Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions (Including Problems)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions (Including Problems)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. Fraction. A rational number in form of — where a and b are integers is called a fraction.
    ‘a’ is called the numerator and Lb’ is called the denominator of the fraction.
  2. Classification of Fractions :
    Decimal fraction : A fraction whose denominator is 10 or multiple of 10.
    Vulgur fraction : A fraction whose denominator is oilier than 10 or multiple of 10.
    Proper fraction : A fraction whose denominator is greater than its numerator.
    Improper fraction : A fraction whose denominator less than its numerator.
    Mixed fraction : A fraction which consists of an integer and a proper fraction.
    Note. If the numerator of a fraction is equal to its denominator, then the fraction is equal to unity i.e. 1.
  3. Equivalent Fractions
    Fractions having the same value are called the equivalent fractions.
  4. Simple and Complex Fractions
    A fraction whose numerator and denominator both are integers, is called a simple fraction.
    A fraction whose numerator or denominator or both are not integers, is called a complex fraction.
  5. Like and Unlike Fractions
    Fractions having the same denominators are called like fractions.
    The fractions with different denominators are called unlike fractions.
  6. Converting unlike fractions into like fractions
    Find the LCM of the denominators of all the give- fractions.
    For each given fraction, multiply its denominator by a suitable numbers so that the product obtained is equal to the LCM in (i).
    Multiply the numerator also by the same number.
  7. To insert a fraction between two given fractions .
    Add the numerators as well as denominators of the given fractions. Then simplify if required.
  8. Addition and Subtraction of fractions
  9. For like fractions, add or subtract (as required) their numerators, keeping the denominator same.
    For unlike fractions, first change all the fractions into like fractions and then add or subtract as above given in (i).
  10. Multiplication
    To multiply two or more fractions, multiply their numerators as well as their denominators.
  11. Division
    To divide on fraction or integer by some other fractions or integer, multiply the first by the reciprocal of the second as given above in multiplication.
  12. Using ‘BODMAS’
    The word ‘BODMAS’ is the abbreviation formed by taking the initial letters of six operations i.e. ‘Bracket’, ;OF, ‘Division’, ‘Multiplication’, ‘Addition’ and ‘Subtraction’. So, according to the rule of ‘BODMAS’, working must be done in the order corresponding to the letters in the word ‘BODMAS’.
  13. Brackets and their removal
    Brackets are four kinds i.e., bar bracket , circular brackets ( ), curly brackets { } and square brackets [ ] and these can be removed in this order i.e. firstly bar, then circular, then curly and lastly square brackets keeping in considerations of the sign given before them.

Fractions Exercise 3A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Classify, each fraction given below, as decimal or vulgar fraction, proper or improper fraction and mixed fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 1

Solution:
(i) Vulgar and Proper
(ii)Decimal and Improper
(iii) Decimal and Proper
(iv) Vulgar and Improper
(v) Mixed
(vi) Decimal
(vii) Mixed and Decimal
(viii) Vulgar and Proper Ans.

Question 2.
Express the following improper fractions as mixed fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 2

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 3

Question 3.
Express the following mixed fractions as improper fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 4

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 5

Question 4.
Reduce the given fractions to lowest terms
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 6
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 8

Question 5.
State : true or false
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 9

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 10
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 11

Question 6.
Distinguish each of the following fractions, given below, as a simple fraction or a complex fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 12

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 13

Fractions Exercise 3B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
For each pair, given below, state whether it forms like fractions or unlike fractions :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 18

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 19

Question 2.
Convert given fractions into fractions with equal denominators :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 21
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 22

Question 3.
Convert given fractions into fractions with equal numerators :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 23

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 24
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 25

Question 4.
Put the given fractions in ascending order by making denominators equal :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 26

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 27
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 28
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 29
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 30

Question 5.
Arrange the given fractions in descending order by making numerators equal :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 31

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 32
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 34

Question 6.
Find the greater fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 35
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 36

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 37
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 38
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 39

Question 7.
Insert one fraction between :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 40

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 41

Question 8.
Insert three fractions between
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 42

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 43
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 44
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 45
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 46
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 47

Question 9.
Insert two fractions between
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 48

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 49

Fractions Exercise 3C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Reduce to a single fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 144

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 51
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 52
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 53
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 54
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 55

Question 2.
Simplify :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 56
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 57

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 58
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 59

Question 3.
Subtract :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 60

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 61

Question 4.
Find the value of
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 62
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 63

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 65

Question 5.
Simplify and reduce to a simple fraction :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 66
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 67
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 68
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 69
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 70
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 71

Question 6.
A bought 3 \(\frac { 3 }{ 4 }\) kg of wheat and 2 \(\frac { 1 }{ 2 }\) kg of rice. Find the total weight of wheat and rice bought.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 72

Question 7.
Which is greater,\(\frac { 3 }{ 5 }\) or \(\frac { 7 }{ 10 }\) and by how much?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 73
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 74

Question 8.
What number should be added to 8 \(\frac { 2 }{ 3 }\) to 12 \(\frac { 5 }{ 6 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 75

Question 9.
What should be subtracted from 8\(\frac { 3 }{ 4 }\) to get 2 \(\frac { 2 }{ 3 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 76

Question 10.
A field is 16 \(\frac { 1 }{ 2 }\) m long and 12 \(\frac { 2 }{ 5 }\) m wide. Find the perimeter of the field.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 77

Question 11.
Sugar costs ₹37 \(\frac { 1 }{ 2 }\)per kg. Find the cost of 8\(\frac { 3 }{ 4 }\) kg sugar.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 78

Question 12.
A motor cycle runs 31\(\frac { 1 }{ 4 }\) km consuming 1 litre of petrol. How much distance will it run consuming 1\(\frac { 3 }{ 5 }\) liter of petrol?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 79

Question 13.
A rectangular park has length = 23 \(\frac { 2 }{ 3 }\) m and breadth = 16 \(\frac { 2 }{ 3 }\) m. Find the area of the park.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 80

Question 14.
Each of 40 identical boxes weighs 4 \(\frac { 4 }{ 5 }\) kg Find the total weight of all the boxes.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 82

Question 15.
Out of 24 kg of wheat, \(\frac { 5 }{ 6 }\) th of wheat is consumed. Find, how much wheat is still left?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 83

Question 16.
A rod of length 2 \(\frac { 2 }{ 5 }\) metre is divided into five equal parts. Find the length of each part so obtained.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 84

Question 17.
IfA = 3\(\frac { 3 }{ 8 }\) and B = 6\(\frac { 5 }{ 8 }\) find :
(i) A+B
(ii) B A
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 85
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 86

Question 18.
Cost of 3 \(\frac { 5 }{ 7 }\) litres of oil is ₹83 \(\frac { 1 }{ 2 }\). Find the
cost of one litre oil.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 87

Question 19.
The product of two numbers is 20 \(\frac { 5 }{ 7 }\). If one of these numbers is 6 \(\frac { 2 }{ 3 }\), find the other.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 88

Question 20.
By what number should 5 \(\frac { 5 }{ 6 }\) be multiplied 1 to get 3\(\frac { 1 }{ 3 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 89

Fractions Exercise 3D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Simplify
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 90

Question 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 91
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 92

Question 3.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 145
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 94

Question 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 95
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 96

Question 5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 97
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 98

Question 6.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 99
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 100

Question 7.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 101
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 102

Question 8.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 103
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 104
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 106

Question 9.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 107
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 108

Question 10.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 109
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 110

Question 11.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 111
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 112

EXERCISE 3 (E)

Question 1.
A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is :
(i) The length of AB to that of CD ?
(ii) \(\frac { 1 }{ 2 }\) the length of AB to that of \(\frac { 1 }{ 3 }\) of CD ?
(iii) \(\frac { 1 }{ 5 }\) of CD to that of AB ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 113

Question 2.
Subtract \(\frac { 2 }{ 7 }\) – \(\frac { 5 }{ 21 }\) from the sum of \(\frac { 3 }{ 4 }\) , \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 12 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 114

Question 3.
From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5\(\frac { 1 }{ 4 }\) kg, 9\(\frac { 1 }{ 2 }\) kg and 9\(\frac { 3 }{ 4 }\) kg respectively.
(i) How many kg did he sell ?
(ii) How many kg are still left in the sack ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 115

Question 4.
If a boy works for six consecutive days for 8 hours, 7\(\frac { 1 }{ 2 }\) hours, 8\(\frac { 1 }{ 4 }\) hours, 6 \(\frac { 1 }{ 4 }\)4 3hours, 6\(\frac { 3 }{ 4 }\) hours and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 116
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 117

Question 5.
A student bought 4 \(\frac { 1 }{ 3 }\) m of yellow ribbon, 6 \(\frac { 1 }{ 6 }\) m of red ribbon and 3\(\frac { 2 }{ 9 }\) m of blue ribbon for decorating a room. How many metres of ribbon did he buy ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 118

Question 6.
In a business, Ram and Deepak invest \(\frac { 3 }{ 5 }\) and \(\frac { 2 }{ 5 }\) of the total investment. IfRs. 40,000 is the total investment, calculate the amount invested by each ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 146
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 120

Question 7.
Geeta had 30 problems for home work. She worked out \(\frac { 2 }{ 5 }\) of them. How many problems were still left to be worked out by her ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 121

Question 8.
A picture was marked at Rs. 90. It was sold at \(\frac { 3 }{ 4 }\) of its marked price. What was the sale price ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 122

Question 9.
Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10\(\frac { 1 }{ 2 }\) kg ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 123

Question 10.
A rope is 25\(\frac { 1 }{ 2 }\) m long. How many pieces , 1 \(\frac { 1 }{ 2 }\) each of length can be cut out from it?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 124

Question 11.
The heights of two vertical poles, above the earth’s surface, are 14 \(\frac { 1 }{ 4 }\) m and 22 \(\frac { 1 }{ 3 }\) respectively. How much higher is the second pole as compared with the height of the first pole ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 125

Question 12.
Vijay weighed 65\(\frac { 1 }{ 2 }\) kg. He gained 1\(\frac { 2 }{ 5 }\) kg during the first week, 1 \(\frac { 1 }{ 4 }\) kg during the second week, but lost \(\frac { 5 }{ 16 }\) kg during the 16 third week. What was his weight after the third week ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 126
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 127

Question 13.
A man spends \(\frac { 2 }{ 5 }\) of his salary on food and \(\frac { 3 }{ 10 }\) on house rent, electricity, etc. What fraction of his salary is still left with him ?
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 128

Question 14.
A man spends \(\frac { 2 }{ 5 }\) of his salary on food and \(\frac { 3 }{ 10 }\) of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 129

Question 15.
Shyam bought a refrigerator for Rs. 5000. He paid \(\frac { 1 }{ 10 }\) of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 130
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 131

Question 16.
A lamp post has half of its length in mud, and \(\frac { 1 }{ 3 }\) of its length in water.
(i) What fraction of its length is above the water ?
(ii) If 3\(\frac { 1 }{ 3 }\) m of the lamp post is above the water, find the whole length of the lamp post.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 132

Question 17.
I spent \(\frac { 3 }{ 5 }\) of my savings and still have Rs. 2,000 left. What were my savings ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 133
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 134

Question 18.
In a school, \(\frac { 4 }{ 5 }\) of the children are boys. If the number of girls is 200, find the number of boys.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 135

Question 19.
If \(\frac { 4 }{ 5 }\) of an estate is worth Rs. 42,000, find the worth of whole estate. Also, find the value of \(\frac { 3 }{ 7 }\) of it.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 136
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 137

Question 20.
After going \(\frac { 3 }{ 4 }\) of my journey, I find that I have covered 16 km. How much Journey is still left ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 138

Question 21.
When Krishna travelled 25 km, he found that \(\frac { 3 }{ 5 }\) of his journey was still left. What was the length of the whole journey.

Solution:Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 139

Question 22.
From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 147
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 140

Question 23.
A boy spent \(\frac { 3 }{ 5 }\) of his money on buying 1 cloth and \(\frac { 1 }{ 4 }\) of the remaining on buying shoes. If initially he has ?2,400; how much did he spend on shoes?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 141

Question 24.
A boy spent \(\frac { 3 }{ 5 }\) of his money on buying cloth and \(\frac { 1 }{ 4 }\) of his money on buying shoes. If initially he has ?2,400; how much did he spend on shoes?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 142
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 3 Fractions image - 143

 

Selina Concise Mathematics class 7 ICSE Solutions – Simple Interest

Selina Concise Mathematics class 7 ICSE Solutions – Simple Interest

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

POINTS TO REMEMBER
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 1

Question 1.
Find the S.I. and amount on :
(i) Rs. 150 for 4 years at 5% per year.
(ii) Rs. 350 for 3\(\frac {1 }{ 2 }\) years at 8% p.a.
(iii) Rs. 620 for 4 months at 8 p. per rupee per month.
(iv) Rs. 3,380 for 30 months at 4 \(\frac { 1 }{ 2 }\) % p.a.
(v) 600 from July 12 to Dec. 5 at 10% p.a.
(vi) Rs. 850 from 10th March to 3rd August at 2 \(\frac { 1 }{ 2 }\) % p.a.
(vii) Rs. 225 for 3 years 9 months at 16% p.a.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 3
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 5
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 6

Question 2.
On what sum of money does the S.I. for 10 years at 5% become Rs. 1,600 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 7

Question 3.
Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a. ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 8

Question 4.
In what time will a sum of money double it self at 8% p.a ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 9

Question 5.
In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2\(\frac { 1 }{ 2 }\) % p.a ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 10

Question 6.
Find the rate percent if the S.I. on ₹275 is 2 years is ₹22.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 11

Question 7.
Find the sum which will amount to ₹700 in 5 years at 8% rate p.a.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 12
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 13

Question 8.
What is the rate of interest, if ₹3,750 amounts to ₹4,650 in 4 years ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 14

Question 9.
In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 15

Question 10.
The interest on a sum of money at the end of 2\(\frac { 1 }{ 2 }\) years is \(\frac { 4 }{ 5 }\) of the sum. What is the rate percent ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 16

Question 11.
What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 17

Question 12.
A sum of Rs. 1,780 become Rs. 2,136 in 4 years,
Find :
(i) the rate of interest.
(ii) the sum that will become Rs. 810 in 7 years at the same rate of interest ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 18

Question 13.
A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest.
Find :
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 19

Question 14.
P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q; find the rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 20

Question 15.
A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 21

Question 16.
The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Simple Interest image - 22

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions
APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Pythagoras Theorem Exercise 16 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.

Solution:
Given : ∆ABC right angled at A and AB = 18 cm, AC = 24 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 1

To find : Length of BC.
According to Pythagoras Theorem,
BC2 = AB2 + AC2
= 182 + 242 = 324 + 576 = 900
∴BC = \(\sqrt { 900 }\) = \(\sqrt { 30 x 30 }\)= 30 cm

Question 2.
Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.

Solution:
Given : ∆XYZ right angled at Z and XY = 13 cm, XZ = 12 cm.
To find : Length of YZ.
According to Pythagoras Theorem,
XY2 = XZ2 + YZ2
132 = 122 + YZ2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 2

169= 144 +YZ2
169- 144 = YZ2
25 = YZ2
∴YZ = \(\sqrt { 25 }\)cm \(\sqrt { 5×5 }\) = 5 cm

Question 3.
Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:
PQ = 34 cm and QR = 33.6 cm.

Solution:
Given : ∆PQR right angled at R and PQ = 34 cm, QR = 33.6 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 3

To find : Length of PR.
According to Pythagoras Theorem,
PR2 + QR2 = PQ2
PR2 + 33.62 = 342
PR2+ 1128.96= 1156
PR2 = 1156- 1128.96
∴ PR = \(\sqrt { 27.04 }\) = 5.2 cm

Question 4.
The sides of a certain triangle are given below. Find, which of them is right-triangle
(i) 16 cm, 20 cm and 12 cm
(ii) 6 m, 9 m and 13 m

Solution:
(i) 16 cm, 20 cm and 12 cm
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (20)2 = (16)2 = (12)2
(20)2 = (16)2 + (12)2
400 = 256 + 144
400 = 400
So, the given triangle is right angled.
(ii) 6 m, 9 m and 13 m
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (13)2 = (9)2 + (6)2
169 = 81+36 169 ≠ 117
So, the given triangle is not right angled.

Question 5.
In the given figure, angle BAC = 90°, AC = 400 m and AB = 300 m. Find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 4
Solution:
AC = 400 m
AB = 300 m
BC = ?
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 5
According to Pythagoras Theorem,
BC2 = AB2 + AC2
BC2 = (300)2 + (400)2
BC2 = 90000 + 160000
BC2 = 250000
BC = \(\sqrt { 250000 }\)= 500 m

Question 6.
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:

(i) CP
(ii) PD
(iii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 6

Solution:
Given : AC = 12 m
BD = 9 m
PA = PB= 15 m
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 7
(i) In right angle triangle ACP
(AP)2 = (AC)2 + (CP)2
152 = 122 + CP2
225 = 144 + CP2
225 – 144 = CP2
81 =CP
\(\sqrt { 81 }\) =CP
∴ CP = 9 m
(ii) In right angle triangle BPD
(PB)2 = (BD)2 + (PD)2
(15)2 = (9)2 + PD2
225 = 81 + PD2
225-81 = PD2
144 = PD2
\(\sqrt { 144 }\)=PD                                               ‘
∴ PD = 12 m
(iii) CP = 9 m
PD = 12 m
∴ CD = CP + PD
= 9+ 12 = 21 m

Question 7.
In triangle PQR, angle Q = 90°, find :
(i) PR, if PQ = 8 cm and QR = 6 cm
(ii) PQ, if PR = 34 cm and QR = 30 cm

Solution:
(i) Given:
PQ = 8 cm
QR = 6 cm
PR = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 8
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
PR2 = 82 + 62
PR2 = 64 + 36
PR2 = 100
∴ PR = \(\sqrt { 100 }\)= 10 cm
(ii) Given :
PR = 34 cm
QR = 30 cm
PQ = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 9
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
(34)2 = PQ2 + (30)2
1156 = PQ2 + 900
1156-900 = PQ2
256 = PQ2
∴ PQ = 16 cm

Question 8.
Show that the triangle ABC is a right-angled triangle; if:
AB = 9 cm, BC = 40 cm and AC = 41 cm

Solution:
AB = 9 cm
CB = 40 cm
AC = 41 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 10
The given triangle will be a right angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
According to Pythagoras Theorem,
(AC)2 = (BC)2 + (AB)2
(41)2 = (40)2 + (9)2
1681 = 1600 + 81
1681 = 1681
Hence, it is a right-angled triangle ABC.

Question 9.
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 11

Solution:
Given:
∆ABD
∠ACB = ∠ACD = 90°
and AB = 10 cm, BC = 6 cm and AD = 17 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 12
To find:
(i) Length of AC
(ii) Length of CD
Proof:
(i) In right-angled triangle ABC
BC = 6 cm, AB = 110 cm
According to Pythagoras Theorem,
AB2 = AC2 + BC2
(10)2 = (AC)2 + (6)2
100 = (AC)2 + 36
AC2 = 100-36 = 64 cm
AC2 = 64 cm
∴ AC = \(\sqrt { 8×8 }\) = 8 cm
(ii) In right-angle triangle ACD
AD = 17 cm, AC = 8 cm
According to Pythagoras Theorem,
(AD)2 = (AC)2 + (CD)2
(17)2 = (8)2 + (CD)2
289 – 64 = CD2
225 = CD2
CD =\(\sqrt { 15×15 }\) = 15 cm

Question 10.
In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 13

Solution:
Given:
∆ABC
∠ADB = 90° and AC = AB = 26 cm
AD = 24 cm
To find : Length of BC In right angled ∆ADC
AB = 26 cm, AD = 24 cm
According to Pythagoras Theorem,
(AC)2 = (AD)2 + (DC)2
(26)2 = (24)2 + (DC)2
676 = 576 + (DC)2
⇒ (DC)2 = 100
⇒ DC = \(\sqrt { 100 }\) = 10 cm
∴ Length of BC = BD + DC
= 10 + 10 = 20 cm

Question 11.
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 14

Solution:
Given :
∆ACD = ∠ABC = 90°
and AD = 13 cm, BC = 12 cm, AB = 3 cm
To find : Length of DC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 15

(i)In right angled ∆ABC
AB = 3 cm, BC = 12 cm
According to Pythagoras Theorem,
(AC)2 = (AB)2 + (BC)2
(AC)2 = (3)2 + (12)2
(AC) = \(\sqrt { 9+144 }\) = \(\sqrt { 153 }\) cm

 (ii) In right angled triangle ACD
AD = 13 cm, AC =\(\sqrt { 153 }\)
According to Pythagoras Theorem,
DC2 = AB2-AC2
DC2= 169-153
DC = \(\sqrt { 16 }\) = 4 cm
∴ Length of DC is 4 cm

Question 12.
A ladder, 6.5 m long, rests against a vertical wall. Ifthe foot of the ladcler is 2.5 m from the foot of the wall, find upto how much height does the ladder reach?

Solution:
Given :
Length of ladder = 6.5 m
Length of foot of the wall = 2.5 m
To find : Height AC According to Pythagoras Theorem,
(BC)2 = (AB)2 + (AC)2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 16

(6.5)2 = (2.5)2 + (AC)2
42.25 = 6.25 + AC2
AC2 = 42.25 – 6.25 = 36 m
AC = \(\sqrt { 6×6 }\) = 6 m
∴ Height of wall = 6 m

Question 13.
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.

Solution:
Given : Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 17

To find: BC
According to Pythagoras Theorem,
In right angled AABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = \(\sqrt { 169 }\) = \(\sqrt { 13×13 }\) = 13 m

Question 14.
Use the information given in the figure to find the length AD.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 18

Solution:
Given :
AB = 20 cm
∴AO =\(\frac { AB }{ 2 }\) = \(\frac { 20 }{ 2 }\) =10cm
BC = OD = 24 cm
To find : Length of AD
In right angled triangle
AOD (AD)2 = (AO)2 + (OD)2
(AD)2 = (10)2 + (24)2
(AD)2 = 100 + 576
(AD)2 = 676
∴ AD =  \(\sqrt { 26×26 }\)
AD = 26 cm

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. The Cost Price (C.P.) of an article is the price at which the article is bought.
  2. The Selling Price (S.P.) of an article is the price at which the article is sold.
  3. If Selling Price of an article is more than its cost price ; it is sold at a profit (gain)
    Profit = Selling Price-Cost Price
    i.e., Profit (gain) = S.P. – C.P. and S.P. = C.P. + Gain
  4. If Selling Price of an article is less than its cost price ; it is sold at a loss.
    Loss = Cost Price – Selling Price
    i.e., Loss = C.P. – S.P. and S.P. = C.P. — Loss
  5. Profit percent and loss percent are always calculated on cost price (C.P.) only.
    i.e., (i) Profit % = \(\frac { Profit }{ C.P. }\) x 100% and
    (ii)
    Loss % = \(\frac {Loss }{ C.P. }\) x 100%
  6. Selling Price = Marked price – Discount
    i.e., S.P. = M.P.—(piscount
    Note : (i) Discount is calculated on marked price (M.P.)
    (ii) Marked price is also written as List price.

EXERCISE 9 (A)

Question 1.
Find the gain or loss percent, if
(i) C.P. = Rs. 200 and S.P.: = Rs. 224
(ii) C.P. = Rs. 450 and S.P. = Rs. 400
(iii) C.P. = Rs. 550 and gain = Rs . 22
(iv) CP. = Rs. 216 and loss = Rs. 72
(v) S.P. = Rs. 500 and loss : = Rs. 100
(vi) S.P. = Rs. 12 and profit = Rs. 4
(vii) C.P. = Rs. 5 and gain = 60 P

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 1
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 3

Question 2.
Find the selling price, if:
(i) C.P. = Rs. 500 and gain = 25%
(ii) C.P. = Rs. 60 and loss = 12 1/2%
(iii) C.P. = Rs. 150 and loss = 20%
(iv) C.P. = Rs. 80 and gain = 2.5%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 5

Question 3.
Rohit bought a tape-recorder for Rs. 1,500 and sold it for Rs. 1,800. Calculate his profit or loss percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 6

Question 4.
An article bought for Rs. 350 is sold at a profit of 20%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 7

Question 5.
An old machine is bought for Rs. 1,400 and is sold at a loss of 15%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 8

Question 6.
Oranges are bought at 5 for Rs. 10 and sold at 6 for Rs. 15. Find profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 53

Question 7.
A certain number of articles are bought at 3 for Rs. 150 and all of them are sold at 4 for Rs. 180. Find the loss or gain as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 10

Question 8.
A vendor bought 120 sweets at 20 p each. In his house, 18 were consumed and he sold the remaining at 30 p each. Find his profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 11

Question 9.
The cost price of an article is Rs. 1,200 and selling price is \(\frac { 5 }{ 4 }\) times of its cost price. Find:
(i) selling price of the article
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 12

Question 10.
The selling price of an article is Rs. 1,200 and cost price is \(\frac { 5 }{ 4 }\) times of its selling price,
find :
(i) cost price of the article ;
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 13

EXERCISE 9 (B)

Question 1.
Find the cost price, if:
(i) S.P. = Rs. 21 and gain = 5%
(ii) S.P. = Rs. 22 and loss = 12%
(iii) S.P. = Rs. 340 and gain = Rs. 20
(iv) S.P. = Rs. 200 and loss = Rs. 50
(v) S.P. = Re. 1 and loss = 5 p.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 14

Question 2.
By selling an article for Rs. 810, a loss of 10 percent is suffered. Find its cost price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 15

Question 3.
By selling a scooter for Rs. 9,200, a man gains 15%. Find the cost price of the scooter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 16

Question 4.
On selling an article for Rs. 2,640, a profit of 10 percent is made. Find
(i) cost price of the article
(ii) new selling price of it, in order to gain 15%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 17

Question 5.
A T.V. set is sold for Rs. 6800 at a loss of 15%. Find
(i)cost price of the T.V. set.
(ii)new selling price of it, in order to gain 12%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 18

Question 6.
A fruit seller bought mangoes at Rs. 90 per dozen and sold them at a loss of 8 percent. How much will a customer pay for.
(i) one mango
(ii) 40 mangoes

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 19

Question 7.
By selling two transistors for Rs. 00 each, a shopkeeper gains 20 percent on one transistor and loses 20 percent on the other.
Find :
(i) C.P. of each transistor

(ii) total C.P. and total S.P. of both the transistors
(iii) profit or loss percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 20
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 21

Question 8.
Mangoes are bought at 20 for Rs. 60. If
they are sold at 33\(\frac { 1 }{ 3 }\) percent profit.
Find:

(i) selling price of each mango.
(ii) S.P. of 8 mangoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 22

Question 9.
Find the cost price of an article, which is sold for Rs. 4050 at a loss of 10%. Also, find the new selling price of the article which must give a profit of 8%.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 23
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 24

Question 10.
By selling an article for ₹825, a man loses \(\frac { 1 }{ 3 }\) equal to j of its selling price.
Find :

(i) the cost price of the article,
(ii) the profit percent or the loss percent made, if the same article is sold for ₹1265.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 25

Question 11.
Find the loss or gain as percent, if the C.P. of 10 articles, all of the same kind, is equal to S.P. of 8 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 26
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 27

Question 12.
Find the loss or gain as percent, if the C.P. of 8 articles, all of the same kind, is equal to S.P. of 10 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 28

Question 13.
The cost price of an article is 96% of its selling price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 30

Question 14.
The selling price of an article is 96% of its cost price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 31

Question 15.
Hundred oranges are bought for ₹350 and all of them are sold at the rate of ₹48 per dozen. Find the profit percent or loss percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 32

Question 16.
Oranges are bought at 100 for ?80 and all of them are sold at ₹80 for ₹100. Find the loss or gain as percent in this transaction.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 33

Question 17.
An article is bought for ₹5,700 and ₹1,300 is spent on its repairing, transportion, etc. For how much should this article be sold in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 34

EXERCISE 9 (C)

Question 1.
A machine is marked at ₹5000 and is sold at a discount of 10%. Find the selling price of the machine.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 35

Question 2.
shopkeeper marked a dinner set for ₹1000. He sold it at ₹900, what percent discount did he give ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 36

Question 3.
A pair of shoes marked at ₹320, are sold at a discount of 15 percent.
Find :
(i) discount
(ii) selling price of the shoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 37

Question 4.
The list price of an article is ₹450 and it is sold for ₹360.
Find :
(i) discount
(ii) discount percent

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 38

Question 5.
A shopkeeper buys an article for₹300. He increases its price by 20% and then gives 10% discount on the new price. Find:
(i) the new price (marked price) of the article.
(ii) the discount given by the shopkeeper.
(iii) the selling price.
(iv) profit percent made by the shopkeeper.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 39
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 40

Question 6.
A car is marked at Rs. 50,000. The dealer gives 5% discount on first Rs. 20,000 and 2% discount on the remaining Rs. 30,000.
Find :
(i) the total discount.
(ii) the price charged by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 41

Question 7.
A dealer buys a T.V. set for Rs. 2500. He marks it at Rs. 3,200 and then gives a discount of 10% on it.
Find :
(i) the selling price of the T.V. set
(ii) the profit percent made by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 42

Question 8.
A sells his goods at 15% discount. Find the price of an article which is sold for Rs. 680.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 43
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 44

Question 9.
A shopkeeper allows 20% discount on the marked price of his articles. Find the marked price of an article for which he charges Rs. 560.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 45

Question 10.
An article is bought for Rs. 1,200 and Rs. 100 is spent on its transportation, etc.
Find :
(i) the total C.P. of the article.
(ii) the selling price of it in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 46

Question 11.
40 pens are bought at 4 for Rs. 50 and all of them are sold at 5 for Rs. 80
Find :
(i) C.P. of one pen.
(ii) S/P. of one pen.
(iii) Profit made by selling one pen.
(iv) Profit percent made by selling one pen.
(v) C.P. of 40 pens
(vi) S.P. of 40 pens.
(vii) Profit made by selling 40 pens.
(viii) Profit percent made by selling 40 pens. Are the results of parts (iv) and (viii) same? What conclusion do you draw from the above result ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 47
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 48

Question 12.
The C.P. of 5 identical articles is equal to S.P. of 4 articles. Calculate the profit percent or loss percent made if all the articles bought are sold.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 49

Question 13.
The C.P. of 8 pens is same as S.P. of 10 pens. Calculate the profit or loss percent made, if all the pens bought are considered to be sold

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 50

Question 14.
A certain number of articles are bought at Rs. 450 per dozen and all of them are sold at a profit of 20%. Find the S.P. of:
(i) one article
(ii)seven articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 51

Question 15.
An article is marked 60% above the cost price and sold at 20% discount. Find the profit percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 52

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Recognition of solids Exercise 18 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 1
Solution:
Nets for a cube are (ii) , (iii) and (v).

Question 2.
Draw at least three different nets for making cube.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 2

Question 3.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 3
In each case, name the object.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 4

Question 4.
Using Euler’s formula, find the values of a, b, c and d.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 5
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 6
(i) a + 6 – 12 = 2 ⇒ a = 2 – 6 + 12 = 14 – 6 = 8
(ii) b + 5- 9 = 2 ⇒6 = 2 + 9- 5 = 6
(iii) 20+ 12 — c = 2 ⇒32 – c = 2 ⇒ c = 32-2 ⇒ c = 30
(iv) 6 + d-12=2 ⇒ d – 6 = 2 ⇒ d = 2 + 6 = 8

Question 5.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 7
Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 8

Question 6.
The following figures represent nets of some solids. Name the solids
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 9
Solution:
The given nets are of the solid as given below :
(i) Cube
(ii) Cuboid

Question 7.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 10

Question 8.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 11

Question 9.
In the map of India, the distance between two cities is 13.8 cm.
Taking scale : 1 cm = 12 km, find the actual distance between these two cities.
Solution:
The scale for a map is given to be 1 cm = 12 km
The distance between these two cities = 13.8 cm on the map
∴ Actual distance between these two cities
= 12 x 13.8 km = 165.6 km

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER
1. Symmetry : A geometrical figure is said to be symmetric about a line if on folding about that line, the two parts of the figure exactly concide each other. The given figure is symmetric about the line PQ. The line is said to be a line of symmetry or an axis of symmetry
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 1

2. Lines of symmetry of given geometrical figures :
It is not necessary that every figure under consideration will definitely have a line symmetry. If we consider different types of triangle ; we find :
1. A scalene triangle has no line of Symmetry : i.e. we can not have a line in a scalene triangle about which if the figure (triangle) is folded, the two parts of the figure will coincide.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 2
2. An isosceles triangle has only one line of symmetry. The bisector of angle of vertex which is also the perpendicular bisector of its base.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 3
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 4

The bisectors of the angle of vertices which are also the perpendicular bisectors of its sides.
4. Line/lincs of symmetry of differed types of quadrilaterals are shown below by dotted lines :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 5
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 6

5. In each of the following, the dotted line/lines are the line/lines of symmetry of the given figure:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 7

6. As shown below:
(i) a circle has infinite lines of symmetry ; every line through its centre is line of symmetry’.
(ii) a semi-circle has one line of symmetry.
(iii) a quadrant (one-fourth) of a circle has one line of symmetry’ and so on.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 8
Note : It is clear from the question numbers 4 and 5, given above that:
1. The largest number of lines of symmetry’ of a triangle is three (3).
2. The largest number of lines of symmetry of a quadrilateral is four (4).
i. e. as the number of sides in a triangle is 3 ; the largest number of lines of symmetry in it is 3 and as the number of sides in a quadrilateral is 4 ; the largest number of lines of symmetry is 4.
In the same way :
1. The largest number of lines of symmetry of a pentagon is 5, as a pentagon has 5 sides.
2. The hexagon has 6 sides and so the largest number of lines of symmetry’ of a hexagon is 6.
In general, we can say, that if a polygon has n sides ; M
the largest number of lines of symmetry, it can have, is n.
3. Reflection (Image): Tire given figure shows a candle place ‘d’ distance before a plane mirror MM’, the image of the candle is obtained in the mirror at the same distance ‘d’ behind the mirror. Geometrically, the line joining the candle (c) and its reflection c’ is (Candle) perpendicular bisector of the mirror line MM’.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 9
4. Reflection in x-axis : Reflection if x-axis means the x-axis is considered as the plane mirror, the given point as the object and then to find its image.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 10
Let P (x, y) be a point and. as shown in the figure, when it is reflected in x-axis to point P’; the co-ordinates of image point P’ are (x, – y).
i. e. reflection of P (x, y) in x-axis = P’ (x, – y)
In other words :
Image of P (x, y) in x-axis = P’ (x, – y)
We can say, when a point (x, y) is reflected in x-axis, the sign of its second component (ordinate) changes i.e. tire sign of y changes and so the image of (x, y) in x-axis is (x, – y).
5. Reflection in y-axis : As is clear from the figure, given alongside, the reflection P (x, y) in y-axis is point P’ (- x, y).
We can say, when a point (x,y) is reflected in y-axis, the sign of its first component (abscissa) change i.e. the sign of x changes and so the image of (x, y) in y- axis is (-x,y).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 11
6. Reflection in Origin : When a point P (x, y) is reflected in origin, the sign of both of its components change i.e. the image of P (x, y) is P’ (- x, -y) as shown along side in the figure.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 12

Symmetry Exercise 17A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
For each figure, given below, draw the line (s) of symmetry, if possible :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 13
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 14

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 15

Question 2.
Write capital letters A to Z of English alphabet ; and in each case, if possible, draw the largest number of lines of symmetry.
Solution:
Line or lines of symmetry’ is possible in the following alphabets. For others alphabets it is not possible
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 16
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 17

Question 3.
By drawing a free hand sketch of each of the following, draw in each case, the line (s) of symmetry,
if any:
(i) a scalene triangle
(ii) an isosceles right angled triangle
(iii) a rhombus
(iv) a kite shaped figure triangle.
(v) a rectangle
(vi) a square
(vii) an isosceles

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 18.

Question 4.
Draw a triangle with :
(i) no line of symmetry,
(ii) only one line of symmetry,
(iii) exactly two lines of symmetry,
(iv) exactly three lines of symmetry,
(v) more than three lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the triangle drawn.

Solution:
(i) Scalene triangle : It has no line of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 19
(ii) Isosceles Triangle : It has one line of symmetry as shown.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 20
(iii) It is not possible.
(iv) Equilateral Triangle : It has three lines of symmetry as shown
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 21
(v) It is not possible.

Question 5.
Draw a quadrilateral with :
(i) no line of symmetry.
(ii) only one line of symmetry.
(iii) exactly two lines of symmetry.
(iv) exactly three lines of symmetry.
(v) exactly four lines of symmetry.
(vi) more than four lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the quadrilateral drawn.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 22

Question 6.
Construct an equilateral triangle with each side 6 cm. In the triangle drawn, draw all the possible lines of symmetry.

Solution:
Steps of Construction :
(i) Draw a line segment BC = 6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 23
(ii) With centres B and C and radius 6 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC
∆ ABC is the required equilateral triangle,
(iv) Draw the angle bisectors of ∠A, ∠B and ∠C.
These are the lines of symmetry which are three in numbers as the triangle is equilateral.

Question 7.
Construct a triangle ABC in which AB = AC = 5cin and BC = 5.6 cm. If possible, draw its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5.6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 24
(ii) With centres B and C and radius 5 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC.
∆ ABC is an isosceles triangle.
(iv) Draw the bisector of ∠A. This is the only one line of symmetry as the triangle is an isosceles.

Question 8.
Construct a triangle PQR such that PQ = QR = 5 .5 cm and angle PQR = 90°. If possible, draw its lines of symmetry.
Solution:
∴ ∠PQR = 90°, and ∠P = ∠R
(opposite sides are equal)
∴∠P + ∠R = 90°
Hence ZP = ZR = \(\frac { 99\circ }{ 2 }\)  = 45°
Steps of Construction:
(i) Draw a line segment QR = 5.5 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 25
(ii) At Q, draw a ray making an angle of 90° and cut off QP = 5.5 cm.
(iii) Join PR.
∆PQR is an isosceles triangle.
(iv) Draw the angle bisector of ∠PQR. It is the line of symmetry. Since the triangle is an isosceles.
∴It has only one line of symmetry.

Question 9.
If possible, draw a rough sketch of a quadrilateral which has exactly two lines of symmetry.

Solution:
Since the quadrilateral has exactly two lilies of symmetry
∴It must be a rectangle or a rhombus
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 26

Question 10.
A quadrilateral ABCD is symmetric about its diagonal AC. Name tire sides of this quadrilateral which are equal.

Solution:
The quadrilateral ABCD is symmetric about its diagonal AC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 27
∴It must be a kite shaped.
Hence side AB = AD and BC = DC.

Symmetry Exercise 17B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
In each figure, given below, find the image of the point P in the line AB :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 28
Solution:
Steps of Construction : Fig. (i) and (ii)
(i) From P, draw a perpendicular to the given line AB meeting it at O.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 29
(ii) Produce PO to P’ such that OP’ = PO.
P’ is the required image of P in AB.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 30

Question 2.
In each figure, given below, find the image of the line segment AB in the line PQ :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 31
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 32
Steps of Construction :
(i) From A and B, draw perpendiculars on PQ intersecting PQ at L and M.
(ii) Produce AL to A’ such that AL = LA’ and produce BM to B’ such that BM = MB’ A’B’ is the image of the line segment AB in PQ.

Question 3.
Complete the following table :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 33
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 34

Question 4.
A point P (7,3)|is reflected in x-axis to point P’. The point P’ is further reflected in v-axis to point P” Find :
(i) the co-ordinates of P’
(ii) the co-ordinates of P”
(iii) the image of P (7, 3) in origin.
Solution:
(i) Image of point P (7,3) when reflected in x-axis is P’ whose co-ordinates will be (7,-3)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 35
(ii) Image of point P’ (7,-3) when reflected in y-axis, is P” whose co-ordinates will be (- 7,-3)
(iii) The image of P (7, 3) in origin is P” whose co-ordinates are (- 7, – 3).

Question 5.
A point A (- 5, 4) is reflected in y-axis to point B. The point B is further reflected in origin to point C. find :
(i) the co-ordinates of B
(ii) the co-ordinates of C
(iii) the image of A (- 5, 4) in x-axis.
Solution:
(i) Image of point A (- 5,4) when reflected in y-axis is B whose co-ordinates will be (5,4)
(ii) Image of B (5, 4) when reflected in origin is C whose co-ordinates will be (- 5, – 4)
(iii) Image of A (- 5,4) in x-axis is C whose co-ordinates are (- 5, – 4)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 36

Question 6.
The point P (3, – 8) is reflected in origin to point Q. The Point Q is further reflected in x-axis to point R. Find :
(i) the co-ordinates of Q
(ii) the co-ordinates of R
(iii) the image of P (3, – 8) in y-axis.
Solution:
(i) The image of the given point P (3, – 8) when reflected in origin is Q whose co-ordinates will be (- 3, 8).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 37
(ii) The image of Q (- 3, 8) when reflected in x-axis is R whose co-ordinates will be (-3,-8)
(iii) Tlie image of P (3, 8) in y-axis is R whose co-ordinates are (- 3, – 8).

Question 7.
Each of the points A (3, 0), B (7, 0), C (- 8, 0), D (- 7, 0) and E (0, 0) is reflected in x-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points are given :
A (3, 0), B (1, 0), C (-8, 0), D (- 7, 0) and E (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 38
This images will be when reflected in x-axis. A’ (3, 0), B’ (7, 0), C’ (- 8, 0) D’ (- 7, 0) and E’ (0, 0) as the given points lie on x-axis.

Question 8.
Each of the points A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) is reflected in y-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The given points
A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) are reflected in y-axis. The co – ordinates of their images will be A’ (0, 4), B’ (0, 10), C’ (0, – 4) D’ (0, – 6) and E’ (0, 0) as they all lie on y-axis.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 39

Question 9.
Each of the points A (0, 7), B (8. 0), C (0, -5), D (- 7, 0) and E (0, 0) are reflected in origin to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points A (0, 7), B (8, 0), C (0, – 5). D (- 7.0) and E (0,0) are reflected in origin.
So, the co-ordinates of their images will be A’ (0,-7), B’ (- 8, 0), C’ (0,5), D’ (7, 0) and E’ (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 40

Question 10.
Mark points A (4, 5) and B (- 5, 4) on a graph paper. Find A’, the image of A in x-axis and B’, the image of B in x-axis.
Mark A’ and B’ also on the same graph paper.
(ii) Join AB and A’ B’ and
find if AB = A’ B’ ?
Solution:
The given points :
A (4, 5) and B (- 5, 4) have been marked on the graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 41
The image of A in x-axis A (4, – 5) and image of B in x-axis is B’ (-5, -4) which have been also plotted on the same graph.
AB and A’ B’ are joined. We see that AB = A’ B’.

Question 11.
Mark points A (6, 4) and B (4, – 6) on a graph paper.
Find A’, the image of A in y-axis and B’, the image of B in y-axis. Mark A’ and B’ also on the same graph paper.
Solution:
The given points are
A (6, 4) and B (4, – 6)
The images of A and B is y-axis are A’ (- 6, 4) and B’ (- 4, – 6) respectively as shown in the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 42

Question 12.
Mark points A (- 6, 5) and B (- 4, – 6) on a graph paper. Find A’, the image of A in origin and B’, the image of B in origin. Mark A’ and B’ also on the same graph paper. Join AB and A’ B’. Is AB = A’ B’ ?
Solution:
The given points are A (- 6, 5) and B (- 4, – 6). The images of A and B in the origin are A’ and B’ where co-ordinates are A’ (6, – 5) and B’ (4, 6) which have been plotted on the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 43
AB and A’ B’ are joined we see that AB = A’ B’.

Symmetry Exercise 17C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
How many lines of symmetry does a rhombus have?
Solution:
It has two lines of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 44

Question 2.
What is the order of rotational symmetry of a rhombus?
Solution:
The order of rotational symmetry can be defined as the number of times that a shape appears exactly the same during a full 360° rotation. Order of rotational symmetry of a rhombus is 2.

Question 3.
Show that each of the following figures has two lines of symmetry and a rotational.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 45
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 46

Question 4.
Name a figure that has a line of symmetry but does not have any roational symmetry.
Solution:
Isosceles triangle has only line symmetry and no rotational symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 47

Question 5.
In each of the following figures, draw all possible lines of symmetry and also write the order of rotational symmetry:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 48
Solution:
(i) It has no line of symmetry and order of rotational symmetry is 3
(ii) It has 2 line of symmetry and order of rotational symmetry is 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 49
(iii) It has 2 line of symmetry and order of rotational symmetry is 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 50
(iv) It has no line of symmetry and order of rotational symmetry is 0.
(v) It has 1 line of symmetry and order of rotational symmetry is 0.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 51

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER
1. Definition of a Set : In our day to day life, different collective nouns are used to describe collection of objects ; such as : a group of students playing cricket, a pack of cards, a bunch of flowers, etc
In mathematics, such collections of objects are named as sets.
A set is a collection of well-defined objects, things or symbols, etc.
The phrase ‘well-defined’ means ; it must be possible to know, without any doubt, whether a given object (thing or symbol) belongs to the set under consideration or not.
For example :
“The set of tall boys of Class 10” is not well-defined ; since it is not possible to know that which boys are to be included and exactly what is the limit.
But when we say, “The set of boys of Class 10, which are taller than Peter”, now we can compare the heights of different boys with the height of Peter and can know exactly, that which boys are to be included in the required set. Thus, the objects are well-defined.
2. Elements of a set: The objects (things, symbols, etc.) used to form a set are called elements or members of the set.
In general, a set is denoted by a capital letter of English alphabet with its elements written inside curly braces and separated by commas.
e.g., Set A = {5, 10, 12, 15}
3. Use of Symbol ‘∈’ or Symbol ‘∉ ‘ : The symbol ‘∈’ stands for ‘belongs to’ or ‘ is an element of’ or ‘is a member of’; whereas the symbol ‘∉’ stands for ‘does not belong to’ or ‘is not an element of or ‘is not a member of’.
e.g., For set P = {3, 6, 8, 13, 18} ; 3 ∈ P, 5∉ P and so on.
(i) The elements in a set can be written in any order.
Thus, {a, b, c, d] is the same set as {b, d, a, c} or {c, b, d, a}, etc.
(ii) The elements in a set should not be repeated, i.e. if any element occurs many times, it should be written only once.
Thus, set of letters of the word ‘crook’ = {c, r, o, k}.
There are two os in the given word “crook” ; but in the set, it is written only once.
4. Representation of A Set: A set, in general, is represented in :
(i) Description method (form)
(ii) Tabular or Roster method (form)
(iii) Set-builder or Rule method
For example :
N is the set of natural numbers [Description method]
N = {1, 2, 3, 4, 5, ……. } [Roster or Tabular method]
N = {x : x is a natural number}, or {x : x ∈ N} [Set-builder or Rule method]
[The symbol ‘ : ’ stands for such that and the set {x : x ∈ N} is read as, “the set of x such that x is a natural number”].
It is clear from the example given above that:
(i) in description method a well-defined description about the set is given.
(ii) in roster or tabular method the elements of the set are written inside a pair of curly braces and are separated by commas.
(iii) in set-builder or ruler method the actual elements of the set are not written, but a rule or a statement or a formula is written in the briefest possible way.
5. Cardinal Number : The cardinal number of a set is the number of elements in it.
Thus, if a set A has 5 elements ; its cardinal number is 5 and we represent it by writing n (A) = 5 .
Similarly, if set B = Set of even natural numbers less than 10 then, B = {2, 4, 6, 8} and n (B) = 4.
If B = {0}, then n (B) = 1. Since, 0 is an element of set B.
6. Types of Sets :
(i) Finite Set: A set is said to be a finite set, if it has a limited (countable) number of elements in it.
For example :
(a) S = Set of natural numbers between 10 and 15 = {11, 12, 13, 14}
(b) P = {0, 1,2, , 20} = {x : x ∈ W and x ≤ 20} and so on.
(ii) Infinite set: A set is said to be an infinite set, if it has an unlimited (uncountable) number of elements in it.
For example :
(a) P = Set of prime numbers = {2, 3, 5,…. }
(b) B = (x : x ∈ N and x ≥ 21} = (21, 22, 23, } ans so on.
(iii) Empty set or Null set: The set, with no element in it, is called the empty set or the null set.
The empty set is represented by a pair of braces with no element in it or by the Danish letter Φ, which is pronounced as ‘oe’
Thus, the empty set = { } = Φ
Note : For empty set, it is wrong to call ‘an empty set’ or ‘a null set’ as there is one and only one empty set though it may have many descriptions.
Therefore, it is always called “the empty set or the null set”.
Some examples of the empty set :
(a) Let A = {a man of age more than 400 years}.
Since there can not be any man with the age more than 400 years; the set A will have no element in it i.e. It is the empty set. And we write : A = { } or Φ
(b) If B = (Triangles with 4 sides} ; it is clear that B = Φ
Note :
1. Φ ≠{0}, since {0} is a set with 0 as its element whereas Φ has no element.
2. {Φ} ≠ {0}, since both the sets have different elements.
3. The cardinal number of the empty set is 0 i.e. n (Φ) = 0.
(iv) Disjoint sets : Sets having no element in common are called disjoint sets.
For example :
Sets P = (5, 7, 9} and O = (4, 6, 10, 12} are disjoint; as they do not have any element in common.
(v) Joint (overlapping) sets : Sets having atleast one element in common are called joint or overlapping sets.
For example:
Set B = (4,6, 8,10,12} and set C = (3,6,9,12,15} are joint sets; as they have elements 6 and 12 common.
(vi) Equal sets : Two sets are said to be equal, if the elements of both the sets are the same.
For example :
If set A = {x, y, z} and set B = {last three letters of English alphabet}.
Clearly, sets A and B have the same elements and so set A = set B.
(vii) Equivalent sets : Two sets are said to be equivalent, if they have equal number of elements in them, i.e. the cardinal numbers of both the sets are equal.
For example :
Let A = (3, 6, 9} and B = {a, b, c}.
Since, set A has 3 elements and set B also has 3 elements i.e., n (A) = n (B); therefore, sets A and B are equivalent and for this, we write : A ↔ B.
Note:
1. Equal sets are always equivalent; but the converse is not always true (i.e. it is not necessary that equivalent sets are equal also).
2. In equivalent sets, the number of elements (cardinal number) are equal, whereas in equal sets the element are the same.
3. Two infinite sets are always equivalent;

Set Concepts Exercise 13A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find, whether or not, each of the following collections represent a set:
(i) The collection of good students in your school.
(ii) The collection of the numbers between 30 and 45.
(iii) The collection of fat-people in your colony.
(iv) The collection of interesting books in your school library.
(v) The collection of books in the library and are of your interest.
Solution:
(i) It is not a set as it is not well defined.
(ii) It is a set
(iii) It is not a set as it is not well defined.
(iv) It is not a set as it is not well defined.
(v) It is a set.

Question 2.
State whether true or false :
(i) Set {4, 5, 8} is same as the set {5, 4, 8} and the set {8, 4, 5}
(ii) Sets {a, b, m, n} and {a, a, m, b, n, n) are same.
(iii) Set of letters in the word ‘suchismita’ is {s, u, c, h, i, m, t, a}
(iv) Set of letters in the word ‘MAHMOOD’ is {M, A, H, O, D}.
Solution:
(i) True
(ii) True
(iii) True as it has the same elements
(iv) True as it has the same elements.

Question 3.
Let set A = {6, 8, 10, 12} and set B = {3, 9, 15, 18}.
Insert the symbol ‘ ∈ ’ or ‘ ∉ ’ to make each of the following true :
(i) 6 …. A
(ii) 10 …. B
(iii) 18 …. B
(iv) (6 + 3) …. B
(v) (15 – 9) …. B
(vi) 12 …. A
(vii) (6 + 8) …. A
(viii) 6 and 8 …. A
Solution:
(i) 6 ∈ A
(ii) 10 ∉ B
(iii) 18 ∈B
(iv) (6 + 3) or 9 ∈ B
(v) 15 – 9 or 6 g B
(vi) 12 ∈ A
(vii) 6 + 8 or 14 ∉ A
(viii) 6 and 8 ∈ A

Question 4.
Express each of the following sets in
roster form :
(i) Set of odd whole numbers between 15 and 27.
(ii) A = Set of letters in the word “CHITAMBARAM”
(iii) B = {All even numbers from 15 to 26}
(iv) P = {x : x is a vowel used in the word ‘ARITHMETIC’}
(v) S = {Squares of first eight whole numbers}
(vi) Set of all integers between 7 and 94; which are divisible by 6.
(vii) C = {All composite numbers between 2 and 20}
(viii) D = Set of Prime numbers from 2 to 23.
(ix) E = Set of natural numbers below 30 which are divisible by 2 or 5.
(x) F = Set of factors of 24.
(xi) G = Set of names of three closed figures in Geometry.
(xii) H = {x : x eW and x < 10}
(xiii) J = {x: x e N and 2x – 3 ≤17}
(xiv) K = {x : x is an integer and – 3 < x < 5}
Solution:
(i) {17, 19, 21, 23, 25}
(ii) A = (C, H, I, T, A, M, B, R}
(iii) B = {16, 18, 20, 22, 24, 26}
(iv) P = {a, e, i}
(v) S = {0, 1, 4, 9, 16, 25, 36, 49}
(vi) {12, 18, 24; 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90}
(vii) C = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}
(viii) D = {2, 3, 5, 7, 11, 13, 17, 19,23}
(ix) E = {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28}
(x) F={l,2, 3, 4, 6, 8, 12, 24}
(xi) G = {Triangle, quadrilateral, circle}
(xii) H = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(xiii) 2x – 3 ≤ 17
⇒ 2x ≤ 17 + 3 2 x ≤ 20
⇒ x ≤ \(\frac { 20 }{ 2 }\)
x ≤ 10
∴ J = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(xiv) ∵ – 3 < x < 5
∴x lies between – 3 and 5
∴K = {- 2, – 1, 0, 1, 2, 3, 4}

Question 5.
Express each of the following sets in set- builder notation (form) :
(i) {3, 6, 9, 12, 15}
(ii) {2, 3, 5, 7, 11, 13 …. }
(iii) {1, 4, 9, 16, 25, 36}
(iv) {0, 2, 4, 6, 8, 10, 12, …. }
(v) {Monday, Tuesday, Wednesday}
(vi) {23, 25, 27, 29, … }
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
(viii) {42, 49, 56, 63, 70, 77}
Solution:
(i) {3, 6, 9, 12, 15}
= {x: x is a natural number divisible by 3 ;x< 18}
(ii) {2, 3, 5, 7, 11, 13, }
= {x : x is a prime number}
(iii) {1,4,9, 16,25,36}
= {x : x is a perfect square ; x < 36}
(iv) {0, 2, 4, 6, 8, 10, 12, }
= {x : x is a whole number divisible by 2}
(v) {Monday, Tuesday, Wednesday}
= {x : x is one of the first three days of 3 week}
(vi) {23, 25, 27, 29, }
= {x : x is an odd natural number; x ≥ 23}
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
= {x: x = \(\frac { 1 }{ n }\) when n is a natural number: 3 ≤ n ≤ 8}
(viii) {42, 49, 56, 63, 70, 77}
= (x: x is a natural number divisible by 7 ; 42 ≤x ≤ 77}

Question 6.
Given : A = {x : x is a multiple of 2 and is less than 25}
B = {x : x is a square of a natural number and is less than 25}
C = {x : x is a multiple of 3 and is less than 25}
D = {x: x is a prime number less than 25}
Write the sets A, B, C and D in roster form.
Solution:
A = {x: x is a multiple of 2 and is less than 25}
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
B = {x : x is a square of natural number and is less than 25}
= {1,4,9,16}
C = {x : x is a multiple of 3 and is less than 25}
= {3, 6, 9, 12, 15, 18,21,24}
D = {x : x is a prime number less than 25}
= {2, 3, 5, 7, 11, 13, 17, 19, 23}

Set Concepts Exercise 13B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Write the cardinal number of each of the following sets:
(i) A = Set of days in a leap year.
(ii) B = Set of numbers on a clopk-face.
(iii) C = {x : x ∈ N and x ≤ 7}
(iv) D = Set of letters in the word “PANIPAT”.
(v) E = Set of prime numbers between 5 and 15.
(vi) F = {x : x ∈ Z and – 2 < x ≤ 5}
(vii) G = {x : x is a perfect square number, x ∈N and x ≤ 30}.
Solution:
(i) n A = 366
(ii) n B = 12
(iii) n C =7
(iv) n D = 5
(v) n E = 3
(vi) n F =7
(vii) n G = 5

Question 2.
For each set, given below, state whether it is finite set, infinite set or the null set :
(i) {natural numbers more than 100}
(ii) A = {x : x is an integer between 1 and 2}
(iii) B = {x : x ∈ W ; x is less than 100}.
(iv) Set of mountains in the world.
(v) {multiples of 8}.
(vi) {even numbers not divisible by 2}.
(vii) {squares of natural numbers}.
(viii) {coins used in India}
(ix) C = {x | x is a prime number between 7 and 10}.
(x) Planets of the Solar system.
Solution:
(i) {Natural numbers more than 100}
= It is an infinite set
(ii) A = {x : x is an integer between 1 and 2}
It is a null set
(iii) B = {x : x ∈ W, x is less than 100}
It is finite set as it has 100 elements i.e. from 0 to 99.
(iv) Set of mountains in the world.
∴ It is an infinite set
(v) {Multiples of 8}
It is an infinite set
(vi) {Even numbers not divisible by 2}
It is a null set
(vii) {Squares of natural numbers}
∴ It is an infinite set
(viii) {Coins used in India}
∴It is a finite set as these are countable
(ix) {x | x is a prime number between 7 and 10}
As there is not such prime number between 7 and 10.
Hence it is null set
(x) Planets of two Solar system.
It is finite set as there are countable.

Question 3.
State, which of the following pairs of sets are disjoint :
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10.
(ii) {birds} and {tress}
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}.
(iv) A = {natural numbers less than 10} and B = {x : x is a multiple of 5}.
(v) {people living in Calcutta} and {people living in West Bengal}.
Solution:
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10}
⇒ {0, 1,2, 6, 8} and {1,3, 5, 7, 9}
∴There sets are not disjoint sets as there is one element (1) is common.
(ii) {Birds} and {trees}
These are disjoint sets as there is no common element in term
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}
These are not disjoint sets as there can be a person who is fan of both the games.
(iv) A = {Natural numbers less than 10} and B = {x : x is a multiple of 5}
⇒ A = {1, 2, 3, 4, 5, 6, 7, 8, 9} and B = {5, 10, 15 }
These are hot disjoint sets as there is one element 5, which is common.
(v) {People living in Calcutta} and {People living in West Bengal}.
These are not disjoint sets as people of Calcutta are the people of West Bengal as Calcutta is a city of West Bengal.
So, only (ii) is a pair of disjoint sets.

Question 4.
State whether the given pairs of sets are equal or equivalent.
(i) A = {first four natural numbers} and B = {first four whole numbers}.
(ii) A = Set of letters of the word “FOLLOW” and B = Set of letters of the word “WOLF”.
(iii) E = {even natural numbers less than 10} and O = {odd natural numbers less than 9}
(iv) A = {days of the week starting with letter S} and B = {days of the week starting with letter T}.
(v) M = {multiples of 2 and 3 between 10 and 20} and N = {multiples of 2 and 5 between 10 and 20}.
(vi) P = {prime numbers which divide 70 exactly} and Q = {prime numbers which divide 105 exactly}
(vii) A = {0², 1², 2², 3², 4²} and = {16, 9,4, 1, 0}.
(viii) E = {8,JO, 12, 14, 16} and F = {even natural numbers between 6 and 18}.
(ix) A = {letters of the word SUPERSTITION} and B = {letters of the word JURISDICTION}.
Solution:
(i) A = {first four natural numbers}
= {1,2, 3, 4}
B = {first first whole number)
= {0, 1,2,3}
These are equivalent sets as both have equal number of elements but not same.
(ii) A = Set of letters of the word ‘FOLLOW’
= {F, O, L, W}
B = Set of letters of the word ‘WOLF’
= {W, O, L, F}
These are equal sets as these have same and equal elements.
(iii) E = {even natural numbers less than 10}
= {2, 4, 6, 8}
O = {odd natural numbers less than 9}
= {L3, 5, 7}
These are equivalent sets as both have equal number of elements but not the same.
(iv) A = {Days of the week starting with letter S}
= {Sunday, Saturday}
B = {Days of the week starting with letter T}
= {Tuesday, Thursday}
These are equivalent sets as both have equal number of elements.
(v) M = {Multiples of 2 and 3 between 10 and 20}
= {12, 14, 15, 16, 18}
N = {Multiples of 2 and 5 between 10 and 20}
= {12, 14, 15, 16, 18}
These are equal sets as these have same and equal number of elements.
(vi) P = {Prime numbers which divide 70 exactly}
= {2, 5, 7}
Q = {Prime numbers which divide 105 exactly}
= {3, 5, 7}
These are equivalent sets as these have equal number of elements.
(vii) A = {02, l2, 22, 32, 42} = {0, 1, 4, 9, 16} B = {16, 9, 4, 1,0}
These are equal sets as these have same and equal number of elements.
(viii) E = {8, 10, 12, 14, 16}
F = {even natural numbers between 6 and 18}
= {8, 10, 12, 14, 16}
These sets are equal as these have same and equal number of elements
(ix) A = {Letters of the word SUPERSTITION}
= {S, U, P, E, R, T, I, O, N}
B = Letters of the word JURISDICTION.
= (J, U, R, I, S, D, C, T, O, N}
These are neither equal nor equivalent sets as these have different and unequal elements.

Question 5.
Examine which of the following sets are the empty sets :
(i) The set of triangles having three equal sides.
(ii) The set of lions in your class.
(iii) {x 😡 + 3 = 2 and xeN}
(iv) P = {x : 3x = 0}
Solution:
(i) The set of triangle having three equal sides. This is not an empty set
(ii) The set of lions in your class This is an empty set
(iii) {x : x +3 = 2 and x ∈N}
x ≠ 3 = 2 ⇒ x = 2-3= -l
which is not a natural number.
∴ It is an empty set.
(iv) P = {x : 3x = 0} = {0} which is not an empty set.
Hence (ii) and (iii) are empty sets.

Question 6.
State true or false :
(i) All examples of the empty set are equal.
(ii) All examples of the empty set are equivalent.
(iii) If two sets have the same cardinal number, they are equal sets.
(iv) If n (A) = n (B) then A and B are equivalent sets.
(v) If B = {x : x + 4 = 4}, then B is the empty set.
(vi) The set of all points in a line is a finite set.
(vii) The set of letters in your Mathematics book is an infinite set.
(viii) If M = {1, 2, 4, 6} and N = {x : x is a factor of 12} ; then M = N.
(ix) The set of whole numbers greater than 50 is an infinite set.
(x) If A and B are two different infinite sets, then n (A) = n (B).
Solution:
(i) True
(ii) True
(iii) False
(iv) True
(v) False
(vi) False
(vii) False
(viii) False
( ix )True
(x) False

Question 7.
Which of the following represent the null set ?
φ, {0}, 0, { }, {φ}

Solution:
φ and { } are the null sets other are not as there have same element.

Set Concepts Exercise 13C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Fill in the blanks :
(i) If each element of set P is also an element of set Q, then P is said to be …… of Q and Q is said to be of P.
(ii) Every set is a ….. of itself.
(iii) The empty set is a …… of every set.
(iv) If A is proper subset of B, then n (A) …. n (B).
Solution:
(i) If each element of set P is also an element of set Q then P is said to be subject of Q ; and Q is said to be super set of P.
(ii) Every set is a subset of itself.
(iii) The empty set is subset of every set.
(iv) If A is proper subset of B, then n (A) is less than n (B)

Question 2.
If A = {5, 7, 8, 9} ; then which of the following are subsets of A ?
(i) B = {5, 8}
(ii) C = {0}
(iii) D = {7, 9, 10}
(iv) E = { }
(V) F = {8, 7, 9, 5}
Solution:
(i) B = {5, 8,}
∴B ⊂ A
(ii) C = {0}
∴ C φ A
(iii) D = {7, 9, 10}
∴D ⊄ A
(iv) E = { }
∴E ⊂A (An empty set is subset of every set)
(v) F = (8, 7, 9, 5}
∴F ⊂A
∵ Every set is subset of it self.
Hence (i), (iv)
and (v) are subsets of A.

Question 3.
If P = {2, 3, 4, 5} ; then which of the following are proper subsets of P ?
(i) A = {3, 4}
(ii) B = { }
(iii) C = {23, 45}
(iv) D = {6, 5, 4}
(v) E = {0}
Solution:
P = {2, 3, 4, 5}
(i) A = {3,4},
(ii) B = { }, C = {23, 45},
D = {6, 5, 4} and E = {0}.
We see that only A and B are the proper subset of P.

Question 4.
If A = {even numbers less than 12},
B = {2, 4},
C = {1, 2, 3},
D = {2, 6} and E = {4}
State which of the following statements are true :
(i) B⊂A
(ii) C⊆A
(iii) D⊂C
(iv) D ⊄ A
(v)E⊇B
(vi) A⊇B⊇E
Solution:
A = {Even number less than 12} = {2, 4, 6, 8, 10}
B = {2, 4}, C = {1, 2, 3},
D = {2, 6} and E = {4}
(i) B ⊂ A: It is true
(ii) C ⊆ A: It is false
(iii) D ⊂ C : It is false
(iv) D ⊄ A
(v) E ⊇ B : It is false
(vi) A ⊇ B ⊇ E : It is true

Question 5.
Given A = {a, c}, B = {p, q, r} and C = Set of digits used to form number 1351.
Write all the subsets of sets A, B and C.
Solution:
(i) A = {a, c}
∴ Subsets are : { } or φ, {a}, {c} and {a, c}
(ii) B = {p,q, r)
∴ subsets are : { } or φ, {p}, {q}, {r}, {p, q}, ip, r}, {q, r} and {p, q, r}
(iii) C = Set of digits used in 135, = {1,3,5}
∴ Subsets are = { }
or φ, {1}, {3}, {5}, {1,3}, {1,5}, {2,5} and {1, 3, 5}

Question 6.
(i) If A = {p, q, r}, then number of subsets of A = ……
(ii) If B = {5, 4, 6, 8}, then number of proper subsets of B = ……
(iii) If C = {0}, then number of subsets of C = …..
(iv) If M = {x : x ∈ N and x < 3}, then M has …… proper subsets.
Solution:
(i) If A = {p, q, r},
then number of subsets of A = 2³ = 2×2×2 = 8
(ii) If B = {5, 4, 6, 8},
then number of proper subsets of B = 24 – 1 = 2 × 2 × 2× 2 – 1 = 16 – 1 = 15
(iii) If C = {0},
then number of subsets of C = 21 = 2
(iv) If M = {x: x ∈ N and x < 3}, = {1, 2}
Then M has proper subsets = 22—1 = 4 — 1 = 3

Question 7.
For the universal set {4, 5, 6, 7, 8, 9, 10, 11,12,13} ; find its subsets A, B, C and D such that
(i) A = {even numbers}
(ii) B = {odd numbers greater than 8}
(iii) C = {prime numbers}
(iv) D = {even numbers less than 10}.
Also, find compliments of each set i.e., find A’, B’, C’ and D’.
Solution:
(i) A = {even numbers}
= {4, 6, 8, 10, 12}
(ii) B = {odd numbers greater then 8}
= {9, 11, 13}
(iii) C = {Prime numbers}
= {5, 7, 11, 13}
(iv) D = {even numbers less than 10}
= {4, 6, 8}
A’= {5, 7, 9, 11, 13},
B’ = {4, 5, 6, 7, 8, 10, 12}
C’ = { 4, 6, 8, 9, 10, 12}
and D’ = {5, 7, 9, 10, 11, 12, 13}

Set Concepts Exercise 13D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
If A = {4, 5, 6, 7, 8} and B = {6, 8, 10, 12}, find :
(i) A∪B
(ii) A∩B
(iii) A-B
(iv) B-A
Solution:
(i) A∪B
= [All the elements from set A and all the elements from set B]
= {4, 5, 6, 7, 8, 10, 12}
(ii) A∩B
= [elements common to both the sets A and B]
= {6, 8}
(iii) A-B
= [elements of set A which are not in set B]
= {4, 5, 7}
(iv) B-A
= [elements of set B which are not in set A]
= {10, 12}

Question 2.
If A = {3, 5, 7, 9, 11} and B = {4, 7, 10}, find:
(i) n(A)
(ii) n(B)
(iii) A∪B and n(A∪B)
(iv) A∩B and n(A∩B)
Solution:
(i) n(A) = (3, 5, 7, 9, 11) = 5
(ii) n(B) = (4, 7, 10) = 3
(iii) A ∪ B = {3, 4, 5, 7, 9, 10, 11} n(A ∪ B) = 7
(iv) A∩B = {6} n(A∩B)=l

Question 3.
If A = {2, 4, 6, 8} and B = {3, 6, 9, 12}, find:
(i) (A ∩ B) and n(A ∩ B)
(ii) (A – B) and n(A – B)
(iii) n(B)
Solution:
(i) (A ∩ B) = {2, 4, 8} n( A ∩ B) = 3
(ii) (A – B) and n(A – B)
⇒ (A – B) = (2, 4, 8)
⇒ n(A-B) = 3
(iii) n(B) = {3, 6, 9, 12} = 4

Question 4.
If P = {x : x is a factor of 12} and Q = {x: x is a factor of 16}, find :
(i) n(P)
(ii) n(Q)
(iii) Q – P and n(Q – P)
Solution:
(i) n(P) = Factors of 12 are
= 1, 2, 3, 4, 6, 12
∴ n(P) = 6
(ii) n(Q) = Factors of 16 are = 1. 2, 4, 8, 16
∴n(Q) = 5
(iii) Q – P and n(Q – P)
Elements of set P = {1, 2, 3, 4, 6, 12}
Elements of set Q = {1, 2, 4, 8, 16}
∴ Q – P = 8, 16
n(Q-P) = 2

Question 5.
M = {x : x is a natural number between 0 and 8) and N = {x : x is a natural number from 5 to 10}. Find :
(i) M – N and n(M – N)
(ii) N – M and n(N – M)
Solution:
Natural numbers between 0 and 8 M = {0, 1, 2, 3, 4, 5, 6, 7} and Natural numbers between 5 to 10 N = {6, 7, 8, 9, 10}
(i) M – N = {1, 2, 3, 4} and n(M – N) = 4
(ii) N – M = {8, 9, 10} and n (N – M) = 3

Question 6.
If A = {x: x is natural number divisible by 2 and x< 16} and B = {x:x is a whole number divisible by 3 and x < 18}, find :
(i) n(A)
(ii) n(B)
(iii) A∩B and n(A∩B)
(iv) n(A – B)
Solution:
(i) A = {x : x is natural number divisible-by 2 and x < 16}
A = {2, 4, 6, 8, 10, 12, 14}
n(A) = 7
(ii) B = {x: x is a whole number divisible by 3 and x < 18}
B = {3, 6, 9, 12, 15, 18}
n(B) = 6
(iii) A n B = {2, 4, 6, 8, 10, 12, 14} n {3, 6, 9, 12, 15, 18}
A∩B = {6,12} n(A ∩ B) = 2
(iv) A – B = {2, 4, 6, 8, 10, 12, 14} – {3, 6, 9, 12, 15, 18}
A-B = {2,4, 8, 10, 14} n(A – B) = 5

Question 7.
Let A and B be two sets such that n(A) = 75, M(B) = 65 and n(A ∩ B) = 45, find :
(i) n(A∪ B)
(ii) n(A – B)
(iii) n(B – A)
Solution:
n(A ∩ B)
n(A) = 75, n(B) = 65 and n(A ∩ B) = 45
(i) We know that,
n( A ∪B) = n(A) + n(B) – n( A ∩ B)
n(A ∪B) =75 + 65 – 45
n(A∪B) = 140-45 = 95
(ii) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A – B) = 75 – 45 = 30
(iii) We know that,
n(B – A) = n(B) – n(A ∩ B)
n(B – A) = 65 – 45 = 20

Question 8.
Let A and B be two sets such that n(A) = 45, n(B) = 38 and n(A ∪B) = 70, find :
(i) n(A∩B)
(ii) n(A-B)
(iii) n(B – A)
Solution:
n(A) = 45, n(B) = 38 and n(A∪ B) = 70
(i) We know that,
n(A ∩ B) = n(A) + M(B) – n(A ∪B)
n(A ∩ B) = 45 + 38 – 70 = 83 – 70 = 13
(ii) We know that,
n(A-B) = n(A ∪B)-n(B)
n(A – B) = 70 – 38 = 32
(iii) We know that,
n(B – A) = n(A ∪ B) – n(A)
n(B – A) = 70 – 45 = 25

Question 9.
Let n(A) 30, n(B) = 27 and n(A∪B) = 45, find :
(i) n(A∩B)
(ii) n(A-B)
Solution:
n(A) = 30, n(B) = 27 and n(A ∪ B) = 45
(i) We know that,
n(A ∩ B) = n( A) + n(B) – n( A∪ B)
n(A ∩ B) = 30 + 27 – 45
n(A ∩ B) = 57 – 45 = 12
(ii) We know that,
n(A-B) = n(A ∪B) – n(B)
n(A – B) = 45 – 27 = 18

Question 10.
Let n(A) = 31, n(B) = 20 and n(A ∩ B) = 6, find:
(i) n(A-B)
(ii) n(B – A)
(iii) n(A ∪B)
Solution:
n(A) = 31, n(B) = 20 and n(A ∩ B) = 6
(i) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A -B) = 31 -6 = 25
(ii) We know that,
n(B – A) = n(B) – n(A n B)
n(B – A) = 20 – 6 = 14
(iii) We know that,
n(A ∪B) = n(A) + n(B) – n(A ∪ B) n(A∪B) = 31 +20-6 = 45

 

Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

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POINTS TO REMEMBER

  1. Constants and Variables : The numbers which has fixed value is called constant and same at English alphabet which can be assigned any value according to the requirement is called variables.
  2. Term : A term is a number, (constant), a variable or a combination of numbers and variables.
  3. Algebraic Expression : An algebraic expression is a collection of one or more terms, which are separated from each other by addition (+) or subtraction (-) signs.
  4. Types of algebraic expressions :
    (i) Monomial : It has only one term
    (ii) Binomial : It has two terms
    (iii) Trinomial : It has three terms
    (iv) Multinomial : It has more than three terms
    (v) Polynomial : It has two or more than two terms.
    Note : An expression of the type \(\frac { 2 }{ 5 }\) does not form a monomial unless JC is not equal to zero.
  5. Product: When two or more quantities are multiplied together, the result is called their product.
  6. Factors : Each of the quantities (numbers or variables) multiplied together to form a term is called a factor of the given term.
  7. Co-efficient: In a monomial, any factor or group of factors of a term is called the co-efficient of the remaining part of the monomial.
  8. Degree of a monomial: The degree of a monomial is the exponent of its variable or the sum of the exponents of its variables.
  9. Degree of a polynomial: The degree of a polynomial is the degree of its highest degree term.
  10. Like and unlike terms : Terms having the same literal co-efficients or alphabetic letters are called like terms ; whereas the terms with different literal co-efficients are called unlike terms.
  11. Addition and subtraction : Addition and subtraction of only like terms is possible by adding or subtracting the numerical co-efficients.
  12. Multiplication and division :
    (A) Multiplication :
    (i) Multiplications of monomials.
    (a) Multiply the numerical co-efficient together
    (ii) Multiply the literal co-efficients separately together.
    (iii) Combine the like terms.
    (B) Division :
    (i) Dividing a polynomial by a monomial Divide each term of the polynomial by monomial and simplify each fractions.
    (ii) While dividing one polynomial by another polynomial ; arrange the terms of both the dividend and the divisior both in descending or in ascending order of their powers and then divide.

SOME IMPORTANT POINTS

TYPES OF BRACKETS:
The name of different types of brackets and the order in which they are removed is shown below:
(a) ____ ; Bar (Vinculum) bracket
(b) ( ); Circular bracket .
(c) { } ; Curly bracket and then
(d) [ ]; square bracket

EXERCISE 11 (A)

Question 1.
Separate constant terms and variable terms from tile following :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 1

Solution:

Constant is only 8 others are variables

Question 2.
Constant is only 8 others are variables
(i) 2x ÷ 15
(ii) ax+ 9
(iii) 3x2 × 5x
(iv) 5 + 2a-3b
(v) 2y – \(\frac { 7 }{ 3 }\) z÷x
(vi) 3p x q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a3 – 3ab2 x c

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 3

Question 3.
Write the coefficient of:
(i) xy in – 3axy
(ii) z2 in p2yz2
(iii) mn in -mn
(iv) 15 in – 15p2

Solution:
(i) Co-efficient of xy in – 3 axy = – 3a
(ii) Co-efficient of z2 in p2yz2 = p2y
(iii) Co-efficient of mn in – mn = – 1
(iv) Co-efficient of 15 in – 15p2 is -p2

Question 4.
For each of the following monomials, write its degree :
(i) 7y
(ii) – x2y
(iii) xy2z
(iv) – 9y2z3
(v) 3 m3n4
(vi) – 2p2q3r4

Solution:
(i) Degree of 7y = 1
(ii) Degree of – x2y = 2+1=3
(iii) Degree of xy2z = 1 + 2 + 1 = 4
(iv) Degree of – 9y2z3 = 2 + 3 = 5
(v) Degree of 3m3n4 = 3 + 4 = 7
(vi) Degree of – 2p2q3r4 = 2 + 3 + 4 = 9

Question 5.
Write the degree of each of the following polynomials :
(i) 3y3-x2y2 + 4x
(ii) p3q2 – 6p2q5 + p4q4
(iii) – 8mn6+ 5m3n
(iv) 7 – 3x2y + y2
(v) 3x – 15
(vi) 2y2z + 9yz3

Solution:
(i) The degree of 3y3 – x2y2+ 4x is 4 as x2
y2 is the term which has highest degree.
(ii) The degree of p3q2 – 6p2q5-p4q4 is 8 as p4 q4 is the term which has highest degree.
(iii) The degree of- 8mn6 + 5m3n is 7 as – 8mx6 is the term which has the highest degree.
(iv) The degree of 7 – 3x2 y + y2 is 3 as – 3x2y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1 as 3x is the term which is highest degree.
(vi) The degree of 2y2 z + 9y z3 is 4 as 9yz3 has the highest degree.

Question 6.
Group the like term together :
(i) 9x2, xy, – 3x2, x2 and – 2xy
(ii) ab, – a2b, – 3ab, 5a2b and – 8a2b
(iii) 7p, 8pq, – 5pq – 2p and 3p

Solution:
(i) 9x2, – 3x2 and x2 are like terms
xy and – 2xy are like terms
(ii) ab, – 3ab, are like terms,
– a2b, 5a2b, – 8a2b are like terms
(iii) 7p, – 2p and 3p are like terms,
8pq, – 5pq are like terms.

Question 7.
Write numerical co-efficient of each of the followings :
(i) y
(ii) -y
(iii) 2x2y
(iv) – 8xy3
(v) 3py2
(vi) – 9a2b3

Solution:
(i) Co-efficient of y = 1
(ii) Co-efficient of-y = – 1
(iii) Co-efficient of 2x2y is = 2
(iv) Co-efficient of – 8xy3 is = – 8
(v) Co-efficient of Ipy2 is = 3
(vi) Co-efficient of – 9a2b3 is = – 9

Question 8.
In -5x3y2z4; write the coefficient of:
(i) z2
(ii) y2
(iii) yz2
(iv) x3y
(v) -xy2
(vi) -5xy2z
Also, write the degree of the given algebraic expression.

Solution:
-5x3y2z4
(i) Co-efficient of z2 is -5x3y2z2
(ii) Co-efficient of y2 is -5x3z4
(iii) Co-efficient of yz2 is -5x3yz2
(iv) Co-efficient of x3y is -5yz4
(v) Co-efficient of -xy2 is 5x2z4
(vi) Co-efficient of -5xy2z is x2z3
Degree of the given expression is 3 + 2 + 4 = 9

EXERCISE 11 (B)

Question 1.
Fill in the blanks :
(i) 8x + 5x = ………
(ii) 8x – 5x =……..
(iii) 6xy2 + 9xy2 =……..
(iv) 6xy2 – 9xy2 = ………
(v) The sum of 8a, 6a and 5b = ……..
(vi) The addition of 5, 7xy, 6 and 3xy = …………
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ………
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = ……..
(x) 16x2 – 9x2 = and 25xy2 – 17xy2=………

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 4

Question 2.
Add :
(i)- 9x, 3x and 4x
(ii) 23y2, 8y2 and – 12y2
(iii) 18pq – 15pq and 3pq

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 5

Question 3.
Simplify :
(i) 3m + 12m – 5m
(ii) 7n2 – 9n2 + 3n2
(iii) 25zy—8zy—6zy
(iv) -5ax2 + 7ax2 – 12ax2
(v) – 16am + 4mx + 4am – 15mx + 5am

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 6

Question 4.
Add : 
(i) a + i and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii)- 3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 7

Question 5.
Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z- 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b-c and a + b + c.
(iii) 4x2+ 8xy – 2y2 and 8xy – 5y2 + x2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
(v) 5x2 – 2xy + 3y2 and – 2x2 + 5xy + 9y2
and 3x2 -xy- 4y2
(vi) a2 + b2 + 2ab, 2b2 + c2 + 2bc
and 4c2-a2 + 2ac
(vii) 9ax – 6bx + 8, 4ax + 8bx – 7
and – 6ax – 46x – 3
(viii) abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca
and 2ab – 3abc – 6ac
(ix) 4a2 + 5b2 – 6ab, 3ab, 6a2 – 2b2
and 4b2 – 5 ab
(x) x2 + x – 2, 2x – 3x2 + 5 and 2x2 – 5x + 7
(xi) 4x3 + 2x2 – x + 1, 2x3 – 5x2– 3x + 6, x2 + 8 and 5x3 – 7x

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 8
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 9
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 10
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 11

Question 6.
Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) – 4xand +7x
(iv) +4a and -7b
(v) x3+3x2y and 2y2
(vi) 11 and -by

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 12

Question 7.
The sides of a triangle are 2x + 3y, x + 5y and 7x – 2y, find its perimeter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 13

Question 8.
The two adjacent sides of a rectangle are 6a + 96 and 8a – 46. Find its, perimeter.

Solution
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 14

Question 9.
Subtract the second expression from the first:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 15

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 16
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 17

Question 10.
Subtract:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 18

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 19
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 20

Question 11.
Subtract – 5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 21
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 22

Question 12.
By how much does 8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x -3 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 23

Question 13.
What must be added to 2a3 + 5a – a2 – 6 to get a2 – a – a3 + 1 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 24

Question 14.
What must be subtracted from a2 + b2 + lab to get – 4ab + 2b2 ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 25

Question 15.
Find the excess of 4m2 + 4n2 + 4pover m2+ 3n2 – 5p2

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 26

Question 16.
By how much is 3x3 – 2x2y + xy2 -y3 less than 4x3 – 3x2y – 7xy2 +2y3

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 27

Question 17.
Subtract the sum of 3a2 – 2a + 5 and a2 – 5a – 7 from the sum of 5a2 -9a + 3 and 2a – a2 – 1

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 28

Question 18.
The perimeter of a rectangle is 28x3+ 16x2 + 8x + 4. One of its sides is 8x2 + 4x. Find the other side

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 30

Question 19.
The perimeter of a triangle is 14a2 + 20a + 13. Two of its sides are 3a2 + 5a + 1 and a2 + 10a – 6. Find its third side.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 31.

Question 20.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 32

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 33

Question 21.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 34

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 35

Question 22.
Simplify:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 36

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 37
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 38
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 39

EXERCISE 11 (C)

Question 1.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 40

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 41
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 42
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 43

Question 2.
Copy and complete the following multi-plications :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 44

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 45
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 46

Question 3.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 47
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 49

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 50
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 51
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 52
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 53
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 54

Question 4.
Evaluate:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 55

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 56

Question 5.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 57

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 58
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 59

Question 6.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 60
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 61
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 62

Question 7.
Multiply:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 63

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 64
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 65

EXERCISE 11 (D)

Question 1.
Divide:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 66
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 67

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 68
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 69
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 70

Question 2.
Divide :
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 71

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 72
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 73
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 74
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 75
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 76
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 77
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 78
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 79

Question 3.
The area of a rectangle is 6x2– 4xy – 10y2 square unit and its length is 2x + 2y unit. Find its breadth

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 80
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 81

Question 4.
The area of a rectangular field is 25x2 + 20xy + 3y2 square unit. If its length is 5x + 3y unit, find its breadth, Hence find its perimeter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 83

Question 5.
Divide:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 84

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 85
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 86

EXERCISE 11 (E)

Simplify
Question 1.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 87

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 88

Question 2.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 89

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 90

Question 3.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 91

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 92

Question 4.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 93

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 94

Question 5.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 95

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 96

Question 6.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 97

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 98

Question 7.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 99

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 100

Question 8.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 101

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 102

Question 9.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 104

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 105

Question 10.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 106

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 107

Question 11.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 108

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 109

Question 12.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 110

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 111
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 112

Question 13.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 113

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 114

Question 14.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 115

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 116

Question 15.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 118

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 119

Question 16.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 120

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 121

Question 17.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 122

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 123

Question 18.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 124

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 125

Question 19.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 126

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 127

Question 20.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 128

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 129

Question 21.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 130

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 131

Question 22.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 132

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 133

Question 23.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 134

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 135

Question 24.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 136

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 137

Question 25.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 138

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 139

Question 26.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 140

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 141
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 142

EXERCISE 11 (F)

Enclose the given terms in brackets as required :

Question 1.
 x – y – z = x-{…….)

Solution:
x – y – z = x – (y + z)

Question 2.
x2 – xy2 – 2xy – y2 = x2 – (…….. )

Solution:
x– xy– 2xy – y2
= x2 – (xy2 + 2xy + y2)

Question 3.
4a – 9 + 2b – 6 = 4a – (…….. )

Solution:
4a – 9 + 2b – 6
= 4a – (9 – 2b + 6)

Question 4.
x2 -y2 + z2 + 3x – 2y = x2 – (…….. )

Solution:
x2 – y2 + z2 + 3x – 2y
= x2 – (y2 – z2 – 3x + 2y)

Question 5.
– 2a2 + 4ab – 6a2b2 + 8ab2 = – 2a (……… )

Solution:
 – 2a2 + 4ab – 6a2b2 + 8ab2
= – 2a (a – 2b + 3ab2 – 4b2)

Simplify :

Question 6.
2x – (x + 2y- z)

Solution:
2x-(x + 2y-z) = 2x – x – 2y + z
= x – 2y + z

Question 7.
p + q – (p – q) + (2p – 3q)

Solution:
p + q – (p – q) + (2p- 3q)
= p + q – p + q + 2p – 3q = 2p – q

Question 8.
9x – (-4x + 5)

Solution:
9x – (-4x + 5) = 9x + 4x – 5
= 13x- 5

Question 9.
6a – (- 5a – 8b) + (3a + b)

Solution:
6a – (- 5a – 8b) + (3a + b)
= 6a + 5a + 8b + 3a + b
= 6a + 5a + 3a + 8b + b
= 14a + 9b

 Question 10.
(p – 2q) – (3q – r)

Solution:
(p-2q) – (3q – r) =p – 2q – 3q + r =p – 5q + r

Question 11.
9a (2b – 3a + 7c)

Solution:
9a (2b – 3a + 7c)
= 18ab – 27a2 + 63ca

Question 12.
-5m (-2m + 3n – 7p)

Solution:
-5m (-2m + 3n- 7p)
= – 5m x (-2m) + (-5m) (3n) – (-5m) (7p)
= 10m2 – 15mn + 35 mp.

Question 13.
-2x (x + y) + x2

Solution:
– 2x (x + y) + x2
= -2x x x + (-2x)y + x2
= – 2x2 – 2xy + x2
= – 2x2 + x2 – 2xy = – x2 – 2xy

Question 14.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 143

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 144

Question 15.
8 (2a + 3b – c) – 10 (a + 2b + 3c)

Solution:
8 (2a + 3b -c)- 10 (a + 2b + 3c)
= 16a + 24b – 8c – 10a – 20b- 30c
= 16a – 10a + 24b – 20b – 8c – 30c
= 6a + 4b – 38c

Question 16.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 145

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 146

Question 17.
5 x (2x + 3y) – 2x (x – 9y)

Solution:
5x (2x + 3y) – 2x (x – 9y)
= 10x2 + 15xy – 2x2 + 18xy
= 10x– 2x2+ 15xy+ 18xy
= 8x2 + 33 xy

Question 18.
a + (b + c – d)

Solution:
a + (b + c – d) = a + (b + c – d)
= a + b + c – d

Question 19.
5 – 8x – 6 – x

Solution:
5 – 8x – 6 – x
= 5 – 6 –  8x – x
= -1 -7x

Question 20.
2a + (6- \(\overline { a-b }\) )

Solution:
2a + (6 – \(\overline { a-b }\) )
= 2a + (b – a + b)
= 2a + b – a + b
= a + 2b

Question 21.
3x + [4x – (6x – 3)]

Solution:
3x + [4x – (6x – 3)]
= 3x + [4x – 6x + 3]
= 3x + 4x – 6x + 3
= 3x + 4x – 6x + 3
= 7x – 6x + 3= x + 3

Question 22.
5b – {6a + (8 – b – a)}

Solution:
5b- {6a + 8- 6-a}
= 5b – 6a – 8 + b + a
= -6a + a + 5b +b – 8
= -5a + 6b-8

Question 23.
2x-[5y- (3x -y) + x]

Solution:
2x – [5y- (3x – y) + x]
= 2x – {5y – 3x +y + x}
= 2x – 5y + 3x -y – x
= 2x + 3x – x – 5y – y
= 4x – 6y

Question 24.
6a – 3 (a + b – 2)

Solution:
6a – 3 (a + b – 2)
=
6a – 3a – 3b + 6
= 3a -3b + 6

Question 25.
8 [m + 2n-p – 7 (2m -n + 3p)]

Solution:
8 [m + 2n-p -1 (2m – n + 3p)]
8 [m + 2n-p- 14m + 7n-21p]
= 8m+ 16n -8p- 112m + 56n – 168p
= 8m – 112m + 16n + 56n -8p – 168p
= -104m + 72n – 176p

Question 26.
{9 – (4p – 6q)} – {3q – (5p – 10)}

Solution:
{9 – {4p – 6q)} – {3q – (5p – 10)}
{9 – 4p + 6q} – {3q -5p+ 10}
= 9 – 4p + 6q – 3q + 5p – 10
= 9 – 4p +
5p + 6q – 3q – 10
= p + 3q – 1

Question 27.
2 [a – 3 {a + 5 {a – 2) + 7}]

Solution:
2 [a – 3 {a + 5 {a – 2) + 7}]
= 2 [a- 3 {a + 5a- 10 + 7}]
= 2 [a -3a- 15a + 30 -21]
= 2a-6a- 30a + 60-42
= 2a- 36a + 60-42
= -34a + 18

Question 28.
5a – [6a – {9a – (10a – \(\overline { 4a-3a }\)  )}]

Solution:
5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – 10a + 4a – 3a}]
= 5a- [6a – 9a + 10a – 4a + 3a]
= 5a – 6a + 9a – 10a + 4a – 3a
= 5a + 9a + 4a – 6a – 10a – 3a
= 18a – 19a = – a

Question 29.
9x + 5 – [4x – {3x – 2 (4x – 3)}]

Solution:
9x + 5 – [4x – {3x – 2 (4x – 3)}]
= 9x + 5 – [4x – {3x – 8x + 6}]
= 9x + 5 – [4x – 3x + 8x – 6]
= 9x + 5-4x + 3x-8x + 6
= 9x + 3x-4x-8x + 5 + 6
= 12x- 12x+ 11 = 11

Question 30.
(x + y – z)x + (z + x – y)y – (x + y – z)z

Solution:
(x + y – z)x + (z + x -y )y – (x + y -z)z
= x+ xy – zx + yz + xy -y– zx – yz + z2
= x2 -y2 + z2 + 2xy – 2zx

Question 31.
-1 [a-3 {b -4 (a-b-8) + 4a} + 10]

Solution:
– 1 [a – 3 {b – 4(a – b – 8) + 4a} + 10]
= -1 [a-3 {b-4{a-b-8) + 4a} + 10]
= -1[a-3 {b-4a + Ab +32 + 4a} + 10]
= -1 [a-3b+ 12a- 126-96- 12a + 10]
= -a + 3b – 12a + 12b + 96 + 12a – 10
= -a-12a + 12a+ 3b+ 12b-96-10
= – a + 15b – 106

Question 32.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 148

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 149

Question 33.
10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}

Solution:

10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}
= 10 – {4a – (7 – a + 5) – (5a – 1 – a)}
= 10- {4a -(12 -a) -(4a- 1)}
= 10 – {4a – 12 + a- 4a + 1}
= 10 – 4a + 12 – a + 4a- 1
= 10 + 12 – 1 – 4a – a + 4a
= 21 -a

Question 34.
7a- [8a- (11a-(12a- \(\overline { 6a-5a }\))}]

Solution:
7a – [8a – {1 la – (12a \(\overline { 6a-5a }\))}]
= 7a-[8a-{11a-(12a-6a + 5a)}]
= 7a -[8a -{11a -(17a -6a)}]
= 7a- [8a- {11a-(11a)}]
= 7a- [8a- {11a- 11a}]
= 7a – 8a = -a

Question 35.
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 150

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Fundamental Concepts (Including Fundamental Operations) image - 151

Question 36.
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))

Solution:
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))
= x – (3y – 4z + 3x  + 2z -5y + 7x)
= x-(-2y-2z+10x)
= x + 2y + 2z- 10x
= -9x + 2y + 2z

 

Selina Concise Mathematics class 7 ICSE Solutions – Percent and Percentage

Selina Concise Mathematics class 7 ICSE Solutions – Percent and Percentage

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. The cent means hundred. Therefore percent means after hundred and notation % is used for it.
  2. To express an ordinary given statement as percent.
    (i) Express the given statement as a fraction.
    (ii) Convert this fraction into an equivalent fraction with denominator 100.
    Therefore to express a fraction or a decimal as percent, multiply it by 100.
  3. To Express-One quantity as a percent of the other.
    (i) If necessary, convert with the quantitities into the same units.
    (ii) From the fraction with the number to be compared as numerator and the number with which it is to be compared as denominator.
    (iii) Multiply the fraction obtained by 100 and at the same time write the percent sign (%).

EXERCISE 8 (A)

Question 1.
Express each of the following as percent :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 1

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 2

Question 2.
Express the following percentages as fractions and as decimal numbers :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 3

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 5

Question 3.
What percent is :
(i) 16 hours of 2 days ?
(ii) 40 paisa of Rs. 2 ?
(iii) 25 cm of 4 metres
(iv) 600 gm of 5 kg ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 6
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 7

Question 4.
Find the value of:
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 8

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 10

Question 5.
In a class of 60 children, 30% are girls. How many boys are there ?

Solution :
Total children = 60,
Girls = 30%
∴Total girls = 30% of 60 = 60 x \(\frac { 30 }{ 100 }\)= 18
∴ No. of boys = 60 – 18 = 42

Question 6.
In an election, two candidates A and B contested. A got 60% of the votes. The total votes polled were 8000. How many votes did each get ?

Solution :
Total number of votes polled = 8000
A got 60% of the votes
A got total votes = 60% of 8000 = 8000 x \(\frac { 60 }{ 100 }\) = 4800
∴ B got total votes = 8000 – 4800 = 3200

Question 7.
A person saves 12% of his salary every month. If his salary is ₹2,500, find his expenditure.

Solution :
Total salary = ₹2500
Saving = 12% of the salary
∴ Total savings = 12% of ₹2500
= ₹2500 x\(\frac { 12 }{ 100 }\) = ₹300
∴Total expenditure = ₹2500 – ₹300 = ₹2200

Question 8.
Seeta got 75% marks out of a total of 800. How many marks did she lose ?

Solution :
Total marks = 800
Marks Seeta got = 75% of total marks
∴ Total marks Seeta got = 75% of 800
= 800 x \(\frac { 75 }{ 100 }\) = 600
∴ Marks Seeta lose = 800 – 600 = 200

Question 9.
A shop worth ₹25,000 was insured for 95% of its value. How much would the owner get in case of any mishappening ?

Solution :
Value of shop =₹25,000
Insured amount = 95% of total value
=95% of ₹25,000
= ₹25,000 x \(\frac { 95 }{ 100 }\)
= ₹ 23,750

Question 10.
A class has 30 boys and 25 girls. What is the percentage of boys in the class ?

Solution :
No. of boys = 30
No. of girls = 25
Total number of children = 30 + 25 = 55
∴Percentage of boys in the class
= \(\frac { 30 }{ 55 }\) x 100
= \(\frac { 600 }{ 11 }\)=54\(\frac { 6 }{ 11 }\) %

Question 11.
Express :
(i) 3 \(\frac { 2 }{ 5 }\) as a percent
(ii) 0.0075 as percent
(iii) 3 : 20 as percent
(iv) 60 cm as percent of 1 m 25 cm
(v) 9 hours as a percent of 4 days.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 11
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 12

Question 12.
(i) Find 2% of 2 hours 30 min.
(ii) What percent of 12 kg is 725 gm?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 13

EXERCISE 8 (B)

Question 1.
Deepak bought a basket of mangoes containing 250 mangoes 12% of these were found to be rotten. Of the remaining, 10% got crushed. How many mangoes were in good condition ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 14

Question 2.
In a Maths Quiz of 60 questions, Chandra got 90% correct answers and Ram got 80% correct answers. How many correct answers did each give ?
What percent is Ram’s correct answers to Chandra’s correct answers ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 15
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 16

Question 3.
In an examination, the maximum marks are 900. A student gets 33% of the maximum marks and fails by 45 marks. What is the passing mark ? Also, find the pass percentage.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 17

Question 4.
In a train, 15% people travel in first class, 35% travel in second class. The balance travel in the A.C. class ? Calculate the percentage of A.C. class travellers ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 18

Question 5.
A boy eats 25% of the cake and gives away 35% of it to his friends. What percent of the cake is still left with him ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 19

Question 6.
What is the percentage of vowels in the English alphabet ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 20

Question 7.
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 21

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 22

Question 8.
The money spent on the repairs of a house was 1% of its value. If the repair, costs Rs. 5,000, find the cost of the house.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 23

Question 9.
In a school out of300 students, 70% are girls and 30% are boys. If 30 girls leave and no new boy is admitted, what is the new percentage of girls in the school ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 24

Question 10.
Kumar bought a transistor for Rs. 960. He paid 12 \(\frac { 1 }{ 2 }\) % cash money. The rest he agreed to pay in 12 equal monthly instalments. How much will he pay each month ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 25
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 26

Question 11.
An ore contains 20% zinc. How many kg of ore will be required to get 45 kg of zinc ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 27

EXERCISE 8 (C)

Question 1.
The salary of a man is increased from Rs. 600 per month to Rs. 850 per month. Express the increase in salary as percent.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 28

Question 2.
Increase :
(i) 60 by 5%
(ii) 20 by 15%
(iii) 48 by 121 %
(iv) 80 by 140%
(v) 1000 by 3.5%

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 30

Question 3.
Decrease :
(i)80 by 20%
(ii) 300 by 10%
(iii) 50 by 12.5%

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 31

Question 4.
What number :
(i) When increased by 10% becomes 88 ?
(ii) When increased by 15% becomes 230 ?
(iii) When decreased by 15% becomes 170 ?
(iv) When decreased by 40% becomes 480 ?
(v) When increased by 100% becomes 100 ?
(vi) When decreased by 50% becomes 50 ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 32
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 33
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 34

Question 5.
The price of a car is lowered by 20% to Rs. 40,000. What was the original price ? Also, find the reduction in price.

Solution :
Let original price of the car = Rs. 100
Reduction = 20%’ = Rs. 20
∴ Reduced price = Rs. 100 – 20 = Rs. 80
If reduced price is Rs. 80, then original price = Rs. 100
and if reduced price is Rs. 40,000 then original price = \(\frac { Rs.100 x 40000 }{ 80 }\)
= Rs. 50,000
and reduction = Rs. 50000 – Rs. 40000
= Rs. 10,000

Question 6.
If the price of an article is increased by 25%, The increase is Rs. 10. Find the new price.

Solution :
Let the price of an article = Rs. 100
Increase = 25%
∴Increase = Rs. 25
If an increased price = Rs. 100 + 25 = Rs. 125
If increase is Rs. 25 then new price = Rs. 125
and if increase is Rs. 10, then new price = Rs. \(\frac { 125 x 10 }{ 25 }\)
= Rs. 50

Question 7.
If the price of an article is reduced by 10%, the reduction is Rs. 40. What is the old price ?

Solution :
Let the original (old) price = Rs. 100
Reduction = 10% = Rs. 10
∴If reduction is Rs. 10, then old price = Rs. 100
and if reduction is Rs. 40, then old price = Rs.\(\frac { 100 x 40 }{ 10 }\) = Rs. 400

Question 8.
The price of a chair is reduced by 25%. What is the ratio of:
(i) Change in price to the old price.
(ii) Old price to the new price.

Solution :
Let old (original) price of a chair = Rs. 100
Reduction = 25% = Rs. 25
∴Reduced price = Rs. 100 – Rs. 25 = Rs. 75
(i) Ratio between change in price and old price = 25 : 100
= 1:4 (Dividing by 25)
(ii) Ratio between old price and new price = 100 : 75
= 4:3 (Dividing by 25)

Question 9.
If x is 20% less than y, find :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 35

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 36

Question 10.
If x is 30% more than y; find :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 37

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 38

Question 11.
The weight of a machine is 40 kg. By mistake it was weighed as 40.8 kg. Find the error percent.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 39

Question 12.
From a cask, containing 450 litres of petrol, 8% of the petrol was lost by leakage and evaporation. How many litres of petrol was left in the cask ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 40

Question 13.
An alloy consists of 13 parts of copper, 7 parts of zinc and 5 parts of nickel. What is the percentage of each metal in the alloy?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 49

Question 14.
In an examination, first division marks are 60%. A student secures 538 marks and misses the first division by 2 marks. Find the total marks of the examination.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 42

Question 15.
Out of 1200 pupils in a school, 900 are boys and the rest are girls. If 20% of the boys and 30% of the girls wear spectacles, find :
(i) how many pupils in all, wear spectacles ?
(ii) what percent of the total number of pupils wear spectacles ?

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 43

Question 16.
Out of 25 identical bulbs, 17 are red, 3 are black and the remaining are yellow. Find the difference between the numbers of red and yellow bulbs and express this difference as percent.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 44

Question 17.
A number first increases by 20% and then decreases by 20%. Find the percentage increase or decrease on the whole.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 45
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 46

Question 18.
A number is first decreased by 40% and then again decreased by 60%. Find the percentage increase or decrease on the whole.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 47

Question 19.
If 150% of a number is 750, find 60% of this number.

Solution :
Selina Concise Mathematics class 7 ICSE Solutions - Percent and Percentage image - 48

Selina Concise Mathematics class 7 ICSE Solutions – Triangles

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 15 Triangles

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 15 Triangles

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

POINTS TO REMEMBER
1. Definition of a triangle : A closed figure, having 3 sides, is called a triangle and is usually denoted by the Greek letter ∆ (delta).
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -1
The figure, given alongside, shows a triangle ABC (∆ABC) bounded by three sides AB, BC and CA.
Hence it has six elements : 3 angles and 3 sides.

2. Vertex : The point, where any two sides of a triangle meet, is called a vertex.
Clearly, the given triangle has three vertices; namely : A, B and C. [Vertices is the plural of vertex]

3. Interior angles : In ∆ABC (given above), the angles BAC, ABC and ACB are called its interior angles as they lie inside the ∆ ABC. The sum of interior angles of a triangle is always 180°.

4. Exterior angles : When any side of a triangle is produced the angle so formed, outside the triangle and at its vertex, is called its exterior angle.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -2
e.g. if side BC is produced to the point D; then ∠ACD is its exterior angle. And, if side AC is produced to the point E, then the exterior angle would be ∠BCE.
Thus. at every vertex, two exterior angles can be formed and that these two angles being vertically opposite angles, are always equal.
Make the following figures clear :
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -3
5. Interior opposite angles : When any side of a triangle is produced; an exterior angle is formed. The two interior angles of this triangle, that are opposite to the exterior angle formed; are called its interior opposite angles.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -4
In the given figure, side BC of ∆ABC is produced to the point D, so that the exterior ∠ACD is formed. Then the two interior opposite angles are ∠B AC and ∠ABC.
6. Relation between exterior angle and interior opposite angles :
Exterior angle of a triangle is always equal to the
sum of its two interior opposite angles.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -5
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B

7. CLASSIFICATION OF TRIANGLES
(A) With regard to their angles :
1. Acute angled triangle : It is a triangle, whose each angle is acute i.c. each angle is less than 90°.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -6

2. Right angled triangle : It is a triangle, whose one angle is a right angle i.e. equal to 90”.
The figure, given alongside, shows a right angled triangle XYZ as ∠XYZ = 90°
Note : (i) One angle of a right triangle is 90° and the other two angles of it are acute; such that their sum is always 90”.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -7
In ∆XYZ, given above, ∠Y = 90° and each of ∠X and ∠Z is acute such that ∠X + ∠Z = 90°. .
(ii)In a right triangle, the side opposite to the right angle is largest of all its sides and is called the hypotenuse. In given right angled ∆ XYZ side XZ is its hypotenuse

3.Obtuse angled triangle : If one angle of a triangle is 1
obtuse, it is called an obtuse angled triangle.
Note : In case of an obtuse angled triangle, each of the other two angles is always acute and their sum is less than 90”.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -8
(B) With regard to their sides :
(1) Scalene triangle: If all the sides of a triangle are unequal, it is called a scalene triangle.
In a scalene triangle; all its angles are also unequal.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -9
(2) Isosceles triangle : If atleast two sides of a triangle are equal, it is called an isosceles triangle.
In ∆ ABC, shown alongside, side AB = side AC.
∴∆ ABC is an isosceles triangle.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -10
Note : (i) The angle contained by equal sides i.e. ∠BAC is called the vertical angle or the angle of vertex.
(ii) The third side (i.e. the unequal side) is called the base of the isosceles triangle.
(iii) The two other angles (i.e. other than the angle of vertex) are called the base angles of the triangle.

IMPORTANT PROPERTIES OF AN ISOSCELES TRIANGLE
The base angles i.e. the angles opposite to equal sides of an isosceles triangle are always equal.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -11
In given triangle ABC,
(i) If side AB = side BC; then angle opposite to AB = angle opposite to BC i.e. ∠C = ∠A.
(ii) If side BC = side AC; then angle opposite to BC = angle opposite to AC i.e. ∠A = ∠B and so on.
Conversely : If any two angles of a triangle are equal; the sides opposite to these angles are also equal i.e. the triangle is isosceles.
Thus in ∆ ABC,
(i) If ∠B = ∠C => side opposite to ∠B = side opposite to ∠C i.e. side AC = side AB.
(ii) If ∠A = ∠B => side BC = side AC and so on.

(3) Equilateral triangle :
If all the sides of a triangle are equal, it is called an equilateral triangle.
In the given figure, A ABC is equilateral, because AB = BC = CA.
Also, all the angles of an equilateral triangle are equal to each other and so each angle = 60°. [∵60° + 60° + 60° = 180°]
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -13
Since, all the angles of an equilateral triangle are equal, it is also known as equiangular triangle. Note : An equilateral triangle is always an isosceles triangle, but its converse is not always true.

(4) Isosceles right angled triangle : If one angle of an isosceles triangle is 90°, it is called an isosceles right angled triangle.
In the given figure, ∆ ABC is an isosceles right angled triangle, because : ∠ ACB = 90° and AC = BC.
Here, the base is AB, the vertex is C and the base angles are ∠BAC and ∠ABC, which are equal.
Since, the sura of the angles of a triangle = 180″
∴∠ABC = ∠BAC = 45 [∵45° + 45° + 90° = 180°]

Triangles Exercise 15A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Stale, if the triangles are possible with the following angles :
(i) 20°, 70° and 90°
(ii) 40°, 130° and 20°
(iii) 60°, 60° and 50°
(iv) 125°, 40° and 15°
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -14
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -15
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -16

Question 2.
If the angles of a triangle are equal, find its angles.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -17

Question 3.
In a triangle ABC, ∠A = 45° and ∠B = 75°, find ∠C.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -18

Question 4.
In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -19

Question 5.
Calculate the unknown marked angles in each figure :
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -20
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -21

Question 6.
Find the value of each angle in the given figures:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -22
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -23
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -24

Question 7.
Find the unknown marked angles in the given figure:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -25
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -26
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -27

Question 8.
In the given figure, show that: ∠a = ∠b + ∠c
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -28
(i) If ∠b = 60° and ∠c = 50° ; find ∠a.
(ii) If ∠a = 100° and ∠b = 55° : find ∠c.
(iii) If ∠a = 108° and ∠c = 48° ; find ∠b.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -29
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -30

Question 9.
Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -31

Question 10.
One angle of a triangle is 60°. The, other two angles are in the ratio of 5 : 7. Find the two angles.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -32
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -38

Question 11.
One angle of a triangle is 61° and the other two angles are in the ratio 1\(\frac { 1 }{ 2 }\) : 1 \(\frac { 1 }{ 3 }\). Find these angles.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -39

Question 12.
Find the unknown marked angles in the given figures :
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -40
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -41
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -42
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -43.

Triangles Exercise 15B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the unknown angles in the given figures:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -44
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -45
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -46
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -47
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -48

Question 2.
Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures :
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -49
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -50
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -51
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -52
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -53

Question 3.
The angle of vertex of an isosceles triangle is 100°. Find its base angles.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -54
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -55

Question 4.
One of the base angles of an isosceles triangle is 52°. Find its angle of vertex.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -56

Question 5.
In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -57
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -58

Question 6.
The vertical angle of an isosceles triangle is 15° more than each of its base angles. Find each angle of the triangle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -59

Question 7.
The base angle of an isosceles triangle is 15° more than its vertical angle. Find its each angle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -60

Question 8.
The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -61
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -62

Question 9.
The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -63

Question 10.
In the given figure, BI is the bisector of∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -64
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -65
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -66

Question 11.
In the given figure, express a in terms of b.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -67
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -68
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -68Selina Concise Mathematics class 7 ICSE Solutions - Triangles ima

Question 12.
(a) In Figure (i) BP bisects ∠ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects ∠ABC and∠ADB = 70°.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -70
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -71
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -72
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -73

Question 13.
In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.
Find, in each case : (i) ∠ABE(ii) ∠BAE
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -74
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -75
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -76

Question 14.
In ∆ ABC, BA and BC are produced. Find the angles a and h. if AB = BC.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -77
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -78

Triangles Exercise 15C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Construct a ∆ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm
Solution:
(i) Steps of Construction :
(i) Draw a line segment BC = 4 cm.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -79
(ii) With centre B and radius 6 cm draw an arc.
(iii) With centre C and radius 5.5 cm, draw another arc intersecting the First are at A.
(iv) Join AB and AC. ∆ABC is the required triangle.
(ii) Steps of Construction :
(i) Draw a line segment CB = 6 5 cm
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -80
(ii) With centre C and radius 4.2 cm draw an arc.
(iii) With centre B and radius 5.1 cm draw another arc intersecting the first arc at A.
(iv) Join AC and AB.
∆ ABC is the required triangle.
(iii) Steps of Construction :
(i) Draw a line segment BC = 4 cm.
(ii) With centre B and radius 3.5 cm, draw an arc
(iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -81
(iv) Join AB and AC.
∆ ABC is the required triangle.

Question 2.
Construct a A ABC such that:
(i) AB = 7 cm, BC = 5 cm and ∠ABC = 60°
(ii) BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°
(iii) AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°
Solution:
(i) Steps of Construction :
(i) Draw a line segment AB = 7 cm.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -82
(ii) At B, draw a ray making an angle of 60° and cut off BC = 5 cm
(iii) Join AC,
∆ABC is the required triangle.
(ii) Steps of Construction :
(i) Draw a line segment BC = 6 cm.
(ii) At C, draw a ray making an angle of 75° and cut off CA = 5.7 cm.
(iii) JoinAB
∆ ABC is the required triangle.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -83
(iii) Steps of Construction :
(i) Draw a line segment AB = 6.5 cm
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -84
(ii) At A, draw a ray making an angle of 45° and cut off AC = 5.8 cm
(iii) JoinCB.
∆ ABC is the required triangle.

Question 3.
Construct a ∆ PQR such that :
(i) PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.
(ii) QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR.
(iii) PR = 5.8 cm, ∠P = 60° and ∠R = 45°.
Measure ∠Q and verify it by calculations
Solution:
(i) Steps of Construction:
(i) Draw a line segment PQ = 6 cm.
(ii) At P, draw a ray making an angle of 45°
(iii) At Q, draw another ray making an angle of 60° which intersects the first ray at R.
∆ PQR is the required triangle.
On measuring ∠R, it is 75°.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -85
(ii) Steps of Construction :
(i) Draw a line segment QR = 44 cm.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -86
(ii) At Q, draw a ray making an angle of 75°
(iii) At R, draw another arc making an angle of 30° ; which intersects the first ray at R
∆ PQR is the required triangle.
On measuring the lengths of PQ and PR, PQ = 2.1 cm and PR = 4. 4 cm.
(iii) Steps of Construction :
(i) Draw a line segment PR = 5.8 cm
(ii) At P, construct an angle of 60°
(iii) At R, draw another angle of 45° meeting each other at Q.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -87
∆ PQR is the required triangle. On measuring ∠Q, it is 75°
Verification : We know that sum of angles of a triangle is 180°
∴∠P + ∠Q + ∠R = 180°
⇒ 60° + ∠Q + 45° = 180°
⇒ ∠Q + 105° = 180°
⇒ ∠Q = 180° – 105° = 75°.

Question 4.
Construct an isosceles A ABC such that:
(i) base BC = 4 cm and base angle = 30°
(ii) base AB = 6-2 cm and base angle = 45°
(iii) base AC = 5 cm and base angle = 75°.
Measure the other two sides of the triangle.
Solution:
(i) Steps of Construction :
We know that in an isosceles triangle base angles are equal.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -88
(i) Draw a line segment BC = 4 cm.
(ii) At B and C, draw rays making an angle of 30° each intersecting each other at A.
∆ ABC is the required triangle.
On measuring the equal sides each is 2.5 cm (approx.) in length.
(ii) Steps of Construction :
We know that in an isosceles triangle, base angles are equal.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -89
(i) Draw a line segment AB = 6.2 cm
(ii) At A and B, draw rays making an angle of 45° each which intersect each other at C.
∆ABC is the required triangle.
On measuring the equal sides, each is 4.3 cm (approx.) in length.
(iii) Steps of Construction :
We know that base angles of an isosceles triangles are equal.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -90
(i) Draw a line segment AC = 5cm.
(ii) At A and C, draw rays making an angle of 75° each which intersect each other at B.
∆ ABC is the required triangle.
On measuring the equal sides, each is 9.3 cm in length.

Question 5.
Construct an isosceles ∆ABC such that:
(i) AB = AC = 6.5 cm and ∠A = 60°
(ii) One of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30°. Measure ∠A and ∠C.
Solution:
(i) Steps of Construction :
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -91
(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 60°.
(iii) Cut off AC = 6.5 cm
(iv) JoinBC.
∆ABC is the required triangle.
(ii) Steps of Construction :
(i) Draw a line segment AB = 6 cm
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -92
(ii) At A, construct an angle equal to 45°
(iii) Cut off AC = 6 cm
(iv) JoinBC.
∆ ABC is the required triangle.
On measuring, ∠B and ∠C, each is equal 1° to, 67\(\frac { 1 }{ 2 }\)°
(iii) Steps of Construction :
(i) Draw a line segment BC = 5.8 cm
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -93
(ii) At B, draw a ray making an angle of 30°.
(iii) Cut off BA = 5.8 cm
(iv) Join AC.
∆ ABC is the required triangle On measuring ∠C and ∠A, each is equal to 75°.

Question 6.
Construct an equilateral A ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.
Solution:
(i) Steps of Construction :
(i) Draw a line segment AB = 5 cm.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -94
(ii) With centres A and B and radius 5 cm each, draw two arcs intersecting each other at C.
(iii) Join AC and BC ∆ABC is the required triangle.
(iv) Draw the perpendicular bisectors of sides AC and BC which intersect each other at P-
(v) Join PA, PB and PC.
On measuring, each is 2.8 cm.
(ii) Steps of Construction :
(i) Draw a line segment AB = 6 cm.
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -95
(ii) At A and B as centre and 6 cm as radius draw two arcs intersecting each other at C.
(iii) Join AC and BC.
∆ABC is the required triangle.

Question 7.
(i) Construct a ∆ ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles ∆PQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -96
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -97
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -98
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -99

Question 8.
(i) Construct a ∆ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles ∆ MNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral ∆DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a ∆ PQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -100
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -101
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -102
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -103
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -104
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -105
Selina Concise Mathematics class 7 ICSE Solutions - Triangles image -106

 

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations (Including Word Problems)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 12 Simple Linear Equations (Including Word Problems)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. Equation: An equation is a statement which states that two expressions are equal.
  2. To solve an equation means to find the value of the variable (unknown quantity) used in it.
    Note : An equation remains unchanged if
    (i) the same number is added to each side of the equation. .
    (ii) the same number is subtracted from each side of the equation.
    (iii) the same number is multiplied to each side of the equation.
    (iv) Each side of the equation is divided by the same non-zero number.
    (v) In transposing any term of an equation from one side to another, then its sign is reversed is
    (a) from positive to negative and from negative to positive
    (b) from multiplication to division and from division to multiplication.
  3. In equation :
    It is a statement of inequality between two expressions involving a single variable with the highest power one.
  4. Replacement set
    For a given inequation, the set from which the values of its variable are taken is called the replacement set or domain of the variable.
  5. Solution set
    It is the subset of the replacement set, consisting of those values of the variable which satisfy the given inequation
  6. Properties of inequations
    Adding, subtracting, multiplying or dividing by the same positive number to each side of an inequation does not change the inequality but multiplying or dividing by a negative number to each side of an inequation, it changes the inequality.

Simple Linear Equations Exercise 12A – Selina Concise Mathematics Class 7 ICSE Solutions

Solve the following equations :

Question 1.
x + 5 = 10

Solution:
x + 5 = 10
⇒ x=10 -5 = 5

Question 2.
2 + y=7

Solution:
2 + y = 7
⇒ = 7- 2 = 5

Question 3.
a – 2 = 6

Solution:
a -2 =6
⇒a = 6 + 2 = 8

Question 4.
x – 5 = 8

Solution:
x-5 =8
⇒ x = 8 +5 = 13

Question 5.
5 – d= 12

Solution:
5-d = 12
⇒ -d = 12-5 =7
⇒ d = – 7

Question 6.
3p = 12

Solution:
3p = 12
⇒ P =\(\frac { 12 }{ 3 }\) = 4 Ans.

Question 7.
14 = 7m

Solution:
14 = 7m
⇒ m = \(\frac { 14 }{ 7 }\) = 2

Question 8.
2x = 0

Solution:
2x = 0 ⇒ x = \(\frac { 0 }{ 2 }\) = 0

Question 9.
\(\frac { x }{ 9 }\) = 2

Solution:
\(\frac { x }{ 9 }\) = 2
⇒x = 2 ×9 = 18
∴ x = 18

Question 10.
\(\frac { y }{ -12 }\) = -4

Solution:
\(\frac { y }{ -12 }\) = -4
⇒ \(\frac { y }{ -12 }\) = -4
⇒ y = (-4) × (-12)
∴ y= 48

Question 11.
8x-2 =38

Solution:
8x-2 =38
8x = 38 + 2 = 40
⇒ x = \(\frac { 40 }{ 8 }\) = 5
∴ x = 5

Question 12.
2x + 5 = 5

Solution:
2x + 5 = 5
⇒ 2x = 5 – 5 = 0
x = \(\frac { 0 }{ 2 }\) = 0
∴x = 0

Question 13.
5x – 1 = 74

Solution:
5x- 1 = 74
⇒ 5x = 74 + 1 = 75
⇒ x =\(\frac { 75 }{ 5 }\) = 15

Question 14.
14 = 27-x

Solution:
14 = 27 -x
⇒ x = 27- 14
⇒ x = 13
∴ x= 13

Question 15.
10 + 6a = 40

Solution:
10 + 6a = 40
⇒ 6a = 40 -10 = 30
⇒ a = \(\frac { 30 }{ 6 }\) = 5
∴ a= 5

Question 16.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 1

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 2

Question 17.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 3

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 4

Question 18.
12 = c – 2

Solution:
12 = c – 2
⇒ 12 + 2 =c
⇒ 14 = c
∴c = 14

Question 19.
4 = x- 2.5

Solution:
4 = x – 2.5
⇒4 + 2.5=x
⇒ 6.5 =x
∴ x = 6.5

Question 20.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 5

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 6

Question 21.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 8

Question 22.
p + 0.02 = 0.08

Solution:
p + 0.02 = 0.08
⇒ p = 0.08 – 0.02 = 0.06
∴ p = 0 06

Question 23.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 9

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 10

Question 24.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 11

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 12

Question 25.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 13

Question 26.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 14

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 15
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 16

Question 27.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 17

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 18

Question 28.
2a – 3 =5

Solution:
2a – 3 = 5
⇒2a = 5 +3
⇒ 2a = 8
⇒ a = \(\frac { 8 }{ 2 }\) = 4
∴a = 4

Question 29.
3p – 1 = 8

Solution:
3p – 1 = 8
⇒3p = 8 + 1 = 9
⇒ p = \(\frac { 9 }{ 3 }\) = 3
∴p = 3

Question 30.
9y -7 = 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 20

Question 31.
2b – 14 = 8

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 21

Question 32.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 22

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 23
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 24

Question 33.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 25

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 26

Simple Linear Equations Exercise 12B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
8y – 4y = 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 27

Question 2.
9b – 4b + 3b = 16

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 28

Question 3.
5y + 8 = 8y – 18

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 29

Question 4.
6 = 7 + 2p -5

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 30

Question 5.
8 – 7x = 13x + 8

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 31

Question 6.
4x – 5x + 2x  = 28 + 3x

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 32

Question 7.
9 + m = 6m + 8 – m

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 33

Question 8.
24 = y + 2y + 3 + 4y

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 34

Question 9.
19x -+ 13 -12x + 3 = 23

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 35

Question 10.
6b + 40 = – 100 – b

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 36

Question 11.
6 – 5m – 1 + 3m = 0 

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 38

Question 12.
0.4x – 1.2  = 0.3x + 0.6

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 39

Question 13.
6(x+4) = 36

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 40

Question 14.
9 ( a+ 5) + 2 = 11

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 42

Question 15.
4 ( x- 2 ) = 12

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 43

Question 16.
-3 (a- 6 ) = 24

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 44

Question 17.
7 ( x-2) = 2 (2x -4)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 45

Question 18.
(x-4) (2x +3 ) = 2x²

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 46
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 47

Question 19.
21 – 3 ( b-7 ) = b+ 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 48

Question 20.
x (x +5 ) = x² +x + 32

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 49

Simple Linear Equations Exercise 12C – Selina Concise Mathematics Class 7 ICSE Solutions

Solve
Question 1.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 50

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 51

Question 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 52

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 53
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 54

Question 3.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 55

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 56

Question 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 58
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 59

Question 5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 60
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 61

Question 6.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 62

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 63

Question 7.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 64

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 65

Question 8.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 66

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 67

Question 9.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 68

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 69

Question 10.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 70

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 71

Question 11.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 72

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 73
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 74

Question 12.
0.6a +0.2a = 0.4 a +8

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 75

Question 13.
p + 104p= 48

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 76

Question 14.
10% of x = 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 77

Question 15.
y + 20% of y = 18

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 78

Question 16.
x – 13% of x = 35

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 79
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 80

Question 17.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 81

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 82

Question 18.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 83

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 84

Question 19.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 85

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 86

Question 20.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 87

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 88
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 89

Question 21.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 90

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 91

Question 22.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 92

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 93

Question 23.
15 – 2 (5-3x ) = 4 ( x-3 ) + 13

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 94

Question 24.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 95

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 96
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 97

Question 25.
21 – 3 (x – 7) = x + 20

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 98

Question 26.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 99

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 100

Question 27.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 101

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 102

Question 28.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 103

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 104
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 105

Question 29.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 106

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 107

Question 30.
2x + 20% of x = 12.1

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 108

Simple Linear Equations Exercise 12D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
One-fifth of a number is 5, find the number.

Solution:
Let the number = x
According to the condition
\(\frac { 1 }{ 5 }\)x = 5 ⇒ x = 5 x 5
⇒ x = 25
∴ Number = 25

Question 2.
Six times a number is 72, find the number.

Solution:
Let the number = x
According to the condition
6x = 72
⇒ x = \(\frac { 72 }{ 6 }\)
⇒x= 12
∴ Number = 12

Question 3.
If 15 is added to a number, the result is 69, find the number.

Solution:
Let the number = x
According to the condition
x+ 15 = 69
⇒ x = 69 – 15 x = 54
∴Number = 54

Question 4.
The sum of twice a number and 4 is 80, find the number.

Solution:
Let the number = x
According to the condition
2x + 4 = 80
⇒2x = 80 – 4
⇒ 2x = 76
⇒ x = \(\frac { 76 }{ 2 }\) = 38
Number = 38

Question 5.
The difference between a number and one- fourth of itself is 24, find the number.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 109

Question 6.
Find a number whose one-third part exceeds its one-fifth part by 20.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 110
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 111

Question 7.
A number is as much greater than 35 as is less than 53. Find the number.

Solution:
Let the number = x
According to the condition
x – 35 = 53 – x
⇒ x + x = 53 + 35
88
⇒2x = 88
⇒ x = \(\frac { 88 }{ 2 }\) = 44
∴Number = 44

Question 8.
The sum of two numbers is 18. If one is twice the other, find the numbers.

Solution:
Let the first number = x
and the second number = y
According to the condition
x + y= 18 …(i)
and x = 27 ….(ii)
Substitute the eq. (ii) in eq. (i), we get
2y + y= 18
x= 2y = 18
⇒ 3y= 18 ⇒y= \(\frac { 18 }{ 3 }\) = 6
Now, substitute the value of y in eq. (ii), we get
x = 2 x 6= 12
∴ The two numbers are 12, 6

Question 9.
A number is 15 more than the other. The sum of of the two numbers is 195. Find the numbers.

Solution:
Let the First number = x
and the Second number = y
According to the condition
x = y+ 15 …(i)
x + 7=195 …(ii)
Substitute the eq. (i) in eq. (ii), we get
y+15+7=195
⇒2y= 195- 15
⇒ y = \(\frac { 180 }{ 2 }\) = 90
Now, substitute the value of y in eq. (i), we get
x = 90+ 15 = 105
∴ The two numbers are 105 and 90

Question 10.
The sum of three consecutive even numbers is 54. Find the numbers.

Solution:
Let the first even number = x
second even number = x + 2
and third even number = x + 4
According to the condition,
x + x + 2 + x + 4 = 54
⇒ 3x + 6 = 54
⇒ 3x = 54 – 6
⇒ x =\(\frac { 48 }{ 3 }\) = 16
∴ First even number = 16
Second even number = 16 + 2 = 18
and third even number = 16 + 4 = 20

Question 11.
The sum of three consecutive odd numbers is 63. Find the numbers.

Solution:
Let the first odd number = x
second odd number = x + 2
and third odd number = x + 4
According to the condition,
x+ x + 2 + x+4 = 63
3x + 6 = 63 ⇒ 3x = 63 – 6
⇒3x = 57 ⇒ x = \(\frac { 57 }{ 3 }\) =19
∴ First odd number = 19
Second odd number = 19 + 2 = 21
third odd number = 19 + 4 = 23

Question 12.
A man has ₹ x from which he spends ₹6. If twice of the money left with him is ₹86, find x.

Solution:
Let the total amount be x
According to the condition
2x = 86
⇒x = \(\frac { 86 }{ 2 }\)
⇒ x = 43
Amount spent by him = 6
∴Total money he have = ₹43 + ₹6 = ₹49

Question 13.
A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.

Solution:
Let the present age of the son = x years
Present age of the father = 4x years
After 20 years,
Son’s age will be (x + 20) years
and Father’s age will be (4x + 20) years
According to the condition,
4x + 20 = 2 (x + 20)
4x + 20 = 2x + 40
4x – 2x = 40 – 20
2x = 20
⇒ x = 10
∴Present age of the son = 10 years and Present age of the father = 4×10 years = 40 years

Question 14.
If 5 is subtracted from three times a number, the result is 16. Find the number.

Solution:
Let the number = x
According to the condition,
3x – 5 = 16
⇒ 3x = 16 + 5
⇒ 3x = 21
⇒ x = \(\frac { 21 }{ 3 }\)
⇒ x = 7
∴The number = 7

Question 15.
Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.

Solution:
Let the first conscutive number = x,
Second consecutive number = x + 1
and Third consecutive number = x + 2
According to the condition,
x + x + 1 = 15 + x + 2
⇒ 2x + 1 = 17 +x
⇒ 2x -x = 17 – 1
⇒ x= 16
∴ The first consecutive number = 16
Second consecutive number =16+1 = 17
Third consecutive number =16 + 2=18

Question 16.
The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

Solution:
Let the smallest number = x
and the largest number = y
According to the condition,
y-x = 7 …(i)
and 6x + y = 77 ….(ii)
From eq. (i)
y = 7 + x …(iii)
Substitute the eq. (iii) in eq. (ii)
6x + 7 + x = 77
⇒ 7x = 77-7
⇒ x = \(\frac { 70 }{ 7 }\) = 10
Now, substitute the value of x in eq. (iii)
y = 7+ 10= 17
∴The smallest number 10 and the largest number is 17.

Question 17.
The length of a rectangular plot exceeds its breadth by 5 metre. If the perimeter of the plot is 142 metres, find the length and the breadth of the plot.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 112
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 113

Question 18.
The numerator of a fraction is four less than its denominator. If 1 is added to both, is numerator and denominator, the fraction becomes \(\frac { 1 }{ 2 }\) Find the fraction.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 114
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 12 Simple Linear Equations image - 115

Question 19.
A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time. Find their present ages.

Solution:
Let the present age of the son = x years
and the present age of the father = 3x years
After 12 years,
Son’s age will be (x + 12) years
and father’s age will be (3x + 12) years
According to the condition,
3x + 12 = 2 (x + 12)
3x + 12 = 2x+ 24
3x – 2x = 24 – 12
x= 12
∴Present age of the son = 12 years
and Present age of the father = 3×12 years
= 36 years

Question 20.
A sum of ₹ 500 is in the form of notes of denominations of ₹ 5 and₹ 10. If the total number of notes is 90, find the number of notes of each type.

Solution:
Let the number of ₹ 5 notes = x
∴ The number of ₹10 notes = 90 – x
Value of ₹10 notes = x ×₹ 5 = ₹3x
and value of ₹10 notes = (90 – x) x ₹ 10 =₹(900 – 10x)
∴Total value of all the notes = ₹500
∴5x+ (900- 10x) = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900
⇒ x = \(\frac { 400 }{ 5 }\)
⇒ x = 80
∴ The number of ₹5 notes = x = 80
and the number of ₹10 notes = 90 – x
= 90 – 80= 10

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Mensuration Exercise 20A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The length and the breadth of a rectangular plot are 135 m and 65 m. Find, its perimeter and the cost of fencing it at the rate of ₹60 per m.
Solution:
Given :
Length (l) = 135 m
Breadth (b) = 65 m
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -1
Perimeter = 2 (l + b)
= 2(135 + 65)
= 2(200) = 400 m
∴Perimeter of rectangular plot is = 400 m
Cost of fencing per m = ₹60
∴Cost of fencing 400 m = ₹60 x 400 m = ₹24000

Question 2.
The length and breadth of a rectangular field are in the ratio 7 : 4. If its perimeter is 440 m, find its length and breadth. Also, find the cost of fencing it @ ₹150 per m.
Solution:
Given : Perimeter = 440 m
Let the length of rectangular field = lx and breadth = 4x
2(l + b) = Perimeter
2(7x + 4x) = 440 m
2(11x) = 440 m
22x = 440 m
x = \(\frac { 440 }{ 22 }\)
x = 11 m
∴Length = 7x = 7 x 11 = 77 m
Breadth = Ax = 4 x 11 = 44 m
Cost of fencing per m = ₹150
Cost of fencing 440 m = ₹150 x 440 = ₹66,000

Question 3.
The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -3

Question 4.
The diagonal of a square is 12\(\sqrt { 2 } \) cm. Find its perimeter.
Solution:
Diagonal of square = Its side x \(\sqrt { 2 } \)
Side \(\sqrt { 2 } \) = \(\sqrt { 2 } \) \(\sqrt { 2 } \)
i.e. side = 12 cm
Perimeter of a square = 4 x Side
= 4 x 12 = 48 cm

Question 5.
Find the perimeter of a rectangle whose length = 22.5 m and breadth = 16 dm.
Solution:
Length = 22.5 m
Breadth = 16 dm = 1.6 m
Perimeter of rectangle = 2(l + b)
– 2(22.5 + 1.6)
– 2(24.1) = 48.2 m

Question 6.
Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm
Solution:
Length of a rectangle (l) = 24 cm Diagonal = 25 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -4
Let breadth of the rectangle = b m
Applying Pythagoras Theorem in triangle ABC,
We get, (AC)2 = (AB)2 + (BC)2
(25)= (24)2 + (b)2
625 = 576 + (b)2
625 – 576 = b2
49 = A2
\(\sqrt { 7 x 7 } \) =b
∴b = 7 cm
Now, perimeter of the rectangle
= 2(1 + b)
= 2(24 + 7)
= 2(31)
= 62 cm

Question 7.
The length and breadth of rectangular piece of land are in the ratio of 5 : 3. If the total cost of fencing it at the rate of ₹48 per metre is ₹19,200, find its length and breadth.
Solution:
Ratio in length and breadth of a rectangular piece of land = 5:3
Cost of fencing =₹ 19,200
and rate = ₹48 per m
∴Perimeter = \(\frac { 19200 }{ 48 }\)= 400 m 48
Let length = 5x.
Then breadth = 3x
∴Perimeter = 2(l + b)
400 = 2(5x + 3x)
400 = 2 x 8x= 16x
∴16x = 400
⇒ x = \(\frac { 400 }{ 16 }\) = 25
∴Length of the land = 5x= 5 x 25 = 125 m and breadth = 3x = 3 x 25 = 75 m

Question 8.
A wire is in the shape of square of side 20 cm. If the wire is bent into a rectangle of length 24 cm, find its breadth.
Solution:
Side of square = 20 cm
Perimeter of square = 4 x 20 = 80 cm
Or perimeter of rectangle = 80 cm
Length of a rectangle = 24 cm
∴ Perimeter of a rectangle = 2(l + b)
b = \(\frac { 80 }{ 2 }\) – 24
b = 40 – 24 = 16 m

Question 9.
If P = perimeter of a rectangle, l= its length and b = its breadth find :
(i) P, if l = 38 cm and b = 27 cm
(ii) b, if P = 88 cm and l = 24 cm
(iii) l, if P = 96 m and b = 28 m
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -5

Question 10.
The cost of fencing a square field at the rate of
Cost of fencing 440 m = ₹150 x 440 = ₹75 per meter is
Cost of fencing 440 m = ₹150 x 440 = ₹67,500. Find the perimeter and the side of the square field.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -6

Question 11.
The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter is equal to the perimeter of a square, find the side of the square.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -7

Question 12.
The radius of a circle is 21 cm. Find the circumference (Take π = 3 \(\frac { 1 }{ 7 }\) ).
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -8

Question 13.
The circumference of a circle is 440 cm. Find its radius and diameter. (Take π = \(\frac { 22 }{ 7 }\)
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -9

Question 14.
The diameter of a circular field is 56 m. Find its circumference and cost of fencing it at the rate of ₹80 per m. (Take n = \(\frac { 22 }{ 7 }\))
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -10

Question 15.
The radii of two circles are 20 cm and 13 cm. Find the difference between their circumferences. (Take π = \(\frac { 22 }{ 7 }\))
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -11
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -12

Question 16.
The diameter of a circle is 42 cm, find its perimeter. If the perimeter of the circle is doubled, what will be the radius of the new circle. (Take π = \(\frac { 22 }{ 7 }\) )
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -13

Question 17.
The perimeter of a square and the circumference of a circle are equal. If the length of each side of the square is 22 cm, find:
(i) perimeter of the square.
(ii) circumference of the circle.
(iii) radius of the circle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -14
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -15

Question 18.
Find the radius of the circle whose circumference is equal to the sum of the circumferences of the circles having radii 15 cm and 8 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -16

Question 19.
Find the diameter of a circle whose circumference is equal to the sum of circumference of circles with radii 10 cm, 12 cm and 18 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -17
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -18

Question 20.
The circumference of a circle is eigth time the circumference of the circle with radius 12 cm. Find its diameter.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -19

Question 21.
The radii of two circles are in the ratio 3 : 5, find the ratio between their circumferences.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -20

Question 22.
The circumferences of two circles are in the ratio 5 : 7, find the ratio between their radii.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -21

Question 23.
The perimeters of two squares are in the ratio 8:15, find the ratio between the lengths of their sides.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -22

Question 24.
The lengths of the sides of two squares are in the ratio 8:15, find the ratio between their perimeters.
Solution:
Let the side of first square = 8x
∴Perimeter of first square = 4 x Side = 4 x 8x = 32 x
and the side of second squares = 15x
∴Perimeter of second square = 4 x Side = 4 x 15s = 60s
Now, the ratio between their perimeter = 32x: 60x= 8: 15

Question 25.
Each side of a square is 44 cm. Find its perimeter. If this perimeter is equal to the circumference of a circle, find the radius of the circle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -23

Mensuration Exercise 20B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the area of a rectangle whose length and breadth are 25 cm and 16 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -24

Question 2.
The diagonal of a rectangular board is 1 m and its length is 96 cm. Find the area of the board.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -25

Question 3.
The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -26
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -27

Question 4.
A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -28

Question 5.
The length and breadth of a rectangular piece of land are in the ratio 5 : 3. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its :
(i) length and breadth
(ii) area
(iii) cost of levelling at the rate of ₹60 per m2.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -29

Question 6.
Find the area of the square whose perimeter is 56 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -30

Question 7.
A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m2 find the area of the lawn.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -31
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -32
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -33

Question 8.
For each figure, given below, find the area of shaded region : (All measurements are in cm)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -34
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -35
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -36

Question 9.
One side of a parallelogram is 20 cm and its distance from the opposite side is 16 cm. Find the area of the parallelogram.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -37

Question 10.
The base of a parallelogram is thrice it height. If its area is 768 cm2, find the base and the height of the parallelogram.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -38
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -39

Question 11.
Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -40

Question 12.
If the area of a rhombus is 112 cm2 and one of its diagonals is 14 cm, find its other diagonal.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -41
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -42

Question 13.
One side of a parallelogram is 18 cm and its area is 153 cm2. Find the distance of the given side from its opposite side.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -43

Question 14.
The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -44
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -45

Question 15.
The area of a rhombus is 84 cm2 and its perimeter is 56 cm. Find its height.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -46

Question 16.
Find the area of a triangle whose base is 30 cm and height is 18 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -47

Question 17.
Find the height of a triangle whose base is 18 cm and area is 270 cm2.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -48

Question 18.
The area of a right-angled triangle is 160 cm2. If its one leg is 16 cm long, find the length of the other leg.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -49
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -50

Question 19.
Find the area of a right-angled triangle whose hypotenuse is 13 cm long and one of its legs is 12 cm long.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -51
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -52

Question 20.
Find the area of an equilateral triangle whose each side is 16 cm. (Take \(\sqrt { 3 } \)= 1.73)
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -53

Question 21.
The sides of a triangle are 21 cm, 17 cm and 10 cm. Find its area.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -54

Question 22.
Find the area of an isosceles triangle whose base is 16 cm and length of each of the equal sides is 10 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -55

Question 23.
Find the base of a triangle whose area is 360 cm2and height is 24 cm.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -56

Question 24.
The legs of a right-angled triangle are in the ratio 4 :3 and its area is 4056 cm2. Find the length of its legs.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -57

Question 25.
The area of an equilateral triangle is (64 x \(\sqrt { 3 } \) ) cm2– Find the length of each side of the triangle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -58
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -59

Question 26.
The sides of a triangle are in the ratio 15 : 13 : 14 and its perimeter is 168 cm. Find the area of the triangle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -60

Question 27.
The diameter of a circle is 20 cm. Taking π = 3.14, find the circumference and its area.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -61
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -62

Question 28.
The circumference of a circle exceeds its diameter by 18 cm. Find the radius of the circle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -63

Question 29.
The ratio between the radii of two circles is 5 : 7. Find the ratio between their :
(i) circumference
(ii) areas
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -64
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -65

Question 30.
The ratio between the areas of two circles is 16 : 9. Find the ratio between their :
(i) radii
(ii) diameters
(iii) circumference
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -66
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -67

Question 31.
A circular racing track has inner circumference 528 m and outer circumference 616 m. Find the width of the track.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -68

Question 32.
The inner circumference of a circular track is 264 m and the width of the track is 7 m. Find:
(i) the radius of the inner track.
(ii) the radius of the outer circumference.
(iii) the length of the outer circumference.
(iv) the cost of fencing the outer circumference at the rate of ₹50 per m.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -69
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -70

Question 33.
The diameter of every wheel of a car is 63 cm. How much distance will the car move during 2000 revolutions of its wheel.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -71

Question 34.
The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel one kilometre?
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -72

Question 35.
A metal wire, when bent in the form of a square of largest area, encloses an area of 484 cm2. Find the length of the wire. If the same wire is bent to a largest circle, find:
(i) radius of the circle formed.
(ii) area of the circle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -73
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -74

Question 36.
A wire is along the boundary of a circle with radius 28 cm. If the same wire is bent in the form of a square, find the area of the square formed.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -75

Question 37.
The length and the breadth of a rectangular paper are 35 cm and 22 cm. Find the area of the largest circle which can be cut out of this paper.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -76
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -77

Question 38.
From each comer of a rectangular paper (30 cm x 20 cm) a quadrant of a circle of radius 7 cm is cut. Find the area of the remaining paper i.e., shaded portion.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -78
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration imagev -79