Selina Class 7 ICSE Solutions

Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

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Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 1 Plant And Animal Tissues

Synopsis

  • The unit of level of organisation is independent in its mode of existence and activity.
  • All multi cellular organisms start their life as a single cell.
  • Plant tissues are basically of two type
    1.  meristematic
    2.  permanent or non-dividing
  • The permanent – plant tissues are further of three types
    1. protective
    2. supportive: parenchyma, collenchyma, sclerenchyma
    3. conductive: xylem, phloem
  • Parenchymal cells have thin-walled cells and usually with a vacuole.
  • Potatoes mainly are composed of parenchymal cells.
  • Collenchyma are parenchymatous cells which are elongated and are thick at the comers. This helps to support the parts of the plant.
  • Sclerenchyma tissue is formed of long, narrow and thick cells. This provides strength to the plant parts.
  • Xylem is formed of thick-walled, tubular and often dead cells. They transport water and minerals absorbed by roots.
  • Old xylem forms the wood.
  • Phloem is formed of living tubular cells which provide a passage for the downward transport of food.
  • The four major groups of animal tissues

    1. epithelial tissue
    2. connective tissue
    3. muscular tissue
    4. nervous tissue
  • The epithelial tissue is further of four types:

    1. squamous epithelium (protective)
    2. cuboidal epithelium (absorption)
    3. columnar epithelium (secretory)
    4. ciliated epithelium (movement of substances)
  • Supportive connective tissue consists of
    1. Cartilage
    2.  Bone
  • Fibrous connective tissue:
    It packs and binds most of the organs. It is of the following types.

    1. areolar tissue: binds skin to underlying tissue.
    2. adipose tissue: filled with fat.
    3. tendon: connect muscles to bones.
    4. ligaments: connect bone to another bone.
  • Fluid connective tissue consists of
    1. Blood
    2. Lymph
  • The liquid part of the blood is called plasma and the cellular part includes:
    1. red blood cells
    2.  white blood cells
    3.  platelets.
  • Three distinct kinds of muscles are
    1. striated or skeletal
    2. unstriated or smooth
    3. cardiac or heart.
  • A nerve cell is formed of a cell body called cyton and one or more elongated hair-like extensions called dendrites. The longest dendrite is called axon.
  • Systems of the body with their primary vital function.Skeletal system: support and protection
    1. Muscular system: movement
    2. Digestive system: nutrition
    3. Respiratory system: exchange of gases
    4. Circulatory system: transport of materials
    5. Excretory system: waste removal
    6. Nervous system: sensation and co-ordination
    7.  Reproductive system: continuation of race.

Review Questions

MULTIPLE CHOICE QUESTIONS

1. Put a tick (✓) against the most appropriate alternative in the following statements.

(i) A group of similar cells to perform a specific function forms a
(a) organ
(b) species
(c) organ system
(d) tissue

(ii) The fine branches given out from the cell body of a nerve cell are
(a) dendrites
(b) cyton
(c) axon
(d) neurons

(iii) Fluid connective tissue of humans is
(a) blood and cartilage
(b) lymph and plasma
(c) blood and lymph
(d) stroma and matrix

Short Answer Questions

Question 1.
1. Define the following terms:

  1. Tissue
  2. Organ

Answer:

  1. Tissue: A group of similar cells which perform a specific function.
    example: Muscular tissue in animals.
  2. Organ: The different type of tissues which group together to function in a co-ordinated manner.
    example: liver

2. Answer the following:

Question 2(i).
What is a meristematic tissue ? How is it different from permanent tissues ?
Answer:
Plant tissues are classified into two types:

  1. Meristematic tissue
  2.  Permanent or non-diving tissue

Meristematic tissues are the plant tissues which are made up of actively dividing cells. These tissues actively divide and lead to the growth of the plant body. They are found at the growth points of the plant like tips of root, stem and branches etc.

  1. Cells are small with thin cell walls.
  2. Cells have large and conspicuous nuclei.
  3. Cells have no vacuoles.
  4. Cells are actively dividing type cells.

Difference between Meristematic and permanent
Meristematic tissue :

  1. Meristematic tissue is present at the tip of the root and stem and in between the xylem and phloem. Form apical meristematic tissue when present at the tips. It is in the form of cambium in between the xylem and phloem.
  2. Meristematic cells divide and form other types of tissues. The cells are thin walled.
  3. Meristematic cells may be intercalary as in case of monocots.
  4. The cells are small and isodiametric, vacuoles are small or absent.
  5. Respiratory and biosynthetic activities maximum.
  6. The cells are immature and mitochondria simple.
  7. Proplastids act as plastids.

Permanent tissue

  1. Permanent tissue may be simple as parenchyma, collenchyma or sclerenchyma and it may be complex as xylem and phloem.
  2. These are made up of more than one kind of cells. These perform a common function Xylem and phloem form vascular system of the plant. These cells do not have the power to divide.
  3. These cells may act as epidermis cortex or grit cells. Sclerenchyma gives strength.
  4. Living cells of permanent tissue have vacuoles. The cells are large and of different shapes.
  5. Both these activities are low.
  6. The cells fully mature, mitochondria fully developed.
  7. Living cells have plastids.

Question 2(ii).
Which living material would you take to demonstrate meristematic tissue ?
Answer:
Green gram seeds can be used to demonstrate meristematic tissue which when soaked in a petridish stuffed with wet cotton and left for 3-4 days would sprout out. These sprouted seeds have roots developing whose root tips have meristematic tissue.

Question 2(iii).
What is the function of meristematic tissue ?
Answer:
The meristematic tissue have the primary role in the growth of the plant tissue as it consists of active dividing cells

Question 3.
State whether the following statements are True or False. 

(i) A tissue is formed of only one type of cells.
Ans. True

(ii) Only one type of tissue forms an organ.
Ans. False.
Correct: Two or more types of tissue form an organ.

(iii) Permanent tissue is made up of undifferentiated and dividing Cells.
Ans. False.
Correct: Meristematic tissue is made up of undifferentiated and dividing cells.

(iv) Meristematic tissue is found at growing tips of a plant.
Ans. True

(v) Phloem is formed of dead tubular cells.
Ans. False.
Correct: Phloem is formed of living tubular cells.

Question 4.
Fill in the blanks by selecting suitable words from the list given below:
“Thin – walled, collenchyma, vascular, tissues, conducting”

  1. A group of different tissues working together to perform a function is called an organ.
  2.  Xylem and phloem form the conducting tissue.
  3.  Conducting tissue is also called vascular tissue.
  4. Cells are elongated and thick at the comers in collenchyma tissue.
  5. Parenchyma is composed of large thin-walled cell

Question 5.
Match the items given is column A with those given in column B:

Column A
(i) Fibrous connective tissue
(ii) Fluid connective tissue
(iii) Supportive connective tissue
(iv) Ligament
(v) Tendon
Column B
(a) blood
(b) cartilage
(iii) Supportive connective tissue
another bone.
(d) areolar tissue
(e) connects a muscle
with a bone.
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 1

Question 6.
How do you rank the following among cells, tissues, organs, or organism ?

  1. Amoeba : organism
  2.  Euglena: organism
  3. Skin : organ
  4. Lungs : organ
  5. Neuron : tissue
  6. Cardiac muscles: Ti1ue

Question 7.
Each of the tissues listed in Column A is related to one of the functions given in Column B. Match the lines correct pairs by drawing
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 2

Question 8.
Name the kind of tissue that

  1.  Carries oxygen around your body — Blood tissue.
  2.  Brings about movements in animals — muscular tissue.
  3.  Transports food to different parts of plant— phloem.
  4.  Transports water in plants — xylem.
  5.  Supports an animal’s body — connective tissue (supportive)
  6.  Binds different tissues together — Fibrous connective tissue.
  7.  Conducts messages from one part of the body to another — nervous tissue.

Question 9.
Based on the following information, identify the three types of epithelial tissue in the figures given below :
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 3

(i) Cuboidal epithelium : It consists of a single layer of cuboidal cells.

(ii) Columnar epithelium: It is composed of tall, cylindrical cells with oval nuclei usually placed at the base of the cells.

(iii) Ciliated epithelium : It consists of cells being hair-like cilia on their free surface.
Answer:
(i) fig. b (ii) fig. a (iii) fig. c

Question 10.
Write three differences between the two principal vascular tissues found in plants.
Answer:
Xylem

  1. Transports water and minerals absorbed by the roots to other plant parts.
  2. Consists mainly of dead cells.
  3. Conduction is unidirectional i.e. only upwards from the roots.

Phloem

  1. Conducts food manufactured in the leaves to other plant parts.
  2. Consists mainly of living cells.
  3. Bidirectional conduction i.e. both upwards and downwards from the leaves.

 

Selina Concise Mathematics class 7 ICSE Solutions – Percent and Percentage

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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.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

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

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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

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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

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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration

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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)2 = (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

Selina Concise Mathematics class 7 ICSE Solutions – Exponents (Including Laws of Exponents)

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Selina Concise Mathematics class 7 ICSE Solutions – Exponents (Including Laws of Exponents)

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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EXERCISE 5 (A)

Question 1.
Find the value of:
(i) 6² 
(ii) 73
(iii) 44
(iv) 55
(v) 83
(vi) 75

Solution:
(i) 62 = 6 x 6 = 36
(ii) 73 = 7 x 7 x 7 = 343
(iii) 44 = 4 x 4 x 4 x 4 = 256
(iv) 55= 5 x 5 x 5 x 5 x 5 = 3125
(v) 83 = 8 x 8 x 8 = 512
(vi) 75 = 7 x 7 x 7 x 7 x 7 =16807

Question 2.
Evaluate:
(i) 23 x 42
(ii) 23 x 52
(iii) 33 x 52
(iv) 22 x 33
(v) 32 x 52
(vi) 53 x 24
(vii) 32 x 42
(ix) (5 x 4)2

Solution:
(i) 23 x 42
= 2 x 2 x 2 x 4 x 4
= 8 x 16
= 128
(ii) 23 x 52
= 2 x 2 x 2 x 5 x 5
= 8 x 25
= 200
(iii) 33 x 52
=3 x 3 x 3 x 5 x 5
= 27 x 25
= 675
(iv) 22 x 33
= 2 x 2 x 3 x 3 x 3
= 4 x 27
= 108
(v) 32 x 53
=3 x3 x 5 x 5 x 5
= 9 x 125
= 1125
(vi) 53 x 24
= 5 x 5 x 5 x 2 x 2 x 2 x 2
= 125 x 16
= 2000
(vii) 32 x 42
=3 x 3 x 4 x 4
= 9 x 16
=144
(viii) (4 x 3)3
=4 x 4 x 4 x 3 x 3 x 3
= 64 x 27
= 1728
(ix) (5 x 4)2
=5 x 5 x 4 x 4
= 25 x 16
= 400

Question 3.
Evaluate:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 39

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 40

Question 4.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 1

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 3

Question 5.
Which is greater :
(i) 23 or 32
(ii) 25 or 52
(iii) 43 or 34
(iv) 54 or 45

Solution:
(i) 23 or 33
Since, 23 = 2 x 2 x 2 = 8
and, 32 = 3 x 3 = 9
∵9 is greater than 8 ⇒ 32 > 23
(ii) 25 or 52
Since, 25 = 2 x 2 x 2 x 2 x 2 = 32
and, 52 = 5 x 5 = 25
∵32 is greater than 25 ⇒ 235 > 532
(iii) 43 or 34
Since, 43 = 4 x 4 x 4 = 64
and, 34 = 3 x 3 x 3 x 3 = 81
∵ 81 is greater than 64 ⇒ 34 > 43
(iv) 54 or 45
Since, 54 = 5 x 5 x 5 x 5 = 625
and, 45 = 4 x 4 x 4 x 4 x 4= 1024
∵ 1024 is greater than 625 ⇒ 45 > 54

Question 6.
Express each of the following in exponential form :
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176

Solution:
(i) 512
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 4

(ii) 1250
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 6

(iii) 1458
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 7

(iv) 3600
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 8

(v) 1350
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 9

(vi) 1176
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 10

Question 7.
If a = 2 and b = 3, find the value of:
(i) (a + b)2
(ii) (b – a)3
(iii) (a x b)a (iv) (a x b)b

Solution:
(i) (a + b)2
= (2 + 3)2 = (5)2 = 5 x 5 = 25

(ii) (b – a)2
= (3 – 2)2= (1)3
= 1 x 1 x 1 = 1

(iii) (a x b)a
= (2 x 3)2 – (6)2
= 6 x 6 = 36

(iv) (a x b)b
= (2 x 3)3 = (6)3 = 6 x 6 x 6 = 216

Question 8.
Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
Solution:
(i) 1024 as a power of 2.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 11

(ii) 343 as a power of 7.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 12

(iii) 729 as a power of 3.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 13

Question 9.
If 27 x 32 = 3x x 2y; find the values of x and y.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 14

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 15

Question 10.
If 64 x 625 = 2a x 5b; find :
(i) the values of a and b.
(ii) 2b x 5a

Solution:
(i) the values of a and b.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 16

(ii) 2b x 5a
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 17

EXERCISE 5 (B)

Question 1.
Fill in the blanks:
In 52 = 25, base = ……… and index = ……….
If index = 3x and base = 2y, the number = ………

Solution:
(i) In 52 = 25, base = 5 and index = 2
(ii) If index = 3x and base = 2y, the number = 2y3x

Question 2.
Evaluate:
(i) 28 ÷ 23
(ii) 23÷ 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 54  x 53 + 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 18
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 19

Question 3.
Simplify, giving Solutions with positive index:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 20
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 41

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 21
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 22
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 23
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 24
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 25
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 26

Question 4.
Simplify and express the Solution in the positive exponent form :
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 27
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 28

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 30
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 31

Question 5.
Evaluate
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 32

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 33
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 34
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 35
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 36

Question 6.
If m2 = -2 and n = 2; find the values of:
(i) m + r2 – 2mn
(ii) mn + nm
(iii) 6m-3 + 4n2
(iv) 2n3 – 3m

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 37
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 38

 

Selina Concise Chemistry Class 7 ICSE Solutions – Metals and Non-metals

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Selina Concise Chemistry Class 7 ICSE Solutions – Metals and Non-metals

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 6 Metals and Non-metals

Points to Remember :

  1. Knowledge of chemistry plays a vital role in the development of human society and civilization.
  2. Metals are known to man from ancient times. Metals are used to make our life comfortable.
  3. Non-metals form another class of elements, e.g. hydrogen, oxygen, carbon, etc. They are used for various purposes.
  4. Alloys are homogeneous solid mixtures containing two or more metals e.g. steel, brass, bronze.
  5. Common salt, hydrocloric acid, carbohydrates, fats, proteins, vitamins, occur naturally and can also be prepared artificially.
  6. Fertilizers are artificially prepared substances, which are necessary for the proper growth of crops.
  7. There are a number of man-made materials that are used in our daily life for various purposes, e.g., cement, plaster of pairs, plastics.
  8. Medicines are used to cure diseases.
  9. Solution is a homogenous mixture of solute and solvent.
  10. Soda water is prepared by dissolving carbon dioxide in water under high pressure.
  11. Syrup is a highly concentrated sugar solutions. It contains a specific flaviour.

EXERCISE

1. Name a metal

  1. that is most malleable : Pure gold
  2. that is brittle : Zinc
  3. as precious as gold : Platinum
  4. that can be cut with knife : Sodium
  5. used in making electric cables : Copper
  6. used as a thermometric liquid : Mercury
  7. that is the best conductor of electricity : Silver

2. Name a non-metal that is :

  1. a good conductor of heat and electricity : Graphite (Carbon)
  2. hardest naturally occurring substance : Diamond (Carbon)
  3. used to kill germs in water : Chlorine
  4. lustrous : Iodine
  5. used for filling into electric bulbs : Argon
  6. used for cancer therapy : Radon
  7. liquid at room temperature : Bromine

3. Mention two uses of the following metals and non-metals

(a) Iron :
It is used to make pipes, tanks, railing, etc.
It is used in the construction of power transmission towers.

(b) Aluminium :
It is used to make electric wires.
It is used to make utensils, cans, window fram’es, etc.

(c) Gold :
It is used for making ornaments and coins.
It is used in the manufacture of electronic devices like computers, telephones, home appliances, etc.

(d) Oxygen :
It is used by all living beings for breathing.
It is important for combustion.

(e) Iodine :
It is used in photographic films in the form of potassium iodide.
It is added to salt to make it iodized salt which is necessary for the growth of human body.

4. Give reasons :

(a) Magnesium is used in fire works.
Ans : Magnesium is used in fire works because it bums with a dazzling light.

(b) Aluminium is used in making aircrafts.
Ans : Aluminium is used in making aircrafts because it is light and strong. It is mixed with other metals to make it stronger.

(c) Copper is used in making electric cables.
Ans : Copper is ductile and a very good conductor of heat and electricity. This is the reason that copper is used in making electric cables.

(d) Graphite is used in the leads of pencils 
Ans : Graphite turns paper black that is why it is used in the leads of pencils.

(e) Impure diamond is used to cut glass
Ans : Impure diamond is used to cut glass becuase it is the hardest substance and can easily exert force required for cutting.

(f) Gold is mixed with copper and nickel.
Ans :  Pure gold is a very soft metal. It cannot be moulded into ornaments so it is mixed with copper and nickel so that it becomes harder and bit cheaper also.

(g) Tungsten is used in electric bulbs.
Ans : It is a shiny grey metal, in solid state at room temperature. It can withstand high temperature because it has highest melting point among metals. Hence, it is used in electric bulbs.

5. Name the metals present in the following alloys

  1. Brass— Copper and zinc
  2. Bronze— Copper and tin
  3. Duralumin— Aluminium and copper
  4. Stainless steel— Iron, chromium, nickel

6. Give four differences between metals and non-metals with reference to their
(a) Melting point and boiling point,
(b) Conductivity of heat and electricity,
(c) Malleability
(d) Solubility

            Metals Non-metals
Melting point and boiling point Metals have both high high melting point and boiling point. Non-metals have both low melting and low boiling point.
Conducti­vity of heat and electricity They are good conductors of heat and electricity. Nofi-metals are bad conductors of heat and electricity.

Malleabi­ lity

 Metals are ususally malleable. All non-metal are non- malleable.

Solubility

 Metals are generally insoluble in water and other organic solvents. They are both soluble and insoluble

7. What are metalloids?
Ans : Metalloids are the elements which show some properties of metals and some properties of non-metals. They all are solids. They are silicon, boron, arsenic, antimony, germanium, tellurium and polonium.

8. Give two uses of

(a) Silicon :

  • Highly pure silicon is used in making microchips for computers, transistors, solar cells, rectifiers and other solid state devices that are used extensively in the electronic and present space age industries.
  • It is used in the manufacture of a waterproof material called “silicone”. Silicone is used to make bags, umbrellas, raincoats, etc.
  • It is an important substance present in steel, an alloy of carbon.

(b) Antimony :

  • Antimony is used in electric industry to make semiconductor devices.
  • It is alloyed with lead to improve its hardness and strength and is used in batteries.
  • It is also used in printing presses as type metal.

(c) Tungsten :

  • It is used in making electrodes.
  • It is used in heating elements.
  • It is used as filaments in electric bulbs and cathode ray tubes.

(d) Germanium:

  • Germanium is used as a semiconductor.
  • It is used as a transistor in many electronic applications when mixed with arsenic, gallium, antiomony, etc.
  • Germanium is also used to form alloys and as a phosphor in fluorescent lamps

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks :

(a) The most ductile metal is silver.
(b) A metal stored in kerosene oil is sodium.
(c) Tungsten metal is a poor conductor of heat.
(d) Pure gold is a soft metal.
(e) Silicon carbide is the hardest compound known to us.
(f) A non-metal used to purify water is phosphorus.
(g) A metal that gives dazzling effect to crackers when they explode is magnesium.
(h) A chemical compound that makes up the striking heads of match sticks is sulphur.

2. Match the following :
Selina Concise Chemistry Class 7 ICSE Solutions - Metals and Non-metals-2
3. Write ‘true’ or ‘false’ for the following statements :
(a) Silver is used to make electric cables : False
(b) Iodine acts as an antiseptic in the form of tincture of iodine : True
(c) Sodium can be cut with a knife : True
(d) Antimony is a metal : False
(e) Sand is an oxide of silicon : True

MULTIPLE CHOICE QUESTIONS

1. The noble gas used in advertising signboards is
(a) Helium
(b) Neon
(c) Argon
(d) Krypton

2. A metal with melting point less than 50°C is
(a) Gallium
(b) Iron
(c) Gold
(d) Aluminium

3. A metal which is neither ductile nor malleable is
(a) Copper
(b) Silver
(c) Zinc
(d) Aluminium

4. Rust is a hydrated oxide of iron which is
(a) Reddish brown
(b) Green
(c) White
(d) Black

5. Aluminium is not used to make :
(a) Foils
(b) Wires
(c) Fireworks
(d) Utensils

6. A metalloid used in the manufacture of microchips used in computer is :
(a) Antimony
(b) Germanium
(c) Silicon
(d) Arsenic

7. A metalloid used to make glass :
(a) Sulphur
(b) Germanium
(c) Silicon
(d) Antimony

 

Selina Concise Chemistry Class 7 ICSE Solutions – Air and Atmosphere

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Selina Concise Chemistry Class 7 ICSE Solutions – Air and Atmosphere

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 7 Air and Atmosphere

Points to Remember :

  1. Air is a mixture of many gases, mainly Nitrogen = 78.1%, Oxygen = 20.9%, Carbon dioxide = 0.03 – 0.04%, Inert gases = 0.9%, [Water vapours, Dust particles and Impurities = Variable].
  2. Nitrogen is a colourless, an odourless and a tasteless gas. It is slightly lighter than air.
  3. The process of conversion of free atmospheric nitrogen into its compounds is called nitrogen-fixation.
  4. Oxygen constitutes about 21% of air by volume. It supports life on earth.
  5. Carbon dioxide is present in air in a very small quantity, e. 0.03 – 0.04%. It is essential for the process of photosynthesis.
  6. Inert gases like neon, argon do not react with any substance.
  7. The harmful substances added to air are called pollutants.
  8. Some pollutants are suspended particles like pollen grains, oxides of sulphur and nitrogen, oxides of carbon, chlorofluorocarbons etc.
  9. Symbol of oxygen = O ; atomic number = 8, relative mass = 16, molecule formula = O2.
  10. Oxygen is available in free and combined state.
  11. A catalyst is a substance that increases or decreases the rate of a chemical reaction without itself undergoing any chemical change.
  12. Oxides are binary compounds formed by the chemical combination of substance with oxygen.
  13. Rusting is the process in which iron slowly reacts with oxygen in the air and produces a flaky brown substance.
  14. Photosynthesis is a process by which C02 and water are used up by green plants in the presence of sunlight to produce glucose and oxygen gases.

A. AIR : A MIXTURE OF GASES

EXERCISE — I

Question 1.
Give one use for each of the following inert gases :
(a) argon
(b) helium
(c) neon
(d) radon
(e) krypton
(f) xenon

Answer:
(a) Argon— Argon is filled into electric bulbs to prevent the oxidation of their filaments.
(b) Helium— It is used in filling up weather observation balloons.
(c) Neon— Neon is used for making advertisement sign boards.
(d) Radon— It is used for treatment of Cancer.
(e) Krypton— It is used in photography.
(f) Xenon— It is also used in photography.

Question 2.
Answer the questions put against each of the following constituents of air :
(a) Nitrogen : Explain its significance for plants and animals.
(b) Oxygen : What is the percentage proportion of oxygen in air ? Why is oxygen called active air.
(c) Carbon dioxide : “Although carbon dioxide plays no role in respiration, all life would come to an end if there is no carbon dioxide in air.” Support this statement with relevant facts.
(d) Water vapours : Explain their role in modifying the earth’s climate.
Answer:
(a) Plants convert nitrogen into protein. It is an important constituent of proteins, which are necessary for the growth of animals, plants and human beings. Plants convert nitrogen into proteins.
(b) 20.9%, oxygen is called active air because it supports life on earth. It is essential for the process of combustion.
(c) Carbon dioxide is essential for photosynthesis by which green plants prepare their food. It minimises heat loss by radiation. Thus, it balances the temperature on earth.
(d) Water vapour determine the earth’s climate conditions. It causes rain. It controls the rate of evaporation from the bodies of plants and animals.

Question 3.
Define the following terms :
(a) pollutants
(b) acid rain
(c) Global warming
(d) smog

Answer:
(a) Pollutants : Air contains substances which are harmful to plants and animals. These harmful substances are called pollutants.
(b) Acid rain : When sulphur trioxide and nitrogen oxide present in the air mix with rainwater they form sulphuric acid and nitric acid respectively. Rainwater containing these acids is called acid rain.
(c) Global warming : An increase in the percentage of carbon dioxide, methane, nitrous oxide and chlorofluorocarbon traps the heat causing the temperature of the earth and its surroundings to rise. This is known as global warming.
(d) Smog : Oxides of nitrogen form a mixture of smoke and fog known as smog which affects our eyes too.

Question 4.
“Air is a mixture”. Support this statement citing at least three evidences.
Answer:
“Air is a mixture” The following are in evidences which prove that air is a mixture.

  1. The composition of air varies from place to place and from time to time.
  2. The components of air retain their individual properties.
  3. Liquid air has no definite boiling point.
  4. No energy exchange occurs when the components of air are mixed with each other.

Question 5.
What is air pollution ? What are the harmful effects of sulphur dioxide, nitrogen dioxide and hydrogen sulphide present in the air ?
Answer:
Air Pollution : Air is polluted by natural processes like volcanic eruption, crop pollination, etc. mostly it is polluted by human activities like burning of coal, wood, diesel oil, kerosene, petrol etc.
Fossil fuels contain sulphur and nitrogen as impurities. When fuels bum these substances combine with air to produce gasses like sulphur dioxide, nitrogen oxide and hydrogen sulphide. They cause many serious respiratory problems. They can destroy the ozone layer, which protects us from the ultra violet radiations of the Sun. They also cause acid rain, which damages crops and buildings.

Question 6.
(a) What are the causes of air pollution ?
(b) Suggest five measures to prevent air pollution.
Answer:
(a) When fuels bum they produce sulphur dioxide, sulphur trioxide, nitrogen dioxide, hydrogen sulphide when these gases mix with rain water. They produce sulphuric and nitric acid. These acids mix with rain water to form acid rain.
(b) Five measures for the prevention of air pollution are:

  1. By using smokeless sources of energy, like solar energy and electrical energy, in place of conventional fossil fuels.
  2.  By using filters for the. smoke coming out of the chimneys of factories and power plants.
  3. By using internal combustion engines in vehicles for complete and efficient burning of fuel.
  4. By locating industries away from residential areas.
  5. By growing more trees.

Question7.
(a) What is nitrogen-fixation ?
(b) What are the two ways in which nitrogen fixation occurs?
(c) Explain the conversion of nitrogen into nitrates during lightning.
Answer:
(a) Nitrogen fixation : Symbiotic bacteria living in the root nodules of leguminous plants like peas, beans, absorb nitrogen directly from air and convert into nitrates. Thereafter, the plants convert it into proteins. Nitrogen is returned to the soil when plant and animal matter decays. This decomposition work is done by organisms called denitrifying bacteria which reconvert dead organic tissue into its constituent nitrogen.
(b) 1. Natural process.
2. Non-biological fixation.
(c) During lightning, temperatures often reach as high as 3000°C. At such high temperatures, nitrogen and oxygen present in the air combine to form nitric oxide, which further react with oxygen to form nitrogen dioxide
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7a
Oxygen constitutes about 21% of air by volume. It is the active part of air.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7b
Nitrogen dioxide then reacts with the water vapour present in air to form nitrous and nitric acids.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7c
Oxygen constitutes about“21% of air by volume. It is the active part of air.
Nitric acid, so formed, reaches the earth along with rain-water, and reacts with metal carbonates to form metal nitrates.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7d
Oxygen constitutes about 21% of air by volume. It is the active part of air.

B. OXYGEN

EXERCISE — II

Question 1.
Name :
(a) The most abundant element in the earth’s crust.
Ans. Oxygen.

(b) A chemical called oxygenated water.
Ans. H2O2 (Hydrogen peroxide)

(c) A metal highly resistant to rusting.
Ans. Tin.

(d) A mixture of oxygen and carbon dioxide used for artificial respiration.
Ans. Carbogen

(e) Two substances from which oxygen can be obtained at a large scale.
Ans. Air, water.

(f) An oxide and a carbonate containing oxygen.
Ans. Mercuric oxide and potassium chlorate.

(g) Two substances which undergo rapid oxidation.
Ans. Sodium, carbon.

Question 2.
(a) Taking hydrogen peroxide, how would you prepare oxygen in the laboratory ?
(b) What is the role of manganese dioxide in the preparation of oxygen ?
(c) Write the balanced chemical equation for the above chemical reaction.
(d) Why is hydrogen peroxide preferred in the preparation of oxygen gas ?
(e) Why is oxygen collected by downward displacement of water ?
(f) What happens when a glowing splinter is introduced in a jar containing oxygen ?
(g) What happens when oxygen gas is passed through alkaline pyrogallol solution ?

Answer:
(a) Take manganese dioxide in a round bottom flask and add hydrogen peroxide drop by drop to it, which acts ; a catalyst as shown in the figure. Collect oxygen by downward displacement of water.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-e2

(b) Manganese dioxides acts as a catalyst.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-e3
(d) H2O2 is preferred for lab preparation of oxygen because of following reasons.

  1. No heating is required.
  2. The rate of evolution of oxygen (O2) is moderate and under control.
    H2O2 is a safe chemical.

(e) Since the water is displaced downward by the gas collecting in the jar, the process is called downward displacement of water. The reasons are :

  1. Oxygen is only slightly soluble in water. Therefpre it can be collected over water without fear of excessive dilution.
  2. Oxygen is slightly heavier than air, so it cannot collected over air.

(f) Introduction of glowing splinter in the jar. The glowing splinter rekindles, but the gas does not catch fire.
(g) Alkaline pyrogallol solution turns brown when oxygen is passed through it.

Question 3.
(a) What happens when

  1. mercuric oxide and
  2. potassium nitrate are heated ?

(b) Why is potassium chlorate not used for laboratory preparation of oxygen ?

Answer: (a) 

  1. When mercuric oxide is heated, it decomposes to give mercury and oxygen.
  2. Potassium nitrate on heating gets converted into molten potassium nitrite with the release of oxygen.

(b) Potassium chlorate needs heating for quite sometime (to a high temperature) before it decomposes.

Question 4.
What are oxides ? Give two examples for each of me – tallic and non-metallic oxides.

Answer:
Oxides are binary compounds formed by the chemical combination of a substance metal or a non-metal with oxygen.
Examples :
Metal:

  1. Sodium oxide (Na2O).
  2. Calcium oxide (CaO).

Non-metal:

  1. Sulphur dioxide (SO2).
  2. Carbon dioxide (CO2).

Question 5.
Name the three types of oxidation processes. In which of these large amount of heat and light energy are produced?

Answer:
Oxidation can be categorised into three types :

  1. Spontaneous oxidation
  2. Rapid oxidation
  3. Slow oxidation

Out of the above said three types, rapid oxidation produces large amount of heat and light energy.

Question 6.
What do you observe when the following substances are heated and then tested with moist blue and red litmus – paper?
(a) Sulphur
(b) Phosphorus
(b) Calcium
(d) Magnesium

Answer:
(a) Sulphur : Blue litmus turns red.
(b) Phosphorus : Blue litmus turns red.
(c) Calcium : Red litmus turns blue.
(d) Magnesium : Red litmus turns blue.

Question 7.
Complete and balance the following chemical equations.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-L

Question 8.
(a) Give four uses of oxygen.
(b) How is oxygen naturally renewed in air ?

Answer:
(a) Uses of oxygen

  1. Oxygen is used by firemen, miners, aviators, sea divers and even by every living being.
  2. Oxygen is necessary for burning of fuels.
  3.  Oxygen mixed with hydrogen as fuel produces.a flame with a very high temperature about 2800°C.
  4. As a fuel in spacecraft.

(b) All living beings use atmospheric oxygen in breathing and burning of fuels and in the formation of oxides of nitrogen. Yet amount of oxygen in the air remains more or less constant. This is because green plants return oxygen to the atmosphere by the process of photosynthesis.

Question 9.
(a) What is rust ?
(b) State at least two ways of prevent rusting.

Answer:
(a) Rust: Rust is hydrated ferric oxide (Fe2O3 . x* H2O), which forms a brownish red coating over iron. (* x can be any number.)

(b) Two ways of prevention of rusting :

  1. Painting with red lead.
  2. Oil paint is applied on doors and windows.
  3. Enamel coating. Enamel is a mixture of iron, and steel with silicates.
  4. Coal tar it is used for coating the lower parts of ships and bridges.

Question 10.
State two differences between : Rusting and burning.

Answer:
Difference between rusting and burning

Rusting Burning
  1. Rusting is the process in which iron slowly reacts with oxygen in the air and produces a flaky substance called rust.
  2. Air and moisture are necessary for rusting.
  1. Burning is fast oxidation process in which large amount of energy is produced.
  2. Only air is necessary for burning.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks :

(a) Argon is the most abundant inert gas present in air.
(b) Oxides of sulphur and nitrogen combine with rain water to form sulphuric acid and nitric acid which cause acid rain.
(c) NO2 and CO are the most common air pollutants.
(d) Joseph Priestly discovered the oxygen gas.
(e) Oxygen occupies about 21% of air by volume.

2. Match the following :
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-o2

MULTIPLE CHOICE QUESTIONS

1. A fuel when used releases least amount of pollutants in the air.
(a) sulphur dioxide
(b) chlorofluorocarbon
(c) smoke
(d) CNG

2. The natural way of adding oxygen to air which involves green plants is called
(a) photosynthesis
(b) respiration
(c) burning
(d) dissolution

3. Which one of the following is most likely to be corroded?
(a) a stainless steel cup-board
(b) a galvanised iron bucket
(c) an iron hammer
(d) a tin plated iron box

4. The process by which oxidation of food in our body takes place is
(a) photosynthesis
(b) respiration
(c) decomposition
(d) combustion

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

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Integers Exercise 1A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Evaluate:

  1. 427 x 8 + 2 x 427
  2. 394 x 12 + 394 x (-2)
  3. 558 x 27 + 3 x 558

Solution:

  1. 427 x 8 + 2 x 427 = 427 x (8 + 2) (Distributive property)
    = 427 x 10
    = 4270
  2. 394 x 12 + 394 x (-2) = 394 x (12-2) (Distributive property)
    = 394 x 10
    = 3940
  3. 558 x 27 + 3 x 558 = 558 x (27 + 3) (Distributive property)
    = 558 x 30
    = 16740

Question 2.
Evaluate:

  1. 673 x 9 + 673
  2. 1925 x 101 – 1925

Solution:

  1. 673 x 9 + 673 = 673 x (9 + 1) (Distributive property) = 673 x 10 = 6730
  2. 1925 x 101 – 1925 = 1925 x (101 – 1) (Distributive property) = 1925 x 100 = 192500

Question 3.
Verify:

  1. 37 x {8 +(-3)} = 37 x 8 + 37 x – (3)
  2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
  3. {7 – (-7)} x 7 = 7 x 7 – (-7) x 7
  4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)

Solution:

  1. 37 x {8 + (-3)} = 37 x 8 + 37 x – (3)
    L.H.S. = 37 x {8 + (-3)}
    = 37 x {8-3}
    = 37 x {5}
    = 37 x 5
    = 185
    R.H.S. = 37 x 8 + 37 – 3
    = 37 x (8 – 3)
    = 37 x 5
    = 185
    Hence, L.H.S. = R.H.S.
  2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
    L.H.S. = (-82) x {(_4) + 19}
    = (-82) x {-4 + 19}
    = (-82)x {15}
    = -82 x 15
    =-1230
    R.H.S. = (-82) x (-4) + (-82) x 19
    = -82 x (-4 + 19)
    = -82 x 15
    =-1230
    Hence, L.H.S. = R.H.S.
  3. {7 – (-7)}. x 7 = 7 x 7 – (-1) x 7
    L.H.S. = {7 – (-7)} x 7
    = {7 + 7} x 7
    = {14} x 7
    = 14 x 7
    = 98
    R.H.S. = 7 x 7 – (-7) x 7
    =7 x 7+7 x 7 =
    7 x (7 + 7)
    = 7 x (14)
    = 98
    Hence, L.H.S. = R.H.S.
  4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)
    L.H.S. = {(-15)-8} x-6
    = {-15-8} x-6
    = {-23} x-6
    = -23 x- 6
    = 138
    R.H.S. = (-15) x (-6) – 8 x (-6)
    = -6 x (-15-8)
    = -6 x -23
    = 138
    Hence, L.H.S. = R.H.S.

Question 4.
Evaluate:

  1. 15 x 8
  2. 15 x (-8)
  3. (-15) x 8
  4. (-15) x -8

Solution:

  1. 15 x 8= 120
  2. 15 x (-8) = -120
  3. (-15) x 8 = -120
  4. (-15) x -8 = 120
    (Since the number of negative integers in the product is even)

Question 5.
Evaluate:

  1. 4 x 6 x 8
  2. 4 x 6 x (-8)
  3. 4 x (-6) x 8
  4. (-4) x 6 x 8
  5. 4 x (-6) x (-8)
  6. (-4) x (-6) x 8
  7. (-4) x 6 x (- 8)
  8. (-4) x (-6) x (-8)

Solution:

  1. 4 x 6 x 8 = 192
  2. 4 x 6 x (-8) = -192
    (It have one negative factor)
  3. 4 x (-6) x 8 = -192
    (It have one negative factor)
  4. (-4 )x 6 x 8 = -192
    (It have one negative factor)
  5. 4 x (-6) x (-8) = 192
    (It have two negative factors)
  6. (-4) x (-6) x 8 = 192
    (It have two negative factors)
  7. (-4) x 6 x (-8) = 192
    (It have two negative factors)
  8. (-4) x (-6) x (-8) = -192
    (It have three negative factors)

Question 6.
Evaluate:

  1. 2 x 4 x 6 x 8
  2. 2 x (-4) x 6 x 8
  3. (-2) x 4 x (-6) x 8
  4. (-2) x (-4) X 6 x (-8)
  5. (-2) x (-4) x (-6) x (-8)

Solution:

  1. 2 x 4 x 6 x 8 = 384
  2. 2 x (-4) x 6 x 8 = -384
    (Number of negative integer in the product is odd)
  3. (-2) x 4 x (-6) x 8 = 384
    (Number of negative integer in the product is even)
  4. (-2) x (-4) x 6 x (-8) = -384
    (Number of negative integer in the product is odd)
  5. (-2) x (-4) x (-6) x (-8) = 384
    (Number of negative integer in the product is even)

Question 7.
Determine the integer whose product with ‘-1’ is:

  1. -47
  2. 63
  3. -1
  4. 0

Solution:

  1. -1 x 47 = -47
    Hence, integer is 47
  2. -1 x -63 = 63
    Hence, integer is -63
  3. -1 x 1 = -1
    Hence, integer is 1
  4. -1 x 0 = 0
    Hence, integer is 0

Question 8.
Eighteen integers are multiplied together. What will be the sign of their product, if:

  1. 15 of them are negative and 3 are positive?
  2. 12 of them are negative and 6 are positive?
  3. 9 of them are positive and the remaining are negative?
  4. all are negative?

Solution:

  1. Since out of eighteen integers, 15 of them are negative, which is odd number. Hence, sign of product will be negative (-).
  2. Since out of eighteen integers 12 of them are negative, which is even number. Hence sign of product will be positive (+).
  3. Since out of eighteen integers 9 of them are negative, which is odd number. Hence, sign of product will be negative (-).
  4. Since all are negative, which is even number. Hence sign of product will be positive (+).

Question 9.
Find which is greater?

  1. (8 + 10) x 15 or 8 + 10 x 15
  2. 12 x (6 – 8) or 12 x 6 – 8
  3. {(-3) – 4} x (-5) or (-3) – 4 x (-5)

Solution:

  1. (8 + 10) x 15 or 8 + 10 x 15
    (8 + 10) x 15 = 18 x 15 = 270
    8 + 10 x 15 = 8 + 150 = 158
    ∴(8 + 10) x 15 > 8 + 10 x 15
  2. 12 x (6 – 8) or 12 x 6 – 8
    12 x (6 – 8) = 12 (-2) = -24
    12 x 6 – 8 = 72 – 8 = 64
    ∴12 x 6 – 8 > 12 x (6-8)
  3. {(-3) – 4} x (-5) or (-3) – 4 x (-5)
    {(-3) – 4} x (-5) = {-3 – 4} x (-5) = -7 x -5 = 35
    (-3) – 4 x (-5) = -7 x (-5) = 35
    ∴{(-3) – 4} x (-5) = (-3) – 4 x (-5)

Question 10.
State, true or false :

  1. product of two integers can be zero.
  2. product of 120 negative integers and 121 positive integers is negative.
  3. a x (b + c) = a x b + c
  4. (b – c) x a=b – c x a

Solution:

  1. False.
  2. False.
    Correct : Since 120 integers are even numbers, hence product will be positive and for 121 integers are positive in numbers, hence product will be positive.
  3. False.
    Correct :a x (b + c) ≠ a x b + c
    ab + ac ≠ ab + c
  4. False.
    Correct: (b – c) x a ≠ b – c x a
    ab – ac ≠ b – ca

Integers Exercise 1B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Divide:
(i) 117 by 9
(ii) (-117) by 9
(iii) 117 by (-9)
(iv) (-117) by (-9)
(v) 225 by (-15)
(vi) (-552) ÷ 24
(vii) (-798) by (-21)
(viii) (-910) ÷ – 26

Solution :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 3

Question 2.
Evaluate:
(i) (-234) ÷ 13
(ii) 234 ÷ (-13)
(iii) (-234) ÷ (-13)
(iv) 374 ÷ (-17)
(v) (-374) ÷ 17
(vi) (-374) ÷ (-17)
(vii) (-728) ÷ 14
(viii) 272 ÷ (-17)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 4
Question 3.
Find the quotient in each of the following divisions:
(i) 299 ÷ 23
(ii) 299 ÷ (-23)
(iii) (-384) ÷ 16
(iv) (-572) ÷ (-22)
(v) 408 ÷ (-17)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 5
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 6

Question 4.
Divide:
(i) 204 by 17
(ii) 152 by-19
(iii) 0 by 35
(iv) 0 by (-82)
(v) 5490 by 10
(vi) 762800 by 100

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 7

Question 5.
State, true or false :

  1. 0 ÷ 32 = 0
  2. 0 ÷ (-9) = 0
  3. (-37) ÷ 0 = 0
  4. 0 ÷ 0 = 0

Solution:

  1. True.
  2. True.
  3. False.
    Correct: It is not meaningful (defined)
  4. False.
    Correct: It is not defined.

Question 6.
Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12+18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 x 3+4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 -12 ÷ 4 x 2
(vii) 16 + 8 ÷ 4- 2 x 3
(viii) 16 ÷ 8 + 4 – 2 x 3
(ix) 16 – 8 + 4 ÷ 2 x 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) x (-1) + 14 – 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 1
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 2

 

Integers Exercise 1C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Evaluate:
18-(20- 15 ÷ 3)
Solution:
18-(20- 15 ÷ 3)
= 18 – \(\left( 20\quad -\frac { 15 }{ 5 } \right)\)
= 18 – (20 – 5)
= 18 – 20 + 5
= 18 + 5 – 20
= 23 – 20
= 3

Question 2.
-15+ 24÷ (15-13)
Solution:
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3

Question 3.
35 – [15 + {14-(13 + \(\overline { 2-1+3 }\))}]
Solution:
35- [15 + {14-(13 + \(\overline { 2-1+3 }\))}]
= 35-[15+ 14-(13+4)]
= 35 — [15 + 14 – (13 + 4}]
= 35-{15 + 14-17]
= 35-15-14+ 17
= 35 + 17-15-14
= 52 – 29
= 23

Question 4.
27- [13 + {4-(8 + 4 – \(\overline { 1+3 }\))}]
Solution:
27- [13 + {4-(8 + 4 – \(\overline { 1+3 }\))}]
= 27-[13 +{4-(8+ 4-4)}]
= 27-[13 + {4-8}]
= 27 – [13 + (-4)]
= 21 – [9]
= 27-9
= 18

Question 5.
32 – [43-{51 -(20 – \(\overline { 18 -7 }\))}]
Solution:
32 – [43 – {51 – (20 – \(\overline { 18 -7 }\))}]
= 32-[43 – {51 -(20- 11)}]
= 32-[43-{51 -9}]
= 32-[43 -42]
= 32-1
=31

Question 6.
46-[26-{14-(15-4÷ 2 x 2)}]
Solution:
46 – [26 – {14 – (15 – 4 ÷ 2 x 2)}]
= 46-[26- {14-(15-2 x 2)}]
= 46-[26- {14-(15 -4)}]
= 46-[26- {14- 11}]
= 46 – [26 – 3]
= 46 – 23
= 23

Question 7.
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
= 45-[38- {60 ÷ 3-(6-3)÷ 3}]
= 45-[38 -{20-3 ÷ 3}]
= 45-[38- {20-1}]
= 45-[38- 19]
= 45-19
= 26

Question 8.
17- [17 — {17 — (17 – \(\overline { 17 -17 }\))}]
Solution:
17- [17-{17-(17 –\(\overline { 17 -17 }\))}]
= 17-[17-{17-(17-0)}]
= 17 – [17 – {17 — 17}]
= 17 — [17 — 0]
= 17-17
= 0

Question 9.
2550 – [510 – {270 – (90 – \(\overline { 80 + 7 }\))}]
Solution:
2550- [510-{270-(90-\(\overline { 80 + 7 }\))}]
= 2550 – [510 – {270 – (90 – 87)}]
= 2550 -[510- {270 -3}]
= 2550-[510-267]
= 2550 – 243
= 2307

Question 10.
30+ [{-2 x (25-\(\overline { 13 -3 }\))}]
Solution:
30+ [{-2 x (25-\(\overline { 13 -3 }\))}]
= 30 + [{-2 x (25 – 10)}]
= 30 + [{-2 x 15}]
= 30 + [-30]
= 30-30
= 0

Question 11.
88-{5-(-48)+ (-16)}
Solution:
88- {5-(-48)+ (-16)}
=88 – \(\left\{ 5-\frac { (-48) }{ -16 } \right\}\)
= 88 – {5-3}
= 88 – 2
= 86

Question 12.
9 x (8-\(\overline { 3 +2 }\)) – 2 (2 + \(\overline { 3 +3 }\))
Solution:
9 x (8-\(\overline { 3 +2 }\)) -2(2 + \(\overline { 3 +3 }\))
= 9 x (8 – 5) – 2(2 + 6)
= 9 x 3 – 2 x 8
= 27- 16
= 11

Question 13.
2 – [3 – {6 – (5 – \(\overline { 4 -3 }\))}]
Solution:
2 – [3 – {6 – (5 – \(\overline { 4 -3 }\))}]
⇒ 2 – [3 – {6 – (5 – 1)}]
⇒ 2 – [3 – {6 – 4}]
⇒2 – (3 – 2)
⇒2-1 = 1

Integers Exercise 1D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The sum of two integers is -15. If one of them is 9, find the other.
Solution:
Sum of two integers = -15
One integer = 9
∴ Second integer = -15 – 9
= -(15 + 9)
= -24

Question 2.
The difference between an integer and -6 is -5. Find the values of x.
Solution:
The difference between an integer
= x-(-6) = -5
∴ Value of
⇒ x – (-6) = -5
⇒ x + 6 = -5
x = -5 – 6
x = -11

Question 3.
The sum of two integers is 28. If one integer is -45, find the other.
Solution:
Sum of two integers = 28
One integer = -45
∴ Second integer = 28 – (-45)
= 28 + 45
= 73

Question 4.
The sum of two integers is -56. If one integer is -42, find the other.
Solution:
Sum of two integers = -56
One integer = -42
∴Second integer = -56 – (-42)
= -56+ 42
=-14

Question 5.
The difference between an integer x and (-9) is 6. Find all possible values ofx.
Solution:
The difference between an integer x – (-9) = 6 or -9 – x = 6
∴ Value of x
⇒ x – (-9) = 6 or ⇒ -9 – x = 6
⇒ x + 9 = 6 or Answer-x = 6 + 9
⇒ x = 6 – 9 or ⇒ -x = 15
⇒x = -3 or ⇒ x = -15
Hence, possible values ofx are -3 and -15.

Question 6.
Evaluate:

  1. (-1) x (-1) x (-1) x  ….60 times.
  2. (-1) x (-1) x (-1) x (-1) x …. 75 times.

Solution:

  1. 1 (because (-1) is multiplied even times.)
  2. -1 (because (-1) is multiplied odd times.)

Question 7.
Evaluate:

  1. (-2) x (-3) x (-4) x (-5) X (-6)
  2. (-3) x (-6) x (-9) x (-12)
  3. (-11) x (-15) + (-11) x (-25)
  4. 10 x (-12) + 5 x (-12)

Solution:

  1. (-2) x (-3) x (-4) x (-5) x (-6)
    ⇒ 6 x 20 x (-6) = 120 x (-6)
    = -720
  2. (-3) x (-6) x (-9) x (-12)
    ⇒ 18 x 108
    = 1944
  3. (-11) x (-15) + (-11) x (-25)
    ⇒ 165 + 275
    = 440
  4. 10 x (-12) + 5 x (-12)
    ⇒ -120-60
    = -180

Question 8.

  1. If x x (-1) = -36, is x positive or negative?
  2. If x x (-1) = 36, is x positive or negative?

Solution:

  1. x x (-1) = -36
    -lx = -36
    x = \(\frac { -36 }{ -1 }\)
    x = 36
    ∵ x = 36
    ∴ It is a positive integer.
  2. x x (-1) = 36
    -1x = 36
    x = \(\frac { 36 }{ -1 }\)
    x = -36
    ∵x = -36
    ∴It is a negative integer.

Question 9.
Write all the integers between -15 and 15, which are divisible by 2 and 3.
Solution:
The integers between -15 and 15 are :
-12, -6, 0, 6 and 12
That are divisible by 2 and 3.

Question 10.
Write all the integers between -5 and 5, which are divisible by 2 or 3.
Solution:
The integers between -5 and 5 are :
-4, -3, -2, 0, 0, 2, 3 and 4
That are divisible by 2 or 3.

Question 11.
Evaluate:

  1. (-20) + (-8) ÷ (-2) x 3
  2. (-5) – (-48) ÷ (-16) + (-2) x 6
  3. 16 + 8 ÷ 4- 2 x 3
  4. 16 ÷ 8 x 4 – 2 x 3
  5. 27 – [5 + {28 – (29 – 7)}]
  6. 48 – [18 – {16 – (5 – \(\overline { 4 +1 }\))}]
  7. -8 – {-6 (9 – 11) + 18 = -3}
  8. (24 ÷ \(\overline { 12 -9 }\) – 12) – (3 x 8 ÷ 4 + 1)

Solution:
We know that, if these type of expressions that has more than one fundamental operations, we use the rule of DMAS i.e., First of all we perform D (division), then M (multiplication), then A (addition) and in the last S (subtraction).

  1. (-20) + (-8) ÷ (-2) x 3
    ⇒ -20 + 4 x 3
    ⇒ -20+ 12
    =-8
  2. (-5) – (-48) ÷ (-16) + (-2) x 6
    ⇒ (-5) – 3 + (-2) x 6
    ⇒ -5 – 3 – 12
    ⇒ -8- 12
    = -20
  3. 16 + 8 ÷ 4 – 2 x 3
    ⇒ 16 + 2 – 2 x 3
    ⇒16 + 2 – 6
    ⇒ 18-6
    = 12
  4. 16 ÷ 8 x 4 – 2 x 3
    ⇒ 2 x 4 – 2 x 3
    ⇒ 8 – 6
    = 2
  5. 27 – [5 + {28 – (29 – 7)}]
    ⇒ 27 – [5 + {28 – 22}]
    ⇒ 27 – [5 + 6]
    ⇒ 27 — 11
    = 16
  6. 48-[18-{16-(5 – \(\overline { 4 +1 }\))}]
    ⇒ 48-[18-{16-(5-5)}]
    ⇒ 48-[18- {16-0)}]
    ⇒ 48-[18- 16]
    ⇒ 48 – 2
    = 46
  7. -8 – {-6 (9 – 11) + 18 ÷ -3}
    ⇒ -8 – {-6 (-2) – 6}
    ⇒ -8- {12-6}
    ⇒ -8 – {6}
    ⇒ -8-6
    = -14
  8. (24 ÷ \(\overline { 12 -9 }\) – 12) – (3 x 8 = 4 + 1)
    ⇒ (24 ÷ 3-12)-(3 x 2 + 1)
    ⇒ (8- 12)-(6+ 1)
    ⇒ —4 — 7
    = —11

Question 12.
Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.
Solution:
Required number = (Sum of all integers between 20 and 30 – Integers between 20 and 30)
(20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 )
⇒ 20 + 30 = 50
∴ Required number = 50

Question 13.
Add the product of (-13) and (-17) to the quotient of (-187) and 11.
Solution:
(-13) x (-17)+ (-187- 11)
⇒ (-13) x (-17) + (-17)
⇒ 221 – 17 = 204

Question 14.
The product of two integers is-180. If one of them is 12, find the other.
Solution:
The product of two integers = -180 One integer = 12
∴ Second integer = -180 – 12 = -15

Question 15.

  1. A number changes from -20 to 30. What is the increase or decrease in the number?
  2. A number changes from 40 to -30. What is the increase or decrease in the number?

Solution:

  1. ∵A number changes from = -20 to 30
    ⇒ -20 – 30 = -50
    ∴-50, it will be increases.
  2. ∵A number changes from = 40 to -30
    ⇒ 40 – (-30)
    40 + 30 = 70
    ∴70, it will be decreases

 

 

Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

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Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

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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The method in which the value of a unit (one) quantity is first calculated to get the value of any other quantity is called the unitary method.
In unitary method, we come across two types of variations :
(i) Direct-variation
(ii) Inverse-variation.

(i) Direct variation : Increase in one quantity causes increase in the other and decrease in one quantity causes decrease in the other.
(ii) Inverse variation : Increase in one quantity causes decrease in the other and decrease in one quantity causes increase in the other.
This is found in the sums of speed, work done etc.

EXERCISE 7 (A)

Question 1.
Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles ?

Answer:
Weight of 8 articles = 4.8 kg
Weight of 1 article = \(\frac { 4.8 }{ 8 }\) kg
and weight of 11 articles =\(\frac { 4.8 }{ 8 }\) x 11 kg
= 0.6 x 11 = 6.6 kg

Question 2.
6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 1

Question 3.
8 men complete a work in 6 hours. In how many hours will 12 men complete the same work ?

Answer:
8 men can complete a work in = 6 hours
1 man can complete the work in = 6×8 hours
12 men can complete the work in = \(\frac { 6 x 8 }{ 12 }\) = 4 hours

Question 4.
If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one, burn ?

Answer:
25 cm long candle burn in = 45 minutes
1 cm long candle will burn in = \(\frac { 45 }{ 25 }\) mintues
25 + 5 = 30 cm long candle will burn in
= \(\frac { 45 x 30 }{ 25 }\)minutes = 54 minutes

Question 5.
A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages ?

Answer:
For typing 24 pages, time is required = 80 minutes
For typing 1 page, time is required =\(\frac { 80 }{ 24 }\) minutes
and for typing 87 pages, time is required
= \(\frac { 80 x 87 }{ 24 }\) minutes = 290 minutes

Question 6.
Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support the same family ?

Answer:
Rs. 750, can support a family for = 15 days
Re. 1 will support for = \(\frac { 15 }{ 750 }\)days
and Rs. 2,500 will support for = \(\frac { 15 }{ 750 }\)x 2500 days = 50 days

Question 7.
400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last ?

Answer:
400 men have provisions for = 23 weeks
1 man will have provisions for = 23 x 400 weeks
and 400 + 60 = 460 men will have provisions for = \(\frac { 23 x 400 }{ 460 }\) weeks = 20weeks

Question 8.
200 men have provisions for 30 days. If 50 men left, the same provisions would last for the remaining men, in how many days?

Answer:
200 men have provisions for = 30 days
1 man will have provisions for = 30 x 200 days
200 – 50 = 150 men will have provisions
for = \(\frac { 30 x 200 }{ 150 }\) days = 40 days

Question 9.
8 men can finish a certain amount of provisions in 40 days. If 2 more men join with them, find for how many days the same amount of provisions be sufficient ?

Answer:
8 men can finish a provision in = 40 days
1 man will finish in = 40 x 8 days
8+2=10 men will finish in =\(\frac { 40 x8 }{ 10 }\)
= 32 days

Question 10.
If interest on Rs. 200 be Rs. 25 in a certain time, what will be the interest on Rs 750 for the same time ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 3

Question 11.
If 3 dozen eggs cost Rs. 90, find the cost of 3 scores of eggs. (1 score = 20)

Answer:
3 dozen = 3 x 12 = 36 eggs,
3 scores = 3 x 20 = 60
The cost of 36 eggs is = Rs. 90
The cost of 1 egg will be = Rs. \(\frac { 90 }{ 36 }\)
∴ Cost of 60 eggs will be = Rs. \(\frac { 90 x60 }{ 36 }\)
= Rs. 150

Question 12.
If the fare for 48 km is Rs. 288, what will be the fare for 36 km ?

Answer:
Fare for 48 km = Rs. 288
fare for 1 km = Rs. \(\frac { 288 x 36 }{ 48 }\) = Rs. 216

Question 13.
What will be the cost of 3.20 kg of an item, if 3 kg of it costs Rs. 360 ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 4

Question 14.
If 9 lines of a print, in a column of a book contains 36 words. How many words will a column of 51 lines cqntain ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 5

Question 15.
125 pupil have food sufficient for 18 days. If 25 more pupil join them, how long will the food last now ? What assumption have you made to come to your answer ?

Answer:
Pupils in the beginning = 125
More pupils joined = 25
Total pupils = 125 + 25 = 150
Food is sufficient for 125 pupils for = 18 days
Food will be sufficient for 1 pupil for = 18 x 125 days (less pupil more days)
and food will be sufficient for 150 pupils = \(\frac { 18 x 125 }{ 150 }\) days (more pupil more days)
= \(\frac { 18 x 5 }{ 6 }\) 15 days

Question 16.
A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how many hours per day must he work in order to make the same chair in 4 days ?

Answer:
A chair is completed in 3 days working per day = 8 hours
Then their will be completed in 1 day working for = 8 x 3 hours per day (less days more hours)
and it will be completed in 4 days working for = \(\frac { 8 x 3 }{ 4 }\)= 6 hours per day.

Question 17.
A man earns ₹5,800 in 10 days. How much will he earn in the month of February of a leap year?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 6

Question 18.
A machine is used for making rubber balls and makes 500 balls in 30 minutes. How many balls will it make in 3\(\frac { 1 }{ 2 }\) hours?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 7

Question 19.
In a school’s hostel mess, 20 children consume a certain quantity of ration in 6 days. However, 5 children did not return to the hostel after holidays. How long will the same amount of ration last now?

Answer:
Total number of children = 20
20 children consume a certain quantity of ration in = 6 days
1 children consume a certain quantity of ration in = 6 x 20 days
As 5 children did not return to the hostel after holidays.
Then number of children in hostel = 20-5 = 15
Hence, 15 children consume certain quantity 6×20
of ration in = \(\frac { 6 x 20 }{ 15 }\) days = 8 days

EXERCISE 7 (B)

Question 1.
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 8

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 9

Question 2.
3\(\frac { 1 }{ 2 }\)m of cloth costs Rs. 168 ; find the cost of 4\(\frac { 1 }{ 3 }\)m of the same cloth.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 10

Question 3.
A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 12

Question 4.
In 2 days and 20 hours, a watch gains 20 sec ; find how much time will the watch take to gain 35 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 13

Question 5.
50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?

Answer:
Land is same in both the cases.
Now 50 men can mow land in = 3 days
∴ 1 man will mow it in = 3 x 50 days
and 15 men will mow it in = \(\frac { 3 x 50 }{ 15 }\) = 10 days

Question 6.
The wages of 10 workers for a six days week are Rs, 1,200. What are the one day wages: (i) of one worker ? (ii) of 4 workers?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 14

Question 7.
If 32 apples weigh 2 kg 800 g. How many apples will there be in a box, containing 35 kg of apples ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 15

Question 8.
A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?

Answer:
For 240 km, diesel is needed = 20 litres
∴ for 1 km, diesel will be needed 20

Question 9.
A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men ?

Answer:
1200 men has provision for = 15 days
1 man will have that provision for = 15 x 1200 days
∴1200 + 600 = 1800 men will has that provisions for =\(\frac { 15 x 1200 }{ 1800 }\)days
= 10 days

Question 10.
A camp has provisions for 60 pupil for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupil ?

Answer:
60 pupil have provision for = 18 days 1 pupil will have provision for = 18 x 60 days (less pupils more days)
and 72 pupils will have provision for = \(\frac { 18 x 60 }{ 72 }\) days
= 15 days.

EXERCISE 7 (C)

Question 1.
A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 16

Question 2.
A and B working together can do a piece of work in 10 days B alone can do the same work in 15 days. How long will A alone take to do the same work ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 17

Question 3.
A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in : (i) one day (ii) 2 days.
What part of work will be left, after they have worked together for 2 days ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 18

Question 4.
A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.
(i) How much work is done in these 3 hours ?
(ii) How much work is still left ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 19

Question 5.
A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 20

Question 6.
Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 21

Question 7.
Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.
Find:
(i) work done by Mohit in 20 days.
(ii) work left after Mohit has worked on it for 20 days.
(iii) time taken by Anuj to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 22
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 23

Question 8.
Joseph and Peter can complete a work in 20 hours and 25 hours respectively.
Find :
(i) work done by both together in 4 hrs.
(ii) work left after both worked together for 4 hrs.
(iii) time taken by Peter to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 24

Question 9.
A is able to complete \(\frac { 1 }{ 3 }\) of a certain work in 10 hrs and B is able to complete\(\frac { 2 }{ 5 }\) of the same work in 12 hrs.
Find:
(i) how much work can A do in 1 hour ?
(ii) how much work can B do in 1 hour ?
(iii) in how much time will the work be completed, if both work together.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 25

Question 10.
Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare :
(i) one chair ?
(ii)14 chairs of the same kind?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 26

Question 11.
A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 27

Selina Concise Mathematics class 7 ICSE Solutions – Data Handling

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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

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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Data Handling Exercise 21A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Consider the following numbers :
68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77 and 71.
(i) Arrange these numbers in ascending order.
(ii) What the range of these numbers?
Solution:
(i) When the above data are written in ascending order. We get,
52, 63, 63, 68, 70, 71, 73, 75, 75, 76, 77, 82, 83, 90, 92, 93, 99, 103, 105, 115
(ii) Range of given numbers = Largest number – Smallest number
= 115-52 = 48

Question 2.
Represent the following data in the form of a frequency distribution table :
16, 17, 21, 20, 16, 20, 16, 18, 17, 21, 17, 18, 19, 17, 15, 15, 19, 19, 18, 17, 17, 15, 15, 16, 17, 17, 19, 18, 17, 16, 15, 20, 16, 17, 19, 18, 19, 16, 21 and 17.
Solution:
The frequency distribution for these data will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 2

Question 3.
A die was thrown 20 times and following scores were recorded.
2, 1, 5, 2, 4, 3, 6, 1, 4, 2, 5, 1, 6, 2, 6, 3, 5, 4, 1 and 3.
Prepare a frequency table for the scores.
Solution:
The frequency table for the scores will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 3

Question 4.
Following data shows the weekly wages (in ₹) of 10 workers in a factory.
3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250 and 4000
(i) Prepare a frequency distribution table.
(ii) What is the range of wages (in ₹)?
(iii) How many workers are getting the maximum wages?
Solution:
(i) The frequency table for the wages of 10 workers will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 4
(ii) Range of wages (₹) = ₹4750 – ₹3500 = ₹1250
(iii) One

Question 5.
The marks obtained by 40 students of a class are given below :
80, 10, 30, 70, 60, 50, 50, 40, 40, 20, 40, 90, 50, 30, 70, 10, 60, 50, 20, 70, 70, 30, 80, 40,20, 80, 90, 50, 80, 60, 70, 40, 50, 60, 90, 60, 40, 40, 60 and 60
(i) Construct a frequency distribution table.
(ii) Find how many students have marks equal to or more than 70?
(iii) How many students obtained marks below 40?
Solution:
(i) The frequency distribution table will be shown as below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 5
(ii)Students have marks equal to or more than 70 = 5 + 4 + 3 = 12
(iii) Students obtained marks below 40 = 2 + 3 + 3 = 8 students

Question 6.
Arrange the following data in descending order:
3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9 and 3.4.
(i) Determine the range.
(ii) How many numbers are less than 3.5?
(iii) How many numbers are 3.8 or above?
Solution:
Descending order : 4.1, 4.0, 3.9, 3.9, 3.8, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2, 3.1
(i) Range = 4.1 – 3.1 = 1
(ii) Number less than 3.5 = 4
i.e., 3.4, 3.3, 3.2, 3.1
(iii) Number are 3-8 or above = 6
i.e., 3.8, 3.8, 3.9, 3.9, 4.0, 4.1

Data Handling Exercise 21B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the mean of 53, 61, 60, 67 and 64.
Solution:
Mean of 53, 6i, 60, 67 and 64
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 6

Question 2.
Find the mean of first six natural numbers.
Solution:
First six natural numbers are : 1, 2, 3, 4, 5, 6
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 7

Question 3.
Find the mean of first ten odd natural numbers.
Solution:
First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 8

Question 4.
Find the mean of all factors of 10.
Solution:
The factor of 10 are 2 and 5
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 9

Question 5.
Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.
Solution:
Mean of x + 3, x + 5, x + 7, x + 9 and x + 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 10

Question 6.
If different values of variable x are 19.8,15.4,13.7,11.71,11.8, 12.6,12.8,18.6,20.5 and 2.1, find the mean.
Solution:
19. +15.4 +13.7 +11.71 +11.8 +12.6 + 12.8 +18.6 + 20.5 +21.1
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 11

Question 7.
The mean of a certain number of observations is 32. Find the resulting mean, if each observation is,
(i) increased by 3
(ii) decreased by 7
(iii) multiplied by 2
(iv) divided by 0.5
(v) increased by 60%
(vi) decreased by 20%
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 12

Question 8.
The pocket expenses (per day) of Anuj, during a certain week, from monday to Saturday were ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25. Find the mean pocket expenses per day.
Solution:
The pocket expenses (per day) during a certain week are : ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 13

∴Anuj expenses per day = ₹85.75

Question 9.
If the mean of 8, 10, 7, x + 2 and 6 is 9, find the value of x.
Solution:
The mean 8, 10, 7, x + 2 and 6 is 9
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 14

Question 10.
Find the mean of first six multiples of 3.
Solution:
The six multiples of 3 are 3, 6, 9, 12, 15, 18
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 15

Question 11.
Find the mean of first five prime numbers.
Solution:
The first five prime numbers are 2, 3, 5, 7, 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 16

Question 12.
The mean of six numbers :x-5,x- 1, x, x + 2, x + 4 and x + 12 is 15. Find the mean of first four numbers.
Solution:
The mean of six numbers are x – 5, x – 1,x,x + 2,x + 4 and x + 12 is 15
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 17
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 18

Question 13.
Find the mean of squares of first five whole numbers.
Solution:
First five whole numbers are 0, 1, 2, 3, 4
Then square the whole prime numbers
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 19

Question 14.
If the mean of 6, 4, 7, p and 10 is 8, find the value of p.
Solution:
The mean of 6, 4, 7, p and 10 is 8
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 20

Question 15.
Find the mean of first six multiples of 5.
Solution:
Six multiples of 5 are :
5, 10, 15, 20, 25 and 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 21

Question 16.
The rainfall (in mm) in a city on 7 days of a certain week is recorded as follows
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 22
Find the total and average (mean) rainfall for the week.
Solution:
The rainfall in a city on 7 days are 0.5, 2.7, 2.6, 0.5, 2, 5.8, 1.5
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 23

Question 17.
The mean of marks scored by 100 students was found to be 40, later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
Mean of 40 observations = 100
Total sum of 40 observations = 100 × 40 = 4000
Incorrect total of 40 observation is = 4000
Correct total of 40 observations = 4000 – 83 + 53 = 3970
∴ Correct mean = \(\frac { 3970 }{ 100 }\) = 39.70

Question 18.
The mean of five numbers is 27. If one number is excluded, the mean of remaining numbers is 25. Find the excluded number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On excluding an observation, the mean of remaining 6 observations = 25
⇒ Total of remaining 4 observations = 25 x 4 = 100
⇒ Included observation = Total mean of 5 observations – Total mean of 4 observations
= 135- 100 = 35

Question 19.
The mean of 5 numbers is 27. If one new number is included, the new mean is 25. Find the included number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 x 5 = 135
On including an observation the mean of 6 observation = 25 x 6 = 150
⇒ Included observations = Total Mean of 6 observations – Total mean of 5 observations = 150- 135 = 15

Question 20.
Mean of 5 numbers is 20 and mean of other 5 numbers is 30. Find the mean of all the 10 numbers taken together.
Solution:
The mean of 5 number = 20
Then, mean of other 5 number = 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 24

Question 21.
Find the median of:
(i) 5,7, 9, 11, 15, 17,2, 23 and 19
(ii) 9, 3, 20, 13, 0, 7 and 10
(iii) 18, 19, 20, 23, 22, 20, 17, 19, 25 and 21
(iv) 3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9 and 7.6.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 25
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 27
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 28

Question 22.
Find the mean and the mode for the following data :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 29
Solution:
We prepare the table given below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 30

Question 23.
Find the mode of:
(i) 5, 6, 9, 13, 6, 5, 6, 7, 6, 6, 3
(ii) 7, 7, 8, 10, 10, 11, 10, 13, 14
Solution:
(i) Arranging the Numbers in ascending order : 3, 5, 5, 6, 6, 6, 6, 6, 7, 9, 13
Mostly repeated term = 6
∴ Mode = 6
(ii) Arranging the Numbers in ascending order = 7, 7, 8, 10, 10, 10, 11, 13, 14
Mostly repeated term =10
∴ Mode = 10

Question 24.
Find the mode of :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 31

Solution:
(i) Since, the frequency of number 18 is maximum
∴Mode = 18
(ii) Since, the frequency of number 41 is maximum
∴ Mode = 41

Question 25.
The heights (in cm) of 8 girls of a class are 140,142,135,133,137,150,148 and 138 respectively. Find the mean height of these girls and their median height.
Solution:
Arranging in ascending order : 133, 135, 137, 138, 140, 142, 148, 150
Here, number of girls = 8 which is even
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 32

Question 26.
Find the mean, the median and the mode of:
(i) 12, 24, 24, 12, 30 and 12
(ii) 21, 24, 21, 6, 15, 18, 21, 45, 9, 6, 27 and 15.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 33.
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 34

Question 27.
The following table shows the market positions of some brands of soap.
Draw a suitable bar graph :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 35
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 36

Question 28.
The birth rate per thousand of different countries over a particular period of time is shown below.
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 37
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 38

Selina Concise Biology Class 7 ICSE Solutions – Classification of Animals

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Selina Concise Biology Class 7 ICSE Solutions – Classification of Animals

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 3 Classification of Animals

Synopsis

  •  The animals can move from one place to another in search of food and shelter and this is called locomotion.
  •  The vertebrates can be classified into five classes:
    1.  Pisces (Fishes)
    2.  Amphibia (Frogs)
    3.  Reptilia (Lizards and Snakes)
    4.  Aves (Birds)
    5. (Mammalia (Milk – nourishing animals)
  • Pisces / Fishes
    1.  have streamlined body shape
    2.  Locomotion with the help of fins
    3.  Body covered with scales.
    4.  Breathe through gills.
    5. Example: Dogfish, Catla
  •  Amphibia / Frogs
    1.  can live in water as well as on land.
    2. always lay their eggs in water.
    3. body covered by a slimy and slippery skin
    4. breathe through lungs and skin.
    5. Example: Frog and toads.
  •  Reptilia
    1.  Mostly live on land
    2.  Skin is dry and scaly
    3.  Breathe through lungs
    4. Females lay eggs on land
    5. Example: Lizards, snakes, crocodiles
  • Aves / Birds
    1. Body covered with feathers.
    2. Have wings to fly.
    3. Scales only on legs.
    4. Have jaws with homy beak and have no teeth.
    5. Example: Pigeon, hen
  • Mammalia / Milk – nourishing animals.
    1. Body covered with hairs.
    2. Posses projecting external ears.
    3. Give birth to young ones.
    4. Mothers suckle their young ones.
    5. Have a tail and four limbs. (Tail may become vestigeal)
    6. Example: dog, tiger, man.
  • Invertebrates can be further divided into nine groups.
    1. Protozoans
    2. Porifera
    3. Coelenterates
    4. platyhelminths
    5. Nemathelminths
    6. Annelids
    7. Molluscs
    8. Arthropoda
    9. Echinoderms
  • Coelenterates
    1. Now called cnidarians
    2. Body is tube like with only one opening called the mouth.
    3. Mouth is surrounded by finger like processes called tentacles for catching food.
    4. Body radially symmetrical
    5. Example: Hydra, Sea-anemone, jelly fish
  • Flatworms / Platyhelminths:
    are usually found as parasites in the bodies of other animals.
    Example: Tapeworm, liver fluke.
  •  Ascaris: The round worm is found in the small intestine of especially those who eat with the unwashed hands.
  •  Annelids:
    1.  are also called segmented worms
    2.  body is composed of rings or segments
    3.  have a body cavity.
    4. have special organs of excretion called nephridia.
      Example: earthworm, leech.
  •  Arthropods can be further divided into
    1.  Crustacea : head and thorax are fused and have many jointed legs.
      Example: crab, lobsters etc.
    2.  Myriagoda: Body is divided into many segments and has one or two pairs of legs on each segment.
      Example: Centipede, millipede.
  •  Insecta: Body is divided into three regions – head, thorax and abdomen.
    — Has three pairs of legs.
    — Have two pairs of wings.
    Example: ant, housefly, butterfly.
  •  Arachnida: Head and thorax fused
    — Have four pairs of legs.
    — Have no wings.
    Example: Spider, Scorpion
  •  Echinoderms
    — also called spiny-skinned animals.
    — Body is star – like or ball – like
    — Have no head or tail.
    — Have no left or right side.
    Example: Starfish, sea urchin.
  • A species can be defined as a group of individuals having common characteristics and which come together to pro¬duce young ones.
  •  Scientific name consists of two parts. The first part is the genus name while the second part is the species name.
    This type of naming is called Binomial nomenclature.
  • The animals can be classified also on the basis of their food habits into as follows.
    (a) Herbivorous:Feed on plants e.g. cow, goat.
    (b) Carnivorous:Feed on the flesh of other animals  e.g. lion, tiger etc.
    (c) Omnivorous:Feed on both plants as well as flesh of other animals, e.g. man, bear etc.
    (d) Parasites:Live either inside or on the outside of the body of other animals and plants and take food from them.
    Example: Leech, mosquitoes etc.

Activity 3
Look at the four animals shown alongside.
Which four classes of vertebrates are represented by them ? Name these classes.
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 1

Answer:
1. Class Mammalia
2. Class Mammalia
3. Class Reptilia
4. Class Pisces

Review Questions

MULTIPLE CHOICE QUESTIONS

1. Tick (✓) the appropriate answer:

(i) Identify the aquatic animal with scaly skin which breathe with gills –
(a) Rohu
(b) Tortoise
(c) Sparrow
(d) Rat

(ii) The unicellular organism causing malaria –
(a) Amoeba
(b) Paramecium
(c) Euglena
(d) Plasmodium

(iii) Identify the animal which is not an Arthropoda —
(a) Prawn
(b) Butterfly
(c) Earthwonn
(d) Spider

(iv) Scientist who introduced binomial nomenclature is —
(a) Charles Darwin
(b) Carolus Linnaeus
(c) Robert Hooke
(d) Gregor Mendel

Short Answer Questions
1. Give two examples of each of the following:
(i) Amphibians:
Ans. Amphibians: 1. Frog 2. Toad
(ii) Segmented worms:
Ans. Segmented worms: 1. Earthworm 2. Leech
(iii) Reptiles:
Ans. Reptiles: 1. Snake 2. Lizard
(iv) Coelenterates:
Ans. Coelenterates : 1. Hydra 2. Jellyfish
(v) Arthropods:
Ans. Arthropods: 1. Crab 2. Centipede
(vi) Flatworms:
Ans. Flatworms: 1. Tapeworm 2. Liverfluke

2. Give names of two animals which are found as parasites inside the human intestine.
Ans. (a) Tapeworm (b)Ascaris ’

3. Name one example each of an animal which shows the following characteristics:
(i) Fixed animals with a pore-bearing body:
Ans. Fixed animals with a pore-bearing body: sponge
(ii) Star-shaped body:
Ans. Star-shaped body: Star-fish
(iii) Can live in water as well as on land:
Ans. Can live in water as well as on land: Frog
(iv) Has a flattened ribbon-like body:
Ans. Has a flattened ribbon-like body: Tapeworm

4. Write one difference each between the following pairs:
(i) Porifera and Coelenterata.
(ii) Arthropoda and mollusca.
(iii) Invertebrates and Vertebrates
(iv) Platyheminthes and Nematoda
Answer:
(i) Porifera and Coelenterata.
 Porifera

  1.  Body is porous i. e. bears many tiny pores to draw water into the body cavity.
  2.  e.g. Sponge

Coelenterata

  1. Sac-like body with only one opening i.e. mouth.
  2. e.g. Jelly fish, hydra,sea-anemone.

(ii) Arthropoda and mollusca.
Arthropoda 

  1.  These are animals with
  2.  They have segmented body.
  3.  They may or may not have wings
    Example: Crab.

Mollusca

  1.  Move with the help of a muscular foot.
  2.  Soft body which is not segmented.
  3.  Body enclosed in a hard shell Example: Octopus

(iii) Invertebrates and Vertebrates

Invertebrates

  1.  The animals which do not have a back bone.
  2.  They are further classified into nine groups.
    Example: Octopus, Starfish.

Vertebrates

  1.  The animals which have a back bone or a vertebral column.
  2.  They are further classified in to five groups.
    Example: Human Being, Lizard.

(iv) Platyheminthes and Nematoda
Platyheminthes

  1.  Body thin and flattened.
  2.  Mostly live as parasites in the bodies of other animals (hosts)
    e.g. Tapeworm.

Nematoda

  1. Body is rounded and unsegmented.
  2. Mostly live as parasites in the body of animals including humans.
    e.g. Roundworm commonly called Ascaris.

5. Match the animals given under column A with their respective classification group given under column B –

Column A                                          Column B
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 2
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 3
Answer:

Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 4
  

6. Write the characteristics of class Aves with reference to their body covering and jaws.
Answer:
The characteristics of class Aves are:

  1.  Body is covered with feathers.
  2.  They have wings to aid flying
  3.  They have scales on legs.
  4.  They have no teeth.
  5. They have jaws provided with homy beaks

7. Categorise the following animals under their appropriate columns of classification.
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 5

Answer:
Worms – Arthropods, Butterfly, Ascaris, Scorpion, Honey bee, Liverfluke, Leech, grasshopper, Eathworm
Molluscs – Snail
Fishes – Rohu
Amphibians – Toad, Frog
Reptiles – Snake, Lizard, Turtle
Birds – Parrot, Pigeon
Mammals – Rat, Bat, Dog, Cattle, Cow, Rabbit, Monkey, Elephant