Prasanna

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 6 Fractions Ex 6.1

Question 1.
Write the following division as fractions:
(i) 3 ÷ 7
(ii) 11 ÷ 78
(iii) 113 ÷ 128
Solution:
(i) 3 ÷ 7 = \(\frac{3}{7}\)
(ii) 11 ÷ 78 = \(\frac{11}{78}\)
(iii) 113 ÷ 128 = \(\frac{113}{128}\)

Question 2.
Write the following fractions in words.
(i) \(\frac{2}{7}\)
(ii) \(\frac{3}{10}\)
(iii) \(\frac{15}{28}\)
Solution:
(i) \(\frac{2}{7}\) = Two – Seventh
(ii) \(\frac{3}{10}\) = Three – Tenth
(iii) \(\frac{15}{28}\) fifteen – Twenty eighth

Question 3.
Write the following fractions in number form:
(i) one – sixth
(ii) three – eleventh,
(iii) seven-fortieth
(iv) thirteen – one hundred twenty fifth
Solution:
(i) One – sixth = \(\frac{1}{6}\)
(ii) Three-eleventh = \(\frac{3}{11}\)
(iiii) seven-forteith = \(\frac{7}{40}\)
(iv) Thirteen-one hundred twenty fifth = \(\frac{13}{125}\)

Question 4.
What fraction of each of the following is shaded part?

Solution:

Question 5.
Shade the parts of the following figures according to given fractions:

Solution:

Question 6.
In the adjoining figure, if we say that the shaded region is \(\frac{1}{4}\) of the whole region, then identify the error in it.

Solution:
The whole rectangle is not divided into four equal parts.

Question 7.
Write the fraction in which
(i) numerator = 5 and denominator = 13
(ii) denominator = 23 and numerator = 17
Solution:
(i) \(\frac{5}{13}\)
(ii) \(\frac{17}{23}\)

Question 8.
Shabana has to stitch 35 dresses. So, ar she has stitched 21 dresses. What fraction of dresses has she stitched?
Solution:
Number of dresses she had to stiches = 35
Number of dresses she has finished = 21
∴ Fraction of dresses she has finished = \(\frac{21}{35}=\frac{3}{5}\)

Question 9.
What fraction of a day is 8 hours ?
Solution:
Number of hours in a day = 24 hours
∴ Required fraction = \(\frac{8}{24}\)

Question 10.
What fraction of an hour is 45 minutes ?
Solution:
An hour (1 hour) = 60 minutes
∴ Required fraction = \(\frac{45}{60}\)

Question 11.
How many natural numbers are there from 87 to 97? What fraction of them are prime numbers?
Solution:
The natural numbers from 87 to 97 are 87, 88, 89, 90, 91, 92, 93, 94, 95, 96 and 97. Total number of natural number = 11 Out of these, the prime numbers are 87 and 97
Total number of these prime numbers = 2
∴ Required fraction = \(\frac{2}{11}\)

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Check Your Progress

Question 1.
State which of the given collections are sets:
(i) Collection of all poor people of Dhanbad.
(ii) Collection of all difficult problems in your maths book.
(iii) Collection of all fools.
(iv) Collection of all countries of Asia.
(v) Collection of four countries of Asia.
(vi) Collection of three cities of India whose name start with the letter ‘J’.
(vii) Collection of all people in this world over 50 year of age.
Solution:
(i) It is not set because elements are not countable.
(ii) It is also not set because problem are different to different students
(iii) It is also not set.
(iv) It is set because the countries are countable.
(v) It is not set because the elements are countable but not defined.
(vi) It is also not set because cities are not defined.
(vii) It is a set.

Question 2.
If A = (3, 5, 7, 9, 11}, then write which of the following statements are true. If a statement is not true, mention why.
(i) 3 ϵ A
(ii) 5, 9 ϵ A
(iii) 8 ∉ A
(iv) 7 ∉ A
(v) {3} ϵ A
(vi) {5, 9} ϵ A
Solution:
(i) 3 ϵ A is true. 3 is element of set.
(ii) 5, 9 ϵ A is true because these are element of set.
(iii) 8 ∉ A is true because 8 is not element of set.
(iv) It is false because 7 is element of set.
(v) It is false because {3} is different set and not element.
(vi) It is not true because (5, 9} is different set.

Question 3.
Write the following sets in the roster farm :
(i) A = (x | x is a month of a year having 30 days}
(ii) B = (x | x = 2n, n ϵ W and n < 5}
(iii) C = (x | x ϵ N and x² < 40}
(iv) D = (all letters in the word PERMISSION}
(v) E = (x : x ϵ I and x² < 10}
(vi) F = (x : x ϵ N, 15 < x < 50 and x is divisible by 6}
(vii) the set of whole numbers which are greater than 14 and divisible by 7.
(viii)the set of signs of four fundamental operation of arithmetic.
Solution:
(i) A= (April, June, September, November}
(ii) B = (0, 2, 4, 6, 8}
(iii) C = (1, 2, 3, 4, 5, 6}
(iv) D = (P, E, R, M, I, S, O, N}
(v) E = {-3, -2, -1, 0, 1, 2, 3}
(vi) F = (18, 24, 30, 36, 42, 48}
(vii) (21, 28, 35, 42 }
(viii) {x, – +, -}

Question 4.
Write the following sets in set builder form :
(i) A = (2, 3, 5, 7, 11, 13, 17, 19}
(ii) B = (all months of a year}
(iii) C = (Monday, Tuesday, Wednesday}
Solution:
(i) A = {x | x is a prime number x < 20}
(ii) B = {x : x is any month of a year}
(iii) C = {x | x is any of the first three days of a week}

Question 5.
Write the following sets in roster form and also in set builder form :
(i) A = {even whole numbers which are less than 50}
(ii) B = {two digit numbers which are perfect square}
(iii) The set of letters in the word MUSSOORIE
Solution:
(i) A = {0, 2, 4, ………. , 48}
A = {x/x ϵ W and x is an even number <50}

(ii) B = {16, 25, 36, 49, 64, 81}
B = {x : x is perfect square and two digit number}

(iii) {M, U, S, O, R, I, E}
{x | x is a letter in the word MUSSORIE.

Question 6.
The sets on the left are in tabular form while the sets on the right are in set builder form. Match them.
(i) {2, 3} – (a) {x/x ϵ N and x < 6}
(ii) {P, A, Y} – (b) {x: x is a prime factor of 6}
(iii) {1, 3, 5} – (c) {x | x is an odd natural number less than 6}
(iv) {1, 2, 3, 4, 5} – (d) {x/x is a letter in word PAPAYA}
Solution:
(i) {2, 3} – (b) – (b) {x: x is a prime factor of 6}
(ii) {P, A, Y} – (d) {x/x is a letter in word PAPAYA}
(iii) {1, 3, 5} – (c) {x | x is an odd natural number less than 6}
(iv) {1, 2, 3, 4, 5} – (a) {x/x ϵ N and x < 6}

Question 7.
Classify the following sets as empty set, finite set or infinite set:
(i) The set of all even prime number > 2.
(ii) The set of even prime numbers.
(iii) The set of prime numbers less than one crore.
(iv) {All points on a line segment of length 3cm}.
Solution:
(i) Empty set.
(ii) Finite set.
(iii) Finite set.
(iv) Infinite set.

Question 8.
Find the cardinal number of the following sets.
(i) A = {x | x is a consonant in the word HUNDRED}
(ii) B = {x/x is a vowel in the word DEHRADOON}
(iii) C = {x | x ϵ W and x² < 50}
(iv) D = {Students of your school having 10 heads}
(v) E = {x | x is a prime between 8 and 30}
Solution:
(i) {HNDR} = 4
(ii) {E, A, O} = 3
(iii) {0, 1, 2, 3, 4, 5, 6, 7} = 8
(iv) 0
(v) {11, 13, 17, 19, 23, 29} = 6

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Objective Type Questions

Mental Maths
Question 1.
Fill in the blanks:
(i) A collection of ……… objects is called a set.
(ii) If x is a member of the set A, we write it as ………
(iii) The order of listing the elements of a set can be ………
(iv) If one or more elements are repeated, the set remains ………
(v) If X is the set of all letters in the word ‘MATHEMATICS’, then the cardinal number of the set X is ………
Solution:
(i) A collection of well defined objects is called a set.
(ii) If x is a member of the set A, we write it as x e A.
(iii) The order of listing the elements of a set can be changed.
(iv) If one or more elements are repeated, the set remains the same.
(v) If X is the set of all letters in the word ‘MATHEMATICS’, then the cardinal number of the set X is 8.

Question 2.
State whether the following statements are true (T) or false (F). Justify your answer.
(i) A collection of stamps is a set.
(ii) A collection of some fruits is a set.
(iii) A group of boys playing cricket is a set.
(iv) Collection of all students of your class taller than you is a set.
(v) Collection of five rivers of India is a set.
Solution:
(i) A collection of stamps is a set. False
Correct :
It is not a set because it is not known which stamps are included in the collection.

(ii) A collection of some fruits is a set. False
Correct :
It is not a set because it is not known which fruits are included in the collection.

(iii) A group of boys playing cricket is a set. False
Correct :
It is not a set because it is not known which students are included in the group.

(iv) Collection of all students of your class taller than you is a set. True
It is a set because every student of your class can be compared with certainly in relation to your height, so it is very easy to select students of your class who are taller than you i.e. it is well defined collection.

(v) Collection of five rivers of India is a set. False
Correct:
It is not a set because it is not known which five rivers of India are included in the collection.

Multiple Choice Questions
Choose the correct answer from the given four options (3 to 9):
Question 3.
Which of the following collections is a set?
(a) Collection of all tasty fruits
(b) Collection of all good football players of your school
(c) Collection of all months of a year
(d) Collection of 5 most intelligent students of your class.
Solution:
Collection of all months of a year is a set,
If we denote the given set, then A = {January, February, March, April, ……., December} (c)

Question 4.
The tabular form of the statement ‘All months of a year whose names begin with the letters J’ is
(a) {January, June, July}
(b) {months of a year whose names begin with the letter J}
(c) {x | x is a month of a year whose name begins with the letter J}
(d) none of these
Solution:
The given set can be written as in Tabular form :
{January, June, July} (a)

Question 5.
The method of representation used in the set A = {x | x is an even natural number less than 15} is called
(a) Description method
(b) Rule method
(c) Roster method
(d) None of these
Solution:
Rule method (b)

Question 6.
The cardinal number of the empty set is
(a) 2
(b) 1
(c) 0
(d) none of these
Solution:
0 (c)

Question 7.
If S = {x | x is a letter in the word AHMEDABAD}, then the cardinal number of S is
(a) 9
(b) 8
(c) 7
(d) 6
Solution:
S = {Letters Of AHMEDABAD} = {A, H, M, E, D, B} has 6 different elements, so n(S) = 6 (d)

Question 8.
If A = {x : x ϵ N and x is an odd prime number less than 17}, then the cardinal number of A is
(a) 8
(b) 6
(c) 5
(d) none of these
Solution:
Set D = {x : x ϵ N and x is an odd prime number less than 17}
⇒ D = {3, 5, 7, 11, 13}
∴ the cardinal number = 5 (c)

Question 9.
{months of a year whose names begin with the letter F} is
(a) an infinite set
(b) empty set
(c) singleton set
(d) none of these
Solution:
Let A= (months of a year whose name begin with the letter F}
⇒ A = {February}
∴ It is a singleton set. (c)

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Ex 5.2

Question 1.
State whether the following sets into empty, finite and infinite sets. In case of (non-empty) finite sets, mention the cardinal number.
(i) {all colours of a rainbow}
(ii) {x | x is a prime number between 7 and 11}
(iii) {multiples of 5}
(iv) {all straight lines drawn in a plane}
(v) {x | x is a digit in the numeral 550131527}
(vi) {x | x is a letter in the word ‘SUFFICIENT’}
(vii) {x | x is a vowel in the word MATHEMATICS}
(viii) {x : x is an even whole number and x ≤ 20}
(ix) {x : x ϵ I and -2 ≤ x ≤ 5}
(x) {x : x is a prime number less than 25}
(xi) {x : x is a prime factor of 180}.
(xii) {x : x ϵ N and x is a composite number < 12}
Solution:
(i) Let A = {all colours of a rainbow}
⇒ A = {Red, Orange, Yellow, Green, Blue, Indigo, Violet}
∴ the given set is finite .’. cardinal number = 7

(ii) Let {x | x is a prime number between 7 and 11}
⇒ B = {ϕ}
∴ the given set is empty

(iii) Let C = {multiples of 5}
⇒ C = {5, 10, 15, }
∴ the given set is infinite

(iv) The given set is infinite
(v) Let D = {x | x is a digit in the numeral 550131527}
⇒ D = {5, 0, 1, 3, 2, 7}, the given set is finite
∴ the cardinal number = 6

(vi) Let E = {x | x is a letter in the word ‘SUFFICIENT’}
⇒ E = {S, U, F, I, C, E, N, T}, the given set is finite
∴ the cardinal number = 8

(vii) Let F = {x | x is a vowel in the word MATHEMATICS}
⇒ F = {A, E, I}, the given set is finite
∴ the cardinal number = 3

(viii)Let F = {x : x is an even whole number and x ≤ 20}
⇒ F = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20}, the given set is finite
∴ the cardinal number = 11

(ix) {x : x ϵ I and -2 ≤ x ≤ 5}
= {-2, -1, 0, 1, 2, 3, 4}
It is a finite set as it has countable element which are 5

(x) G = {x : x is a prime number less than 25}
⇒ G = {2, 3, 5, 7, 11, 13, 17, 19, 23}
∴ the given set is finite
∴ the cardinal number = 9

(xi) H = {x : x is a prime factor of 180}
⇒ H = {2, 3, 5}
the given set is finite the cardinal number = 3

(xii) {x: x e N and x is a composite number < 12}
{4, 6, 8, 9, 10} given set is finite cardinal number is = 5

Question 2.
State whether the following pairs of sets are equal or not:
(i) A = {2, 4, 6, 8, 10}, B = {even natural numbers}
(ii) A = {3, 5, 7, 9, 11, 13}, B = {odd numbers between 2 and 14}
(iii) A = {PUPPET}, B = {P, U, E, T}
(iv) A = {x | x is a letter in the word SOPHIA}
B = {x | x is a letter in the word MUMTAZ}
(v) A = {kids 5 metres tall}, B = {x : x ϵ N and 2x = 3}.
Solution:
(i) A = {2, 4, 6, 8, 10}
B = {0, 2, 4, 6, 8, 10, 12 ………}
∴ A ≠ B

(ii) A = {3, 5, 7, 9, 11, 13}
B = {odd numbers between 2 and 14}
⇒ B = {3, 5, 7, 9, 11, 13}
∴ A = B

(iii) A = {PUPPET}, B= {P, U, E, T}, then A = B because the elements in a set can be repeated or rearranged.

(iv) A = {x | x is a letter in the word SOPHIA}
⇒ A= {S, O, P, H, I, A}
B = {x |x is a letter in the word MUMTAZ}
⇒ B = {M, U, T, A, Z}
∴ A ≠ B

(v) A = {Kids 5 metres tall}
⇒ A = {} ⇒ A is empty set
B = {x : x ϵ N and 2x = 3}
⇒ B = {} ⇒ B is empty set
∴ A = B

Question 3.
Given that A = {2, 5, 7, 8, 10}, B = {5, 7, 2, x, 10} and A = B, write the value of x.
Solution:
A = {2, 5, 7, 8, 10}
B = {5, 7, 2, x, 10}
∵ A = B
∴ x = 8

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Ex 5.1

Question 1.
State which of the following collections are sets:
(i) collection of odd natural numbers less than 50
(ii) collection of four colours of a rainbow
(iii) collection of first three days of a week
(iv) collection of all tall students of your class
(v) collection of all clever students of your school
(vi) collection of all rich people of Bangalore
(vii) collection of some multiples of 5 (viii) collection of all prime numbers
(ix) collection of all even integers which lie between -5 and 15
(x) collection of all good cricket players of India
(xi) collection of three youngest students of your class
(xii) collection of three healthy students of your class
Solution:
(i) It is a set.
If we denote the given set by A, then A = {1, 3, 5, 7, …………. , 47, 49}
(ii) It is not a set because the given collection is not well-defined-people may differ on four colours of a rainbow.
(iii) It is a set.
If we denote the given set by A, then A = {Sunday, Monday, Tuesday}
(iv) It is not a set because the given collection is not well-defined-people may differ on whether a student is tall or not.
(v) It is not a set because the given collection is not well-defined-people may differ on whether a student is clever or not.
(vi) It is not a set because the given collection is not well-defined-people may differ on whether a person is rich or not.
(vii) It is not a set because the given collection is not well defined-people may differ on which are multiples of 5. .
(viii) It is a set because the given collection is well defined.
(ix) It is a set.
If we denote the given set by A, then
A = {-4,-2, 0, 2, 4, 6, 8, 10, 12, 14}
(x) It is not a set because the given collection is not well defined-people may differ on whether a cricket player of India is good or not.
(xi) It is a set because the given collection is well defined-people can choose three youngest students of their classes.
(xii) It is not a set because the given collection is not well defined – people may differ on whether a student is healthy or not.

Question 2.
Let E = {even integers}. Insert the appropriate symbol ϵ or ∉ in the blanks:
(i) 10 …. E
(ii) -8 …. E
(iii) 13 … E
(iv) {6} …. E
(v) a …. E (vi) -4, 12, …. E
Solution:
E = {even numbers}
⇒ E = { ………… , -6, -4, -2, 0, 2, 4, 6, 8, ……………. }
(i) 10 ϵ E
(ii) -8 ϵ E
(iii) 13 ∉ E
(iv) {6} ϵ E
(v) a ∉ E
(vi) -4, 12, ϵ E

Question 3.
Let V = {vowels in English alphabet}. Write which of the following statements are true and which are false :
(i) c ϵ V
(ii) {a} ϵ V
(iii) a, e, i ϵ V
(iv) a, b ϵ V
(v) {a, u} ∉ V
(vi) {a, o, u} ϵ V
Solution:
V = {Vowels of English alphabet}
(i) c ϵ V Which is false.
(ii) {a} ϵ V Which is false.
(iii) a, e, i ϵ V Which is true.
(iv) a, b ϵ V Which is false.
(v) {a, u} ϵ V Which is true.
(vi) {a, o, u} ϵ V Which is true.

Question 4.
Write the following sets in roster form:
(i) the set of first five odd counting numbers
(ii) the set of all even natural numbers less than 101
(iii) {months of year whose names begin with a vowel}
(iv) {one digit natural numbers which are perfect squares}
(v) the set of multiples of 7 which lie between -20 and 25
(vi) {factors of 36}
(vii) {prime factors of 360}
(viii) the set of whole numbers which are multiples of 5
(ix) the set of all letters in the word ‘CHENNAI’
(x) The set of all vowels in the word ‘MUSSOORIE’
(xi) the set of all consonants in the word ‘MATHEMATICS’
Solution:
(i) The given set can be written as in roster form: { 1, 3, 5, 7, 9}
(ii) The given set can be written as in roster form: {2, 4, 6, 8, ……….. , 98, 100}
(iii) The given set can be written as in roster form: {April, August, October}
(iv) The given set can be written as in roster form: {1, 4, 9}
(v) The given set can be written as in roster form: {-14, -7, 0, 7, 14, 21}
(vi) The given set can be written as in roster form: {1, 2, 3, 4, 6, 9, 12, 18, 36}
(vii) The given set can be written as in roster form: {2, 3, 5}
(viii) The given set can be written as in roster form: {0, 5, 10, 15, }
(ix) The given set can be written as in roster form: {C,H,E,N,A,I}
(x) The given set can be written as in roster form: {U, O, I, E}
(xi) The given set can be written as in roster form : {M,T,H,C,S}

Question 5.
Write the following sets in tabular form:
(i) {x : is a natural number and x < 7}
(ii) {x : x e W and x ≤ 5}
(iii) {x : x is a month.of a year having less than 31 days}
(iv) {x | x is a letter in the word ‘CIRCUMFERENCE’}
(v) {x | x is a vowel in the word ‘NOTATION’}
(vi) (x : x is a digit in the numeral 110526715}
(vii) {x : x is a factor of 48}
(viii) (x : x is a multiple of 11 and 0 ≤ x < 80}
(ix) [y : y is a two digit natural number divisible by 10}
Solution:
(i) The given set can written as in Tabular form: {1, 2, 3, 4, 5, 6}
(ii) The given set can be written as in Tabular form: {0, 1, 2, 3, 4, 5}
(iii) The given set can be written as in Tabular form: {February, April, June, September, November}
(iv) The given set can be written as in Tabular form: {C, I, R, U, M, F, E, N}
(v) The given set can be written as in Tabular form: {O, A, I}
(vi) The given set can be written as in Tabular form: {1, 0, 5, 2, 6, 7}
(vii) The given set can be written as in Tabular form: {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
(viii) The given set can be written as in Tabular form: {0, 11, 22, 33, 44, 55, 66, 77}
(ix) {y : y is a two digit natural number divisible by 10} = {10, 20, 30, 40, 50, 60, 70, 80, 90}

Question 6.
Write the following sets in roster form and also in set builder form:
(i) the set of integers which lie between -2 and 3 (both inclusive)
(ii) the set of letters in the word ‘ULTIMATUM’
(iii) {months of a year whose names begin with J}
(iv) the set of single digit whole numbers which are perfect squares
Solution:
(i) The given set can be written as {-2, -1, 0, 1, 2, 3} (roster form)
{x : x ϵ I, -2 ≤ x ≤ 3} (set builder form)

(ii) The given set can be written as { U, L, T, I, M, A}(roster form)
{x : x is a letter in the word ‘ULTIMATUM’ }(set builder form)

(iii) The given set can be written as {January, June, July}(roster form)
{x | x is a month of a year whose names begin with J}(set builder form)

(iv) The given set can be written as {0, 1, 4, 9}(roster form)
{x | x is prefect square one digit number} (set builder form)

Question 7.
Write the following sets in tabular form and also in descriptive form :
(i) {x : x is a prime number less than 30}
(ii) the set of whole numbers which are multiples of 8 and less than 50
(iii) {x | x is a consonant in the word ‘QUESTION PAPER’}
Solution:
(i) The given set can be written as {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} (Tabular form)
{Prime numbers less than 30} (descriptive form)

(ii) The given set can be written as {0, 8, 16, 24, 32, 40, 48}(Tabular form)
{Whole numbers which are multiples of 8 and less than 50} (descriptive form)

(iii) The given set can be written as {Q, S, T, N, P, R} (Tabular form)
{Consonants in the word “QUESTION PAPER’{(descriptive form)

Question 8.
Write the following sets in the set builder form:
(i) A= {0, 1, 2, …, 11}
(ii) B = {7, 14, 21, 28, …}
(iii) C = {1, 4, 9, 16, 25, 36, 49}
(iv) D = {-12, -9, -b, -3, 0, 3, 6, 9, 12, 15, 18}
Solution:
(i) A= {0, 1, 2, …, 11}
= {x : x ϵ W, x ≤ 11}
(ii) B = {7, 14, 21, 28, …}
= {x : x = In, n ϵ N}
(iii) C = {1, 4, 9, 16, 25, 36, 49}
= {x : x = n², n ϵ N and n ≤ 7}
(iv) D = {-12,-9,-6,-3, 0,3, 6, 9, 12, 15, 18}
= {x : x = 3 n, n ϵ 1 and -4 ≤ n ≤ 6}

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Check Your Progress

Question 1.
Write all factors of:
(i) 88
(ii) 105
(iii) 96
Solution:
(i) 88 = {1, 2, 4, 8, 11, 22, 44, 88}
(ii) 105 = {1, 3, 5, 7, 15, 21, 35, 105}
(iii) 96 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96}

Question 2.
Find the common mutliples of 8 and 12.
Solution:
The multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64 72, 80, 88, 96, 104,
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108,
The common multiples of 8 and 12 are 24, 48, 72, 96

Question 3.
Which of the following pairs of numbers are co-prime?
(i) 25 and 105
(ii) 59 and 97
(iii) 161 and 192
Solution:
(i) 25 and 105
The factors of 25 are 1, 5, 25
The factors of 105 are 1,3, 5, 7, 15, 21, 35, 105
The common factors of 25 and 105 are 1, 5
∴ They are not co-prime

(ii) 59 and 97
The factors of 59 are 1, 59
The factors of 97 are 1, 97
The common factors of 59 and 97 is 1
∴ They are co-prime

(iii) 161 and 192
The factors of 161 are 1, 161
The factors of 192 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192
The common factors of 161, 192 is 1
∴ They are co-prime.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 4, 6, 8, 9 or 11:
(i) 197244
(ii) 613440
(iii) 4100448
Solution:
197244: divisible by 4, 6, 9.
It is divisible by 4 as last two digits is divisible by 4.
It is divisible by 6 as last digit of given number is divisible 2 and their sum is also divisible by 3.
Sum of digits 1 + 9 + 7 + 2 + 4 + 4 = 27 Which is divisible by 3.
It is not divisible by 8 as the sum of last three digit 2 + 4 + 4 = 10, is not divisible by 8.
It is divisible by 9 as the sum of its digits 27 is divisible by 9.
It is not divisible by 11 as the difference of the sum of alternate number 1 + 7 + 4 = 12 and 9 + 2 + 4 = 15, (15 – 12) = 3 is not divisible by 11.
∴ 197244 is divisible by 4, 6 and 9.

(ii) 613440 : divisible by 4, 6, 8, 9.
It is divisible by 4 as last two digit is divisible by 4.
It is divisible by 6 as the sum of all digits (6 + 1 + 3 + 4 + 4 + 0)=18 is divisible by 3 and by 2 also as last digit is 0.
It is divisible by 8 as the sum of last three digits (4 + 4 + 0) = 8 is divisible 8.
It is also divisible by 9 as the sum of its digits 18 is divisible by 9.
It is not divisible by 11 as the difference of the sum of alternate number 6 + 3 + 4 = 13 and 1 + 4 + 0 = 5, (13 – 5) = 8 is not divisible by 11.
613440 is divisible by 4, 6, 8 and 9.

(iii) 4100448: divisible by 4, 6, 8, 11.
It is divisible by 4 as last two digit is divisible by 4.
It is divisible by 6 as the sum of all digits 4 + 1 + 0 + 0 + 4 + 4 + 8 = 21 is divisible by 3 and also last digit is divisible by 2.
It is divisible by 8 as the sum of last three digits 4 + 4 + 8 = 16 is divisible by 8.
It is not divisible by 9 as the sum of its digit 21 is not divisible by 9.
It is divisible by 11 as the difference of the sum of alternate number 4 + 0 + 4 + 8= 16 and 1 + 0 + 4 = 5, (16 – 6) = 11 which is divisible by 11.
∴ 4100448 is divisible by 4, 6, 8, 11.

Question 5.
In 92 * 389, replace * by a digit so that the number formed is divisible by 11.
Solution:
The given number is 92 * 389
Here, * occur at odd place.
Sum of digits at odd place = 9 + 8 = 17 (Except *)
Sum of digits at even place = 2 + 3 + 9 = 14
Their difference = 17 – 14 = 3
If ‘*’ is replaced by 8, then sum of digits at odd place = 9 + 8 + 8 = 25
Their difference (Sum of digits at odd places – Sum of digits at even places)
= 25 – 14 = 11
Which is divisible by 11
∴ ‘*’ is to be replaced by the digit 8.

Question 6.
Find the prime factorisation of the following numbers:
(i) 168
(ii) 2304
Solution:
(i) 168

= 2 × 2 × 2 × 3 × 7

(ii) 2304

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Question 7.
Find the G.C.D. of the given numbers by prime factorisation method :
(i) 24,45
(ii) 180, 252, 324
Solution:
(i) 24, 45
24 = 2 × 2 × 2 × 3
45 = 3 × 3 × 5
The greatest common factor is 3.
G.C.D = 3

(ii) 180, 252, 324
180 = 2 × 2 × 3 × 3 × 5

252 = 2 × 2 × 3 × 3 × 7

324 = 2 × 2 × 3 × 3 × 3 × 3

G.C.D = 2 × 2 × 3 × 3 = 36
We notice that 2 and 3 both occurs as the common factor in the given numbers two time each.

Question 8.
Find the H.C.F of the given numbers by division method.
(i) 54, 82
(ii) 84, 120, 156
Solution:
(i) 54, 82
H.C.F = 2

(ii) 84, 120, 156
Solution:
H.C.F = 12

Question 9.
Find the L.C.M of the given numbers by prime factorisation method.
(i) 27, 90
(ii) 36, 48, 210
Solution:
(i) 27, 90
= 2 × 3 × 3 × 3 × 5 = 270

(ii) 36, 48, 210
= 2 × 2 × 3 × 3 × 2 × 2 × 5 × 7 = 5040

Question 10.
Find the L.C.M of the given numbers by division method:
(i) 48, 60
(ii) 112, 168, 266
Solution:
(i) 48, 60
= 2 × 2 × 3 × 4 ×5 = 240

(ii) 112, 168, 266
= 2 × 2 × 2 × 7 × 2 × 3 × 19 = 6384

Question 11.
Find the greatest number which divides 2706, 7041 and 8250 leaving remainder 6, 21 and 42 respectively.
Solution:
When 2706 is divided by the required number, 6 is left as a remainder. So, 2706 – 6 = 2700 i.e. 2700 is exactly divisible by that number.
Similarly, 7041 – 21 = 7020 is exactly divisible by that number.
Similarly, also, 8250 – 42 = 8208 is exactly divisible by that number.
Therefore, 2700, 7020 and 8208 are divisible by that number.
Thus, the required number is the H.C.F. of 2700, 7020 and 8208.
First, we find H.C.F. of 2700 and 7020.

∴ The H.C.F. of 2700, 7020 and 8208 is 108.
Hence the required number is 108

Question 12.
Find the least number which on decreasing by 20 is exactly divisible by 18, 21, 28 and 30.
Solution:
First, we find the least number which is exactly divisible by the numbers 18, 21, 28 and 30. For this, we find the L.C.M. of 18, 21, 28 and 30.

∴ L.C.M. = 2 × 2 × 3 × 3 × 5 × 7 = 1260
According to given, the required number will be 20 more than 1260.
The required number = 1260 + 20 = 1280

Question 13.
There are three heaps of rice weighing 120 kg, 144 kg and 204 kg. Find the maximum capacity of a bag so that the rice of each heap can be packed in exact number of bags.
Solution:
Weights of three heaps = 120 kg, 144 kg and 204 kg

∴ Maximum capacity of a bag, which exactly divides the heaps in exact number HCF of 120, 144, 204 = 12
∴  Required capacity of bag = 12 kg

Question 14.
Three bells are ringing continuously at intervals of 30, 36 and 45 minutes respectively. At what time will they ring together again if they ring simultaneously at 8 a.m.
Solution:
L.C.M = 2 × 3 × 3 × 2 × 5 = 180.
After 180 minute at 11: 00 a.m.

Question 15.
Two numbers are co-prime and their L.C.M. is 4940. If one of the numbers is 65, find the other number.
Solution:
One number = 65
and let the other number = x
We know that,
Two numbers are co-prime if their HCF is 1
Now, H.C.F. × L.C.M. of two numbers = Product of given two numbers
1 × 4940 = 65 × x
⇒ 4940 = 65 × x
⇒ 65 × x = 4940
⇒ x = 4940 ÷ 65 = 76

∴ The other number is 76

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Objective Type Questions

Mental Maths
Question 1.
Fill in the blanks:
(i) The only natural number which has exactly one factor is ……….
(ii) The only prime number which is even is ……….
(iii) The HCF of two co-prime numbers is ……….
(iv) Two perfect numbers are ………. and ……….
(v) The only prime-triplet is ……….
Solution:
(i) The only natural number which has exactly one factor is 1.
(ii) The only prime number which is even is 2.
(iii) The HCF of two co-prime numbers is 1.
(iv) Two perfect numbers are 6 and 28.
(v) The only prime-triplet is 3, 5, 7.

Question 2.
State whether the following statements are true (T) or false (F):
(i) Every natural number has a finite number of factors.
(ii) Every natural number has an infinite number of its multiples.
(iii) There are infinitely many prime numbers.
(iv) If two numbers are separately divisible by a number, then their difference is also divisible by that number.
(v) LCM of two prime numbers equals their product.
(vi) LCM of two co-prime numbers equals their product.
Solution:
(i) Every natural number has a finite number of factors. True
(ii) Every natural number has an infinite number of its multiples. True
(iii) There are infinitely many prime numbers. True
(iv) If two numbers are separately divisible by a number, then their difference is also divisible by that number. True
(v) LCM of two prime numbers equals their product. True
(vi) LCM of two co-prime numbers equals their product. True

Question 3.
State whether the following statements are true or false. If a statement is false, justify your answer.
(i) The sum of two prime numbers is always an even number.
(ii) The sum of two prime numbers is always a prime number.
(iii) The sum of two prime numbers can never be a prime number
(iv) No odd number can be written as the sum of two prime numbers.
(v) If two numbers are co-prime, then atleast one of them must be prime.
(vi) If a number is divisible by 18, it must be divisible by 3 and 6 both.
(vii) If a number is divisible by 2 and 4 both, it must be divisible by 8.
(viii) If a number is divisible by 3 and 6 both, it must be divisible by 18.
(ix) HCF of an even number and an odd number is always 1.
Solution:
(i) The sum of two prime numbers is always an even number. False
Correct:
2 and 7 both are prime numbers but their sum = 2 + 7 = 9, which is an odd number.

(ii) The sum of two prime numbers is always a prime number. False
Correct:
3 and 5 both are prime numbers but their sum = 3 + 5 = 8, which is a composite number.

(iii) The sum of two prime numbers can never
be a prime number. False
Correct:
2 and 5 both are prime numbers but their sum = 2 + 5 = 7, which is a prime number.

(iv) No odd number can be written as the sum of two prime numbers. False
Correct:
13 is an odd number and 13 = 2+11, which is the sum of two prime numbers.

(v) If two numbers are co-prime, then atleast one of them must be prime. False
Correct :
8 and 15 are co-prime numbers but neither 8 is prime nor 15 is prime.

(vi) If a number is divisible by 18, it must be divisible by 3 and 6 both. True

(vii) If a number is divisible by 2 and 4 both, it
must be divisible by 8. False
Correct :
20 is divisible by 2 and 4 both but 20 is not divisible by 8.

(viii)If a number is divisible by 3 and 6 both, it must be divisible by 18. False
Correct:
12 is divisible by 3 and 6 both bu 12 is not divisible by 18.

(ix) HCF of an even number and an odd number is always 1. False
Correct:
6 is even and 9 is odd but HCF of 6 and 9 is 3.

Multiple Choice Questions
Choose the correct answer from the given four options (4 to 28):
Question 4.
All factors of 6 are
(a) 1, 6
(b) 2, 3
(c) 1, 2, 3
(d) 1, 2, 3, 6
Solution:
The factors of 6 are 1, 2, 3, 6 (d)

Question 5.
Which of the following is an odd composite number?
(a) 7
(b) 9
(c) 11
(d) 12
Solution:
9, is an odd composite number. (c)

Question 6.
The number of even numbers between 68 and 90 is
(a) 10
(b) 11
(c) 12
(d) 31
Solution:
The even numbers between 68 and 90 is 70, 72, 74, 76, 78, 80, 82, 84, 86, 88 = 10 numbers (a)

Question 7.
Which of the following is a prime number?
(a) 69
(b) 87
(c) 91
(d) 97
Solution:
Since, the factors of 97 are 1 and 97
97 is a prime number. (d)

Question 8.
Which of the following is a pair of twin- prime number?
(a) 19, 21
(b) 43, 47
(c) 59, 61
(d) 73, 79
Solution:
59, 61
Pairs of prime numbers whose difference is 2 are called twin-prime numbers. (c)

Question 9.
The number of distinct prime factors of the largest 4-digit number is
(a) 2
(b) 3
(c) 5
(d) none of these
Solution:
Largest 4 digit number = 9999

3 is prime factor. (b)

Question 10.
The number of distinct prime factors of the smallest 5-digit number is
(a) 2
(b) 4
(c) 6
(d) 8
Solution:
Smallest 5-digit number = 10000

Number of distinct prime factors of smallest 5-digit number = 2 (a)

Question 11.
The sum of the prime factors of 1729 is
(a) 13
(b) 19
(c) 32
(d) 39
Solution:

Prime factors of 1729 are 7, 13 and 19 Sum of prime factors = 7 + 13 + 19 = 39 (d)

Question 12.
Which of the following is a pair of co-prime numbers?
(a) 8, 45
(b) 3, 18
(c) 5, 35
(d) 6, 39
Solution:
8, 15
The factors of 8 are 1, 2, 4, 8 The factors of 15 are 1, 3, 5, 15 The common factors of 8 and 15 is 1 They are co-prime. (a)

Question 13.
Every natural number has an infinite number of
(a) prime factors
(b) factors
(c) multiples
(d) none of these
Solution:
Multiples
e.g. Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, …….. (c)

Question 14.
Which of the following numbers is divisible by 4?
(a) 308594
(b) 506784
(c) 732106
(d) 9301538
Solution:
506784
Because the number formed by tens and ones digits is divisible by 4 i.e. 84 ÷ 4 = 21 (b)

Question 15.
Which of the following numbers is divisible by 8?
(a) 503786
(b) 505268
(c) 305678
(d) 703568
Solution:
703568
Because the number formed by hundred, tens
and ones digit is divisible by 8
i. e. 568 – 8 = 71 (d)

Question 16.
Which of the following numbers is divisible by 3?
(a) 50762
(b) 42063
(c) 52871
(d) 37036
Solution:
42063
Because sum of its digits is = 4 + 2 + 0 + 6 + 3 = 15 Which is divisible by 3 (b)

Question 17.
Which of the following numbers is divisible by 9?
(a) 972063
(b) 730542
(c) 785423
(d) 5612844
Solution:
972063
Because sum of digits
= 9 + 7 + 2 + 0 + 6 + 3 = 27 Which is divisible by 9 (a)

Question 18.
Which of the following numbers is divisible by 6?
(a) 560324
(b) 650374
(c) 798653
(d) 750972
Solution:
750972
Because sum of its digit
= 7 + 5 + 0 + 9 + 7 + 2 = 30 Which is divisible by 3.
Hence it is divisible by 6. (d)

Question 19.
The digit by which ‘*’ should be replaced in 54 * 281 so that the number formed is divisible by 9 is
(a) 6
(b) 7
(c) 8
(d) 9
Solution:
For a number to be divisible by 9, sum of its digits should be divisible by 9.
Sum of given digits in 54 * 281
= 5 + 4 + 2 + 8 + 1 = 20.
If we add 7, it becomes 27, which is divisible by 9.
∴ * is to be replaced by 7. (b)

Question 20.
The digit by which should be replaced in 7254 * 98 so that the number formed is divisible by 22 is
(a) 0
(b) 1
(c) 2
(d) 6
Solution:
For a number to be divisible by 22, sum of its digits should be divisible by 2 and by 11.
Since, the last digit of 7254 * 98 is 8, which is divisible by 2.
Now,
Sum of the digits at odd places = 7 + 5 + 8 = 20
Sum of the digits at even places = 9 + 4 + 2 = 15
∴ Their Difference = 20 – 15 = 5
Since, 5 is not divisble by 11, so to make a number divisible by 11 we must add 6.
∴ * is to be replaced by 6 (d)

Question 21.
If a number is divisible by 5 and 6 both, then it may not be divisible by
(a) 10
(b) 15
(c) 30
(d) 60
Solution:
60 (d)

Question 22.
The number of common prime factors of 60, 75 and 105 is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
60, 75 and 105

=2 × 2 × 3 × 5 × 5 × 7 = 2 (a)

Question 23.
The H.C.F. of 144 and 198 is
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
H.C.F. of 144 = 2 × 2 × 2 × 2 × 3 × 3
H.C.F. of 198 = 2 × 3 × 3 × 11

∴ H. C. F of 144 and 198
= 2 × 3 × 3 = 18 (d)

Question 24.
The L.C.M. of 30 and 45 is
(a) 15
(b) 30
(c) 45
(d) 90
Solution:
L.C.M. of 30 and 45 is 90

∴ L.C.M. = 3 × 5 × 2 × 3 = 90 (d)

Question 25.
The L.C.M. of 4 and 44 is
(a) 4
(b) 11
(c) 44
(d) 176
Solution:
LCM of 4 and 44 is 44

∴ L.C.M = 4 × 11 =44 (c)

Question 26.
The LCM of 7 and 13 is 1
(a) 1
(b) 7
(c) 13
(d) 91
Solution:
LCM of 7 and 13 is

LCM of 7 and 13 is 91
L.C.M. = 7 × 13 = 91 (d)

Question 27.
If H.C.F. of two numbers is 15 and their product is 1575, then their L.C.M. is
(a) 15
(b) 105
(c) 525
(d) 1575
Solution:
Product of numbers =1575
H.C.F. = 15
We know,
L.C. M = \(\frac{\text { Product of numbers }}{\text { H.C.F. }}\)
\(=\frac{1575}{15}=105\) (b)

Question 28.
If the LCM of two natural numbers is 180, then which of the following is not the HCF of the numbers?
(a) 45
(b) 60
(c) 75
(d) 90
Solution:
L.C.M. of 2 natural numbers = 180
We know that,
L.C.M. of 2 numbers is always exactly divisible by their H.C.F.
∴ Taking (a) 45 as H.C.F.

Here remainder = 0
∴ 45 is H.C.F.
Now, taking (b) 60 as H.C.F.

Here remainder = 0
∴ 60 is also H.C.F.
Now, taking 75 as H.C.F.

Here, remainder = 30
i. e. remainder ≠ 0
Hence, 75 is not the H.C.F. of two natural numbers whose L.C.M. is 180
Hence, answer is (c).

Value Based Questions
Question 1.
To teach the value of gratitude and appreciation to the students, a school organised a ‘Card Making’ activity in which the students were asked to make “THANK YOU CARDS” for the people who helped them in some way. Assorted cards were made with different titles. Their numbers are given below:
T cards for teachers = 120
F cards for friends = 540
S cards for servants = 90
P cards for parents = 240 and
G cards for grandparents = 150
(i) Find the HCF and LCM of all the different number of cards.
(ii) Find HCF and LCM of maximum and minimum number of cards.
(iii) Is the number of T-cards is a factor of number of P-cards?
Solution:
(i) HCF of 120, 540, 90, 240, 150


∴ HCF = 30
Now, LCM of 120, 540, 90, 240 and 150 is

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5
= 10800

(ii) Maximum number of cards = 540
Minimum number of cards = 90
∴ HCF is as follow :

Hence, HCF of 90 and 540 is 90
LCM of 90 and 540 is as follow :

LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540

(iii) Yes.
T cards = 120
P cards = 240
240 = 120 × 2

Higher Order Thinking Skills (Hots)
Question 1.
Write 2-digit odd numbers whose sum of digits is 8.
Solution:
17, 71, 35, 53

Question 2.
Write all pairs of 2-digit twin primes such that on changing the places of their digits, they still remain prime numbers.
Solution:
11, 13, 71, 73

Question 3.
There are just four natural numbers less than 100, which have exactly three factors. One of them is 25, what are the other three? What can be said about these numbers?
Solution:
Four natural numbers less than 100 which have three factors :
One of them is 25 = 1, 5, 25
Second is 49 = 1, 7, 49
Third is 9 = 1, 3, 9
Fourth is 4 = 1, 2, 4

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Ex 4.5

Question 1.
Find the L.C.M. of the given numbers by prime factorisation method :
(i) 28, 98
(ii) 36, 40, 126
(iii) 108, 135, 162
(iv) 24, 28, 196.
Solution:
(i) Prime factorisation of the given numbers are:
28 = 2 × 2 × 7
98 = 2 × 7 × 7
Here 2 and 7 occurs as a prime factor maximum 2 times
∴ L.C.M. = 2 × 2 × 7 × 7 = 196

(ii) Prime factorisation of the given numbers are:
36 = 2 × 2 × 3 × 3
40 = 2 × 2 × 2 × 5
126 = 2 × 3 × 3 × 7
Notice that 2 occurs as a prime factor maximum 3 times, 3 two times, 5 one times and 7 one times
∴ L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 × 7
= 2520

(iii) Prime factorisation of given umbers are
108 = 2 × 2 × 3 × 3 × 3
135 = 3 × 3 × 3 × 5
162 = 2 × 3 × 3 × 3 × 3
Notice that 2 occurs as a prime factor
maximum 2 times, 3, four time and 5, one time
∴ L.C.M.= 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1620

(iv) Prime factorisation of the given numbers are
24 = 2 × 2 × 2 × 3
28 = 2 × 2 × 7
196 = 2 × 2 × 7 × 7
Notice that 2 occurs as a prime factor maximum 3 times, 3 one times, 7 two times
∴ L.C.M. = 2 × 2 × 2 × 3 × 7 × 7 = 1176

Question 2.
Find the L.C.M. of the given numbers by division method :
(i) 480, 672
(ii) 6, 8, 45
(iii) 24, 40, 84
(iv) 20, 36, 63, 67
Solution:
(i) 480, 672

∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7 = 3360

(ii) 6, 8, 45

∴ LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360

(iii) 24, 40, 84

∴ LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840

(iv) 20, 36, 63, 97

∴ L.C.M. = 2 × 2 × 3 × 3 × 5 × 7 × 11 = 13860

Question 3.
Find the least number which when increased by 15 is exactly divisible by 15, 35 and 48.
Solution:
First, we find the least number which is exactly divisible by the numbers 15, 35 and 48. For this, we find L.C.M. of 15, 35 and 48.

∴ L.C.M. = 3 × 5 × 7 × 2 × 2 × 2 × 2 = 1680
According to given condition, the required number will be 15 less than 1680.
∴ The required least number = 1680- 15 = 1665

Question 4.
Find the least number which when divided by 6, 15 and 18 leaves remainder 5 in each case.
Solution:
LCM of 6, 15 and 18

= 2 × 3 × 3 × 5 = 90
Hence, the required number is 90 + 5 i.e. 95 48

Question 5.
Find the least number which when divided by 24, 36, 45 and 54 leaves a remainder of 3 in each case.
Solution:
24, 36, 45 and 54

∴ L.C.M. = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080
According to given condition, the required number will be 3 more than 1080.
∴ The required number = 1080 + 3 = 1083

Question 6.
Find the greatest 3-digit number which is exactly divisible by 8, 20 and 24.
Solution:
First, we find the LCM of 8, 20 and 24

∴ LCM of given numbers = 2 × 2 × 2 × 3 × 5 = 120
Greatest number of 3 digit is 999
We divide 999 by 120 and find the remainder.

According to given condition, we need a greatest 3-digit number which is exactly divisible by 120.
∴ The required number = 999 – 39 = 960

Question 7.
Find the smallest 4-digit number which is exactly divisible by 32, 36 and 48.
Solution:
First, we find the LCM of 32, 36 and 48

∴ LCM of given number
= 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
Smallest number of 4-digit = 1000
We divide 1000 by 288 and find the remainder

According to given condition, we need a least number of 4-digit which is exactly divisible by 288.
∴ The required number = 1000 + (288 – 136) = 1152

Question 8.
Find the greatest 4-digit number which is exactly divisible by each of 8, 12 and 20.
Solution:
First, we find the LCM of 8, 12 and 20

∴ LCM of given numbers = 2 × 2 × 2 × 3 × 5 = 120
According to given condition, we need a greatest number of 4-digit which is exactly divisible by 120.
Greatest number of 4-digit = 9999
We divide 9999 by 120 and find the remainder

∴ The required number = 9999 – 39 = 9960

Question 9.
Find the least number of five digits which is exactly divisible by 32, 36 and 45.
Solution:
First we, find the LCM of 32, 36 and 45

∴ LCM of given numbers
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440
Smallest 5-digit number = 10000
We divided 10000 by 1440 and find the remainder

According to given condition,
We need a least 5-digit number which is exactly divisible by 1440
The required number
= 10000 + 1440 – 1360
= 10080

Question 10.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the same distance in complete steps?
Solution:
The L.C.M. of 63, 70 and 77

⇒ 63 = 3 × 3 × 7
70 = 2 × 5 × 7
77 = 7 × 11
∴ L.C.M. = 3 × 3 × 2 × 5 × 7 × 11 = 6930
∴ The minimum distance each shall cover is 6930 cm i.e. 69 m 30 cm

Question 11.
Traffic lights at three different road crossing change after 48 seconds, 72 seconds and 108 seconds respectively. At what time will they change together again if they change simultaneously at 7 A.M.?
Solution:
LCM of 48, 72 and 108

= 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
432 seconds = 7 minutes 12 seconds past 7 A.M.

Question 12.
If the product of two numbers is 4032 and their HCF is 12, find their LCM.
Solution:
Product of two number = 4032
H.C.F = 12
L.C.M = 4032 ÷ 12 = 336

Question 13.
The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.
Solution:
HCF × LCM = one number × 2nd number
9 × 270 = 45 × 2nd number.
2430 = 45 × 2nd number
2430 = 45 × 2nd number
\(=\frac{2430}{45}=\frac{162}{3}=54\)
∴ 54 is other number.

Question 14.
Find the HCF of 180 and 336. Hence, find their LCM.
Solution:
Division method: HCF of 180 and 336

∴ H.C.R of 180 and 336 = 12
Products of numbers LCM of 180 and 336= \(\frac{\text { Products of numbers }}{\text { their H.C.F. }}\)
\(=\frac{180 \times 336}{12}=15 \times 336=5040\)

Question 15.
Can two numbers have 15 as their HCF and 110 as their LCM? Give reason to justify your answer.
Solution:
On dividing 110 by 15, we get

7 as quotient and 5 as remainder
We find that the remainder ≠ 0
So 110 is not exactly divisible by 15
∴ HCF and L.C.M. of two numbers cannot be 15 and 110 respectively.
As LCM of two numbers is always exactly divisible by their HCF.

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Objective Type Questions

Mental Maths

Question 1.
Fill in the blanks:
(i) A solid having no vertex and no edge is a ………..
(ii) A solid that has congruent and parallel polygons as top and bottom faces and all other faces rectangular is known as ………
(iii) A pyramid having 4 equilateral triangles as its faces is known as ………
(iv) A solid having 3 faces (one curved and two circulars), no vertex and two curved edges are known as ……..
(v) A solid having a circular base and one vertex is called a ………
(vi) A triangular prism has ……… faces, ………… edges and ………. vertices.
(vii) A triangular pyramid has ……… faces, ………. edges and ……… vertices.
(viii) A square pyramid has ……… faces, ……….. edges and ……… vertices.
(ix) The base of a triangular pyramid is a ………
(x) Out of ……….. faces of a triangular prism, ……… are rectangle and ……….. are triangles.
(xi) Out of ……….. faces of a square pyramid, ………. are a triangle and ………. is/are squares.
(xii) Out of ………. faces of a rectangular pyramid, ………. are triangles and the base is a ……….
(xiii) A ……….. is a sort of skeleton – outline in 2-D, which on folding results in a 3-D shape.
(xiv) If the sum of numbers on the two dice thrown together is 9, then the sum of the numbers opposite to these faces is ……….
Solution:
(i) A solid having no vertex and no edge is a sphere.
(ii) A solid that has congruent and parallel polygons as top and bottom faces
and all other faces rectangular is known as a prism.
(iii) A pyramid having 4 equilateral triangles as its faces is known as a tetrahedron.
(iv) A solid having 3 faces (one curved and two circulars),
no vertex and two curved edges are known as a cylinder.
(v) A Solis’having a circular base and one vertex is called a cone.
(vi) A triangular prism has 5 faces, 9 edges, and 6 vertices.
(vii) A triangular pyramid has 4 faces, 6 edges, and 4 vertices.
(viii) A square pyramid has 5 faces, 8 edges, and 5 vertices.
(ix) The base of a triangular pyramid is a triangle.
(x) Out of 5 faces of a triangular prism, 3 are rectangle and 2 are triangles.
(xi) Out of 5 faces of a square pyramid, 4 are triangle and 1 is/are squares.
(xii) Out of 5 faces of a rectangular pyramid, 4 are triangles and the base is a rectangle.
(xiii) A net is a sort of skeleton – outline in 2-D, which on folding results in a 3-D shape.
(xiv) If the sum of numbers on the two dice thrown together is 9,
then the sum of the numbers opposite to these faces is 5.

Question 2.
State whether the following statements are true (T) or false (F):
(i) The faces of a prism are triangular.
(ii) A cube can be treated as a prism.
(iii) A pyramid has only one vertex.
(iv) All the faces, except the base, of a square pyramid are triangular.
(v) A tetrahedron has 3 rectangular faces and 1 rectangle face.
(vi) A square pyramid has 5 faces and one vertex.
(vii) A cone has one vertex, two faces, and one curved edge.
(viii)The shadow of a 3-D object is a 2-D figure.
(ix) A cube can cast a shadow in the shape of a rectangle.
(x) A cube can cast a shadow in the shape of a hexagon.
(xi) In an isometric sketch, the line segments of different lengths can represent the sides of a cube.
(xii) In an oblique sketch of a cuboid, the size of the opposite faces must be different.
(xiii) The top, front and side views of a sphere are different.
(xiv) The adjoining net is of a hexagonal pyramid.

Solution:
(i) The faces of a prism are triangular. (False)
Correct:
Faces are rectangular.
(ii) A cube can be treated as a prism. (True)
(iii) A pyramid has only one vertex. (False)
Correct:
It has three or more than three.
(iv) All the faces, except the base, of a square pyramid are triangular. (True)
(v) A tetrahedron has 3 rectangular faces and 1 rectangle face. (False)
Correct:
It has triangular faces.
(vi) A square pyramid has 5 faces and one vertex. (False)
Correct:
It has five vertices, not one.
(vii) A cone has one vertex, two faces, and one curved edge. (True)
(viii) The shadow of a 3-D object is a 2-D figure. (True)
(ix) A cube can cast a shadow in the shape of a rectangle. (True)
(x) A cube can cast a shadow in the shape of a hexagon. (False)
(xi) In an isometric sketch, the line segments of different lengths
can represent the sides of a cube. (False)
Correct:
A cube has equal length.
(xii) In an oblique sketch of a cuboid, the size of the opposite faces
must be different. (False)
(xiii) The top, front and side views of a sphere are different. (False)
Correct:
All are equal.
(xiv) The adjoining net is of a hexagonal pyramid. (True)

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 11):
Question 3.
A triangular prism has
(a) 4 vertices and 6 edges
(b) 6 vertices and 9 edges
(c) 6 vertices and 6 edges
(d) 9 vertices and 6 edges
Solution:
A triangular prism has 6 vertices and 9 edges. (b)

Question 4.
A square pyramid has
(a) 4 vertices and 4 faces
(b) 4 vertices and 5 faces
(c) 5 vertices and 4 faces
(d) 5 vertices and 5 faces
Solution:
A square pyramid has 5 vertices and 5 faces. (d)

Question 5.
A solid having 4 (plane) faces, 4 vertices and 6 edges is called a
(a) triangular prims
(b) rectangular prism
(c) triangular pyramid
(d) rectangular pyramid
Solution:
A solid having 4 (plane) faces,
4 vertices and 6 edges is called rectangular pyramid. (c)

Question 6.
The number of cubes in the given structure is
(a) 12
(b) 10
(c) 9
(d) 8

Solution:
The number of cubes in the given structure is 12. (a)

Question 7.
An isometric sheet is made up of dots forming
(a) squares
(b) rectangles
(c) right-angled triangles
(d) equilateral triangles
Solution:
An isometric sheet is made up of dots
forming equilateral triangles. (d)

Question numbers 8 to 11 are based on the given figure in which unit cubes are put together to form a structure as shown:

Question 8.
The number of unit cubes in the given structure is
(a) 13
(b) 20
(c) 21
(d) 22
Solution:
The number of unit cubes in the given structure is 21. (c)

Question 9.
The number of unit cubes to be added to make a cuboid of dimensions 4 unit × 4 unit × 2 unit is
(a) 11
(b) 12
(c) 13
(d) 14
Solution:
The number of unit cubes to be added to make a cuboid of dimensions
4 unit × 4 unit × 2 unit is 4 × 4 × 2 = 32 – 21 = 11 (a)

Question 10.
If the structure is painted on the surface everywhere, then the number of unit cubes having no face painted is
(a) 0
(b) 1
(c) 2
(d) 11
Solution:
If the structure is painted on the surface everywhere,
then the number of unit cubes ‘ having no face painted is 1. (b)

Question 11.
The side view of the given structure is

Solution:
The side view of the given structure is (c).

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Ex 16.2

Question 1.
Find the area of each of the following parallelogram:

Solution:
(i) Base of the parallelogram (b) = 8 cm and height (h) = 4.5 cm
Area = b × h = 8 × 4.5 = 36 cm²
(ii) Base of the parallelogram (b) = 2 cm and height (h) = 4.4 cm
Area = b × h = 2 × 4.4 = 8.8 cm²
(iii) Base of the parallelogram (b) = 2.5 cm and height (h) = 3.5 cm
Area = b × h = 2.5 × 3.5 = 8.75 cm²

Question 2.
Find the area of each of the following triangles:

Solution:
(i) Base of the triangle (b) = 6.4 cm and height (b) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 6.4 × 6 cm²
= 19.2 cm²
(ii) Base of triangle (b) = 5 cm and height (h) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 5 × 6 = 15 cm²
(iii) Base of the triangle (b) = 4.5 cm and altitude (h) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 4.5 × 6 cm²
= 13.5 cm²

Question 3.
Find the missing values:

Solution:
Area of ||gm = b × h

Question 4.
Find the missing values:

Solution:

Question 5.
In the given figure, ABCD is a parallelogram whose two adjacent sides are 6 cm and 4 cm. If the height corresponding to the base AB is 3 cm, find:
(i) the area of parallelogram ABCD
(ii) the height corresponding to the base AD.

Solution:
In ||gm ABCD,
Base AB (b) = 6 cm
Altitude (h) = 3 cm

(i) Area = b × h = 6 × 3 = 18 cm²
In second case,
(ii) Base = 4 cm
Area= 18 cm²
Altitude (to AD) = \(\frac { Area }{ Base }\)
= \(\frac { 18 }{ 4 }\) cm
= 4.5 cm

Question 6.
In the given figure, ABC is an isosceles triangle with AB = AC = 7.5 cm and BC = 9 cm. If the height AD from A to BC is 6 cm, find:
(z) the area of ∆ABC
(ii) the height CE from C to AB.

Solution:
In an isosceles ∆ABC
AB = AC = 7.5 cm, BC = 9 cm
Height AD to BC = 6 cm
(i) Area of ∆ABC = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × BC × AD
= \(\frac { 1 }{ 2 }\) × 9 × 6 = 27 cm²
(ii) Area of ∆ABC = 27 cm²
Base AB = 7.5 cm

Question 7.
If the base of a right-angled triangle is 8 cm and the hypotenuse is 17 cm, find its area.
Solution:
Base of a right angled triangle = 8 cm
and hypotenuse = 17 cm
Height² = (Hypotenuse)² – (Base)² = 17² – 8² = 289 – 64 = 225 = (15)²
Height = 15 cm
Now are of ∆ = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 8 × 15 cm² = 60 cm²

Question 8.
In the given figure, ∆ABC is right-angled at B. Its legs are 8 cm and 6 cm. Find the length of perpendicular BN on the side AC.

Solution:
In the given figure,
In ∆ABC,
Base BC = 8 cm
and height AB = 6 cm
Area = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 8 x 6
= 24 cm²
Now, BN ⊥ AC
AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100 = (10)²
AC = 10 cm

Question 9.
In the given figure, the area of ∆ABE is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the length of the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆AMD, where M is mid-point of side DC?

Solution:
In the given figure, M is mid-point of DC
Area of ∆ABE = Area of ||gm ABCD
Now base of ∆ABC = 10 cm and height = 16 cm
Area = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 10 × 16 = 80 cm²
Now area of ||gm = Area of ∆ = 80 cm²
Base = 10 cm
Length of altitude = \(\frac { Area }{ Base }\) = \(\frac { 80 }{ 10 }\) = 8 cm
and Area of ∆AMD = \(\frac { 1 }{ 2 }\) × Base MD × Altitude
= \(\frac { 1 }{ 2 }\) × \(\frac { 10 }{ 2 }\) × \(\frac { 16 }{ 2 }\)
= 20 cm²

Question 10.
In the given figure, ABCD is a rectangle of size 18 cm by 10 cm. In ABEC, ∠E = 90° and EC = 8 cm. Find the area of the shaded region.

Solution:
In the figure,
ABCD is a rectangle in which
Base (b) = 18 cm
Height (h) = 10 cm
A ∆DEC is cut in which
∠E = 90°, EC = 8 cm
Now area of rectangle = l × b = 18 × 10 = 180 cm²
In ∆EBC,
BC = 10 cm, EC = 8 cm
EB² = BC² – EC² (Pythagoras Theorem)
= 10² – 8² = 100 – 64 = 36 = (6)²
EB = 6 cm
Now area of right ∆EBC = \(\frac { 1 }{ 2 }\) × EB × EC
= \(\frac { 1 }{ 2 }\) × 6 × 8 = 24 cm²
Area of shaded portion = 180 – 24 = 156 cm²

Question 11.
In the following figures, find the area of the shaded regions:


Solution:
(i) ABCD is a rectangle in which
Length (l) = 18 cm
Breadth (b) = 10 cm
Area of rectangle = l × b = 18 × 10 = 180 cm²
DE = 10 cm
EC = 18 – 10 = 8 cm
Now area of ∆BCE = \(\frac { 1 }{ 2 }\) × BC × EC
= \(\frac { 1 }{ 2 }\) × 10 × 8
= 40 cm²
and area of ∆FDE = \(\frac { 1 }{ 2 }\) × DC × DF
= \(\frac { 1 }{ 2 }\) × 10 × 6 = 30 cm²
Area of shaded portion = Area of rectangle – Area of ∆BCE – Area of ∆FDE
= 180 – (40+ 30)
= 180 – 70
= 110 cm²
(ii) In the given figure,
ABCD is a square whose each side = 20 cm
E and F are mid-points of AB and AD respectively
EC and FC are joined
Area of square ABCD = (Side)² = (20)² = 400 cm²
Area of ∆EBC = \(\frac { 1 }{ 2 }\) × EB × BC
= \(\frac { 1 }{ 2 }\) × 10 × 20 = 100 cm²
Area of ∆FDC = \(\frac { 1 }{ 2 }\) × FD × DC
= \(\frac { 1 }{ 2 }\) × 10 × 20 = 100 cm²
Area of ∆AEF = \(\frac { 1 }{ 2 }\) × 10 × 10 = 50 cm²
Area of shaded portion = 400 – (100 + 100 + 50) cm² = 400 – 250 = 150 cm²

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Ex 16.1

Question 1.
ABCD is a square of side 24 cm. EF is parallel to BC and AE = 15 cm. By how much does
(i) the perimeter of AEFD exceed the perimeter of EBCF?
(ii) the area of AEFD exceed the area of EBCF?

Solution:
Side of the square ABCD = 24 cm
EF || BC || AB is drawn and AE = 15 cm
EB = 24 – 15 = 9 cm

(i) Now perimeter of AEFD = 2(15 + 24) cm = 2 × 39 = 78 cm
and perimeter of EBCF = 2(9 + 24) = 2 × 33 cm = 66 cm
Difference of perimeter = 78 – 66 = 12 cm
(ii) Now Area of AEFD = l × b = 15 × 24 = 360 sq. cm
and area of EBCF = 9 × 24 = 216 sq. cm
Difference = 360 – 216 = 144 sq. cm

Question 2.
Nagma runs around a rectangular park 180 m long and 120 m wide at the rate of 7.5 km/ hour. In how much time will she complete five rounds?
Solution:
Length of rectangular plot (l) = 180 m
and breadth (b) = 120 m
Perimeter = 2(l + b) = 2(180 + 120) m = 2 × 300 = 600 m
Distance travelled in 5 rounds = 600 × 5 = 3000 m = 3 km
Speed = 7.5 km/hr

Question 3.
The area of a rectangular plot is 540 m². if its length is 27 m, find its breadth and perimeter.
Solution:
Area of a rectangular plot = 540 m²
Length (l) = 27 m

and perimeter = 2(l + b) = 2(27 + 20) m = 2 × 47 = 94 m

Question 4.
The perimeter of a rectangular field is 151 m. If its breadth is 32 m, find its length and area.
Solution:
Perimeter of a rectangular field = 151 m
Breadth = 32 m
Length =\(\frac { Perimeter }{ 2 }\) – Breadth
= \(\frac { 151 }{ 2 }\) – 32
= \(\frac { 87 }{ 2 }\)
= 43.5 m
and area = l × b = 43.5 × 32 m² = 1392 m²

Question 5.
The area of a rectangular plot is 340 m² and its breadth is 17 m. Find the cost of surrounding the plot with a fence at ₹ 5.70 per meter.
Solution:
Area of plot = 340 m²
and breadth (b) = 17 m
Length = \(\frac { A }{ b }\) = \(\frac { 340 }{ 17 }\) = 20 m
Perimeter = 2(l + b) = 2(20 + 17) = 2 × 37 = 74 m
Rate of fencing around it = ₹ 5.70 per m
Total cost = ₹5.70 × 74 = ₹ 421.80

Question 6.
The area of a square park is the same as that of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
Side of a square park = 60 m
Area = (Side)² = 60 × 60 = 3600 m²
Area of rectangular park = 3600 m²
and length (l) = 90 m

Question 7.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also, find which shape encloses more area and by how much?
Solution:
A wire in shape of a rectangle whose length (l) = 40 m
and breadth (b) = 22 m
Perimeter (Length) of wire = 2(l + b) = 2(40 + 22) cm = 2 × 62 cm = 124 cm
Perimeter of square wire = 124 cm
Then side = \(\frac { Perimeter }{ 4 }\) = \(\frac { 124 }{ 4 }\) = 31 m
Now area of rectangle = l × b = 40 × 22 = 880 cm²
and area of square = (Side)² = (31)² cm² = 961 cm²
Difference in area = 961 – 880 = 81 cm²
Area of 81 cm² is more of square shaped wire.

Question 8.
A door of breadth 1 m and height 2 m is fitted in a wall. The length of the wall is 4.5 m and the height is 3.6 m. Find the cost of whitewashing the wall, if the rate of whitewashing the wall is ₹ 20 per m².
Solution:
Breadth of door = 1 m and height = 2 m
Area of door = l × b = 1 × 2 = 2 m²
Length of wall = 4.5 m and height = 3.6 m
Area = 4.5 × 3.6 m² = 16.2 m²
Area of wall excluding area of door = 16.2 – 2 = 14.2 m²
Rate of white washing = ₹ 20 per m²
Total cost = 14.2 × 20 = ₹ 284

Question 9.
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
Solution:
Length of a rectangular park (l) = 45 m
and breadth (b) = 30 m
Width of path outside the park = 2.5 m
Outer length (L) = 45 + 2 × 2.5 m = 45 + 5 = 50 m

and width (B) = 30 + 2 × 2.5 = 30 + 5 = 35 m
Area of park = L × B – l × b
= 50 × 35 – 45 × 30 m²
= 1750 – 1350
= 400 m²

Question 10.
A carpet of size 5 m × 2 m has 25 cm wide red border. The inner part of the carpet is blue in colour. Find the area of the blue portion. What is the ratio of the areas of red portion to blue portion?
Solution:
Length of blue carpet (l) = 5 m
Breadth (b) = 2 m
Width of red border = 25 cm

Inner length = 5 – \(\frac { 2\times 25 }{ 100 }\) = 5 – 0.5 = 4.5 m
and breadth = 2 – 0.5 = 1.5 m
Now area of carpet = 4.5 × 1.5 m² = 6.75 m²
and area of border = 5 × 2 – 6.75 m² = 10 – 6.75 = 3.25 m²
Now ratio between border and carpet (blue part) = 3.25 : 6.75 = 13 : 27

Question 11.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2.
Solution:
Width of a verandah = 2.25 m
Length of room (i) = 5.5 m
and breadth (b) = 4.0 m
Outer length (L) = 5.5 + 2 × 2.25 m = 5.5 + 4.5 = 10 m
and outer breadth = 4 + 4.5 = 8.5 m
(i) Area of verandah = Outer area – Inner area
= 10 × 8.5 – 5.5 × 4 m²
= 85 – 22 m²
= 63 m²
(ii) Rate of cementing the floor of verandah = ₹ 200 per m²
Total cost = ₹ 63 × 200 = ₹ 12600

Question 12.
Two crossroads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of ₹ 105 per m2.
Solution:
Length of rectangular park (l) = 70 m
and breadth (ft) = 45 m
Width of each road = 5 m

(i) Area of roads = 70 × 5 + 45 × 5 – (5)² m²
= 350 + 225 – 25 = 550 m²
(ii) Rate of constructing the roads = ₹105 per m²
Total cost = ₹ 105 × 550 = ₹ 57750

Question 13.
A rectangular room is 10 m long and 7.5 m wide. Find the cost of covering the floor with carpet 1.25 m wide at ₹ 250 per metre.
Solution:
Length of rectangular room (l) = 10 m
and breadth (b) = 7.5 m
Area of floor of the room = l × b = 10 × 7.5 = 75 m²
Width of carpet = 1.25 m

Cost of 1 m carpet = ₹ 250
Total cost = ₹ 250 × 60 = ₹ 15000

Question 14.
Find the cost of flooring a room 6.5 m by 5 m with square tiles of side 25 cm at the rate of ₹ 9.40 per tile.
Solution:
Length of floor of a room (l) = 6.5 m
and breadth (b) = 5 m
Area = 6.5 × 5 m² = 32.5 m²

Cost of tile at the rate of ₹ 9.40 per tile = ₹ 9.40 × 520 = ₹ 4888

Question 15.
The floor of a room is in the shape of a square of side 4.8 m. The floor is to be covered with square tiles of perimeter 1.2 m. Find the cost of covering the floor if each tile costs ₹ 27.
Solution:
Side of square room = 4.8 m
Area = (4.8)² m² = 23.04 m²
Perimeter of one tile = 1.2 m
Side = \(\frac { 1.2 }{ 4 }\) = 0.3 m
Area of one tile = (0.3)² = 0.09 m²

Cost of one tile = ₹ 27
Total cost = 256 × ₹ 27 = ₹ 6912

Question 16.
A rectangular plot of land is 50 m wide. The cost of fencing the plot at the rate of ₹ 18 per metre is ₹ 4680. Find:
(i) the length of the plot.
(ii) the cost of leveling the plot at the rate of ₹ 7.6 per m².
Solution:
Breadth of a plot = 50 m
Cost of fencing around it = ₹ 4680
Rate of fencing = ₹ 18 per m
Perimeter = \(\frac { 4680 }{ 18 }\) = 260 m
Length =\(\frac { Perimeter }{ 2 }\) – Breadth
= \(\frac { 260 }{ 2 }\) – 50
= 130 – 50
= 80 m
(ii) Now area of plot = l × b = 80 × 50 m² = 4000 m²
Rate of leveling the plot = ₹ 7.6 per m²
Total cost = ₹ 4000 × ₹ 7.6 = ₹ 30400

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Check Your Progress

Question 1.
Identify the nets which can be used to make a tetrahedron (cut out copies of the nets and try it):

Solution:
The net/nets for a tetrahedron is (i) and (iii).

Question 2.
If four cubes each with 2 cm edge are placed side by side, what would the dimensions of the resultant cuboid be?
Solution:
Four cubes with 2 cm edge are placed side by side,
then the dimensions of the resulting cuboid will be
Length = 4 × 2 = 8 cm, breadth = 2 cm and height = 2 cm
An isometric sketch and oblique sketches are given below:

Question 3.
Two dice are placed side by side as shown in the given figure. What would be the total on the face opposite to
(i) 5 + 6
(b) 4 + 3?

Solution:
When two dice are placed side by side
as shown in the given figure,
the total would be on face opposite to
(i) 5 + 6 is 3
(ii) 4 + 7 is 7

Question 4.
For the structures given below sketch the front, side, and top view:

Solution:
The front view side view and top view of the given two sketches are given below:

ML Aggarwal Class 7 Solutions for ICSE Maths