Category: CBSE

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

• Class 8 Maths Comparing Quantities Exercise 8.1
• Class 8 Maths Comparing Quantities Exercise 8.2
• Class 8 Maths Comparing Quantities Exercise 8.3
• Comparing Quantities Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1

Ex 8.1 Class 8 Maths Question 1.
Find the ratio of the following:
(a) speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
(a) Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour
= \(\frac { 15 }{ 30 }\) = \(\frac { 1 }{ 2 }\)
Hence, the ratio = 1 : 2
(b) 5 m to 10 km
= 5 m : 10 × 1000 m [∵ 1 km = 1000 m]
= 5 m : 10000 m
= 1 : 2000
Hence, the ratio = 1 : 2000
(c) 50 paise to ₹ 5
= 50 paise : 5 × 100 paise
= 50 paise : 500 paise
ratio = 1 : 10

Ex 8.1 Class 8 Maths Question 2.
Convert the following ratios to percentages:
(a) 3 : 4
(b) 2 : 3
Solution:

Ex 8.1 Class 8 Maths Question 3.
72% of 25 students are good in mathematics. How many are not good in mathematics?
Solution:
Number of students who are good in mathematics = 72% of 25

Number of students who are not good in mathematics = 25 – 18 = 7

Ex 8.1 Class 8 Maths Question 4.
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution:
40 matches the team won out of 100 matches
1 match was won out of \(\frac { 100 }{ 40 }\) matches
10 matches the team will won out of \(\frac { 100 }{ 40 }\) × 10 = 25 matches
Hence, the total number of matches played by the team = 25

Ex 8.1 Class 8 Maths Question 5.
If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Let the money with Chameli be ₹ 100
Money spent by her = 75% of 100
= \(\frac { 75 }{ 100 }\) × 100 = ₹ 75
The money left with her = ₹ 100 – ₹ 75 = ₹ 25
₹ 25 are left with her out of ₹ 100
₹ 1 is left with her out of ₹ \(\frac { 100 }{ 25 }\)
₹ 600 will be left out of ₹ \(\frac { 100 }{ 25 }\) × 600 = ₹ 2400
Hence, she had ₹ 2400 in beginning.

Ex 8.1 Class 8 Maths Question 6.
If 60% of people in a city like a cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.
Solution:
Total number of people = 50,00,000
Number of people who like cricket = 60% of 50,00,000
= \(\frac { 60 }{ 100 }\) × 50,00,000
= 30,00,000
Number of people who like football = 30% of 50,00,000
= \(\frac { 30 }{ 100 }\) × 50,00,000
= 15,00,000
Number of people who like other games = 50,00,000 – (30,00,000 + 15,00,000)
= 50,00,000 – 45,00,000
= 5,00,000
Percentage of the people who like other games = \(\frac { 500000 }{ 5000000 }\) × 100 = 10%
Hence, 10% of people like other game.

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit
• NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2

• Class 8 Maths Cubes and Cube Roots Exercise 7.1
• Class 8 Maths Cubes and Cube Roots Exercise 7.2
• Cubes and Cube Roots Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2

Ex 7.2 Class 8 Maths Question 1.
Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Solution:

Ex 7.2 Class 8 Maths Question 2.
State True or False.
(i) Cube of an odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution:
(i) False – Cube of any odd number is always odd, e.g., (7)³ = 343
(ii) True – A perfect cube does not end with two zeros.
(iii) True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5)² = 25 and (5)³ = 625
(iv) False – (12)³ = 1728 (ends with 8)
(v) False – (10)³ = 1000 (4-digit number)
(vi) False – (99)³ = 970299 (6-digit number)
(vii) True – (2)³ = 8 (1-digit number)

Ex 7.2 Class 8 Maths Question 3.
You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
The given perfect cube = 1331
Forming groups of three from the rightmost digits of 1331

IInd group = 1
1st group = 331
One’s digit in first group = 1
One’s digit in the required cube root may be 1.
The second group has only 1.
Estimated cube root of 1331 = 11
Thus \(\sqrt [ 3 ]{ 1331 }\) = 11
(i) Given perfect cube = 4913
Forming groups of three from the right most digit of 4913

IInd group = 4
1st group = 913
One’s place digit in 913 is 3.
One’s place digit in the cube root of the given number may be 7.
Now in IInd group digit is 4
1³ < 4 < 2³
Ten’s place must be the smallest number 1.
Thus, the estimated cube root of 4913 = 17.
(ii) Given perfect cube = 12167
Forming group of three from the rightmost digits of 12167

We have IInd group = 12
1st group = 167
The ones place digit in 167 is 7.
One’s place digit in the cube root of the given number may be 3.
Now in Ilnd group, we have 12
2³ < 12 < 3³
Ten’s place of the required cube root of the given number = 2.
Thus, the estimated cube root of 12167 = 23.
(iii) Given perfect cube = 32768
Forming groups of three from the rightmost digits of 32768, we have

IInd group = 32
1st group = 768
One’s place digit in 768 is 8.
One’s place digit in the cube root of the given number may be 2.
Now in IInd group, we have 32
3³ < 32 < 4³
Ten’s place of the cube root of the given number = 3.
Thus, the estimated cube root of 32768 = 32.

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit
• NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

• Class 8 Maths Cubes and Cube Roots Exercise 7.1
• Class 8 Maths Cubes and Cube Roots Exercise 7.2
• Cubes and Cube Roots Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.1

Ex 7.1 Class 8 Maths Question 1.
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Solution:
(i) Prime factorisation of 216 is:
216 = 2 × 2 × 2 × 3 × 3 × 3
In the above factorisation, 2 and 3 have formed a group of three.
Thus, 216 is a perfect cube.

(ii) Prime factorisation of 128 is:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, 2 is left without making a group of three.
Thus 128 is not a perfect cube.

(iii) Prime factorisation of 1000, is:
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, no number is left for making a group of three.
Thus, 1000 is a perfect cube.

(iv) Prime factorisation of 100, is:
100 = 2 × 2 × 5 × 5
Here 2 and 5 have not formed a group of three.
Thus, 100 is not a perfect cube.

(v) Prime factorisation of 46656 is:
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Here 2 and 3 have formed the groups of three.
Thus, 46656 is a perfect cube.

Ex 7.1 Class 8 Maths Question 2.
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Solution:
(i) Prime factorisation of 243, is:
243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3 × 3
Here, number 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3
Thus, the required smallest number to be multiplied is 3.

(ii) Prime factorisation of 256, is:
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2 × 2
Here, a number 2 is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2
Thus, the required smallest number to be multiplied is 2.

(iii) Prime factorisation of 72, is:
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3 × 3
Here, a number 3 is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3
Thus, the required smallest number to be multiplied is 3.

(iv) Prime factorisation of 675, is:
675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5 × 5
Here, a number 5 is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5
Thus, the required smallest number is 5.

(v) Prime factorisation of 100, is:
100 = 2 × 2 × 5 × 5
Here, number 2 and 5 are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube, i.e., 2 × 2 × 2 × 5 × 5 × 5
Thus, the required smallest number to be multiplied is 2 × 5 = 10.

Ex 7.1 Class 8 Maths Question 3.
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 92
(v) 704
Solution:
(i) Prime factorisation of 81, is:
81 = 3 × 3 × 3 × 3 = 3³ × 3
Here, a number 3 is the number by which 81 is divided to make it a perfect cube,
i.e., 81 ÷ 3 = 27 which is a perfect cube.
Thus, the required smallest number to be divided is 3.

(ii) Prime factorisation of 128, is:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2
Here, a number 2 is the smallest number by which 128 is divided to make it a perfect cube,
i.e., 128 ÷ 2 = 64 which is a perfect cube.
Thus, 2 is the required smallest number.

(iii) Prime factorisation of 135 is:
135 = 3 × 3 × 3 × 5 = 3³ × 5
Here, 5 is the smallest number by which 135 is divided to make a perfect cube,
i.e., 135 ÷ 5 = 27 which is a perfect cube.
Thus, 5 is the required smallest number.

(iv) Prime factorisation of 192 is:
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2³ × 2³ × 3
Here, 3 is the smallest number by which 192 is divided to make it a perfect cube,
i.e., 192 ÷ 3 = 64 which is a perfect cube.
Thus, 3 is the required smallest number.

(v) Prime factorisation of 704 is:
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2³ × 2³ × 11
Here, 11 is the smallest number by which 704 is divided to make it a perfect cube,
i.e., 704 ÷ 11 = 64 which is a perfect cube.
Thus, 11 is the required smallest number.

Ex 7.1 Class 8 Maths Question 4.
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?
Solution:
The sides of the cuboid are given as 5 cm, 2 cm and 5 cm.
Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm³
For the prime factorisation of 50, we have
50 = 2 × 5 × 5
To make it a perfect cube, we must have
2 × 2 × 2 × 5 × 5 × 5
= 20 × (2 × 5 × 5)
= 20 × volume of the given cuboid
Thus, the required number of cuboids = 20.

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit
• NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

• Class 8 Maths Squares and Square Roots Exercise 6.1
• Class 8 Maths Squares and Square Roots Exercise 6.2
• Class 8 Maths Squares and Square Roots Exercise 6.3
• Class 8 Maths Squares and Square Roots Exercise 6.4
• Squares and Square Roots Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.4

Ex 6.4 Class 8 Maths Question 1.
Find the square root of each of the following numbers by Long Division method.
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
Solution:

Ex 6.4 Class 8 Maths Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation)
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution:
We know that if n is number of digits in a square number then
Number of digits in the square root = \(\frac { n }{ 2 }\) if n is even and \(\frac { n+1 }{ 2 }\) if n is odd.
(i) 64
Here n = 2 (even)
Number of digits in √64 = \(\frac { 2 }{ 2 }\) = 1
(ii) 144
Here n = 3 (odd)
Number of digits in square root = \(\frac { 3+1 }{ 2 }\) = 2
(iii) 4489
Here n = 4 (even)
Number of digits in square root = \(\frac { 4 }{ 2 }\) = 2
(iv) 27225
Here n = 5 (odd)
Number of digits in square root = \(\frac { 5+1 }{ 2 }\) = 3
(iv) 390625
Here n = 6 (even)
Number of digits in square root = \(\frac { 6 }{ 2 }\) = 3

Ex 6.4 Class 8 Maths Question 3.
Find the square root of the following decimal numbers.
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:

Ex 6.4 Class 8 Maths Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution:
(i)

Here remainder is 2
2 is the least required number to be subtracted from 402 to get a perfect square
New number = 402 – 2 = 400
Thus, √400 = 20

(ii)

Here remainder is 53
53 is the least required number to be subtracted from 1989.
New number = 1989 – 53 = 1936
Thus, √1936 = 44

(iii)

Here remainder is 1
1 is the least required number to be subtracted from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
Thus, √3249 = 57

(iv)

Here, the remainder is 41
41 is the least required number which can be subtracted from 825 to get a perfect square.
New number = 825 – 41 = 784
Thus, √784 = 28

(v)

Here, the remainder is 31
31 is the least required number which should be subtracted from 4000 to get a perfect square.
New number = 4000 – 31 = 3969
Thus, √3969 = 63

Ex 6.4 Class 8 Maths Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution:
(i)

Here remainder is 41
It represents that square of 22 is less than 525.
Next number is 23 an 23² = 529
Hence, the number to be added = 529 – 525 = 4
New number = 529
Thus, √529 = 23

(ii)

Here the remainder is 69
It represents that square of 41 is less than in 1750.
The next number is 42 and 42² = 1764
Hence, number to be added to 1750 = 1764 – 1750 = 14
Require perfect square = 1764
√1764 = 42

(iii)

Here the remainder is 27.
It represents that a square of 15 is less than 252.
The next number is 16 and 16² = 256
Hence, number to be added to 252 = 256 – 252 = 4
New number = 252 + 4 = 256
Required perfect square = 256
and √256 = 16

(iv)

The remainder is 61.
It represents that square of 42 is less than in 1825.
Next number is 43 and 43² = 1849
Hence, number to be added to 1825 = 1849 – 1825 = 24
The required perfect square is 1848 and √1849 =43

(v)

Here, the remainder is 12.
It represents that a square of 80 is less than in 6412.
The next number is 81 and 81² = 6561
Hence the number to be added = 6561 – 6412 = 149
The require perfect square is 6561 and √6561 = 81

Ex 6.4 Class 8 Maths Question 6.
Find the length of the side of a square whose area = 441 m²
Solution:
Let the length of the side of the square be x m.
Area of the square = (side)² = x² m²
x² = 441 ⇒ x = √441 = 21

Thus, x = 21 m.
Hence the length of the side of square = 21 m.

Ex 6.4 Class 8 Maths Question 7.
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Solution:
(a) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]
⇒ AC² = (6)² + (8)² = 36 + 64 = 100
⇒ AC = √100 = 10
Thus, AC = 10 cm.
(b) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]
⇒ (13)² = AB² + (5)²
⇒ 169 = AB² + 25
⇒ 169 – 25 = AB²
⇒ 144 = AB²
AB = √144 = 12 cm
Thus, AB = 12 cm.

Ex 6.4 Class 8 Maths Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this.
Solution:
Let the number of rows be x.
And the number of columns also be x.
Total number of plants = x × x = x²
x² = 1000 ⇒ x = √1000

Here the remainder is 39
So the square of 31 is less than 1000.
Next number is 32 and 32² = 1024
Hence the number to be added = 1024 – 1000 = 24
Thus the minimum number of plants required by him = 24.
Alternative method:

The minimum number of plants required by him = 24.

Ex 6.4 Class 8 Maths Question 9.
There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
Let the number of children in a row be x. And also that of in a column be x.
Total number of students = x × x = x²
x² = 500 ⇒ x = √500

Here the remainder is 16
New Number 500 – 16 = 484
and, √484 = 22
Thus, 16 students will be left out in this arrangement.

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit
• NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

• Class 8 Maths Squares and Square Roots Exercise 6.1
• Class 8 Maths Squares and Square Roots Exercise 6.2
• Class 8 Maths Squares and Square Roots Exercise 6.3
• Class 8 Maths Squares and Square Roots Exercise 6.4
• Squares and Square Roots Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.3

Ex 6.3 Class 8 Maths Question 1.
What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i) One’s digit in the square root of 9801 maybe 1 or 9.
(ii) One’s digit in the square root of 99856 maybe 4 or 6.
(iii) One’s digit in the square root of 998001 maybe 1 or 9.
(iv) One’s digit in the square root of 657666025 can be 5.

Ex 6.3 Class 8 Maths Question 2.
Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408
(iv) 441
Solution:
We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares.
(i) 153 is not a perfect square number. (ending with 3)
(ii) 257 is not a perfect square number. (ending with 7)
(iii) 408 is not a perfect square number. (ending with 8)
(iv) 441 is a perfect square number.

Ex 6.3 Class 8 Maths Question 3.
Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:
Using the method of repeated subtraction of consecutive odd numbers, we have
(i) 100 – 1 = 99, 99 – 3 = 96, 96 – 5 = 91, 91 – 7 = 84, 84 – 9 = 75, 75 – 11 = 64, 64 – 13 = 51, 51 – 15 = 36, 36 – 17 = 19, 19 – 19 = 0
(Ten times repetition)
Thus √100 = 10

(ii) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69, 69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0
(Thirteen times repetition)
Thus √169 = 13

Ex 6.3 Class 8 Maths Question 4.
Find the square roots of the following numbers by the prime factorisation Method.
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution:
(i) We have 729
Prime factors of 729
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3² × 3² × 3²
√729 = 3 × 3 × 3 = 27

(ii) We have 400
Prime factors of 400
400 = 2 × 2 × 2 × 2 × 5 × 5 = 2² × 2² × 5²
√400 = 2 × 2 × 5 = 20

(iii) 1764
1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²
√1764 = 2 × 3 × 7 = 42

(iv) 4096
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) Prime factorisation of 7744 is
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
√7744 = 2 × 2 × 2 × 11 = 88

(vi) Prime factorisation of 9604 is
9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2² × 7² × 7²
√9604 = 2 × 7 × 7 = 98

(vii) Prime factorisation of 5929 is
5929 = 7 × 7 × 11 × 11 = 7² × 11²
√5929 = 7 × 11 = 77

(viii) Prime factorisation of 9216 is
9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

(ix) Prime factorisation of 529 is
529 = 23 × 23 = 23²
√529 = 23

(x) Prime factorisation of 8100 is
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2² × 3² × 3² × 5²
√8100 = 2 × 3 × 3 × 5 = 90

Ex 6.3 Class 8 Maths Question 5.
For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.
(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768
Solution:
(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factorisation is not in pair. 7 has no pair.
Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.
The new square number is 252 × 7 = 1764
Square root of 1764 is
√1764 = 2 × 3 × 7 = 42

(ii) Primp factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
New square number = 180 × 5 = 900
The square root of 900 is
√900 = 2 × 3 × 5 = 30
Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) Prime factorisation of 1008 is
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, 7 has no pair.
New square number = 1008 × 7 = 7056
Thus, 7 is the required number.
Square root of 7056 is
√7056 = 2 × 2 × 3 × 7 = 84

(iv) Prime factorisation of 2028 is
2028 = 2 × 2 × 3 × 13 × 13
Here, 3 is not in pair.
Thus, 3 is the required smallest whole number.
New square number = 2028 × 3 = 6084
Square root of 6084 is
√6084 = 2 × 13 × 3 = 78

(v) Prime factorisation of 1458 is
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, 2 is not in pair.
Thus, 2 is the required smallest whole number.
New square number = 1458 × 2 = 2916
Square root of 1458 is
√2916 = 3 × 3 × 3 × 2 = 54

(vi) Prime factorisation of 768 is
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
Thus, 3 is the required whole number.
New square number = 768 × 3 = 2304
Square root of 2304 is
√2304 = 2 × 2 × 2 × 2 × 3 = 48

Ex 6.3 Class 8 Maths Question 6.
For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Solution:
(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here 7 has no pair.
7 is the smallest whole number by which 252 is divided to get a square number.
New square number = 252 ÷ 7 = 36
Thus, √36 = 6

(ii) Prime factorisation of 2925 is
2925 = 3 × 3 × 5 × 5 × 13
Here, 13 has no pair.
13 is the smallest whole number by which 2925 is divided to get a square number.
New square number = 2925 ÷ 13 = 225
Thus √225 = 15

(iii) Prime factorisation of 396 is
396 = 2 × 2 × 3 × 3 × 11
Here 11 is not in pair.
11 is the required smallest whole number by which 396 is divided to get a square number.
New square number = 396 ÷ 11 = 36
Thus √36 = 6

(iv) Prime factorisation of 2645 is
2645 = 5 × 23 × 23
Here, 5 is not in pair.
5 is the required smallest whole number.
By which 2645 is multiplied to get a square number
New square number = 2645 ÷ 5 = 529
Thus, √529 = 23

(v) Prime factorisation of 2800 is
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, 7 is not in pair.
7 is the required smallest number.
By which 2800 is multiplied to get a square number.
New square number = 2800 ÷ 7 = 400
Thus √400 = 20

(vi) Prime factorisation of 1620 is
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, 5 is not in pair.
5 is the required smallest prime number.
By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324
Thus √324 = 18

Ex 6.3 Class 8 Maths Question 7.
The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Total amount of money donated = ₹ 2401
Total number of students in the class = √2401
= \(\sqrt { { 7 }^{ 2 }\times { 7 }^{ 2 } }\)
= \(\sqrt { { 7 }\times { 7\times 7\times 7 } }\)
= 7 × 7
= 49

Ex 6.3 Class 8 Maths Question 8.
2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Total number of rows = Total number of plants in each row = √2025
= \(\sqrt { 3\times 3\times 3\times 3\times 5\times 5 }\)
= \(\sqrt { { 3 }^{ 2 }\times { 3 }^{ 2 }\times { 5 }^{ 2 } }\)
= 3 × 3 × 5
= 45
Thus the number of rows and plants = 45

Ex 6.3 Class 8 Maths Question 9.
Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution:
LCM of 4, 9, 10 = 180
The least number divisible by 4, 9 and 10 = 180
Now prime factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
The required smallest square number = 180 × 5 = 900

Ex 6.3 Class 8 Maths Question 10.
Find the smallest number that is divisible by each of the numbers 8, 15 and 20.
Solution:
The smallest number divisible by 8, 15 and 20 is equal to their LCM.
LCM = 2 × 2 × 2 × 3 × 5 = 120
Here, 2, 3 and 5 have no pair.
The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit
• NCERT Solutions