Chapter 5 Squares and Square Roots Ex 5.3 NCERT Solutions for Class 8 Maths

Latest Updated : October 2023

If you are in search of NCERT solutions for Class 8 Maths Chapter 5 Squares and Square Roots, then you are at the right place. When it comes to Mathematics, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed NCERT solutions for Class 8 Maths of all the chapters, exercise wise with updated syllabus by NCERT. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark ncertforclass8.com to boost your preparation.

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.3

NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roo ts Ex 5.3

Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.3

Exercise 5.3

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) The unit’s digit of the square of a number having digit as unit’s place 1 is 1 and also 9 is 1 [ 9 square = 81 and 1 square = 1]
∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

(ii) The unit’s digit of the square of a number having digit as unit’s place of 6 is 6 and also 4 is 6 [6 square =36 and 4 square =16, both the squares have unit digit 6].
∴ Unit’s digit of the square root of number 99856 is equal to 4 or 6.

(iii) The unit’s digit of the square of a number having digit as unit’s place 1 is 1 and also 9 is 1 [9 square =81 whose unit place is 1].
∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

(iv) The unit’s digit of the square of a number having digit as unit’s place 5 is 5.
∴ Unit’s digit of the square root of number 657666025 is equal to 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153
(ii) 257
(iii) 408
(iv) 441

Solution:

Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution:

Square root of 100 =

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

Here, we have performed subtraction ten times.
∴ √100 = 10

Square root of 169 =

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

Here, we have performed subtraction thirteen times.
∴ √169 = 13

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) 729 = 3×3×3×3×3×3×1
⇒ 729 = (3×3×3)×(3×3×3)

\[⇒729=(3\times3\times3)^{^2}\]

\[\sqrt{729\ }=\ 3\times3\times3\ =\ 27\]

(ii) 400 = 2×2×2×2×5×5×1
⇒ 400 = (2×2×5)×(2×2×5)

\[⇒400=(2\times2\times5)^2\]

\[\sqrt{400}\ =\ 2\times2\times5=20\]

(iii) 1764 = 2×2×3×3×7×7
⇒ 1764 = (2×3×7)×(2×3×7)

\[⇒1764=(2\times3\times7)^2\ \]

\[⇒\ \sqrt{1764}=2\times3\times7=42\]

(iv) 4096 = 2×2×2×2×2×2×2×2×2×2×2×2
⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)

\[⇒4096=(2\times2\times2\times2\times2\times2)^2\]

\[\sqrt{4096\ }\ =\ 2\times2\times2\times2\times2\times2=64\]

(v) 7744 = 2×2×2×2×2×2×11×11×1
⇒ 7744 = (2×2×2×11)×(2×2×2×11)

\[⇒7744=(2\times2\times2\times11)^2\]

\[\sqrt{7744}=\ 2\times2\times2\times11=88\]

(vi) 9604 = 62 × 2 × 7 × 7 × 7 × 7
⇒ 9604 = ( 2 × 7 ×7 ) × ( 2 × 7 ×7 )

\[⇒9604=(2\times7\times7)^2\]

\[\sqrt{9604}\ =\ 2\times7\times7=98\]

(vii) 5929 = 7×7×11×11
⇒ 5929 = (7×11)×(7×11)

\[⇒5929=(7\times11)^2\]

\[\sqrt{5929\ }\ =\ 7\times11=77\]

(viii) 9216 = 2×2×2×2×2×2×2×2×2×2×3×3×1
⇒ 9216 = ( 2 × 2 × 2 × 2 × 2 × 3) × ( 2 × 2 × 2 × 2 × 2 × 3)
⇒ 9216 = 96 × 96

\[⇒9216=(96)^2\]

\[\sqrt{9216\ }=\ 96\]

(ix) 529 = 23×23

\[529=(23)^2\]

\[\sqrt{529}\ =\ 23\]

(x) 8100 = 2×2×3×3×3×3×5×5×1
⇒ 8100 = (2×3×3×5)×(2×3×3×5)
⇒ 8100 = 90×90

\[⇒8100=(90)^2\]

\[\sqrt{8100}\ =\ 90\]

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) 252 = 2 × 2 × 3 x 3 x 7
Here, prime factor 7 has no pair.
Therefore 252 must be multiplied by 7 to make it a perfect square.
∴ 252 x 7 = 1764
And, √1764 = 2 × 3 × 7= 42

(ii) 180 = 2 × 2 × 3 x 3x 5
Here, prime factor 5 has no pair.
Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
And, √900 = 2 x 3 x 5 = 30

(iii) 1008 = 2 × 2 x 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair.
Therefore 1008 must be multiplied by 7 to make it a perfect square.
∴ 1008 x 7 = 7056
And, √7056 = 2x2x3x7= 84

(iv) 2028 = 2 × 2 × 3 × 13 x 13
Here, prime factor 3 has no pair.
Therefore 2028 must be multiplied by 3 to make it a perfect square.
∴ 2028 x 3 = 6084
And, √6084 = 2 × 2 × 3 × 3 × 13×13=78

(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, prime factor 2 has no pair.
Therefore 1458 must be multiplied by 2 to make it a perfect square.
∴1458 x 2 = 2916
And, √2916 = 2x3x3x3= 54

(vi) 768 = 2 × 2 × 2× 2 × 2 × 2 × 2 × 2 × 3
Here, prime factor 3 has no pair.
Therefore 768 must be multiplied by 3 to make it a perfect square.
∴ 768 x 3 = 2304
And, √2304 = 2 x 2 x 2 x 2 x 3 = 48

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) 252 = 2×2×3×3×7
Here, prime factor 7 has no pair.
Therefore 252 must be divided by 7 to make it a perfect square.
∴ 252 ÷ 7= 36
And, √36 = 2 x 3 = 6

(ii) 2925 = 3 × 3 × 5 × 5 × 13
Here, prime factor 13 has no pair.
Therefore 2925 must be divided by 13 to make it a perfect square.
∴ 2925 ÷ 13 = 225
And, √225= 3 x 5 = 15

(iii) 396 = 2 × 2×3×3×11
Here, prime factor 11 has no pair.
Therefore 396 must be divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36
And, √36 = 2×3=6

(iv) 2645 = 5 x 23 x 23
Here, prime factor 5 has no pair.
Therefore 2645 must be divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529
And, √529 = 23 x 23 = 23

(v) 2800 = 2 × 2x 2 x 2 x 5 x 5 x 7
Here, prime factor 7 has no pair.
Therefore 2800 must be divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400
And, √400 = 2 x 2 x 5=20

(vi) 1620 = 2 x 2 x3 x 3 x 3 x 3 x 5
Here, prime factor 5 has no pair.
Therefore 1620 must be divided by 5 to make it a perfect square.
∴1620 ÷ 5 = 324
And, √324 = 2 x 3 x 3 = 18

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:

Here, Donated money = ₹2401
Let the number of students be x.
Therefore donated money = x x x
According to question,

\[x^{2\ }=2401\ \]

\[⇒x\ =\ \sqrt{2401}=\ \sqrt{7\ x\ 7\ x\ 7\ x\ 7}\]

⇒x= 7 x7 = 49

Hence, the number of students is 49.

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:

Here, Number of plants = 2025
Let the number of rows of planted plants be x.
And each row contains number of plants = x
According to question,

\[x^2=2025\]

\[⇒x=\sqrt{2025}\ =\ \sqrt{3\ x\ 3\ x\ 3\ x\ 3\ x\ 5\ x\ 5}\]

⇒x= 3x3x5= 45

Hence, each row contains 45 plants.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

L.C.M. of 4, 9 and 10 is 180.
Prime factors of 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair.
Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

L.C.M. of 8, 15 and 20 is 120.
Prime factors of 120 = 2 x 2 x 2 x 3 x 5
Here, prime factor 2, 3 and 5 has no pair.
Therefore 120 must be multiplied by 2 × 3 × 5 to make it a perfect square.
∴ 120 x 2 x 3 x 5 = 3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.

Using Class 8 NCERT solutions offers several key advantages for students:

  1. Accurate Answers: NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.
  2. Aligned with Curriculum: NCERT Solutions for Class 8 Maths are specifically designed to go along with the CBSE curriculum, making them the ideal resource for students studying in CBSE affiliated schools.
  3. Crisp an Clear Explanation: The NCERT solutions for Class 8 Maths provide step by step explanations, making complex concepts easier to understand. This helps students to grasp the concepts easily and build a better foundation.
  4. Time-Saver: Instead of spending extra time searching for answers, students can quickly access these NCERT solutions for Class 8 Maths , saving time and allowing them to focus on other aspects of their studies.
  5. Extra Questions: NCERT Solutions for Class 8 Maths also has extra questions that offer additional opportunities for students to build better problem solving skills.
  6. Free Forever: We provide answers which will always be free to access and share.
  7. Self-Study Support: These are beneficial for students who prefer self study. They can independently work through the solutions and improve their understanding of the subject at their own pace.

Summary

On our platform ncertforclass8.com, you will get all the NCERT Solutions for Class 8 Maths expertly solved, ensuring utmost care and accuracy. The NCERT book for Class 8 Maths is strictly followed while creating the NCERT solutions for Class 8 Maths, adhering to the latest CBSE guidelines. We make sure to provide you quality content which will help you to understand the concepts in the best way possible. Also our solutions are completely free for you to access. You have the freedom to copy the NCERT Solutions for Class 8 Maths, download them and even create PDF’s, making it convenient for your offline use.

Leave a Comment

Scroll to Top