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NCERT Solutions

Class 8

Maths

Chapter 5 Squares And Square Roots Exercise 5.3

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

NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - FREE PDF Download

NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 are provided by Vedantu. Our updated PDF solutions guide answers all the questions in the latest NCERT exercise book for the subject. Class 8 Math 5.3 dives deeper into the concepts introduced earlier in the chapter, helping students understand how to find square roots using various methods. This exercise is crucial as it builds foundational knowledge that is applicable in higher-level math and various real-life scenarios. The PDF can be freely downloaded from Vedantu’s website and can go a long way to improve your chances of scoring well in your CBSE Class 8 Maths exam.

Table of Content

Access NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - Squares and Square Roots

1. What could be the possible ‘one’s digits of the square root of each of the following numbers?

i. $\text{9801}$

Ans: We know that the one’s digit of the square root of the number ending with $1$ can be $1$ or $9$.

Thus, the possible one’s digit of the square root of $9801$ is either $1$ or $9$.

ii. $\text{99856}$

Ans: We know that the one’s digit of the square root of the number ending with $6$ can be $6$ or $4$.

Thus, the possible one’s digit of the square root of $99856$ is either $6$ or $4$.

iii. $\text{998001}$

Ans: We know that the one’s digit of the square root of the number ending with $1$ can be $1$ or $9$.

Thus, the possible one’s digit of the square root of $998001$ is either $1$ or $9$.

iv. $\text{657666025}$

Ans: We know that the one’s digit of the square root of the number ending with $5$ will be $5$.

Thus, the only possible one’s digit of the square root of $657666025$ is $5$.

2. Find the numbers which are surely not perfect squares without doing any calculations.

i. $\text{153}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $153$ has its unit place digit as $3$.

Hence, $153$ cannot be a perfect square.

ii. $\text{257}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $257$ has its unit place digit as $7$.

Hence, $257$ cannot be a perfect square.

iii. $\text{408}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $408$ has its unit place digit as $8$.

Hence, $408$ cannot be a perfect square.

iv. $\text{441}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $441$ has its unit place digit as $1$.

Hence, $441$ is a perfect square.

3. Find the square roots of $\text{100}$ and $\text{169}$ by the method of repeated subtraction.

Ans: It is already known to us that the sum of the first n odd natural numbers is n2.

For $\sqrt{100}$

$100-1=99$
$99-3=96$
$96-5=91$
$91-7=84$
$84-9=75$
$75-11=64$
$64-13=51$
$51-15=36$
$36-17=19$
$19-19=0$

After subtracting successive odd numbers from $1$ to $100$ , we are getting a $0$ at the 10th step.

Hence, $\sqrt{100}=10$

For $\sqrt{169}$

$169-1=168$
$168-3=165$
$165-5=160$
$160-7=153$
$153-9=144$
$144-11=133$
$133-13=120$
$120-15=105$
$105-17=88$
$88-19=69$
$69-21=48$
$48-23=25$
$25-25=0$

After subtracting successive odd numbers from $1$ to $169$, we are getting a $0$ at the 13th step.

Hence, $\sqrt{169}=13$

4. Find the square roots of the following numbers by the Prime Factorisation Method.

i. $\text{729}$

Ans:

The factorization of $729$ is as follows:

$3$	$729$
$3$	$243$
$3$	$81$
$3$	$27$
$3$	$9$
$3$	$3$
	$1$

$729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

$\sqrt{729}=3\times 3\times 3$

So, $\sqrt{729}=27$

ii. $\text{400}$

Ans: The factorization of $400$ is as follows:

$2$	$400$
$2$	$200$
$2$	$100$
$2$	$50$
$5$	$25$
$5$	$5$
	$1$

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

iii. $\text{1764}$

Ans:

The factorization of $1764$ is as follows:

$2$	$1764$
$2$	$882$
$3$	$441$
$3$	$147$
$7$	$49$
$7$	$7$
	$1$

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

iv. $\text{4096}$

Ans: The factorization of $4096$ is as follows:

$2$	$4096$
$2$	$2048$
$2$	$1024$
$2$	$512$
$2$	$256$
$2$	$128$
$2$	$64$
$2$	$32$
$2$	$16$
$2$	$8$
$2$	$4$
$2$	$2$
	$1$

$4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}$

$\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$

So, $\sqrt{4096}=64$

v. $\text{7744}$

Ans: The factorization of $7744$ is as follows:

$2$	$7744$
$2$	$3872$
$2$	$1936$
$2$	$968$
$2$	$484$
$2$	$242$
$11$	$121$
$11$	$11$
	$1$

$7744=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{11\times 11}$

$\sqrt{7744}=2\times 2\times 2\times 11$

So, $\sqrt{7744}=88$

vi. $\text{9604}$

Ans: The factorization of $9604$ is as follows:

$2$	$9604$
$2$	$4802$
$7$	$2401$
$7$	$343$
$7$	$49$
$7$	$7$
	$1$

$9604=\underline{2\times 2}\times \underline{7\times 7}\times \underline{7\times 7}$

$\sqrt{9604}=2\times 7\times 7$

So, $\sqrt{9604}=98$

vii. $\text{5929}$

Ans: The factorization of $5929$ is as follows:

$7$	$5929$
$7$	$847$
$11$	$121$
$11$	$11$
	$1$

$5929=\underline{7\times 7}\times \underline{11\times 11}$

$\sqrt{5929}=7\times 11$

So, $\sqrt{5929}=77$

viii. $\text{9216}$

Ans: The factorization of $9216$ is as follows:

$2$	$9216$
$2$	$4608$
$2$	$2304$
$2$	$1152$
$2$	$576$
$2$	$288$
$2$	$144$
$2$	$72$
$2$	$36$
$2$	$18$
$3$	$9$
$3$	$3$
	$1$

$9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$

$\sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3$

So, $\sqrt{9216}=96$

ix. $\text{529}$

Ans: The factorization of $529$ is as follows:

$23$	$529$
$23$	$23$
	$1$

$529=\underline{23\times 23}$

So, $\sqrt{529}=23$

x. $\text{8100}$

Ans: The factorization of $8100$ is as follows:

$2$	$8100$
$2$	$4050$
$3$	$2025$
$3$	$675$
$3$	$225$
$3$	$75$
$5$	$25$
$5$	$5$
	$1$

$8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}$

$\sqrt{8100}=2\times 3\times 3\times 5$

So, $\sqrt{8100}=90$

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. $\text{252}$

Ans: The factorization of $252$ is as follows:

$2$	$252$
$2$	$126$
$3$	$63$
$3$	$21$
$7$	$7$
	$1$

Here, $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to multiply $252$ by $7$ to get a perfect square.

The new number will be $252\times 7=1764$

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

ii. $\text{180}$

Ans: The factorization of $180$ is as follows:

$2$	$180$
$2$	$90$
$3$	$45$
$3$	$15$
$5$	$5$
	$1$

Here, $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$

We can see that $5$ is not paired

So, we have to multiply $180$ by $5$ to get a perfect square.

The new number will be $180\times 5=900$

$900=\underline{2\times 2}\times \underline{3\times 3}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{900}=2\times 3\times 5$

So, $\sqrt{900}=30$

iii. $\text{1008}$

Ans: The factorization of $1008$ is as follows:

$2$	$1008$
$2$	$504$
$2$	$252$
$2$	$126$
$3$	$63$
$3$	$21$
$7$	$7$
	$1$

Here, $1008=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to multiply $1008$ by $7$ to get a perfect square.

The new number will be $1008\times 7=7056$

$7056=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ which is a perfect square

$\sqrt{7056}=2\times 2\times 3\times 7$

So, $\sqrt{7056}=84$

iv. $\text{2028}$

Ans: The factorization of $2028$ is as follows:

$2$	$2028$
$2$	$1014$
$3$	$507$
$13$	$169$
$13$	$13$
	$1$

Here, $2028=\underline{2\times 2}\times 3\times \underline{13\times 13}$

We can see that $3$ is not paired

So, we have to multiply $2028$ by $3$ to get a perfect square.

The new number will be $2028\times 3=6084$

$6084=\underline{2\times 2}\times \underline{3\times 3}\times \underline{13\times 13}$ which is a perfect square

$\sqrt{6084}=2\times 3\times 13$

So, $\sqrt{6084}=78$

v. $\text{1458}$

Ans: The factorization of $1458$ is as follows:

$2$	$1458$
$3$	$729$
$3$	$243$
$3$	$81$
$3$	$27$
$3$	$9$
$3$	$3$
	$1$

Here, $1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

We can see that $2$ is not paired

So, we have to multiply $1458$ by $2$ to get a perfect square.

The new number will be $1458\times 2=2916$

$2916=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{2916}=2\times 3\times 3\times 3$

So, $\sqrt{2916}=54$

vi. $\text{768}$

Ans: The factorization of $768$ is as follows:

$2$	$768$
$2$	$384$
$2$	$192$
$2$	$96$
$2$	$48$
$2$	$24$
$2$	$12$
$2$	$6$
$3$	$3$
	$1$

Here, $768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3$

We can see that $3$ is not paired

So, we have to multiply $768$ by $3$ to get a perfect square.

The new number will be $768\times 3=2304$

$2304=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{2304}=2\times 2\times 2\times 2\times 3$

So, $\sqrt{2304}=48$

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. $\text{252}$

Ans: The factorization of $252$ is as follows:

$2$	$252$
$2$	$126$
$3$	$63$
$3$	$21$
$7$	$7$
	$1$

Here, $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to divide $252$ by $7$ to get a perfect square.

The new number will be $252\div 7=36$

$36=\underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

ii. $\text{2925}$

Ans: The factorization of $2925$ is as follows:

$3$	$2925$
$3$	$975$
$5$	$325$
$5$	$65$
$13$	$13$
	$1$

Here, $2925=\underline{3\times 3}\times \underline{5\times 5}\times 13$

We can see that $13$ is not paired

So, we have to divide $2925$ by $13$ to get a perfect square.

The new number will be $2925\div 13=225$

$225=\underline{3\times 3}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{225}=3\times 5$

So, $\sqrt{225}=15$

iii. $\text{396}$

Ans: The factorization of $396$ is as follows:

$2$	$396$
$2$	$198$
$3$	$99$
$3$	$33$
$11$	$11$
	$1$

Here, $396=\underline{2\times 2}\times \underline{3\times 3}\times 11$

We can see that $11$ is not paired

So, we have to divide $396$ by $11$ to get a perfect square.

The new number will be $396\div 11=36$

$36=\underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

iv. $\text{2645}$

Ans: The factorization of $2645$ is as follows:

$5$	$2645$
$23$	$529$
$23$	$23$
	$1$

Here, $2645=5\times \underline{23\times 23}$

We can see that $5$ is not paired

So, we have to divide $2645$ by $5$ to get a perfect square.

The new number will be $2645\div 5=529$

$529=\underline{23\times 23}$ which is a perfect square

So, $\sqrt{529}=23$

v. $\text{2800}$

Ans: The factorization of $2800$ is as follows:

$2$	$2800$
$2$	$1400$
$2$	$700$
$2$	$350$
$5$	$175$
$5$	$35$
$7$	$7$
	$1$

Here, $2800=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\times 7$

We can see that $7$ is not paired

So, we have to divide $2800$ by $7$ to get a perfect square.

The new number will be $2800\div 7=400$

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

vi. $\text{1620}$

Ans: The factorization of $1620$ is as follows:

$2$	$1620$
$2$	$810$
$3$	$405$
$3$	$135$
$3$	$45$
$3$	$15$
$5$	$5$
	$1$

Here, $1620=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times 5$

We can see that $5$ is not paired

So, we have to divide $1620$ by $5$ to get a perfect square.

The new number will be $1620\div 5=324$

$324=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{324}=2\times 3\times 3$

So, $\sqrt{324}=18$

7. The students of Class VIII of a school donated Rs 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.

Ans: According to the question, each student donated as many rupees as the number of students in the class.

We can find the number of students in the class by doing the square root of the total amount donated by the students of Class VIII.

The total amount donated by students is Rs. $2401$

Then, the number of students in the class will be $\sqrt{2401}$

$\sqrt{2401}=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}$

$=7\times 7$

$=49$

Thus, there are total $49$ students in the class.

8. Around 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.

Ans: According to the question, the plants are being planted in a garden in such a way that each row contains as many plants as the number of rows.

So, the number of rows will be equal to the number of plants in each row.

Hence,

The number of rows $\times $ Number of plants in each row $=$Total number of plants

The number of rows $\times $ Number of plants in each row $=$ $2025$

The number of rows $\times $ The number of rows $=$ $2025$

The number of rows $=$$\sqrt{2025}$

$\sqrt{2025}=\sqrt{\underline{5\times 5}\times \underline{3\times 3}\times \underline{3\times 3}}$

$=5\times 3\times 3$

$=45$

Thus, the number of rows $=45$ and the number of plants in each row $=45$.

9. Find the smallest square number that is divisible by each of the numbers $\text{4,9}$ and $\text{10}$.

Ans: We know that the number that is perfectly divisible by each one of $4,9$ and $10$ is their L.C.M

So, taking the L.C.M of these numbers

$2$	$4,9,10$
$2$	$2,9,5$
$3$	$1,9,5$
$3$	$1,3,5$
$5$	$1,1,5$
	$1,1,1$

L.C.M$=2\times 2\times 3\times 3\times 5$

$=180$

It can be clearly seen that $5$ cannot be paired.

Therefore, we have to multiply $180$ by $5$ in order to get a perfect square.

Thus, the smallest square number divisible by $4,9$ and $10$$=180\times 5=900$

10. Find the smallest square number that is divisible by each of the numbers $\text{8,15}$ and $\text{20}$.

Ans: We know that the number that is perfectly divisible by each one of $8,15$ and $20$ is their L.C.M

So, taking the L.C.M of these numbers

$2$	$8,15,20$
$2$	$4,15,10$
$2$	$2,15,5$
$3$	$1,15,5$
$5$	$1,5,5$
	$1,1,1$

L.C.M$=2\times 2\times 2\times 3\times 5$

$=120$

It can be clearly seen that the prime factors $2$, $3$ and $5$ cannot be paired.

Therefore, we have to multiply $120$ by $2$, $3$ and $5$ in order to get a perfect square.

Thus, the smallest square number divisible by $8, 15$ and $20$ is $120\times 2\times 3\times 5 = 3600$

Conclusion

NCERT Solutions for Class 8 Maths Chapter Square and Square Roots Exercise 5.3 by Vedantu helps students master various methods of finding square roots, including prime factorization, division, and estimation methods. This exercise is crucial for developing a strong foundation in understanding the properties of numbers and improving problem-solving skills. Students should focus on mastering each method and practicing a variety of problems to gain confidence. Vedantu's detailed, step-by-step solutions ensure that students can follow along easily, making complex concepts more understandable and aiding in thorough exam preparation.

Class 8 Maths Chapter 5: Exercises Breakdown

Exercise	Number of Questions
Exercise 5.1	9 Questions and Solutions
Exercise 5.2	2 Questions and Solutions

CBSE Class 8 Maths Chapter 5 Other Study Materials

S. No	Important Links for Chapter 5 Squares and Square Roots
1	Class 8: Squares and Square Roots Revision Notes
2	Class 8: Squares and Square Roots Important Questions
3	Class 8: Squares and Square Roots Formula
4	Class 8: Squares and Square Roots NCERT Exemplar Solution
5	Class 8: Squares and Square Roots RD Sharma Solutions
6	Class 8: Squares and Square Roots RS Aggarwal Solutions

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

S. No	NCERT Solutions Class 8 Maths Chapter-wise List
1	Chapter 1 - Rational Numbers Solutions
2	Chapter 2 - Linear Equations in One Variable Solutions
3	Chapter 3 - Understanding Quadrilaterals Solutions
4	Chapter 4 - Data Handling Solutions
5	Chapter 5 - Squares and Square Roots Solutions
6	Chapter 6 - Cubes and Cube Roots Solutions
7	Chapter 7 - Comparing Quantities Solutions
8	Chapter 8 - Algebraic Expressions and Identities Solutions
9	Chapter 9 - Mensuration Solutions
10	Chapter 10 - Exponents and Powers Solutions
11	Chapter 11 - Direct and Inverse Proportions Solutions
12	Chapter 12 - Factorisation Solutions
13	Chapter 13 - Introduction to Graphs Solutions

Important Related Links for CBSE Class 8 Maths

S. No	Other Study Materials for CBSE Class 8 Maths
1	CBSE Class 8 Maths NCERT Textbook
2	CBSE Class 8 Maths Important Questions
3	CBSE Class 8 Maths Sample Papers
4	CBSE Class 8 Maths NCERT Exemplar Solutions
5	CBSE Class 8 Maths RD Sharma Solutions
6	CBSE Class 8 Maths RS Aggarwal Solutions
7	CBSE Class 8 Maths Formulas
8	CBSE Class 8 Maths NCERT Solutions
9	CBSE Class 8 Maths Syllabus
10	CBSE Class 8 Maths Revision Notes

FAQs on NCERT Solutions for Class 8 Maths Chapter 5 Squares And Square Roots Ex 5.3

1. What is the correct stepwise method to solve Exercise 5.1 of NCERT Class 8 Maths Chapter 5 Squares and Square Roots?

To solve Exercise 5.1 as per the updated CBSE 2025–26 NCERT pattern, start by identifying the basic square numbers as required in each question. For finding if a number is a perfect square, use factorization and grouping in pairs. Always arrange steps clearly: state the number, list its prime factors, group them into pairs, and check for unpaired factors. The answer must conclude if the number is or isn't a perfect square as shown in the NCERT solution format.

2. How do I write stepwise NCERT solutions for questions involving square roots by prime factorization in Chapter 5?

The NCERT-approved stepwise method for finding square roots using prime factorization requires: writing the number, breaking it down into its prime factors, grouping the factors in pairs, selecting one factor from each pair and multiplying, then providing the final square root. Each step should be shown clearly in line with the format in the official Class 8 Maths NCERT textbook answers.

3. Are the answers for Exercise 5.2 and 5.3 in NCERT Solutions for Class 8 Maths Chapter 5 provided in a CBSE-approved format?

Yes, all solved NCERT Solutions for Exercises 5.2 and 5.3, including the stepwise explanations for estimating square roots and using division methods, strictly follow the official CBSE 2025–26 textbook answer style. Each answer presents all calculation steps, procedures, and reasons exactly as expected for CBSE exams.

4. How can I check if my answer matches the NCERT answer key for Class 8 Maths Chapter 5?

You can match your answer by comparing each step with the detailed solutions provided for Class 8 Maths Chapter 5 in the NCERT Solutions. Ensure your method of factorization, pairing, and final answers follow the same NCERT structure, especially for questions in Exercises 5.1 to 5.4. Pay special attention to the layout and stepwise reasoning as per the CBSE guidelines.

5. What is square root in Class 8 NCERT Solutions and how is it calculated for non-perfect squares?

In Class 8 NCERT Solutions, the square root of a number is the value which, when multiplied by itself, gives the original number. For non-perfect squares, the standard CBSE format uses the long division method. This process is shown in a stepwise manner in the solutions, with each step of division and digit grouping made clear, as per the official 2025–26 NCERT answer structure.

6. Where can I download the PDF of Class 8 Maths Chapter 5 NCERT Solutions for ‘Squares and Square Roots’?

You can download the full PDF of NCERT Solutions for Class 8 Maths Chapter 5 ‘Squares and Square Roots’ from Vedantu’s website. The solutions are free and provided in the latest CBSE 2025–26 format, covering all exercises including 5.1, 5.2, 5.3, and 5.4 in both English and Hindi medium as per NCERT textbook answers.

7. Which NCERT pattern is used to solve Class 8 Maths Chapter 5 Exercise 5.4?

Exercise 5.4 in Class 8 Maths Chapter 5 is solved using the systematic CBSE NCERT method involving stepwise calculations for estimating square roots to the required number of decimal places. The solution shows digit grouping and division, each step explicitly indicated, matching the format set out in the official NCERT textbook for the 2025–26 syllabus.

8. Is Class 8 Maths Chapter 5 ‘Squares and Square Roots’ considered difficult and how do solutions help?

While Class 8 Maths Chapter 5 can appear challenging due to new concepts like prime factorization and division methods, the NCERT Solutions provide clear, stepwise answers that make it easier to understand. Each question is solved as per CBSE and NCERT expectations, helping students follow logical steps and avoid common mistakes, thus simplifying the chapter.

9. Which is the hardest chapter in maths class 8 according to most students?

Many students find ‘Algebraic Expressions and Identities’ and ‘Mensuration’ more challenging than Chapter 5, but ‘Squares and Square Roots’ can be tricky due to unfamiliar methods. Using the detailed NCERT Solutions for Chapter 5 ensures concepts become clear and the stepwise approach is easier to follow for all CBSE questions.

10. How can I get full marks in Class 8 Maths Chapter 5 ‘Squares and Square Roots’ in board exams?

To score full marks in CBSE and school tests for Chapter 5, solve all NCERT exercises using the exact answer structure shown in the solutions, show every calculation step, practice the division and prime factorization methods, and check your answers against the NCERT Solution PDF. Consistency in the CBSE-approved NCERT answer format is key.

11. Do the NCERT Solutions for Class 8 Maths Chapter 5 cover all types of exam questions?

Yes, the Class 8 Maths Chapter 5 NCERT Solutions are comprehensive and cover every question in the textbook, including intext and exercise problems. The answers are stepwise, as per the CBSE format, and suitable for all major school exams as well as Olympiad preparation based on the NCERT syllabus.

12. What approach should I use if my prime factorization does not result in perfect pairs?

If, after prime factorization, a number does not have every factor paired, it means the number is not a perfect square. In your stepwise answer, point out the unpaired factor and conclude as per the NCERT answer key instructions for Class 8 Maths Chapter 5.

13. How should I solve exercise questions if my school follows Hindi-medium textbooks?

NCERT Solutions for Class 8 Maths Chapter 5 are available for both English and Hindi mediums. Regardless of language, the stepwise solution pattern, methods for prime factorization, and division technique remain the same according to the official NCERT format, ensuring you get the correct answer for CBSE evaluation.