SEBA Class 10 Maths Revision Exercise R-1 Solutions: Square and Square Root

Get free solutions to SEBA Class 10 Maths Chapter R-1: Square and Square Root from the new SCERT textbook 2026. In this article, we have solved all questions of this revision chapter, R-1, in a simple, step-by-step manner. 

See More:

• SEBA Class 10 Maths Chapter R-1 Solutions: Square and Square Root
• SEBA Class 10 Maths Chapter R-2 Solutions: Cube and Cube Root
• SEBA Class 10 Maths Chapter R-3 Solutions: Indices and Power
• SEBA Class 10 Maths Chapter R-4 Solutions: Factorisation
• SEBA Class 10 Maths Chapter R-5 Solutions: Congruence of Triangles

1. What will be the digits in the unit place of the squares of the following numbers?

(i) 272

Solution:-

Unit digit of $272$ is $2$.

$2^2 = 4$

Therefore, the unit digit of $ 272^2$ is $4$.

(ii) 79

$9^2 = 81$ (unit digit is $1$)

Therefore, the unit digit of $79^2$ will be $1$.

(iii) 400

Unit digit of $400$ is $0$.

$0^2 = 0$

Therefore, the unit digit of $400^2$ is $0$.

(iv) 2637

$7^2 = 49$ (unit digit is $9$)

Therefore, the unit digit of $2637^2$ will be $9$.

(v) 640

$0^2 = 0$

Therefore, the unit digit of $640^2$ is $0$.

2. Why do the following numbers are not perfect square?

A perfect square number can only end with digits $0, 1, 4, 5, 6,$ or $9$ and must end with an even number of zeroes.

(i) 1057

Solⁿ:

Since no square number can end with $7$

So, $1057$ is not a perfect square.

(ii) 7928

Solⁿ:

Since no square number can end with $8$

So, $7928$ is not a perfect square.

(iii) 222

Solⁿ:

Since no square number can end with $2$, $222$ is not a perfect square.

3. What are the squares of the following numbers?

(i) 19

Solⁿ:

$19^2 = 19 \times 19 = 361$

(ii) 37

Solⁿ:

$37^2 = 37 \times 37 = 1369$

(iii) 53

Solⁿ:

$53^2 = 53 \times 53 = 2809$

(iv) 78

Solⁿ:

$78^2 = 78 \times 78 = 6084$

4. Find the square roots of the following numbers.

(i) 1764

Solⁿ:

By prime factorisation, we get:

$1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7$

$= 2^2 \times 3^2 \times 7^2$

$\therefore \sqrt{1764} = 2 \times 3 \times 7 = 42$

(ii) 9216

Solⁿ:

By prime factorisation, we get:

$9216 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$

$= 2^{10} \times 3^2$

$= (2^5 \times 3)^2 $

$= (32 \times 3)^2 $

$= 96^2$

$\therefore \sqrt{9216} = 96$

(iii) 7744

Solⁿ:

By prime factorisation, we get:

$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11$

$= 2^6 \times 11^2$

$= (2^3 \times 11)^2 $

$= (8 \times 11)^2 $

$= 88^2$

$\therefore \sqrt{7744} = 88$

(iv) 9801

Solⁿ:

By prime factorisation, we get:

$9801 = 3 \times 3 \times 3 \times 3 \times 11 \times 11$

$= 3^4 \times 11^2$

$= (3^2 \times 11)^2 $

$= (9 \times 11)^2 $

$= 99^2$

$\therefore \sqrt{9801} = 99$

5. Find the least numbers (integer) with which the following numbers are to be multiplied so that they become perfect squares.

(i) 1525

Solⁿ:

By prime factorisation:

$1525 = 5 \times 5 \times 61 = 5^2 \times 61$

Here, the prime factor $61$ does not have a pair.

Therefore, the least number to be multiplied is $61$.

(ii) 1008

Solⁿ:

By prime factorisation:

$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 $

$= 2^4 \times 3^2 \times 7$

Here, the prime factor $7$ does not have a pair.

Therefore, the least number to be multiplied is $7$.

(iii) 2028

Solⁿ:

By prime factorisation:

$2028 = 2 \times 2 \times 3 \times 13 \times 13 $

$= 2^2 \times 13^2 \times 3$

Here, the prime factor $3$ does not have a pair.

Therefore, the least number to be multiplied is $3$.

(iv) 768

Solⁿ:

By prime factorisation:

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 $

$= 2^8 \times 3$

Here, the prime factor $3$ does not have a pair.

Therefore, the least number to be multiplied is $3$.

6. With what least numbers (integer) the following numbers are to be divided so that they become perfect squares.

(i) 468

Solⁿ:

By prime factorisation:

$468 = 2 \times 2 \times 3 \times 3 \times 13 $

$= 2^2 \times 3^2 \times 13$

Here, the prime factor $13$ is left unpaired.

Therefore, the least number to be divided is $13$.

(ii) 1584

Solⁿ:

By prime factorisation:

$1584 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 11 $

$= 2^4 \times 3^2 \times 11$

Here, the prime factor $11$ is left unpaired.

Therefore, the least number to be divided is $11$.

(iii) 2645

Solⁿ:

By prime factorisation:

$2645 = 5 \times 23 \times 23 $

$= 5 \times 23^2$

Here, the prime factor $5$ is left unpaired.

Therefore, the least number to be divided is $5$.

(iv) 1620

Solⁿ:

By prime factorisation:

$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 $

$= 2^2 \times 3^4 \times 5$

Here, the prime factor $5$ is left unpaired.

Therefore, the least number to be divided is $5$.

7. Find the square root of the following decimal numbers:

(i) 12.25

Solⁿ:

$12.25 = \frac{1225}{100}$
$\sqrt{12.25} = \sqrt{\frac{1225}{100}} $
$= \frac{\sqrt{1225}}{\sqrt{100}} $
$= \frac{35}{10} = 3.5$

(ii) 24.01

Solⁿ:

$24.01 = \frac{2401}{100}$
$\sqrt{24.01} = \sqrt{\frac{2401}{100}} $
$= \frac{\sqrt{2401}}{\sqrt{100}} $
$= \frac{49}{10} = 4.9$

(iii) 146.41

Solⁿ:

$146.41 = \frac{14641}{100}$

$\sqrt{146.41} = \sqrt{\frac{14641}{100}} $
$= \frac{\sqrt{14641}}{\sqrt{100}} $
$= \frac{121}{10} = 12.1$

(iv) 102.01

Solⁿ:

$102.01 = \frac{10201}{100}$

$\sqrt{102.01} = \sqrt{\frac{10201}{100}} $
$= \frac{\sqrt{10201}}{\sqrt{100}} $
$= \frac{101}{10} = 10.1$

8. Four options are given for each of the following. Find the correct option.

(a) Which of the following is a square of an odd natural number?

(i) 256

(ii) 169

(iii) 546

(iv) 754

Solⁿ:

The square of an odd natural number is always odd.

Among the given options, $169 = 13^2$ is an odd number.

So the correct option is (ii) 169

(b) Which of the following will have 1 (one) in the unit place?

(i) $19^2$

(ii) $34^2$

(iii) $18^2$

(iv) $20^2$

Solⁿ:

Unit digit of $19$ is $9$, and $9^2 = 81$ (unit digit is $1$).

So the correct option is (i) $19^2$

(c) Between $18^2$ and $19^2$ how many natural numbers are there?

(i) 38

(ii) 36

(iii) 42

(iv) 40

Solⁿ:

Between the squares of $n$ and $(n+1)$, there are $2n$ natural numbers.

Here, $n = 18$.

$\text{Number of natural numbers} = 2 \times 18 = 36$

So the correct option is (ii) 36

(d) Which of the following is not a perfect square?

(i) 441

(ii) 572

(iii) 576

(iv) 729

Solⁿ:

A perfect square number never ends with the digit $2$.

Here, $572$ ends with $2$, so it is not a perfect square.

So the correct option is (ii) 572

(e) If $\sqrt{2025} = 45$, then $\sqrt{20.25}$ is equal to:

(i) 45

(ii) 4.5

(iii) 0.45

(iv) 0.045

Solⁿ:

$\sqrt{20.25} $
$= \sqrt{\frac{2025}{100}} $
$= \frac{\sqrt{2025}}{\sqrt{100}} $
$= \frac{45}{10} = 4.5$

So the correct option is (ii) 4.5