# Class 8 ICSE Maths Solutions Chapter 3

## Class 8 ICSE Maths Solutions Chapter 3 Squares and Square Roots (Selina Concise)

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 3, Squares and Square Roots. Here students can easily find step by step solutions of all the problems for Squares and Square Roots, Exercise 3A, 3B and 3C Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Squares and Square Roots Exercise 3(A) Solution :

Question no – (1)

Solution :

(i) 59

= 59²

= 3481

Thus, the square of 59 is 3481

(ii) 6.3

= (6.3)²

= 39.69

Hence, the square of 6.3 is 39.69

(iii) 15

= (15)²

= 225

Therefore, the square of 15 is 225.

Question no – (2)

Solution :

(i) 11025

= √11025

= √5 × 5 × 7 × 7 × 3 × 3

= 5 × 7 × 3

= 105

So, the square root of 11025 is 105.

(ii) 396900

= √3969900

= √2 × 2 × 3 × 3 × 7 × 5 × 5 × 3 ×3

= √2 × 3 × 7 × 5 × 3

= 630

Hence, the square root of 396900 is 630.

(iii) 194481

= √194481

= √3 × 3 × 3 × 3 × 7 × 7 × 7 × 7

= 3 × 3 × 7 × 7

= 441

Therefore, the square root of 194481 is 441.

Question no – (3)

Solution :

(i) As per the question,

2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

Hence, to make it perfect square the given number should be multiplied by 2.

(ii) According to the question,

12748 = 2 × 2 × 3187

Thus, to make it perfect square, the given number should multiplied by 3187.

Question no – (4)

Solution :

As per the question,

10368 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 2

The give number should be divided by 2.

10368/2 = 5184

√5184

= 72

Question no – (5)

Solution :

(i) 0.1764

= √0.1764

= √1764/10000

= 42/100

= 0.42

So, the square root of 0.1764 is 0.42

(ii) 96 1/25

= √96 1/25

= √2401/25

= √7× 7 × 7 × 7

= 7 × 7/5

= 49/5

Hence, the square root of 96 1/25 is 49/5

(iii) 0.0169

= √0.0169

= √169/10000

= 13/100

= 0.13

Therefore, the square root of 0.0169 is 0.13.

Question no – (6)

Solution :

(i) √14.4/22.5

= √144/225

= 12/15

(ii) √0.225/28.9

= √225/29900

= √ 15 × 15/17 × 17 × 10 × 10

= 15/17× 10

= 15/170

= 3/34

(iii) √25/32 × 2 13/18 × 0.25

= √25/32 × 49/18 × 25/100

= √25 × 49/32 × 18 × 4

= 5 × 7/2 × 24

= 35/48

(iv) √1 4/5 × 14 21/44 × 2 7/55

= √9/5 × 637/44 × 117/55

= √9 × 7 × 7 × 13 × 13 × 9/5 × 11 × 2 × 2 × 11 × 5

= √9 × 7 × 13/5 × 11 × 2

= 819/110

= 7 49/110

Question no – (7)

Solution :

(i) √3² × 6³ × 24

= √3² × 6³ × 2 × 2 × 6

= √3² × 6² × 2 × 2 × 6 × 6

= 3 × 6 × 2 × 6

= 216

(ii) √(0.5)³ × 6 × 3⁵

= √(0.5)² × 0.5 × 3 × 2 × 3⁵

= √(0.5) ² × 1 × 3.6

= 0.5 × 3³

= 0.5 × 27

= 13.5

(iii) √(5 + 2 21/25) × 0.169/1.6

= √(5 + 71/25) × 0.169/1.600

= √196/25 × 169/1600

= √14 × 14/5 × 5 × 13 × 13/40 × 40

= 14 × 13/5 × 40

= 91/100

= 0.91

(iv) √5(2 3/4 – 3/10)

= √5(11/4 – 3/10)

= √5(53 – 6/20)

= √5(49/20)

= √49/4

= 7/2

(v) √248 + √52 + √144

= √248 + √52 + 12

= √248 + 64

= √248 + 8

= √356

= 16

Question no – (8)

Solution :

Let, the no of days = x

Rupees spent = x

Total spent = (x × x) = x²

x² = 1296

= x = √1296

= 36

Therefore, His tour will last 36 Days.

Question no – (9)

Solution :

Total students = 745

Left = 16

Rows,

= (745 – 16)

= 729

Number of rows,

= √729

= 27

Hence, the number of rows will be 27.

Question no – (10)

Solution :

Here, 169 = (13)²

961 = (31)²

Similarly, two such no’s are = 12 and 21

(12)² = 144

(21)² = 441

And the no’s are 102 and 201

(102)² = 10404

(201)² = 40401

Question no – (11)

Solution :

As per the given question,

L.C.M = 2 × 3 × 2 × 5 = 60

To perfect square, we should multiply by 3 x 5 = 15

Perfect square,

= (60 × 15)

= 900

Question no – (12)

Solution :

(i) √784 = 28

√7.84 = 2.8

and √√78400 = 280

Now, √7.84 + √78400

= 2.8 + 280

= 282.8

Therefore, the value is 282.8

(ii) √0.0784 + √0.000784

Now, √0.000784 = 8.028

√0.0784 = 0.28

So, √0.0784 + √0.000784

= 0.28 + 0.028

= 0.308

Hence, the value is 0.308

Squares and Square Roots Exercise 3(B) Solution :

Question no – (1)

Solution :

(i) 4761

= √4761

= 69

Thus, the square root of 4761 is 69.

(ii) 7744

√7744

= 88

Hence, the square root of 7744 is 88.

(iii) 15129

∴ √15129

= 123

Thus, the square root of 15129 is 123.

(iv) 0.2916

√0.2916

= 0.54

Therefore, the square root of 0.2916 is 0.54

(v) 0.001225

√0.001225

= 0.035

Hence, the square root of 0.001225 is 0.035

(vi) 0.023104

√0.023104

= 0.152

Thus, the square root of 0.023104 is 0.152

(vii) 27.3529

√27.3529

= 5.23

So, the square root of 27.3529 is 5.23

Question no – (2)

Solution :

(i) 4.2025

√4.2025 = 2.05

Thus, the square root of 4.2025 is 2.05.

(ii) 531.7636

√531.7636 = 23.06

Thus, the square root of 531.7636 is 23.06

(iii) 0.007225

√0.007225 = 0.085

Hence, the square root of 0.007225 is 0.085

Question no – (3)

Solution :

(i) 245

Square root = 15.65 (Correct up to 2 decimal)

(iii) In the given question, 82.6

Square root = 9.09 (Correct up to 2 decimal)

(iv) 0.065

√0.0650 = 0.254 (Correct in 3decimal)

Question no – (4)

Solution :

(i) 3 4/5

= 19/5

= 3.80

√3 4/5

= 1.96 (correct to 2 decimal)

(ii) 6 7/8

= 6.875

√6 7/8

= 2.62

Question no – (5)

Solution :

(i) 796

Least no to be subtracted is 12

(ii) 1886

Least no subtracted to make perfect square is 37

(iii) 23497

Least no subtracted is 88 to make perfect square.

Question no – (6)

Solution :

(i) Given number, 511

(511 + 27)

= 529

= 23²

Required number,

= (529 – 27)

= 18

Therefore, the required number will be 18

(ii) 7172

= (84)² = 7056

Required number,

= (7225 – 7172)

= 53

Therefore, the required number will be 53

(iii) 55078

Now, (234)² = 54756

= (235)² = 53225

Now, required number,

= (55225 – 53078)

= 147

Hence, the required number will be 147.

Question no – (7)

Solution :

Given in the question, √4 + √7/√4 – √7

Now, √4 + √7/√4 – √7

= 4 + (√7)(4 + (√7)/(4 -(√7) (4 + √7)

= √(4 + √7)²/4² – (√7)

= √(4 + √7)²/9

= 4 + √7/3

= 4 + 2.65/3

= 2.22

Question no – (8)

Solution :

In the given question, 3 – √5/3 + √5

Now, √3 – √5/3 + √5

= (3 – √5) (3 – √5)/(3 + √5) ( 3 – √5)

= √3 – √5)/9 – 5

= √(3 – √5)/4

= 3 – 2.24/4

= 0 76/4

= 0.38

Question no – (9)

Solution :

(i) 1764/2809

= √1764/2809

= 42/53

Hence, the square root of 1764/2809 is 42/53.

(ii) 507/4107

507/4107

= 169/1369

= √169/1369

= 13/37

Thus, the square root of 507/4107 is 13/37.

(iii) √108 × 2028

= 468

So, the square root of √108 × 2028 is 468

(iv) 0.01 + √0.0064

= 0.01 + 0.08

= 0.09

Therefore, the square root of 0.01 + √0.0064 is 0.09

Question no – (10)

Solution :

(i) In the given question, 7.832

√7.832 = 2.80 (Correct up to 2 decimal)

Question no – (11)

Solution :

As per the question,

(1205 – 44) = 1158

√1156 = 34

Thus, the least number is 34.

Question no – (12)

Solution :

According to the question,

So, now, 34 = 1156

and 35 = 1205

Required number,

= (1225 – 1205)

= 20

Therefore, the least number is 20.

Question no – (13)

Solution :

Now, (2037 – 12)

= 2025

√2025

= 45

Therefore, the least number will be 45.

Question no – (14)

Solution :

So, 74² = 5476

75² = 5625

The required number,

= (5625 – 5483)

= 142

Therefore, the least number will be 142.

Squares and Square Roots Exercise 3(C) Solution :

Question no – (4)

Solution :

The square of the numbers will not have 6 at their unit’s place are – 35, 23, 98

Because, (4)² = 16, (6)² = 36

Question no – (5)

Solution :

The numbers have 6 at their unit’s place are

(i) 26²

(ii) 34²

(iii) 244²}

Because, (4)² = 16, (6)² = 36

Question no – (6)

Solution :

If a number ends with 3 zeroes,

(2 × 3 = 6), then its square will have 6 zeroes.

Question no – (7)

Solution :

If the square of a number ends with 10 zeroes,

(10/2 = 5) then the number will have 5 zeroes.

Question no – (8)

Solution :

No, it is not possible for the square of a number to have 5 zeroes that is odd because the no of zeroes of the square must be 2 zeroes.

Question no – (9)

Solution :

None of the given numbers is a perfect square, because number having 2, 3, 7, 8 at the unit place is have a perfect square.

Question no – (10)

Solution :

(i) 23 – odd number.

(ii) 54 – even number.

(iii) 76 – even number.

(iv) 75 – odd number.

Question no – (11)

Solution :

Because, the given numbers has not a even number of zeroes.

Question no – (12)

Solution :

(i) 37² – 36²

= (37 + 36) (37 – 36)

= 73 × 1

= 73

(ii) 85² – 84²

= (85 + 84) (85 – 84)

= 169 × 1

= 169

(iii) 101² – 100²

= (101 + 100) (101 – 100)

= 201 × 1

= 201

Question no – (13)

Solution :

(i) Sum of 1st 12 odd natural number is 12²

= 144

Thus, sum is 144.

(ii) 1 + 3 + 5 + 7 + 9 + ___ + 39 + 41

= Sum of 1st 21 odd natural numbers

= 21² = 441

Therefore, the sum is 441.

(iii) 1 + 3 + 5 + 7 + 9 + ___ + 51 + 53

= Sum of 1st 27 odd natural numbers

= 27²

= 729

Therefore, the sum is 729

Question no – (14)

Solution :

So, 3² + 4² = 5

= 9 + 16 = 25

= 25 = 25

And, 6² + 8² = 10

= 36 + 64 = 100

= 100 = 100

And, 5² + 12² = 169

= 25 + 144 = 169

= 169 = 169

Next Chapter Solution :

Updated: June 20, 2023 — 10:31 am