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# Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.4 | Set 1

### Question 1.Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots?

(i) 9801

Solution:

Unit digit of 9801 is 1

Unit digit of square root is 1 or 9

9801 is an odd number. Therefore, square root is also odd

(ii) 99856

Solution:

Unit digit of 99856 = 6

Unit digit of square root is 4 or 6

99856 is an even number. Therefore, square root is also even

(iii) 998001

Solution:

Unit digit of 998001 = 1

Unit digit of square root is 1 or 9

998001 is an odd number. Therefore, square root is also odd

(iv) 657666025

Solution:

Unit digit of 657666025 = 5

Unit digit of square root is 5

657666025 is an odd number. Therefore, square root is also odd

### Question 2. Find the square root of each of the following by prime factorization.

(i) 441

Solution:

Prime factorization of

441 = 3×3×7×7 (Pairing of 3 and 7)

= 32×72

√441 = 3×7

= 21

(ii) 196

Solution:

Prime factorization of

196 = 2×2×7×7 (Pairing of 2 and 7)

= 22×72

√196 = 2×7

= 14

(iii) 529

Solution:

Prime factorization of

529 = 23×23 (Pairing of 23)

= 232

√529 = 23

(iv) 1764

Solution:

Prime factorization of

1764 = 2×2×3×3×7×7 (Pairing of 2, 3 and 7)

= 22×32×72

√1764 = 2×3×7

= 42

(v) 1156

Solution:

Prime factorization of

1156 = 2×2×17×17 (Pairing of 2 and 17)

= 22×172

√1156 = 2×17

= 34

(vi) 4096

Solution:

Prime factorization of

4096 = 2×2×2×2×2×2×2×2×2×2×2×2 (Pairing of 2)

= 212

√4096 = 26

= 64

(vii) 7056

Solution:

Prime factorization of

7056 = 2×2×2×2×21×21 (Pairing of 2 and 21)

= 22×22×212

√7056 = 2×2×21

= 84

(viii) 8281

Solution:

Prime factorization of

8281 = 91×91 (Pairing of 91)

= 912

√8281=91

(ix) 11664

Solution:

Prime factorization of

11664 = 2×2×2×2×3×3×3×3×3×3 (Pairing of 2 and 3)

= 22×22×32×32×32

√11664 = 2×2×3×3×3

= 108

(x) 47089

Solution:

Prime factorization of

47089 = 217×217 (Pairing of 217)

= 2172

√47089 = 217

(xi) 24336

Solution:

Prime factorization of

24336 = 2×2×2×2×3×3×13×13 (Pairing of 2,3 and 13)

= 22×22×32×132

√24336 = 2×2×3×13

= 156

(xii) 190969

Solution:

Prime factorization of

190969 = 23×23×19×19 (Pairing of 23 and 19)

= 232×192

√190969 = 23×19

= 437

(xiii) 586756

Solution:

Prime factorization of

586756 = 2×2×383×383 (Pairing of 2 and 383)

= 22×3832

√586756 = 2×383

= 766

(xiv) 27225

Solution:

Prime factorization of

27225 = 5×5×3×3×11×11 (Pairing of 5,3 and 11)

= 52×32×112

√27225 = 5×3×11

= 165

(xv) 3013696

Solution:

Prime factorization of

3013696 = 2×2×2×2×2×2×217×217 (Pairing of 2 and 17)

= 26×2172

√3013696 = 23×217

= 1736

### Question 3.Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.

Solution:

Prime factorization of

180 = (2 × 2) × (3 × 3) × 5 (Pairing of 2 and 3)

=22 × 32 × 5

5 is left out

Therefore, multiplying the number with 5

180 × 5 = 22 × 32 × 52

Therefore, square root of √ (180 × 5) = 2 × 3 × 5

= 30

### Question 4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.

Solution:

Prime factorization of

147 = (7 × 7) × 3 (Pairing of 7)

=72 × 3

3 is left out

Therefore, multiplying the number with 3

147 × 3 = 72 × 32

Therefore, square root of √ (147 × 3) = 7 × 3

= 21

### Question 5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.

Solution:

Prime factorization of

3645 = (3 × 3) × (3 × 3) × (3 × 3) × 5 (Pairing of 3)

=32 × 32 × 32 × 5

5 is left out

Therefore, by dividing with 5

3645 ÷ 5 = 32 × 32 × 32

Therefore, square root of √ (3645 ÷ 5) = 3 × 3 × 3

= 27

### Question 6. Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained.

Solution:

Prime factorization of

1152 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3) (Pairing of 2 and 3)

=22 × 22 × 22 × 32 × 2

2 is left out

Therefore, by dividing with number 2

1152 ÷ 2 = 22 × 22 × 22 × 32

Therefore, square root of √ (1152 ÷ 2) = 2 × 2 × 2 × 3

= 24

### Question 7. The product of two numbers is 1296. If one number is 16 times the other, find the numbers.

Solution:

Let us consider two numbers x and y

y =16x

x × y= 1296

x × 16x = 1296

16x2 = 1296

x2 = 1296/16 = 81

x = 9

y = 16x

= 16(9)

= 144

Therefore, y =144 and x =9

### Question 8. A welfare association collected Rs 202500 as donation from the residents. If each paid as many rupees as there were residents, find the number of residents.

Solution:

Let us consider total residents as x

So, each paid Rs. x

Total collection = x (x) = x2

Total Collection = 202500

x = √ 202500

x = √(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)

= 2 × 3 × 3 × 5 × 5

= 450

Therefore, total residents = 450

### Question 9. A society collected Rs 92.16. Each member collected as many paisas as there were members. How many members were there and how much did each contribute?

Solution:

Let there be x members and x paisa collected by each member

Therefore,

x2=Total amount= 9216 paisa

x2 = 9216

x = √9216

= 2 × 2 × 2 × 12

= 96

Therefore, there were 96 members in the society and each contributed 96 paisas.

### Question 10. A society collected Rs 2304 as fees from its students. If each student paid as many paisas as there were students in the school, how many students were there in the school?

Solution:

Let there be x number of students and each contributed Rs.x

Total money obtained = x2 = 2304 paisa

x = √2304

x = √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

x = 2 × 2 × 2 × 2 × 3

x = 48

Therefore, there were 48 students in the school

### Chapter 3 Squares and Square Roots – Exercise 3.4 | Set 2

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