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#### EXERCISE 3.1 PAGE NO: 3.4

**1. Which of the following numbers are perfect squares?**

**(i) 484**

**(ii) 625**

**(iii) 576**

**(iv) 941**

**(v) 961**

**(vi) 2500**

**Solution:**

**(i) **484

First find the prime factors for 484

484 = 2×2×11×11

By grouping the prime factors in equal pairs we get,

= (2×2) × (11×11)

By observation, none of the prime factors are left out.

∴ 484 is a perfect square.

**(ii)** 625

First find the prime factors for 625

625 = 5×5×5×5

By grouping the prime factors in equal pairs we get,

= (5×5) × (5×5)

By observation, none of the prime factors are left out.

∴ 625 is a perfect square.

**(iii)** 576

First find the prime factors for 576

576 = 2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3)

By observation, none of the prime factors are left out.

∴ 576 is a perfect square.

**(iv)** 941

First find the prime factors for 941

941 = 941 × 1

We know that 941 itself is a prime factor.

∴ 941 is not a perfect square.

**(v)** 961

First find the prime factors for 961

961 = 31×31

By grouping the prime factors in equal pairs we get,

= (31×31)

By observation, none of the prime factors are left out.

∴ 961 is a perfect square.

**(vi)** 2500

First find the prime factors for 2500

2500 = 2×2×5×5×5×5

By grouping the prime factors in equal pairs we get,

= (2×2) × (5×5) × (5×5)

By observation, none of the prime factors are left out.

∴ 2500 is a perfect square.

**2. Show that each of the following numbers is a perfect square. Also find the number whose square is the given number in each case:(i) 1156(ii) 2025(iii) 14641(iv) 4761**

**Solution:**

**(i)** 1156

First find the prime factors for 1156

1156 = 2×2×17×17

By grouping the prime factors in equal pairs we get,

= (2×2) × (17×17)

By observation, none of the prime factors are left out.

∴ 1156 is a perfect square.

To find the square of the given number

1156 = (2×17) × (2×17)

= 34 × 34

= (34)^{2}

∴ 1156 is a square of 34.

**(ii)** 2025

First find the prime factors for 2025

2025 = 3×3×3×3×5×5

By grouping the prime factors in equal pairs we get,

= (3×3) × (3×3) × (5×5)

By observation, none of the prime factors are left out.

∴ 2025 is a perfect square.

To find the square of the given number

2025 = (3×3×5) × (3×3×5)

= 45 × 45

= (45)^{2}

∴ 2025 is a square of 45.

**(iii)** 14641

First find the prime factors for 14641

14641 = 11×11×11×11

By grouping the prime factors in equal pairs we get,

= (11×11) × (11×11)

By observation, none of the prime factors are left out.

∴ 14641 is a perfect square.

To find the square of the given number

14641 = (11×11) × (11×11)

= 121 × 121

= (121)^{2}

∴ 14641 is a square of 121.

**(iv)** 4761

First find the prime factors for 4761

4761 = 3×3×23×23

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23)

By observation, none of the prime factors are left out.

∴ 4761 is a perfect square.

To find the square of the given number

4761 = (3×23) × (3×23)

= 69 × 69

= (69)^{2}

∴ 4761 is a square of 69.

**3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square:(i) 23805(ii) 12150(iii) 7688**

**Solution:**

**(i)** 23805

First find the prime factors for 23805

23805 = 3×3×23×23×5

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

23805 × 5 = (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)^{2}

∴ Product is the square of 345.

**(ii)** 12150

First find the prime factors for 12150

12150 = 2×3×3×3×3×3×5×5

By grouping the prime factors in equal pairs we get,

= 2×3 × (3×3) × (3×3) × (5×5)

By observation, prime factor 2 and 3 are left out.

So, multiply by 2×3 = 6 we get,

12150 × 6 = 2×3 × (3×3) × (3×3) × (5×5) × 2 × 3

= (2×3×3×3×5) × (2×3×3×3×5)

= 270 × 270

= (270)^{2}

∴ Product is the square of 270.

**(iii)** 7688

First find the prime factors for 7688

7688 = 2×2×31×31×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (31×31) × 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

7688 × 2 = (2×2) × (31×31)× (2×2)

= (2×31×2) × (2×31×2)

= 124 × 124

= (124)^{2}

∴ Product is the square of 124.

**4. Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:(i) 14283(ii) 1800(iii) 2904**

**Solution:**

**(i)** 14283

First find the prime factors for 14283

14283 = 3×3×3×23×23

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23) × 3

By observation, prime factor 3 is left out.

So, divide by 3 to eliminate 3 we get,

14283/3 = (3×3) × (23×23)

= (3×23) × (3×23)

= 69 × 69

= (69)^{2}

∴ Resultant is the square of 69.

**(ii)** 1800

First find the prime factors for 1800

1800 = 2×2×5×5×3×3×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (5×5) × (3×3) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

1800/2 = (2×2) × (5×5) × (3×3)

= (2×5×3) × (2×5×3)

= 30 × 30

= (30)^{2}

∴ Resultant is the square of 30.

**(iii)** 2904

First find the prime factors for 2904

2904 = 2×2×11×11×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (11×11) × 2 × 3

By observation, prime factor 2 and 3 are left out.

So, divide by 6 to eliminate 2 and 3 we get,

2904/6 = (2×2) × (11×11)

= (2×11) × (2×11)

= 22 × 22

= (22)^{2}

∴ Resultant is the square of 22.

**5. Which of the following numbers are perfect squares?11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121**

**Solution:**

11 it is a prime number by itself.

So it is not a perfect square.

12 is not a perfect square.

16= (4)^{2}

16 is a perfect square.

32 is not a perfect square.

36= (6)^{2}

36 is a perfect square.

50 is not a perfect square.

64= (8)^{2}

64 is a perfect square.

79 it is a prime number.

So it is not a perfect square.

81= (9)^{2}

81 is a perfect square.

111 it is a prime number.

So it is not a perfect square.

121= (11)^{2}

121 is a perfect square.

**6. Using prime factorization method, find which of the following numbers are perfect squares?189, 225, 2048, 343, 441, 2961, 11025, 3549**

**Solution:**

189 prime factors are

189 = 3^{2}×3×7

Since it does not have equal pair of factors 189 is not a perfect square.

225 prime factors are

225 = (5×5) × (3×3)

Since 225 has equal pair of factors. ∴ It is a perfect square.

2048 prime factors are

2048 = (2×2) × (2×2) × (2×2) × (2×2) × (2×2) × 2

Since it does not have equal pair of factors 2048 is not a perfect square.

343 prime factors are

343 = (7×7) × 7

Since it does not have equal pair of factors 2048 is not a perfect square.

441 prime factors are

441 = (7×7) × (3×3)

Since 441 has equal pair of factors. ∴ It is a perfect square.

2961 prime factors are

2961 = (3×3) × (3×3) × (3×3) × (2×2)

Since 2961 has equal pair of factors. ∴ It is a perfect square.

11025 prime factors are

11025 = (3×3) × (5×5) × (7×7)

Since 11025 has equal pair of factors. ∴ It is a perfect square.

3549 prime factors are

3549 = (13×13) × 3 × 7

Since it does not have equal pair of factors 3549 is not a perfect square.

**7. By what number should each of the following numbers by multiplied to get a perfect square in each case? Also find the number whose square is the new number.**

**(i) 8820**

**(ii) 3675(iii) 605**

**(iv) 2880(v) 4056**

**(vi) 3468(vii) 7776**

**Solution:**

**(i)** 8820

First find the prime factors for 8820

8820 = 2×2×3×3×7×7×5

By grouping the prime factors in equal pairs we get,

= (2×2) × (3×3) × (7×7) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

8820 × 5 = (2×2) × (3×3) × (7×7) × (5×5)

= (2×3×7×5) × (2×3×7×5)

= 210 × 210

= (210)^{2}

∴ Product is the square of 210.

**(ii)** 3675

First find the prime factors for 3675

3675 = 5×5×7×7×3

By grouping the prime factors in equal pairs we get,

= (5×5) × (7×7) × 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3675 × 3 = (5×5) × (7×7) × (3×3)

= (5×7×3) × (5×7×3)

= 105 × 105

= (105)^{2}

∴ Product is the square of 105.

**(iii)** 605

First find the prime factors for 605

605 = 5×11×11

By grouping the prime factors in equal pairs we get,

= (11×11) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

605 × 5 = (11×11) × (5×5)

= (11×5) × (11×5)

= 55 × 55

= (55)^{2}

∴ Product is the square of 55.

**(iv)** 2880

First find the prime factors for 2880

2880 = 5×3×3×2×2×2×2×2×2

By grouping the prime factors in equal pairs we get,

= (3×3) × (2×2) × (2×2) × (2×2) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

2880 × 5 = (3×3) × (2×2) × (2×2) × (2×2) × (5×5)

= (3×2×2×2×5) × (3×2×2×2×5)

= 120 × 120

= (120)^{2}

∴ Product is the square of 120.

**(v)** 4056

First find the prime factors for 4056

4056 = 2×2×13×13×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (13×13) × 2 × 3

By observation, prime factors 2 and 3 are left out.

So, multiply by 6 we get,

4056 × 6 = (2×2) × (13×13) × (2×2) × (3×3)

= (2×13×2×3) × (2×13×2×3)

= 156 × 156

= (156)^{2}

∴ Product is the square of 156.

**(vi)** 3468

First find the prime factors for 3468

3468 = 2×2×17×17×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (17×17) × 3

By observation, prime factor 3 is left out.

So, multiply by 3 we get,

3468 × 3 = (2×2) × (17×17) × (3×3)

= (2×17×3) × (2×17×3)

= 102 × 102

= (102)^{2}

∴ Product is the square of 102.

**(vii)** 7776

First find the prime factors for 7776

7776 = 2×2×2×2×3×3×3×3×2×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (3×3) × (3×3) × 2 × 3

By observation, prime factors 2 and 3 are left out.

So, multiply by 6 we get,

7776 × 6 = (2×2) × (2×2) × (3×3) × (3×3) × (2×2) × (3×3)

= (2×2×3×3×2×3) × (2×2×3×3×2×3)

= 216 × 216

= (216)^{2}

∴ Product is the square of 216.

**8. By What numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.(i) 16562(ii) 3698(iii) 5103(iv) 3174(v) 1575**

**Solution:**

**(i)** 16562

First find the prime factors for 16562

16562 = 7×7×13×13×2

By grouping the prime factors in equal pairs we get,

= (7×7) × (13×13) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

16562/2 = (7×7) × (13×13)

= (7×13) × (7×13)

= 91 × 91

= (91)^{2}

∴ Resultant is the square of 91.

**(ii)** 3698

First find the prime factors for 3698

3698 = 2×43×43

By grouping the prime factors in equal pairs we get,

= (43×43) × 2

By observation, prime factor 2 is left out.

So, divide by 2 to eliminate 2 we get,

3698/2 = (43×43)

= (43)^{2}

∴ Resultant is the square of 43.

**(iii)** 5103

First find the prime factors for 5103

5103 = 3×3×3×3×3×3×7

By grouping the prime factors in equal pairs we get,

= (3×3) × (3×3) × (3×3) × 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

5103/7 = (3×3) × (3×3) × (3×3)

= (3×3×3) × (3×3×3)

= 27 × 27

= (27)^{2}

∴ Resultant is the square of 27.

**(iv)** 3174

First find the prime factors for 3174

3174 = 2×3×23×23

By grouping the prime factors in equal pairs we get,

= (23×23) × 2 × 3

By observation, prime factor 2 and 3 are left out.

So, divide by 6 to eliminate 2 and 3 we get,

3174/6 = (23×23)

= (23)^{2}

∴ Resultant is the square of 23.

**(v)** 1575

First find the prime factors for 1575

1575 = 3×3×5×5×7

By grouping the prime factors in equal pairs we get,

= (3×3) × (5×5) × 7

By observation, prime factor 7 is left out.

So, divide by 7 to eliminate 7 we get,

1575/7 = (3×3) × (5×5)

= (3×5) × (3×5)

= 15 × 15

= (15)^{2}

∴ Resultant is the square of 15.

**9. Find the greatest number of two digits which is a perfect square.**

**Solution:**

We know that the two digit greatest number is 99

∴ Greatest two digit perfect square number is 99-18 = 81

**10. Find the least number of three digits which is perfect square.**

**Solution:**

We know that the three digit greatest number is 100

To find the square root of 100

∴ the least number of three digits which is a perfect square is 100 itself.

**11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.**

**Solution:**

First find the prime factors for 4851

4851 = 3×3×7×7×11

By grouping the prime factors in equal pairs we get,

= (3×3) × (7×7) × 11

∴ The smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.

**12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.**

**Solution:**

First find the prime factors for 28812

28812 = 2×2×3×7×7×7×7

By grouping the prime factors in equal pairs we get,

= (2×2) × 3 × (7×7) × (7×7)

∴ The smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.

**13. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.**

**Solution:**

First find the prime factors for 1152

1152 = 2×2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3) × 2

∴ The smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.

The number after division, 1152/2 = 576

prime factors for 576 = 2×2×2×2×2×2×3×3

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (2×2) × (3×3)

= 2^{6} × 3^{2}

= 24^{2}

∴ The resulting number is the square of 24.

**All Chapter RD Sharma Solutions For Class 8 Maths**

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