The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

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(i) 9801

(ii) 99856

(iii) 998001

(iv) 657666025

(ii) The unit digit of the number 99856 is 6. So, the possible unit digits are 4 or 6 (Table 3.4). Since its last digit is 6 (an even number), it cannot have an odd number as its square root.

(iii) The unit digit of the number 998001 is 1. So, the possible unit digits are 1 or 9. Note that 998001 is equal to (3

(iv) The unit digit of the number 657666025 is 5. So, the only possible unit digit is 5. Note that 657666025 is equal to (5 x 23 x 223)

Hence, among the given numbers, (i), (iii) and (iv) have odd numbers as their square roots.

(i) 441

(ii) 196

(iii) 529

(iv) 1764

(v) 1156

(vi) 4096

(vii) 7056

(viii) 8281

(ix) 11664

(x) 47089

(xi) 24336

(xii) 190969

(xiii) 586756

(xiv) 27225

(xv) 3013696

441 = 3 x 3 x 7 x 7

441 = (3 x 3) x (7 x 7)

Taking one factor for each pair, we get the square root of 441:

3 x 7 = 21

(ii) Resolving 196 into prime factors:

196 = 2 x 2 x 7 x 7

196 = (2 x 2) x (7 x 7)

Taking one factor for each pair, we get the square root of 196:

2 x 7 = 14

(iii) Resolving 529 into prime factors:

529 = 23 x 23

529= (23 x 23)

Taking one factor for each pair, we get the square root of 529 as 23.

(iv) Resolving 1764 into prime factors:

1764 = 2 x 2 x 3 x 3 x 7 x 7

1764 = (2 x 2) x (3 x 3) x (7 x 7)

Taking one factor for each pair, we get the square root of 1764:

2 x 3 x 7 = 42

(v) Resolving 1156 into prime factors:

1156 = 2 x 2 x 17 x 17

1156 = (2 x 2) x (17 x 17)

Taking one factor for each pair, we get the square root of 1156:

2 x 17 = 34

(vi) Resolving 4096 into prime factors:

4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

4096 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2)

Taking one factor for each pair, we get the square root of 4096:

(2 x 2) x (2 x 2) x (2 x 2) = 64

(vii) Resolving 7056 into prime factors:

7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7

7056 = (2 x 2) x (2 x 2) x (3 x 3) x (7 x 7)

Taking one factor for each pair, we get the square root of 705:

2 x 2 x 3 x 7 = 84

(viii) Resolving 8281 into prime factors:

8281 = 7 x 7 x 13 x 13

8281 = (7 x 7) x (13 x 13)

Taking one factor for each pair, we get the square root of 8281:

7 x 13 = 91

(ix) Resolving 11664 into prime factors:

11664 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

11664 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3)

Taking one factor for each pair, we get the square root of 11664:

2 x 2 x 3 x 3 x 3 = 108

(x) Resolving 47089 into prime factors:

47089 = 7 x 7 x 31 x 31

47089 = (7 x 7) x (31 x 31)

Taking one factor for each pair, we get the square root of 47089:

7 x 31 = 217

(xi) Resolving 24336 into prime factors:

24336 = 2 x 2 x 2 x 2 x 3 x 3 x 13 x 13

24336 = (2 x 2) x (2 x 2) x (3 x 3) x (13 x 13)

Taking one factor for each pair, we get the square root of 24336:

2 x 2 x 3 x 13 = 156

(xii) Resolving 190969 into prime factors:

190969 = 19 x 19 x 23 x 23

190969 = (19 x 19) x (23 x 23)

Taking one factor for each pair, we get the square root of 190969:

19 x 23 = 437

(xiii) Resolving 586756 into prime factors:

586756 = 2 x 2 x 383 x 383

586756 = (2 x 2) x (383 x 383)

Taking one factor for each pair, we get the square root of 586756:

2 x 383 = 766

(xiv) Resolving 27225 into prime factors:

27225 = 3 x 3 x 5 x 5 x 11 x 11

27225 = (3 x 3) x (5 x 5) x (11 x 11)

Taking one factor for each pair, we get the square root of 27225:

3 x 5 x 11 = 165

(xv) Resolving 3013696 into prime factors:

3013696 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 31 x 31

3013696 = (2 x 2) x (2 x 2) x (2 x 2) x (7 x 7) x (31 x 31)

Taking one factor for each pair, we get the square root of 3013696:

2 x 2 x 2 x 7 x 31 = 1736

180 = 2 x 2 x 3 x 3 x 5

Grouping the factors into pairs of equal factors, we get:

180 = (2 x 2) x (3 x 3) x 5

The factor, 5 does not have a pair. Therefore, we must multiply 180 by 5 to make a perfect square. The new number is:

(2 x 2) x (3 x 3) x (5 x 5) = 900

Taking one factor from each pair on the LHS, the square root of the new number is 2 x 3 x 5, which is equal to 30.

147 = 3 x 7 x 7

Grouping the factors into pairs of equal factors, we get:

147 = 3 x (7 x 7)

The factor, 3 does not have a pair. Therefore, we must multiply 147 by 3 to make a perfect square. The new number is:

(3 x 3) x (7 x 7) = 441

Taking one factor from each pair on the LHS, the square root of the new number is 3 x 7, which is equal to 21.

3645 = 3 x 3 x 3 x 3 x 3 x 3 x 5

Grouping the factors into pairs of equal factors, we get:

3645 = (3 x 3) x (3 x 3) x (3 x 3) x 5

The factor, 5 does not have a pair. Therefore, we must divide 3645 by 5 to make a perfect square. The new number is:

(3 x 3) x (3 x 3) x (3 x 3) = 729

Taking one factor from each pair on the LHS, the square root of the new number is 3 x 3 x 3, which is equal to 27.

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping the factors into pairs of equal factors, we get:

1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2

The factor, 2, at the end, does not have a pair. Therefore, we must divide 1152 by 2 to make a perfect square. The new number is:

(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = 576

Taking one factor from each pair on the LHS, the square root of the new number is 2 x 2 x 2 x 3, which is equal to 24.

From the first statement, we have:

If one number is 16 times the other, then we have:

Substituting this value in the first equation, we get:

By simplifying both sides, we get:

Hence,

To find

Since

So, the two numbers satisfying the question are 9 and 144.

Let

Total donation =

Since the money received as donation is the same as the number of residents:

∴

Substituting this in the first equation, we get:

So, the number of residents is 450.

Let

The total contribution can be expressed as follows:

Since the amount received as donation is the same as the number of members:

∴

Substituting this in the first equation, we get:

To find

Let

Let

The total contribution can be expressed as follows:

Since the amount received as donation is the same as the number of members:

∴

So, there are 96 members and each paid 96 paise.

Let

The total contribution can be expressed by

Since each student paid as many paise as the number of students, then

So, there are 48 students in total in the school.

The area of the square is

Then, we have the equation as follows:

Taking the square root, we get

Hence the perimeter of the square is 4 x

Now let

Let

The perimeter is equal to the perimeter of square.

Hence, we have:

2(

Moreover, since the length is twice the width:

Substituting this in the previous equation, we get:

2 x (2 x

3 x

To find

∴ Area of the rectangular field =

(i) 6, 9, 15 and 20

(ii) 8, 12, 15 and 20

Factorising 60 into its prime factors:

60 = 2 x 2 x 3 x 5

Grouping them into pairs of equal factors:

60 = (2 x 2) x 3 x 5

The factors 3 and 5 are not paired. To make 60 a perfect square, we have to multiply it by 3 x 5, i.e . by 15.

The perfect square is 60 x 15, which is equal to 900.

(i) The smallest number divisible by 8, 12, 15 and 20 is their L.C.M., which is equal to 120.

Factorising 120 into its prime factors:

120 = 2 x 2 x 2 x 3 x 5

Grouping them into pairs of equal factors:

120 = (2 x 2) x 2 x 3 x 5

The factors 2, 3 and 5 are not paired. To make 120 into a perfect square, we have to multiply it by 2 x 3 x 5, i.e. by 30.

The perfect square is 120 x 30, which is equal to 3600.

121 − 1 = 120

120 − 3 = 117

117 − 5 = 112

112 − 7 = 105

105 − 9 = 96

96 − 11 = 85

85 − 13 = 72

72 − 15 = 57

57 − 17 = 40

40 − 19 = 21

21 − 21 = 0

In total, there are 11 numbers to subtract from 121. Hence, the square root of 121 is 11.

To find the square root of 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

In total, there are 13 numbers to subtract from 169. Hence, the square root of 169 is 13.

(i) 7744

(ii) 9604

(iii) 5929

(iv) 7056

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

Grouping them into pairs of equal factors, we get:

7744 = (2 × 2) × (2 × 2) × (2 × 2) × (11 × 11)

Taking one factor from each pair, we get :

9604 = 2 × 2 × 7 × 7 × 7 × 7

Grouping them into pairs of equal factors, we get:

9604 = (2 × 2) × (7 × 7) × (7 × 7)

Taking one factor from each pair, we get:

5929 = 7 x 7 x 11 x 11

Grouping them into pairs of equal factors, we get:

5929 = (7 × 7) × (11 × 11)

Taking one factor from each pair, we get:

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

Grouping them into pairs of equal factors, we get:

7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)

Taking one factor from each pair, we get:

Let

The total donation can be expressed by:

Since the total amount in rupees is equal to the number of students,

Substituting this in the first equation: