RD Sharma Class 8 Math 3rd Chapter “Squares & Square Roots” Exercise 3.4 Solution

Chapter: “Squares & Square Roots Exercise 3.4”

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

(ii) 99856 

(iii) 998001

(iv) 657666025

Solution:

(i) The unit digit of the number 9801 is 1. So, the possible unit digits are 1 or 9. Note that 9801 is equal to 992. Hence, the square root is an odd number.

(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 (3³ × 3⁷)². Hence, the square root is an odd number.

(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 × 23 × 223)². Hence, the square root is an odd number.

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

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

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:

The prime factorization of 180:

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.

 

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:

We know that the prime factorization of 147 is

147 = 3 x 7 x 7

Now, group the factors into pairs of equal factors.

147 = 3 x (7 x 7)

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

So, when you multiply 3 by 147, the perfect square number is obtained.

This means that,

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

So, to get the square root of 441, take one factor from each pair.

3 x 7= 21.

Hence, the square root of the perfect square number is 21.

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:

The prime factorization of 3645:

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.

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

Solution:

We know that the prime factorization of 1152 is

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

Now, group the factors into pairs of equal factors.

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

Here, the factor, 2 does not have a pair.

So, when you divide 1152 by 2, the perfect square number is obtained.

This means that,

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

So, to get the square root of 576, take one factor from each pair.

2 x 2 x 2 x 3 = 24.

Hence, the square root of the perfect square number is 24.

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

Solution:

Let the two numbers be a and b.

From the first statement, we have:

a x b = 1296

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

b = 16 x a.

Substituting this value in the first equation, we get:

a x (16 x a) = 1296

By simplifying both sides, we get:

a² = 1296/16 = 81

Hence, a is the square root of 81, which is 9.

To find b, use equation b = 16 x a.

Since a = 9: b = 16 x 9 = 144

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

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

Solution:

Given that, Let N be the number of residents.

Let r the money in rupees donated by each resident.

Therefore, Total donation = N x r = 202500 ….(1)

It is given that, the money received is the same as the number of residents.

i.e.,

r = N ….(2)

Substitute (2) in (1), we get

N x N = 202500

N² = 202500

N² = (2 x 2) x (5 x 5) x (5 x 5) x (3 x 3)²

Now, take square root on both the sides, we get

N =2 x 5 x 5 x 3 x 3 = 450

Therefore, the number of residents is 450.

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

Solution:

Let M be the number of members.

Let r be the amount in paise donated by each member.

The total contribution can be expressed as follows:

M x r = Rs 92.16 = 9216 paise

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

r = M

Substituting this in the first equation, we get:

M x M = 9216

M² = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2  x 3 x 3

M² = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3)

M = 2 x 2 x 2 x 2 x 2 x 3 = 96

To find r, we can use the relation r = M.

Let M be the number of members.

Let r be the amount in paise donated by each member.

The total contribution can be expressed as follows:

M x r = Rs 92.16 = 9216 paise

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

r = 96

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

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

Solution:

Let the number of students be S.

Let the money donated by each student be r

Therefore, the total contribution is

(S)(r) = Rs 2304  …..(1)

It is given that each student paid as many paise as the number of students

So, r = S …..(2)

Substitute (2) in (1), we get:

S x S = 2304

S²= 2 x 2 x  2 x 2  x 2 x 2 x 2 x  2 x 3 x 3

S² = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3)

take square roots on both the sides, it becomes

S =2 x 2 x 2  x 2 x 3 = 48

Therefore, there are total of 48 students in the school.

 11.The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.

Solution:

First, we have to find the perimeter of the square.

The area of the square is r², where r is the side of the square.

Then, we have the equation as follows:

r² = 5184 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)

Taking the square root, we get r = 2 x 2 x 2 x 3 x 3 = 72

Hence the perimeter of the square is 4 x r = 288 m

Now, let L be the length of the rectangular field.

Let W be the width of the rectangular field.

The perimeter is equal to the perimeter of square.

Hence, we have:

2(L + W) = 288

Moreover, since the length is twice the width:

L = 2 x W.

Substituting this in the previous equation, we get:

2 x (2 x W+ W) = 288 3 x W= 144 W= 48

To find L: L = 2 x W = 2 x 48 = 96

Area of the rectangular field = L x W = 96 x 48 = 4608 m²

12. Find the least square number, exactly divisible by each one of the numbers:

(i) 6, 9, 15 and 20                           

(ii) 8, 12, 15 and 20

Solution:

(i)  Given Number: 6, 9, 15 and 20

It is known that the smallest number that divisible by 6, 9, 15 and 20 is their L.C.M.,

So, LCM of 6, 9, 15 and 20 is 60.

Now factorize the number 60 into its prime factors.

60 =2 x2 x 3 x 5

Now, group the factors into pairs of equal factors

60 =(2 x 2)x 3 x 5

Here, factors 3 and 5 are not paired.

So, when you multiply 3×5 by 60, the perfect square number is obtained.

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

Therefore, the least square number is 900.

(ii) Given Number: 8, 12, 15 and 20

It is known that the smallest number that divisible by 8, 12, 15 and 20 is their L.C.M.,

So, LCM of 8, 12, 15 and 20 is 120.

Now factorize the number 120 into its prime factors.

120 =2 x 2 x 2 x 3 x 5

Now, group the factors into pairs of equal factors

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

Here, factors 2,  3 and 5 are not paired.

So, when you multiply 2 x3x5 by 120, the perfect square number is obtained.

(2 x 2) x 2 x 3 x 5 x 2 x 3 x5 = 3600.

Therefore, the least square number is 3600.

13.Find the square roots of 121 and 169 by the method of repeated subtraction.

Solution:

To find the square root of 121:

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

14. Write the prime factorization of the following numbers and hence find their square roots:

(i) 7744                    

(ii) 9604                            

(iii) 5929                     

(iv) 7056

Solution:

(i) Given number: 7744

We know that the prime factorization of 7744 is

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

Now, group the factors into pairs of equal factors.

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

Here, all the factors are paired and there are no factors left.

So, to get the square root of 7744, take one factor from each pair.

2 × 2 × 2 × 11 = 88

Hence, the square root of the perfect square number 7744 is 88.

(ii) Given number: 9604

We know that the prime factorization of 9604 is

9604 = 2 × 2 × 7 × 7 × 7 × 7

Now, group the factors into pairs of equal factors.

9604 = ( 2 × 2 ) × ( 7 × 7 ) × ( 7 × 7  )

Here, all the factors are paired and there are no factors left.

So, to get the square root of 9604, take one factor from each pair.

2 × 7 × 7 = 98

Hence, the square root of the perfect square number 9604 is 98.

(iii)  Given number: 5929

We know that the prime factorization of 5929 is

5929 = 7 × 7 × 11 × 11

Now, group the factors into pairs of equal factors.

5929 = ( 7 × 7 ) × ( 11 × 11 )

Here, all the factors are paired and there are no factors left.

So, to get the square root of 5929, take one factor from each pair.

7 × 11 = 77

Hence, the square root of the perfect square number 5929 is 77.

(iv) Given number: 7056

We know that the prime factorization of 7056 is

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

Now, group the factors into pairs of equal factors.

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

Here, all the factors are paired and there are no factors left.

So, to get the square root of 7056, take one factor from each pair.

2 × 2 × 3 × 7 = 84

Hence, the square root of the perfect square number 7056 is 84.

15.The student of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student denoted as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

Let S be the number of students.

Let r be the amount in rupees denoted by each student.

The total donation can be expressed by:

S × r = Rs. 2401

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

So, r is equal to S.

Substituting this in the first equation:

S × S = 2401

S² = (7 × 7) × (7 × 7)

S = 7 × 7 = 49

So, there are 49 students in the class.

16. A PT teacher wants to arrange a maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.

Solution:

Given data: Maximum possible number of students = 6000

71 students were left out,

So, there are only 5929, i.e., (6000 – 72) students remaining.

Hence, the number of rows and columns is simply the square root of 5929.

Factorize the number 5929 into its prime factors:

5929 = 7 × 7 × 11 × 11

Now, group the factors into pairs of equal factors:

5929 = ( 7 × 7 ) × ( 11 × 11 )

Here, all the factors are paired and there are no factors left.

So, to get the square root of 5929, take one factor from each pair.

7 × 11 = 77

Therefore, there were 77 rows of students in the arrangement.