RD Sharma Solutions Class 8 Squares And Square Roots Exercise 3.4

RD Sharma Class 8 Solutions Chapter 3 Ex 3.4 PDF Free Download

RD Sharma Solutions Class 8 Chapter 3 Exercise 3.4

Q-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) Given: 9801

From the given number, the unit digit of 9801 is 1.

From table 3.4, we can say that the possible unit digits are 1 or 9

It is noted that 9801 is equal to 992, the square root of a number is an odd number.

(ii) Given: 99856

From the given number, the unit digit of 99856 is 6.

From table 3.4, we can say that the possible unit digits are 4 or 6

Since the last digit of a number is an even number (6), the number cannot have an odd number as its square root.

(iii)  Given: 998001

From the given number, the unit digit of 998001 is 1.

From table 3.4, we can say that the possible unit digits are 1 or 9

It is noted that 998001 is equal to (33 x 37)2, the square root of a number is an odd number.

(iv) Given: 657666025

From the given number, the unit digit of 657666025 is 5.

We can say that the only possible unit digit is 5

It is noted that 657666025 is equal to (5 x 23 x 223)2, the square root of a number is an odd number.

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

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

(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

Solution:

(i) Given number:441

Factorize the given numbers into prime factors, we get

441=3 x 3 x 7 x 7

1

Now, group the factors into pairs of equal factors.

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

To get the square root of 441, take one factors from each pair.

3 x 7 = 21

Therefore, the square root of 441using prime factorization method is 21.

(ii) Given number:196

Factorize the given numbers into prime factors, we get

196 = 2 x 2 x 7 x 7

 

2

Now, group the factors into pairs of equal factors.

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

To get the square root of 196, take one factor from each pair.

2 × 7 = 14

Therefore, the square root of 196 using prime factorization method is 14.

(iii) Given number: 529

Factorize the given numbers into prime factors, we get

529 = 23 x 23

 

3

Now, group the factors into pairs of equal factors.

529= (23 x 23)

To get the square root of 529, take one factor from each pair as 23.

Therefore, the square root of 529 using prime factorization method is 23.

(iv)  Given number: 1764

Factorize the given numbers into prime factors, we get

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

 

4

Now, group the factors into pairs of equal factors.

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

To get the square root of 1764, take one factor from each pair.

2 x 3 x 7 = 42

Therefore, the square root of 1764 using prime factorization method is 42.

(v) Given number: 1156

Factorize the given numbers into prime factors, we get

1156 = 2 x 2 x 17 x 17

5

Now, group the factors into pairs of equal factors.

1156 = ( 2 × 2 ) × ( 17 × 17 )

To get the square root of 1156, take one factor from each pair.

2 × 17 = 34

Therefore, the square root of 1156 using prime factorization method is 34.

(vi)Given number: 4096

Factorize the given numbers into prime factors, we get

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

6

Now, group the factors into pairs of equal 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)

To get the square root of 4096, take one factor from each pair.

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

Therefore, the square root of 4096 using prime factorization method is 64.

(vii) Given number: 7056

Factorize the given numbers into prime factors, we get

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

7

Now, group the factors into pairs of equal factors.

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

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

2 × 2 × 3 × 7 = 84

Therefore, the square root of 7056 using prime factorization method is 84.

(viii)Given number: 8281

Factorize the given numbers into prime factors, we get

8281 = 7 × 7 × 13 × 13

8

Now, group the factors into pairs of equal factors.

8281 = ( 7 × 7 ) × ( 13 × 13 )

To get the square root of 8281, take one factor from each pair.

7  × 13 = 91

Therefore, the square root of 8281 using prime factorization method is 91.

(ix)Given number:11664

Factorize the given numbers into prime factors, we get

11664 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

9

Now, group the factors into pairs of equal factors.

11664 = ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 ) × ( 3 × 3 ) × ( 3 × 3 )

To get the square root of 11664, take one factor from each pair.

2 × 2 × 3 × 3 × 3 = 108

Therefore, the square root of 11664 using prime factorization method is 108.

(x)Given number: 47089

Factorize the given numbers into prime factors, we get

47089 = 7 × 7 × 31 × 31

10

Now, group the factors into pairs of equal factors.

47089 = ( 7 × 7 ) × ( 31 × 31 )

To get the square root of 47089, take one factor from each pair.

7 × 31 = 217

Therefore, the square root of 47089 using prime factorization method is 217.

(xi) Given number: 24336

Factorize the given numbers into prime factors, we get

24336 = 2 × 2 × 2 × 2 × 3 × 3 × 13 × 13

11

Now, group the factors into pairs of equal factors.

24336 = ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 ) × ( 13 × 13 )

To get the square root of 24336, take one factor from each pair.

2 × 2 × 3 × 13 = 156

Therefore, the square root of 24336 using prime factorization method is 156

(xii) Given number: 190969

Factorize the given numbers into prime factors, we get

190969 = 19 × 19 × 23 × 23

12

Now, group the factors into pairs of equal factors.

190969 = ( 19 × 19 ) × ( 23 × 23 )

To get the square root of 190969, take one factor from each pair.

19 × 23 = 437

Therefore, the square root of 190969 using prime factorization method is 437

(xiii) Given number: 568756

Factorize the given numbers into prime factors, we get

568756 = 2 × 2 × 383 × 383

13

Now, group the factors into pairs of equal factors.

568756 = ( 2 × 2 ) × ( 383 × 383 )

To get the square root of 568756, take one factor from each pair.

2 × 383 = 766

Therefore, the square root of 568756 using prime factorization method is 766.

(xiv) Given number: 27225

Factorize the given numbers into prime factors, we get

27225 = 3 × 3 × 5 × 5 × 11 × 11

14

Now, group the factors into pairs of equal factors.

27225 = ( 3 × 3 ) × ( 5 × 5 ) × ( 11 × 11 )

To get the square root of 27225, take one factor from each pair.

3 × 5 × 11 = 165

Therefore, the square root of 27225 using prime factorization method is 165.

(xv) Given number: 3013696

Factorize the given numbers into prime factors, we get

3013696 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 31 × 31

15

Now, group the factors into pairs of equal factors.

3013696 = ( 2 × 2 ) × ( 2 × 2 ) × ( 2 × 2 ) × ( 7 × 7 ) × ( 31 × 31 )

To get the square root of 3013696, take one factor from each pair.

2 × 2 × 2 × 7 × 31 = 1736

Therefore, the square root of 3013696 using prime factorization method is 1736.

Q-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:

We know that the prime factorization of 180 is

180 = 2 x 2 x 3 x 3 x 5

Now, group the factors into pairs of equal factors.

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

Here, the factor, 5 does not have a pair.

So, when you multiply 5 by 180, the perfect square number is obtained.

This means that,

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

So, to get the square root of 900, take one factor from each pair.

2 x 3 x 5 = 30.

Hence, the square root of the perfect square number is 30.

Q-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.

Q-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:

We know that the prime factorization of 3645 is

3645=3 x3x3x3x3x3x5

Now, group the factors into pairs of equal factors.

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

Here, the factor, 5 does not have a pair.

So, when you divide 3645 by 5, the perfect square number is obtained.

This means that,

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

So, to get the square root of 729, take one factor from each pair.

3 x 3 x 3 = 27.

Hence, the square root of the perfect square number is 27.

Q-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.

Q-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 p and q.

From the given data, the first statement defines

p x q = 1296 ….(1)

Also given that, if one number is 16 times the other, then we have:

q = 16 x p …..(2)

Substitute (2) in (1), we get

p x ( 16 x p ) = 1296

Simplify both the sides of an equation

p=1296/16 = 81

Therefore, the square root of 81 = 9

To find the value of  q, substitute the value of  “a” in (2)

q= 16 x p.

q =16 x 9 = 144

Hence, the numbers are 9 and 144.

Q-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.

Q-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 N be the number of members.

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

Therefore, the total contribution

N x r = Rs 92.16 = 9216 paise …..(1)

It is given that the amount received is the same as the number of members:

r = N …..(2)

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

N x N = 9216

N2= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2  x 3 x 3

N2 = (2 x 2) x (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

N=2 x 2 x 2 x 2 x 2 x 3 = 96

To find r, we can use the relation r = N = 96

Therefore, there are 96 members and each contributed 96 paise.

Q-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.

Q-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:

Given: Area of a square field =5184 m2.

To find the area of a rectangular field.

so, first, find the perimeter of the square.

We know that,

The area of a square = a2

where “a” is the side of a square.

Therefore,

a2=5184 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)

take sqaure roots on both the sides

Therefore, the a = 2 x 2 x  2 x 3  x 3 = 72

Using “a” value, find the perimeter of the square

4 x a = 288 m

It is given that, the perimeter of a rectangular field is equal to the perimeter of the square field.

So,  let l be the length of the rectangular field.

Let w be the width of the rectangular field.

2(l + w) = 288

Moreover, it is given that length is twice the width:

l = 2 x w.

Substitute the “l” value in the above equation, we get

2 x (2 x w+ w) = 288

3 x w= 144

Therefore, w= 48m

So, to find L, we know that

l = 2 x w = 2 x 48 = 96m

Therefore, the area of the rectangular field, A = l x w = 96 x 48 = 4608 m2

Q-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.

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

Solution:

To find the square root of 121:

By using repeated subtraction method,

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

From above method, it is observed that,there are 11 numbers to subtract from 121.

Therefore, the square root of 121 is 11.

To find the square root of 169:

By using repeated subtraction method,

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

From above method, it is observed that there are 13 numbers to subtract from 169.

Therefore, the square root of 169 is 13.

Q-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.

Q-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 the number of students be S and r be the amount in rupees denoted by each student.

The total donation is given as

S × r = Rs. 2401 ….(1)

It is given that the total amount in rupees is equal to the number of students

So, r = S ….(2)

Substitute (2) in (1), we get

S × S = 2401

S2 = ( 7 × 7 ) × ( 7 × 7 )

take square roots on both the sides, we get

S = 7 × 7 = 49

Therefore, the number of students in the class is 49.

Q-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.

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