RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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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
(ii) 99856
(iii) 998001
(iv) 657666025

Answer 1: (i) The unit digit of the number 9801 is 1. So, the possible unit digits are 1 or 9 (Table 3.4). 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 x 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 x 23 x 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.

Question 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

Answer 2: (i) Resolving 441 into prime factors:
441 = 3 x 3 x 7 x 7
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping 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)
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

Grouping the factors into pairs of equal factors:
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 

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.

Answer 3: The prime factorisation 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. 

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.

Answer 4: The prime factorisation of 147:
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. 

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.

Answer 5: The prime factorisation 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. 

Question 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.

Answer 6: The prime factorisation of 1152:
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. 

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

Answer 7: Let the two numbers be a and b.
From the first statement, we have:
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 = 16 x a.
Since = 9:
= 16 x 9 = 144
So, the two numbers satisfying the question are 9 and 144. 

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.

Answer 8: Let R be the number of residents.
Let r be the money in rupees donated by each resident.
Total donation = R x r = 202500
Since the money received as donation is the same as the number of residents:
 r = R.
Substituting this in the first equation, we get:
R x R = 202500
R2 = 202500
R2 = (2 x 2) x (5 x 5) x (5 x 5) x (3 x 3)2
R = 2 x 5 x 5 x 3 x 3 = 450
So, the number of residents is 450. 

Question 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?

Answer 9: Let be the number of members.
Let 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 be the number of members.
Let 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. 

Question 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?

Answer 10: Let S be the number of students.
Let be the money donated by each student.
The total contribution can be expressed by (S)(r) = Rs 2304
Since each student paid as many paise as the number of students, then r = S. Substituting this in the first equation, 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)
S = 2 x 2 x 2 x 2 x 3 = 48
So, there are 48 students in total in the school. 

Question 11: The area of a square field is 5184 cm2. 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.

Answer 11: 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 = 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 

Question 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

Answer 12: (i) The smallest number divisible by 6, 9, 15 and 20 is their L.C.M., which is equal to 60.
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. 

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

Answer 13: 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. 

Question 14 Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056

Answer 14: (i) The prime factorisation of 7744:
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 :
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

(ii) The prime factorisation of 9604:
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:
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

(iii) The prime factorisation of 5929:
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:
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

(iv) The prime factorisation of 7056:
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:
RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-3) Class 8 Notes | EduRev

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

Answer 15: Let S be the number of students.
Let r be the amount in rupees donated 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, r is equal to S.
Substituting this in the first equation:
S ××