RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

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Class 8 : RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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QUESTION 7: BY WHAT NUMBER SHOULD EACH OF THE FOLLOWING NUMBERS BE 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

ANSWER 7: Factorising each number. 
(i) 8820 = 2 x 2 x 3 x 3 x 5 x 7 x 7
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
8820 = (2 x 2) x (3 x 3) x (7 x 7) x 5
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 8820 must be multiplied by 5 for it to be a perfect square.
The new number would be (2 x 2) x (3 x 3) x (7 x 7) x (5 x 5).
Furthermore, we have:
(2 x 2) x (3 x 3) x (7 x 7) x (5 x 5) = (2 x 3 x 5 x 7) x (2 x 3 x 5 x 7)
Hence, the number whose square is the new number is:
2 x 3 x 5 x 7 = 210
(ii) 3675 = 3 x 5 x 5 x 7 x 7
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
3675 = (5 x 5) x (7 x 7) x 3
The factor, 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3675 must be multiplied by 3 for it to be a perfect square.
The new number would be (5 x 5) x (7 x 7) x (3 x 3).
Furthermore, we have:
(5 x 5) x (7 x 7) x (3 x 3) = (3 x 5 x 7) x (3 x 5 x 7)
Hence, the number whose square is the new number is:
3 x 5 x 7 = 105
(iii) 605 = 5 x 11 x 11
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
605 = 5 x (11 x 11)
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 605 must be multiplied by 5 for it to be a perfect square.
The new number would be (5 x 5) x (11 x 11).
Furthermore, we have:
(5 x 5) x (11 x 11) = (5 x 11) x (5 x 11)
Hence, the number whose square is the new number is:
5 x 11 = 55
(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
2880 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 5
There is a 5 as the leftover. For a number to be a perfect square, each prime factor has to be paired. Hence, 2880 must be multiplied by 5 to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5) = (2 x 2 x 2 x 3 x 5) x (2 x 2 x 2 x 3 x 5)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 5 = 120
(v) 4056 = 2 x 2 x 2 x 3 x 13 x 13
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
4056 = (2 x 2) x (13 x 13) x 2 x 3
The factors at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 4056 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (3 x 3) x (13 x 13).
Furthermore, we have:
(2 x 2) x (2 x 2) x (3 x 3) x (13 x 13) = (2 x 2 x 3 x 13) x (2 x 2 x 3 x 13)
Hence, the number whose square is the new number is:
2 x 2 x 3 x 13 = 156
(vi) 3468 = 2 x 2 x 3 x 17 x 17
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
3468 = (2 x 2) x (17 x 17) x 3
The factor 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3468 must be multiplied by 3 for it to be a perfect square.
The new number would be (2 x 2) x (17 x 17) x (3 x 3).
Furthermore, we have:
(2 x 2) x (17 x 17) x (3 x 3) = (2 x 3 x 17) x (2 x 3 x 17)
Hence, the number whose square is the new number is:
2 x 3 x 17 = 102
(vii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
7776 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x 2 x 3
The factors, 2 and 3 at the end are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 7776 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3) = (2 x 2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 3 x 3 x 3)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 3 x 3 = 216

Question 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

Answer 8: Factorising each number.
(i) 16562 = 2 x 7 x 7 x 13 x 13
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
16562 = 2 x (7 x 7) x (13 x 13)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 16562 must be divided by 2 for it to be a perfect square.
The new number would be (7 x 7) x (13 x 13).
Furthermore, we have:
(7 x 7) x (13 x 13) = (7 x 13) x (7 x 13)
Hence, the number whose square is the new number is:
7 x 13 = 91
(ii) 3698 = 2 x 43 x 43
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
3698 = 2 x (43 x 43)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3698 must be divided by 2 for it to be a perfect square.
The new number would be (43 x 43).
Hence, the number whose square is the new number is 43.
(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
5103 = (3 x 3) x (3 x 3) x (3 x 3) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 5103 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(3 x 3) x (3 x 3) x (3 x 3) = (3 x 3 x 3) x (3 x 3 x 3)
Hence, the number whose square is the new number is:
3 x 3 x 3 = 27
(iv) 3174 = 2 x 3 x 23 x 23
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
3174 = 2 x 3 x (23 x 23)
The factors, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3174 must be divided by 6 (2 x 3) for it to be a perfect square.
The new number would be (23 x 23). 

Hence, the number whose square is the new number is 23.    
(v) 1575 = 3 x 3 x 5 x 5 x 7
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
1575 = (3 x 3) x (5 x 5) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1575 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (5 x 5).
Furthermore, we have:
(3 x 3) x (5 x 5) = (3 x 5) x (3 x 5)
Hence, the number whose square is the new number is:
3 x 5 = 15
 

QUESTION 9: FIND THE GREATEST NUMBER OF TWO DIGITS WHICH IS A PERFECT SQUARE.
ANSWER 9: WE KNOW THAT 102 IS EQUAL TO 100 AND 92 IS EQUAL TO 81.
SINCE 10 AND 9 ARE CONSECUTIVE NUMBERS, THERE IS NO PERFECT SQUARE BETWEEN 100 AND 81. SINCE 100 IS THE FIRST PERFECT SQUARE THAT HAS MORE THAN TWO DIGITS, 81 IS THE GREATEST TWO-DIGIT PERFECT SQUARE.

Question 10: Find the least number of three digits which is perfect square.
Answer 10:
 Let us make a list of the squares starting from 1.
12 = 1
22 = 4
32 = 9
42 = 16
52 = 25
62 = 36
72 = 49
82 = 64
92 = 81
102 = 100
The square of 10 has three digits. Hence, the least three-digit perfect square is 100.

QUESTION 11: FIND THE SMALLEST NUMBER BY WHICH 4851 MUST BE MULTIPLIED SO THAT THE PRODUCT BECOMES A PERFECT SUQARE.
ANSWER 11: Prime factorisation of 4851:
4851 = 3 x 3 x 7 x 7 x 11
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRevGrouping them into pairs of equal factors:
4851 = (3 x 3) x (7 x 7) x 11
The factor, 11 is not paired. The smallest number by which 4851 must be multiplied such that the resulting number is a perfect square is 11. 

QUESTION 12: FIND THE SMALLEST NUMBER BY WHICH 28812 MUST BE DIVIDED SO THAT THE QUOTIENT BECOMES A PERFECT SQUARE.
ANSWER 12: Prime factorisation of 28812:
28812 = 2 x 2 x 3 x 7 x 7 x 7 x 7
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRevGrouping them into pairs of equal factors:
28812 = (2 x 2) x (7 x 7) x (7 x 7) x 3
The factor, 3 is not paired. Hence, the smallest number by which 28812 must be divided such that the resulting number is a perfect square is 3. 

QUESTION 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.
ANSWER 13: 
Prime factorisation of 1152:
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math Class 8 Notes | EduRev

Grouping them into pairs of equal factors:
1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2
The factor, 2 at the end is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1152 must be divided by 2 for it to be a perfect square.

The resulting number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3).
Furthermore, we have:

(2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = (2 x 2 x 2 x 3) x (2 x 2 x 2 x 3)
Hence, the number whose square is the resulting number is:
2 x 2 x 2 x 3 = 24.


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