Class 8 NCERT Solutions – Chapter 6 Squares and Square Roots – Exercise 6.1

Problem 1: What will be the unit digit of the squares of the following numbers?

Solution:

To find the unit digit of the square of any number, find the square of the unit digit of that number and 
its unit digit will be the unit digit of the square of the actual number. 
eg:- if a number is 2007, the square of the unit digit of 2007 is 49 (7 square). and the unit digit of 49 is 9. Therefore, the unit digit of the square of 2007 is 9. 

 (i) 81

Ans: 1

Reason: If a number has 1 or 9 in the units place, then its square ends in 1. 
(or) Unit digit of 81 is 1. and 1² = 1, therefore unit digit of square of 81 is 1.  

(ii) 272

Ans: 4 

Reason: Unit digit of 272 is 2 and 2² = 4. Therefore unit digit of the square of 272 is 4. 

(iii) 799

Ans: 1

Reason: If a number has 1 or 9 in the units place, then its square ends in 1. (or) Unit digit of 799 is 9 and 9² = 81, and unit digit of 81 is 1. Therefore unit digit of square of 799 is 1. 

(iv) 3853

Ans: 9

Reason: Unit digit of 3853 is 3 and 3² = 9, therefore unit digit of square of 3853 is 9. 

(v) 1234

Ans: 6

Reason: Unit digit of 1234 is 4 and 4² = 16, and the unit digit of 16 is 6. Therefore the unit digit of the square of 1234 is 6. 

(vi) 26387

Ans: 9

Reason: Unit digit of 26387 is 7 and 7² = 49, and the unit digit of 49 is 9. Therefore the unit digit of the square of 26387 is 9. 

(vii) 52698

Ans: 4

Reason: Unit digit of 52698 is 8 and 8² = 64, and the unit digit of 64 is 4. Therefore the unit digit of the square of 52698 is 4. 

(viii) 99880

Ans: 0

Reason: If a number has 0 in its unit place then square of its number will also have 0 in its units place (since 0² = 0). 

(ix) 12796

Ans: 6

Reason: Unit digit of 12796 is 6 and 6² = 36, and the unit digit of 36 is 6. Therefore the unit digit of the square of 12796 is 6. 

(x) 55555

Ans: 5

Reason: The unit digit of 55555 is 5 and 5² = 25, and the unit digit of 25 is 5. Therefore the unit digit of the square of 55555 is 5. 

Problem 2: The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222

(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Solution:

• If a number ends with 2 (or) 3 (or) 7 (or) 8 at units place then we can tell it is not a perfect square. From the above rule, we can tell 1057, 23453, 7928, 222222, 89722 are not perfect squares.
• For a number to be a perfect square it should have an even number of zeros at the end. From rule 2 we can tell 64000, 222000, 505050 are not perfect squares as they are having an odd number of zeros at the end.

Problem 3: The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004 

Ans: 431 and 7779

Reason: The square of an odd number will be an odd number, and the square of an even number will be an even number. 

Problem 4: Observe the following pattern and find the missing digits.

11² = 121

101² = 10201

1001² = 1002001

100001² = 1 ……… 2 ……… 1

10000001² = ………………………

Ans:

100001² = 10000200001

10000001² = 100000020000001  

Problem 5: Observe the following pattern and supply the missing numbers.

11² = 121

101² = 10201

10101² = 102030201

1010101² = ………………………

…………² = 10203040504030201

Ans:

1010101² = 1020304030201 
101010101² = 10203040504030201  

Problem 6: Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² +  4² + 12² = 13²

4² + 5² + __²  = 21²

5² + __² + 30² = 31²

6² + 7² +  __²  =  __²

Solution:

To find pattern third number is related to first and second number. How?

If we multiply the first and second numbers we will get the third number.  

The fourth number is related to the third number. How?

Forth number = third number + 1

4² + 5² + __² = 21²

Ans: 4 * 5 = 20

5² + __² + 30² = 31²

Ans: 5 * x = 30

       => x = (30 / 5) = 6

6² + 7² + x² = y²

Ans: x = 6 * 7 = 42

        y = x + 1 = 43

Problem 7: Without adding, find the sum.

Solution:

The Sum of first n odd natural numbers is n²

(i) 1 + 3 + 5 + 7 + 9

Ans: 25 
Sum of first 5 odd natural numbers is 5², and 5² = 25. 

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

Ans: 100

Sum of first 10 odd natural numbers is 10², and 10² = 100.  

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Ans: 144

Sum of first 12 odd natural numbers is 12², and 12² = 144. 

Problem 8

(i) Express 49 as the sum of 7 odd numbers.

Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13 
Explanation: 49 = 7² therefore the sum of the first 7 odd natural numbers is 49.  

(ii) Express 121 as the sum of 11 odd numbers

Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 
Explanation: 121 = 11² therefore sum of first 11 odd natural numbers is 121. 

Problem 9: How many numbers lie between squares of the following numbers?

In between n² and (n + 1)², there will be 2n natural numbers 

(i) 12 and 13

Ans: 24 (as n is 12, 2n will be 24) 

(ii) 25 and 26

Ans: 50 (as n is 25, 2n will be 50) 

(iii) 99 and 100

Ans: 198 (as n is 99, 2n will be 198) 

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