NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1
NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1 are the part of NCERT Solutions for Class 8 Maths (Rationalised Contents). Here you can find the
NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.1.
Ex 5.1 Class 8 Maths Question 1.
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution:
The unit digits of the squares of the given numbers are shown against the numbers in the table given below:
Ex 5.1 Class 8 Maths Question 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:
A number which ends with 2, 3, 7 or 8 cannot be a perfect square. Again, a number which ends with odd number of zero(s) cannot be a perfect square.
(i) Since the number 1057 ends with 7, so it cannot be a perfect square.
(ii) Since the number 23453 ends with 3, so it cannot be a perfect square.
(iii) Since the number 7928 ends with 8, so it cannot be a perfect square.
(iv) Since the number 222222 ends with 2, so it cannot be a perfect square.
(v) Since the number 64000 ends with an odd number of zeros, so it cannot be a perfect square.
(vi) Since the number 89722 ends with 2, so it cannot be a perfect square.
(vii) Since the number 222000 ends with an odd number of zeros, so it cannot be a perfect square.
(viii) Since the number 505050 ends with an odd number of zeros, so it cannot be a perfect square.
Ex 5.1 Class 8 Maths Question 3.
The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
(i) The number 431 is an odd number, so its square would be odd.
(ii) The number 2826 is an even number, so its square would be even.
(iii) The number 7779 is an odd number, so its square would be odd.
(iv) The number 82004 is an even number, so its square would be even.
Hence, the squares of the numbers 431 and 7779 would be odd numbers.
Ex 5.1 Class 8 Maths Question 4.
Observe the following pattern and find the missing digits:
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 ………….. 2 ……….. 1
100000012 = ………………..
Solution:
The missing digits are as follows:
1000012 = 10000200001
100000012 = 100000020000001
Ex 5.1 Class 8 Maths Question 5.
Observe the following pattern and supply the missing numbers:
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………..
……………. 2 = 10203040504030201
Solution:
The missing numbers are as follows:
10101012 = 1020304030201
1010101012 = 10203040504030201
Ex 5.1 Class 8 Maths Question 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:
The missing numbers are as follows:
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
Ex 5.1 Class 8 Maths Question 7.
Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
(i) 1 + 3 + 5 + 7 + 9 = Sum of first 5 odd numbers
= 52 = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
= Sum of first 10 odd numbers
= 102 = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
= Sum of first 12 odd numbers
= 122 = 144
Ex 5.1 Class 8 Maths Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
(i) 49 = 72 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = 112 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Ex 5.1 Class 8 Maths Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Solution:
We know that there are 2n numbers lie between n2 and (n + 1)2.
(i) Therefore, between 122 and 132, there are 24 (i.e., 2 × 12) numbers.
(ii) Therefore, between 252 and 262, there are 50 (i.e., 2 × 25) numbers.
(iii) Therefore, between 992 and 1002, there are 198 (i.e., 99 × 2) numbers.
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