Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Squares and Square Roots

Short Answer Type

41.

(–1) = 1. Is –1, a square root of 1?

(–2)² = 4. Is –2, a square root of 4?

(–9)² = 81. Is –9, a square root of 81?

133 Views

Long Answer Type

42.
By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root.
(i) 121 (ii) 55 (iii) 36

(iv) 49 (v) 90

(i) Subtracting the successive odd numbers from 121, we have

                  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

∴ $space space space square root of 121$ = 11 ( ∵ We had subtract the first 11 odd numbers)

(ii)   ∵                          55 - 1 = 54                       54 - 3 = 51
                51 - 5 = 46                        46 - 7 = 39
                            39 - 9 = 30                      30 - 11 = 19
                     19 - 13 = 6 6 - 15 = -9

and we do not reach to o

∴ 55 is not a perfect square.

(iii) ∵     36 - 1 = 35    35 - 3 = 32
                  32 - 5 = 27                      27 - 7 = 20
20 - 9 = 11    11 - 11 = 0

and we have obtained 0 after subtracting 6 successive odd numbers.

∴ 36 is a perfect square.

Thus, $space space space square root of 36 space end root$ = 6

(iv) We have     49 - 1 = 48    48 - 3 = 45
                 45 - 5 = 40                      40 - 7 = 33
           33 - 9 = 24    24 - 11 = 13
                            13 - 13 = 0

∵ We have obtained o after successive subtraction of 7 odd numbers

∴ 49 is a perfect square.

Thus, $space space space space space space space square root of 49 space equals space 7$

(v) We have     90 - 1 = 89    89 - 3 = 86
                 86 - 5 = 81                      81 - 7 = 74
           74 - 9 = 65    65 - 11 = 54
                           54 - 13 = 41                     41 - 15 = 26
26 - 17 = 9    9 -19 = -10
Since, we can not reach to 0 after subtracting successive odd numbers

∴ 90 is not a perfect square.

695 Views

Short Answer Type

43.

What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

1005 Views

44.

Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 152 (ii) 257 (iii) 408 (iv) 441

356 Views

45.

Find the square roots of 100 and 169 by the method of repeated subtraction.

946 Views

46.

Can we say that if a perfect square is of n-digits, then its square root will have $space space space space space n over 2$ digits if n is even of $space space space space space fraction numerator left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction$ if n is odd?

363 Views

47.

Without calculating square roots, find the number of digits in the square root of the following numbers.

(i) 25600 (ii) 100000000 (iii) 36864

750 Views

48.

Estimate the value of the following to the nearest whole number.

(i) $space space space space space space space square root of 80$ (ii) $space space space space space space square root of 1000$ (iii) $space space space space square root of 350$ (iv) $space space space space space square root of 500$