[1 Mark Question Answer]

MCQ 11 Mark

Give reason to show that none of the numbers, given below, is a perfect square.

✓
$2162$
B
$6843$
C
$9637$
D
$6598$

Answer

Correct option: A.

$2162$

A number having $2,3,7$ or $8$ at the unit place is never a perfect square.

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Question 21 Mark

Is it possible for the square of a number to end with $5$ zeroes $?$ Give reason.

Answer

No, it is not possible for the square of a number, to have $5$ zeroes which is odd because the number of zeros of the square must be $2n$ zeroes i.e., even number of zeroes

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Question 31 Mark

If a number ends with $3$ zeroes, how many zeroes will its square have$?$

Answer

We know that if a number ends with n zeros, then its square will have $2n$ zeroes at their ends
A number ends with $3$ zeroes, then its square will have $3 \times 2 = 6 $ zeroes

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Question 41 Mark

Which of the following numbers will have $6$ at their unit’s place : $(i) \ 262$ $(ii) \ 492$ $(iii) \ 342$ $(iv) \ 432$ $(v) \ 2442$

Answer

The following numbers have $6$ at their units place as $(4)^2=16,(6)^2=36$ has $6$ at their units place $262,342,2442$ i.e.,$ (i), (iii)$ and $(v)$

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Question 51 Mark

Square of which of the following numbers will not have $6$ at their unit’s place :$(i) \ 35\ (ii) \ 23\ (iii) \ 64\ (iv) \ 76\ (v) \ 98$

Answer

The squares of the following numbers, Will not have $6$ at their units place as only $(4)^2=16,(6)^2=36$ has but its units place $35,23$ and $98$ i.e., $(i), (ii),$ and $(v)$

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Question 61 Mark

Squares of which of the following numbers will have $1 ($one$)$ at their unit’s place :$(i) \ 57\
(ii) \ 81\ (iii)\ 139\ (iv)\ 73\ (v) \ 64$

Answer

The square of the following numbers will have $1$ at their units place as $(1)^2=1,(9)^2=81 \ 81$ and $139$ i.e., $(i)$ and $(iii)$

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Question 71 Mark

Without doing the actual addition, find the sum of:$1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53$

Answer

$1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53$
$=$ Sum of first $27$ odd natural number $= 272 = 729$

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Question 81 Mark