Quantitative Aptitude · Mathematics

Squares and Square Roots

196 Questions

Squares and square roots questions assess numerical ability through perfect squares, digit properties, and fast calculation techniques. Many competitive exams feature Vedic mathematics shortcuts to solve these quickly. Mastering these rules improves overall calculation speed.

Perfect square propertiesVedic math squaresFinding square rootsDigit replacement puzzlesLeast perfect squares

Squares and Square Roots Questions

Multiple choice

Square of 9x – 7xy is

  1. 81x2 + 49x2y2

  2. 81x2 – 49x2y2

  3. 81x2 + 49x y2 – 126x2y

  4. 81x 2 + 49x2y2 – 63x2y
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Use identity

Multiple choice

Which of the following is a perfect square? Hint: You can identify the perfect square using the ones' digit of the given number.

  1. 1683

  2. 1087

  3. 7862

  4. 2025

  5. 4298
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

There are couple of clues as follows: 1) As the other options are not perfect squares, this has higher chances to be the perfect square. 2) As the one's digit is 5, there is a chance for it to be a perfect square. 3) Rewriting 2025 as 5 * 405 = 5 * 5 * 81 = 52 * 92 Therefore, this option is correct.

Multiple choice

If a perfect square has 'n' digits in it, then the square root of that number will have how many digits in it?

  1. 2n digits

  2. 2n - 1 digits

  3. n/2 digits

  4. (n + 1)/2 digits

  5. n/2 digits if 'n' is even or (n + 1)/2 digits if 'n' is odd
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Square root(2809) = 53 2809 has 4 digits in it and its square root 53 has only 2 digits in it.  As per this option, number of digits should be 4/2 = 2, which is correct.

Square root(27225) = 165 27225 has 5 digits in it and its square root 165 has only 3 digits in it. As per this option, number of digits should be (5+1)/2 = 3, which is correct. This option is correct.

Multiple choice

Which of the squares (1232, 572, 922, 612, 2092) will end with the digit 1?

  1. Only 61

  2. 61 and 209

  3. 57 and 61

  4. Only 209

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Yes, it is the correct option. For this problem, you don't need to calculate the full value of the square of the given number. Just pick the digit in units place in each number and multiply it with itself. Lets do it for all the given numbers: 1232: 3 x 3 = 9          So, it will not end with 1 572 : 7 x 7 = 49        So, it will not end with 1 922 : 2 x 2 = 4           So, it will not end with 1 612   :  1 x 1 = 1        Yes, it will end with 1 2092 : 9 x 9 = 81       Yes, it will end with 1

Multiple choice statistics moving averages moving average and variation simple moving average uses of average in day-to-day life introduction to time series introduction to time series and forecasting

The total numbers of squares on a chessboard is

  1. $206$
  2. $205$
  3. $204$
  4. $202$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From Permutation and Combination,


no. of squares of 1x1 $=1^2$
no. of squares of 2x2 $=2^2$
no. of squares of 3x3$=3^2$
.
.
.
no. of squares of 8x8 $=8^2$

Hence Total number of squares $=1^2+2^2+......+8^2$
 $\implies 204$

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

 The square of any positive odd integer for some integer $ m$ is of the form 

  1. 7m+1

  2. 8m+1

  3. 8m+3

  4. 7m+2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that any positive odd integer a is of the form 4q + 1 or 4q + 3 where q is some integer.
Case-1: $a=4q+1$
$\Rightarrow a^2=16q^2+8q+1=8(2q^2+q)+1$
$=8m+1$,
where $m=2q^2+q=integer$.
Case-2: $a=4q+3$
$\Rightarrow a^2=16q^2+24q+9$
$=8(2q^2+3q+1)+1=8m+1$,
where $m=2q^2+3q+1=integer$.
Hence square of any positive odd integer is of the form $8m+1$ for some integer m.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

Sum of digits of the smallest number by which $1440$ should be multiplied so that it becomes a perfect cube is

  1. $4$
  2. $6$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\because  1440 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$
The pairs of $2$ and $3$ and $5$ are incomplete to make it perfect cube.
$\therefore$ Smallest  number  to  be  multiplied  $=  2 \times 3 \times 5 \times 5 = 150$

$\therefore$  The  sum  of  its  digits  $= 1+5+0 = 6.$