Squares and Square Roots: MCQ test, Set 1

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

$b^2$ means
A $b + 2$ B $2 \times b$ C $b \times b$ D $b \times b \times b$

Answer and solution

Answer: (C) $b \times b$

$a^2 = a \times a$

So $b^2 = b \times b$ ( $b$ multiplied by itself, not by $2$ ).
The square of $11$ is
A $121$ B $100$ C $22$ D $144$

Answer and solution

Answer: (A) $121$

$11^2 = 11 \times 11 = 121$
The next number in $49, 64, 81, \ldots$ is
A $99$ B $121$ C $98$ D $100$

Answer and solution

Answer: (D) $100$

These are $7^2, 8^2, 9^2$ .

Next is $10^2 = 100$
$80^2 =$
A $64000$ B $640$ C $160$ D $6400$

Answer and solution

Answer: (D) $6400$

$80^2 = 80 \times 80$

$8 \times 8 = 64$ , then write two zeros: $6400$
To find $75^2$ by the rule for numbers ending in 5, we first find
A $7 \times 5$ B $7 \times 8$ C $7 \times 7$ D $7 + 8$

Answer and solution

Answer: (B) $7 \times 8$

Rule: multiply the number before $5$ by the next number, then write $25$ .

$7 \times 8 = 56$ , so $75^2 = 5625$
Given $118^2 = 13924$ , the square root of $13924$ is
A $6962$ B $236$ C $\sqrt{118}$ D $118$

Answer and solution

Answer: (D) $118$

$a^2 = b \Rightarrow \sqrt{b} = a$

$118^2 = 13924 \Rightarrow \sqrt{13924} = 118$

(The square root is not half the number.)
$\sqrt{729} =$
A $26$ B $28$ C $27$ D $364.5$

Answer and solution

Answer: (C) $27$

$27^2 = 27 \times 27 = 729$

$\therefore \sqrt{729} = 27$
The number whose square root is $26$ is
A $676$ B $\sqrt{26}$ C $13$ D $52$

Answer and solution

Answer: (A) $676$

$\sqrt{b} = a \Rightarrow b = a^2$

$26^2 = 26 \times 26 = 676$
$576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$ . So $\sqrt{576} =$
A $24$ B $288$ C $6$ D $48$

Answer and solution

Answer: (A) $24$

Take one factor from each pair: $2 \times 2 \times 2 \times 3 = 24$
$\sqrt{3 \times 3 \times 7 \times 7} =$
A $7$ B $10$ C $441$ D $21$

Answer and solution

Answer: (D) $21$

$\sqrt{3 \times 3 \times 7 \times 7} = \sqrt{(3 \times 3) \times (7 \times 7)}$

$= 3 \times 7 = 21$
By the long division method, $\sqrt{6889} =$
A $83$ B $93$ C $73$ D $87$

Answer and solution

Answer: (A) $83$

Pairs: $\overline{68}\;\overline{89}$ .

$8 \times 8 = 64 \le 68$ : first digit $8$ , and $68 - 64 = 4$

Bring down $89$ : we get $489$ . New divisor: $8 \times 2 = 16$ ; $163 \times 3 = 489$ ( $164 \times 4 = 656$ is too big); $489 - 489 = 0$

Remainder $0$ . $\therefore \sqrt{6889} = 83$
In finding $\sqrt{5776}$ by the long division method, the first digit of the root is $7$ . The new divisor starts with
A $14$ B $49$ C $7$ D $8$

Answer and solution

Answer: (A) $14$

Double the root found so far: $7 \times 2 = 14$

Then $146 \times 6 = 876$ ( $147 \times 7 = 1029$ is too big), so $\sqrt{5776} = 76$
In finding $\sqrt{45369}$ by the long division method, the first digit of the root is
A $6$ B $2$ C $4$ D $3$

Answer and solution

Answer: (B) $2$

Pair from the right: $\overline{4}\;\overline{53}\;\overline{69}$ ; the first group is $4$ .

$2 \times 2 = 4 \le 4$ , so the first digit is $2$
$\sqrt{12} \times \sqrt{3} =$
A $\sqrt{15}$ B $6$ C $36$ D $18$

Answer and solution

Answer: (B) $6$

$\sqrt{12} \times \sqrt{3} = \sqrt{12 \times 3} = \sqrt{36} = 6$
$(\sqrt{29})^2 =$
A $841$ B $\sqrt{29}$ C $29$ D $58$

Answer and solution

Answer: (C) $29$

$(\sqrt{a})^2 = \sqrt{a} \times \sqrt{a} = a$

$\therefore (\sqrt{29})^2 = 29$
$\displaystyle \frac{\sqrt{80}}{\sqrt{5}} =$
A $\sqrt{75}$ B $16$ C $8$ D $4$

Answer and solution

Answer: (D) $4$

$\displaystyle \frac{\sqrt{80}}{\sqrt{5}} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4$
$\displaystyle \frac{23}{\sqrt{23}} =$
A $1$ B $23$ C $\sqrt{23}$ D $23\sqrt{23}$

Answer and solution

Answer: (C) $\sqrt{23}$

$\displaystyle \frac{23}{\sqrt{23}} = \frac{\sqrt{23} \times \sqrt{23}}{\sqrt{23}} = \sqrt{23}$
$\sqrt{36} + \sqrt{64} =$
A $10$ B $100$ C $14$ D $196$

Answer and solution

Answer: (C) $14$

$\sqrt{36} + \sqrt{64} = 6 + 8 = 14$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , so do not add first.
For $m > 0$ and $n > 0$ , which of these is true?
A $\sqrt{m^2 + n^2} = m + n$ B $\sqrt{m} \times \sqrt{n} = \sqrt{m \times n}$ C $\sqrt{m} - \sqrt{n} = \sqrt{m - n}$ D $\sqrt{m} + \sqrt{n} = \sqrt{m + n}$

Answer and solution

Answer: (B) $\sqrt{m} \times \sqrt{n} = \sqrt{m \times n}$

$\sqrt{m} \times \sqrt{n} = \sqrt{m \times n}$ is a property in the notes.

The other three are the 'not equal' rules: $\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , $\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$ , $\sqrt{a^2 + b^2} \ne a + b$ .
$\sqrt{9^2 + 12^2} =$
A $225$ B $15$ C $\sqrt{21}$ D $21$

Answer and solution

Answer: (B) $15$

$\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$

$\sqrt{a^2 + b^2} \ne a + b$
$\sqrt{(17 - 9)^2} =$
A $\sqrt{208}$ B $208$ C $8$ D $64$

Answer and solution

Answer: (C) $8$

$\sqrt{(a - b)^2} = a - b$ when $a - b > 0$

$\sqrt{(17 - 9)^2} = \sqrt{8^2} = 8$
$\sqrt{0.49} =$
A $0.7$ B $0.07$ C $0.245$ D $7$

Answer and solution

Answer: (A) $0.7$

$\displaystyle \sqrt{0.49} = \sqrt{\frac{49}{100}}$

$\displaystyle = \frac{\sqrt{49}}{\sqrt{100}}$

$\displaystyle = \frac{7}{10}$

$= 0.7$
$\sqrt{0.000009} =$
A $0.0000045$ B $0.003$ C $0.03$ D $0.0003$

Answer and solution

Answer: (B) $0.003$

$\displaystyle \sqrt{0.000009} = \sqrt{\frac{9}{1000000}}$

$\displaystyle = \frac{\sqrt{9}}{\sqrt{1000000}}$

$\displaystyle = \frac{3}{1000}$

$= 0.003$
$\sqrt{5.29} =$
A $0.23$ B $2.645$ C $23$ D $2.3$

Answer and solution

Answer: (D) $2.3$

$\displaystyle \sqrt{5.29} = \sqrt{\frac{529}{100}} = \frac{\sqrt{529}}{\sqrt{100}} = \frac{23}{10} = 2.3$
$\sqrt{\square} = 5.5$ . The number in the box is
A $302.5$ B $11$ C $30.25$ D $3.025$

Answer and solution

Answer: (C) $30.25$

$\displaystyle 5.5 = \frac{55}{10}$

$\displaystyle \left(\frac{55}{10}\right)^2 = \frac{3025}{100} = 30.25$

Check: $\displaystyle \sqrt{30.25} = \frac{\sqrt{3025}}{\sqrt{100}} = 5.5$

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MCQ test, Set 1