Squares and Square Roots: MCQ test, Set 7

$a^2$ is read as
A '2 times $a$ ' B ' $a$ plus 2' C ' $a$ cubed' D ' $a$ squared'

Answer and solution

Answer: (D) ' $a$ squared'

$a^2 = a \times a$ is read as ' $a$ squared' (or) ' $a$ power 2'.
The square of $27$ is
A $676$ B $784$ C $54$ D $729$

Answer and solution

Answer: (D) $729$

$27^2 = 27 \times 27 = 729$
The next number in $361, 400, 441, \ldots$ is
A $484$ B $483$ C $529$ D $482$

Answer and solution

Answer: (A) $484$

These are $19^2, 20^2, 21^2$ .

Next is $22^2 = 484$
$70^2 =$
A $140$ B $4900$ C $490$ D $49000$

Answer and solution

Answer: (B) $4900$

$70^2 = 70 \times 70$

$7 \times 7 = 49$ , then write two zeros: $4900$
To find $45^2$ by the rule for numbers ending in 5, we first find
A $4 \times 6$ B $4 \times 5$ C $4 \times 4$ D $4 + 5$

Answer and solution

Answer: (B) $4 \times 5$

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

$4 \times 5 = 20$ , so $45^2 = 2025$
If $m^2 = n$ , then
A $\sqrt{n} = 2 \times m$ B $n^2 = m$ C $\sqrt{m} = n$ D $\sqrt{n} = m$

Answer and solution

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

By the definition, $a^2 = b \Rightarrow \sqrt{b} = a$ .

So $m^2 = n \Rightarrow \sqrt{n} = m$
$\sqrt{361} =$
A $20$ B $18$ C $180.5$ D $19$

Answer and solution

Answer: (D) $19$

$19^2 = 19 \times 19 = 361$

$\therefore \sqrt{361} = 19$
The number whose square root is $24$ is
A $\sqrt{24}$ B $48$ C $576$ D $12$

Answer and solution

Answer: (C) $576$

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

$24^2 = 24 \times 24 = 576$
$5776 = 2 \times 2 \times 2 \times 2 \times 19 \times 19$ . So $\sqrt{5776} =$
A $2888$ B $38$ C $76$ D $152$

Answer and solution

Answer: (C) $76$

Take one factor from each pair: $2 \times 2 \times 19 = 76$
$\sqrt{2 \times 2 \times 5 \times 5 \times 11 \times 11} =$
A $55$ B $110$ C $18$ D $12100$

Answer and solution

Answer: (B) $110$

$\sqrt{2 \times 2 \times 5 \times 5 \times 11 \times 11} = \sqrt{(2 \times 2) \times (5 \times 5) \times (11 \times 11)}$

$= 2 \times 5 \times 11 = 110$
By the long division method, $\sqrt{3481} =$
A $49$ B $59$ C $51$ D $69$

Answer and solution

Answer: (B) $59$

Pairs: $\overline{34}\;\overline{81}$ .

$5 \times 5 = 25 \le 34$ : first digit $5$ , and $34 - 25 = 9$

Bring down $81$ : we get $981$ . New divisor: $5 \times 2 = 10$ ; $109 \times 9 = 981$ ; $981 - 981 = 0$

Remainder $0$ . $\therefore \sqrt{3481} = 59$
In finding $\sqrt{1225}$ by the long division method, the first digit of the root is $3$ . The new divisor starts with
A $3$ B $4$ C $9$ D $6$

Answer and solution

Answer: (D) $6$

Double the root found so far: $3 \times 2 = 6$

Then $65 \times 5 = 325$ ( $66 \times 6 = 396$ is too big), so $\sqrt{1225} = 35$
In finding $\sqrt{39601}$ by the long division method, the first digit of the root is
A $1$ B $6$ C $3$ D $2$

Answer and solution

Answer: (A) $1$

Pair from the right: $\overline{3}\;\overline{96}\;\overline{01}$ ; the first group is $3$ .

$1 \times 1 = 1 \le 3$ , so the first digit is $1$
$\sqrt{2} \times \sqrt{72} =$
A $12$ B $\sqrt{74}$ C $72$ D $144$

Answer and solution

Answer: (A) $12$

$\sqrt{2} \times \sqrt{72} = \sqrt{2 \times 72} = \sqrt{144} = 12$
$(\sqrt{41})^2 =$
A $\sqrt{41}$ B $82$ C $41$ D $1681$

Answer and solution

Answer: (C) $41$

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

$\therefore (\sqrt{41})^2 = 41$
$\displaystyle \frac{\sqrt{128}}{\sqrt{2}} =$
A $32$ B $64$ C $\sqrt{126}$ D $8$

Answer and solution

Answer: (D) $8$

$\displaystyle \frac{\sqrt{128}}{\sqrt{2}} = \sqrt{\frac{128}{2}} = \sqrt{64} = 8$
$\displaystyle \frac{35}{\sqrt{35}} =$
A $35\sqrt{35}$ B $\sqrt{35}$ C $1$ D $35$

Answer and solution

Answer: (B) $\sqrt{35}$

$\displaystyle \frac{35}{\sqrt{35}} = \frac{\sqrt{35} \times \sqrt{35}}{\sqrt{35}} = \sqrt{35}$
$\sqrt{49} + \sqrt{36} =$
A $\sqrt{85}$ B $85$ C $13$ D $169$

Answer and solution

Answer: (C) $13$

$\sqrt{49} + \sqrt{36} = 7 + 6 = 13$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , so do not add first.
$\sqrt{676} - \sqrt{100} =$
A $576$ B $36$ C $24$ D $16$

Answer and solution

Answer: (D) $16$

$\sqrt{676} - \sqrt{100} = 26 - 10 = 16$

$\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$ , so do not subtract first.
$\sqrt{14^2 + 48^2} =$
A $50$ B $62$ C $\sqrt{62}$ D $2500$

Answer and solution

Answer: (A) $50$

$\sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50$

$\sqrt{a^2 + b^2} \ne a + b$
$\sqrt{(15 - 4)^2} =$
A $11$ B $\sqrt{209}$ C $121$ D $209$

Answer and solution

Answer: (A) $11$

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

$\sqrt{(15 - 4)^2} = \sqrt{11^2} = 11$
$\sqrt{0.25} =$
A $0.05$ B $0.125$ C $0.5$ D $5$

Answer and solution

Answer: (C) $0.5$

$\displaystyle \sqrt{0.25} = \sqrt{\frac{25}{100}}$

$\displaystyle = \frac{\sqrt{25}}{\sqrt{100}}$

$\displaystyle = \frac{5}{10}$

$= 0.5$
$\sqrt{0.0625}$ written as a fraction is
A $\displaystyle \frac{25}{1000}$ B $\displaystyle \frac{25}{100}$ C $\displaystyle \frac{25}{10}$ D $\displaystyle \frac{625}{100}$

Answer and solution

Answer: (B) $\displaystyle \frac{25}{100}$

$\displaystyle \sqrt{0.0625} = \sqrt{\frac{625}{10000}} = \frac{\sqrt{625}}{\sqrt{10000}} = \frac{25}{100}$
$\sqrt{4.41} =$
A $21$ B $0.21$ C $2.1$ D $2.205$

Answer and solution

Answer: (C) $2.1$

$\displaystyle \sqrt{4.41} = \sqrt{\frac{441}{100}} = \frac{\sqrt{441}}{\sqrt{100}} = \frac{21}{10} = 2.1$
$\sqrt{\square} = 8.5$ . The number in the box is
A $72.25$ B $17$ C $722.5$ D $7.225$

Answer and solution

Answer: (A) $72.25$

$\displaystyle 8.5 = \frac{85}{10}$

$\displaystyle \left(\frac{85}{10}\right)^2 = \frac{7225}{100} = 72.25$

Check: $\displaystyle \sqrt{72.25} = \frac{\sqrt{7225}}{\sqrt{100}} = 8.5$

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MCQ test, Set 7