Squares and Square Roots: MCQ test, Set 3

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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.

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

Answer and solution

Answer: (B) $n \times n$

$a^2 = a \times a$

So $n^2 = n \times n$ ( $n$ multiplied by itself, not by $2$ ).
The square of $18$ is
A $361$ B $36$ C $289$ D $324$

Answer and solution

Answer: (D) $324$

$18^2 = 18 \times 18 = 324$
The next number in $121, 144, 169, \ldots$ is
A $195$ B $196$ C $194$ D $225$

Answer and solution

Answer: (B) $196$

These are $11^2, 12^2, 13^2$ .

Next is $14^2 = 196$
$60^2 =$
A $3600$ B $120$ C $36000$ D $360$

Answer and solution

Answer: (A) $3600$

$60^2 = 60 \times 60$

$6 \times 6 = 36$ , then write two zeros: $3600$
To find $95^2$ by the rule for numbers ending in 5, we first find
A $9 \times 5$ B $9 \times 9$ C $9 + 10$ D $9 \times 10$

Answer and solution

Answer: (D) $9 \times 10$

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

$9 \times 10 = 90$ , so $95^2 = 9025$
If $p^2 = q$ , then
A $\sqrt{p} = q$ B $q^2 = p$ C $\sqrt{q} = 2 \times p$ D $\sqrt{q} = p$

Answer and solution

Answer: (D) $\sqrt{q} = p$

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

So $p^2 = q \Rightarrow \sqrt{q} = p$
$\sqrt{529} =$
A $22$ B $264.5$ C $23$ D $24$

Answer and solution

Answer: (C) $23$

$23^2 = 23 \times 23 = 529$

$\therefore \sqrt{529} = 23$
The number whose square root is $22$ is
A $\sqrt{22}$ B $44$ C $484$ D $11$

Answer and solution

Answer: (C) $484$

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

$22^2 = 22 \times 22 = 484$
$784 = 2 \times 2 \times 2 \times 2 \times 7 \times 7$ . So $\sqrt{784} =$
A $14$ B $28$ C $56$ D $392$

Answer and solution

Answer: (B) $28$

Take one factor from each pair: $2 \times 2 \times 7 = 28$
$\sqrt{5 \times 5 \times 7 \times 7} =$
A $1225$ B $7$ C $12$ D $35$

Answer and solution

Answer: (D) $35$

$\sqrt{5 \times 5 \times 7 \times 7} = \sqrt{(5 \times 5) \times (7 \times 7)}$

$= 5 \times 7 = 35$
By the long division method, $\sqrt{4761} =$
A $59$ B $69$ C $79$ D $61$

Answer and solution

Answer: (B) $69$

Pairs: $\overline{47}\;\overline{61}$ .

$6 \times 6 = 36 \le 47$ : first digit $6$ , and $47 - 36 = 11$

Bring down $61$ : we get $1161$ . New divisor: $6 \times 2 = 12$ ; $129 \times 9 = 1161$ ; $1161 - 1161 = 0$

Remainder $0$ . $\therefore \sqrt{4761} = 69$
In finding $\sqrt{4096}$ by the long division method, the first digit of the root is $6$ . The new divisor starts with
A $6$ B $7$ C $12$ D $36$

Answer and solution

Answer: (C) $12$

Double the root found so far: $6 \times 2 = 12$

Then $124 \times 4 = 496$ ( $125 \times 5 = 625$ is too big), so $\sqrt{4096} = 64$
By the long division method, $\sqrt{24649} =$
A $157$ B $147$ C $167$ D $153$

Answer and solution

Answer: (A) $157$

Pair the digits from the right: $\overline{2}\;\overline{46}\;\overline{49}$

$1 \times 1 = 1 \le 2$ : first digit $1$ , and $2 - 1 = 1$

Bring down $46$ : we get $146$ . New divisor: $1 \times 2 = 2$ ; $25 \times 5 = 125$ ( $26 \times 6 = 156$ is too big); $146 - 125 = 21$

Bring down $49$ : we get $2149$ . New divisor: $15 \times 2 = 30$ ; $307 \times 7 = 2149$ ( $308 \times 8 = 2464$ is too big); $2149 - 2149 = 0$

Remainder $0$ . $\therefore \sqrt{24649} = 157$
$\sqrt{5} \times \sqrt{80} =$
A $200$ B $400$ C $20$ D $\sqrt{85}$

Answer and solution

Answer: (C) $20$

$\sqrt{5} \times \sqrt{80} = \sqrt{5 \times 80} = \sqrt{400} = 20$
$(\sqrt{31})^2 =$
A $\sqrt{31}$ B $961$ C $31$ D $62$

Answer and solution

Answer: (C) $31$

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

$\therefore (\sqrt{31})^2 = 31$
$\displaystyle \frac{\sqrt{98}}{\sqrt{2}} =$
A $49$ B $7$ C $\sqrt{96}$ D $24.5$

Answer and solution

Answer: (B) $7$

$\displaystyle \frac{\sqrt{98}}{\sqrt{2}} = \sqrt{\frac{98}{2}} = \sqrt{49} = 7$
$\displaystyle \frac{30}{\sqrt{30}} =$
A $\sqrt{30}$ B $1$ C $30\sqrt{30}$ D $30$

Answer and solution

Answer: (A) $\sqrt{30}$

$\displaystyle \frac{30}{\sqrt{30}} = \frac{\sqrt{30} \times \sqrt{30}}{\sqrt{30}} = \sqrt{30}$
$\sqrt{81} + \sqrt{144} =$
A $441$ B $225$ C $15$ D $21$

Answer and solution

Answer: (D) $21$

$\sqrt{81} + \sqrt{144} = 9 + 12 = 21$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , so do not add first.
$\sqrt{289} - \sqrt{225} =$
A $8$ B $2$ C $32$ D $64$

Answer and solution

Answer: (B) $2$

$\sqrt{289} - \sqrt{225} = 17 - 15 = 2$

$\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$ , so do not subtract first.
$\sqrt{6^2 + 8^2} =$
A $10$ B $14$ C $100$ D $\sqrt{14}$

Answer and solution

Answer: (A) $10$

$\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

$\sqrt{a^2 + b^2} \ne a + b$
If $a = 4$ and $b = 5$ , then $a^2 + b^2 =$
A $9$ B $81$ C $41$ D $18$

Answer and solution

Answer: (C) $41$

$a^2 + b^2 = 4^2 + 5^2 = 16 + 25 = 41$

Note: $a^2 + b^2 \ne (a + b)^2$ (here $(a + b)^2 = 81$ ).
$\sqrt{0.64} =$
A $0.8$ B $8$ C $0.08$ D $0.32$

Answer and solution

Answer: (A) $0.8$

$\displaystyle \sqrt{0.64} = \sqrt{\frac{64}{100}}$

$\displaystyle = \frac{\sqrt{64}}{\sqrt{100}}$

$\displaystyle = \frac{8}{10}$

$= 0.8$
$\sqrt{0.0121}$ written as a fraction is
A $\displaystyle \frac{11}{100}$ B $\displaystyle \frac{11}{10}$ C $\displaystyle \frac{11}{1000}$ D $\displaystyle \frac{121}{100}$

Answer and solution

Answer: (A) $\displaystyle \frac{11}{100}$

$\displaystyle \sqrt{0.0121} = \sqrt{\frac{121}{10000}} = \frac{\sqrt{121}}{\sqrt{10000}} = \frac{11}{100}$
$\sqrt{6.76} =$
A $3.38$ B $26$ C $0.26$ D $2.6$

Answer and solution

Answer: (D) $2.6$

$\displaystyle \sqrt{6.76} = \sqrt{\frac{676}{100}} = \frac{\sqrt{676}}{\sqrt{100}} = \frac{26}{10} = 2.6$
$\sqrt{\square} = 2.4$ . The number in the box is
A $5.76$ B $0.576$ C $57.6$ D $4.8$

Answer and solution

Answer: (A) $5.76$

$\displaystyle 2.4 = \frac{24}{10}$

$\displaystyle \left(\frac{24}{10}\right)^2 = \frac{576}{100} = 5.76$

Check: $\displaystyle \sqrt{5.76} = \frac{\sqrt{576}}{\sqrt{100}} = 2.4$

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MCQ test, Set 3