Squares and Square Roots: MCQ test, Set 4

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

Answer and solution

Answer: (C) ' $x$ squared'

$x^2 = x \times x$ is read as ' $x$ squared' (or) ' $x$ power 2'.
Which of these is not a perfect square?
A $784$ B $494$ C $169$ D $484$

Answer and solution

Answer: (B) $494$

$169 = 13^2$ , $484 = 22^2$ , $784 = 28^2$ , so they are perfect squares.

$494$ is not in the table of squares.
The next number in $225, 256, 289, \ldots$ is
A $322$ B $361$ C $324$ D $323$

Answer and solution

Answer: (C) $324$

These are $15^2, 16^2, 17^2$ .

Next is $18^2 = 324$
$90^2 =$
A $81000$ B $8100$ C $810$ D $180$

Answer and solution

Answer: (B) $8100$

$90^2 = 90 \times 90$

$9 \times 9 = 81$ , then write two zeros: $8100$
$55^2 = \square 25$ . The number in the box is
A $25$ B $11$ C $36$ D $30$

Answer and solution

Answer: (D) $30$

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

$5 \times 6 = 30$ , so $55^2 = 3025$
If $137^2 = 18769$ , then $\sqrt{18769} =$
A $274$ B $\sqrt{137}$ C $9384.5$ D $137$

Answer and solution

Answer: (D) $137$

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

$137^2 = 18769 \Rightarrow \sqrt{18769} = 137$

(The square root is not half the number.)
$\sqrt{841} =$
A $28$ B $30$ C $29$ D $420.5$

Answer and solution

Answer: (C) $29$

$29^2 = 29 \times 29 = 841$

$\therefore \sqrt{841} = 29$
$\sqrt{\square} = 11$ . The number in the box is
A $22$ B $\sqrt{11}$ C $5.5$ D $121$

Answer and solution

Answer: (D) $121$

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

$11^2 = 11 \times 11 = 121$
$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$ . So $\sqrt{2025} =$
A $45$ B $135$ C $15$ D $1012.5$

Answer and solution

Answer: (A) $45$

Take one factor from each pair: $3 \times 3 \times 5 = 45$
In the prime factorization method, from each pair of equal prime factors we take
A one factor B no factor C both factors D the sum of the two factors

Answer and solution

Answer: (A) one factor

$\sqrt{2 \times 2} = 2$ : each pair gives one factor to the square root.
By the long division method, $\sqrt{1849} =$
A $47$ B $53$ C $33$ D $43$

Answer and solution

Answer: (D) $43$

Pairs: $\overline{18}\;\overline{49}$ .

$4 \times 4 = 16 \le 18$ : first digit $4$ , and $18 - 16 = 2$

Bring down $49$ : we get $249$ . New divisor: $4 \times 2 = 8$ ; $83 \times 3 = 249$ ( $84 \times 4 = 336$ is too big); $249 - 249 = 0$

Remainder $0$ . $\therefore \sqrt{1849} = 43$
While finding $\sqrt{9025}$ by long division, the first digit of the root is $9$ . The next divisor begins with
A $18$ B $9$ C $10$ D $81$

Answer and solution

Answer: (A) $18$

Double the root found so far: $9 \times 2 = 18$

Then $185 \times 5 = 925$ ( $186 \times 6 = 1116$ is too big), so $\sqrt{9025} = 95$
In finding $\sqrt{60516}$ by the long division method, the first digit of the root is
A $3$ B $7$ C $2$ D $6$

Answer and solution

Answer: (C) $2$

Pair from the right: $\overline{6}\;\overline{05}\;\overline{16}$ ; the first group is $6$ .

$2 \times 2 = 4 \le 6$ , so the first digit is $2$
$\sqrt{7} \times \sqrt{63} =$
A $\sqrt{70}$ B $21$ C $220.5$ D $441$

Answer and solution

Answer: (B) $21$

$\sqrt{7} \times \sqrt{63} = \sqrt{7 \times 63} = \sqrt{441} = 21$
$(\sqrt{22})^2 =$
A $22$ B $44$ C $484$ D $\sqrt{22}$

Answer and solution

Answer: (A) $22$

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

$\therefore (\sqrt{22})^2 = 22$
$\displaystyle \frac{\sqrt{150}}{\sqrt{6}} =$
A $25$ B $12.5$ C $\sqrt{144}$ D $5$

Answer and solution

Answer: (D) $5$

$\displaystyle \frac{\sqrt{150}}{\sqrt{6}} = \sqrt{\frac{150}{6}} = \sqrt{25} = 5$
$\displaystyle \frac{31}{\sqrt{31}} =$
A $31$ B $\sqrt{31}$ C $1$ D $31\sqrt{31}$

Answer and solution

Answer: (B) $\sqrt{31}$

$\displaystyle \frac{31}{\sqrt{31}} = \frac{\sqrt{31} \times \sqrt{31}}{\sqrt{31}} = \sqrt{31}$
$\sqrt{4} + \sqrt{81} =$
A $121$ B $85$ C $11$ D $\sqrt{85}$

Answer and solution

Answer: (C) $11$

$\sqrt{4} + \sqrt{81} = 2 + 9 = 11$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , so do not add first.
$\sqrt{625} - \sqrt{49} =$
A $24$ B $576$ C $18$ D $32$

Answer and solution

Answer: (C) $18$

$\sqrt{625} - \sqrt{49} = 25 - 7 = 18$

$\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$ , so do not subtract first.
$\sqrt{10^2 + 24^2} =$
A $34$ B $676$ C $\sqrt{34}$ D $26$

Answer and solution

Answer: (D) $26$

$\sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26$

$\sqrt{a^2 + b^2} \ne a + b$
$\sqrt{(8 + 9)^2} =$
A $145$ B $17$ C $289$ D $\sqrt{145}$

Answer and solution

Answer: (B) $17$

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

$\sqrt{(8 + 9)^2} = \sqrt{17^2} = 17$
$\sqrt{0.0009} =$
A $0.03$ B $0.3$ C $0.003$ D $0.00045$

Answer and solution

Answer: (A) $0.03$

$\displaystyle \sqrt{0.0009} = \sqrt{\frac{9}{10000}}$

$\displaystyle = \frac{\sqrt{9}}{\sqrt{10000}}$

$\displaystyle = \frac{3}{100}$

$= 0.03$
$\sqrt{0.0225}$ written as a fraction is
A $\displaystyle \frac{15}{100}$ B $\displaystyle \frac{225}{100}$ C $\displaystyle \frac{15}{1000}$ D $\displaystyle \frac{15}{10}$

Answer and solution

Answer: (A) $\displaystyle \frac{15}{100}$

$\displaystyle \sqrt{0.0225} = \sqrt{\frac{225}{10000}} = \frac{\sqrt{225}}{\sqrt{10000}} = \frac{15}{100}$
$\sqrt{2.89} =$
A $17$ B $1.445$ C $1.7$ D $0.17$

Answer and solution

Answer: (C) $1.7$

$\displaystyle \sqrt{2.89} = \sqrt{\frac{289}{100}} = \frac{\sqrt{289}}{\sqrt{100}} = \frac{17}{10} = 1.7$
$\sqrt{\square} = 6.5$ . The number in the box is
A $4.225$ B $42.25$ C $13$ D $422.5$

Answer and solution

Answer: (B) $42.25$

$\displaystyle 6.5 = \frac{65}{10}$

$\displaystyle \left(\frac{65}{10}\right)^2 = \frac{4225}{100} = 42.25$

Check: $\displaystyle \sqrt{42.25} = \frac{\sqrt{4225}}{\sqrt{100}} = 6.5$

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MCQ test, Set 4