MCQ test, Set 5

Answer: (B) $56^2$

$56$ is multiplied by itself $2$ times.

$\therefore 56 \times 56 = 56^2$

Answer: (D) $576$

$24^2 = 24 \times 24 = 576$

Answer: (D) $484$

In the table, $24^2$ and $26^2$ differ by $100$, $23^2$ and $27^2$ by $200$, $22^2$ and $28^2$ by $300$, $21^2$ and $29^2$ by $400$.

$22^2 = 784 - 300 = 484$

Answer: (A) $1600$

$40^2 = 40 \times 40$

$4 \times 4 = 16$, then write two zeros: $1600$

Answer: (B) $6 \times 7$

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

$6 \times 7 = 42$, so $65^2 = 4225$

Answer: (D) $142$

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

$142^2 = 20164 \Rightarrow \sqrt{20164} = 142$

(The square root is not half the number.)

Answer: (C) $12$

$12^2 = 12 \times 12 = 144$

$\therefore \sqrt{144} = 12$

Answer: (D) $625$

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

$25^2 = 25 \times 25 = 625$

Answer: (B) $75$

Take one factor from each pair: $3 \times 5 \times 5 = 75$

Answer: (A) $26$

$\sqrt{2 \times 2 \times 13 \times 13} = \sqrt{(2 \times 2) \times (13 \times 13)}$

$= 2 \times 13 = 26$

Answer: (B) $87$

Pairs: $\overline{75}\;\overline{69}$.

$8 \times 8 = 64 \le 75$: first digit $8$, and $75 - 64 = 11$

Bring down $69$: we get $1169$. New divisor: $8 \times 2 = 16$; $167 \times 7 = 1169$ ($168 \times 8 = 1344$ is too big); $1169 - 1169 = 0$

Remainder $0$. $\therefore \sqrt{7569} = 87$

Answer: (A) $8$

Double the root found so far: $4 \times 2 = 8$

Then $89 \times 9 = 801$, so $\sqrt{2401} = 49$

Answer: (D) $3$

Pairs from the right: $\overline{14}\;\overline{51}\;\overline{61}$, that is $3$ groups.

One digit of the root for each group: $\sqrt{145161} = 381$

Answer: (A) $30$

$\sqrt{6} \times \sqrt{150} = \sqrt{6 \times 150} = \sqrt{900} = 30$

Answer: (D) $37$

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

$\therefore (\sqrt{37})^2 = 37$

Answer: (C) $4$

$\displaystyle \frac{\sqrt{112}}{\sqrt{7}} = \sqrt{\frac{112}{7}} = \sqrt{16} = 4$

Answer: (C) $\sqrt{33}$

$\displaystyle \frac{33}{\sqrt{33}} = \frac{\sqrt{33} \times \sqrt{33}}{\sqrt{33}} = \sqrt{33}$

Answer: (B) $11$

$\sqrt{64} + \sqrt{9} = 8 + 3 = 11$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$, so do not add first.

Answer: (C) $\displaystyle \frac{\sqrt{p}}{\sqrt{q}} = \sqrt{\frac{p}{q}}$

$\displaystyle \frac{\sqrt{p}}{\sqrt{q}} = \sqrt{\frac{p}{q}}$ is a property in the notes.

The other three are the 'not equal' rules: $\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$, $\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$, $\sqrt{a^2 + b^2} \ne a + b$.

Answer: (A) $34$

$\sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1156} = 34$

$\sqrt{a^2 + b^2} \ne a + b$

Answer: (A) $7$

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

$\sqrt{(20 - 13)^2} = \sqrt{7^2} = 7$

Answer: (C) $0.4$

$\displaystyle \sqrt{0.16} = \sqrt{\frac{16}{100}}$

$\displaystyle = \frac{\sqrt{16}}{\sqrt{100}}$

$\displaystyle = \frac{4}{10}$

$= 0.4$

Answer: (C) $0.009$

$\displaystyle \sqrt{0.000081} = \sqrt{\frac{81}{1000000}}$

$\displaystyle = \frac{\sqrt{81}}{\sqrt{1000000}}$

$\displaystyle = \frac{9}{1000}$

$= 0.009$

Answer: (B) $1.5$

$\displaystyle \sqrt{2.25} = \sqrt{\frac{225}{100}} = \frac{\sqrt{225}}{\sqrt{100}} = \frac{15}{10} = 1.5$

Answer: (B) $3.24$

$\displaystyle 1.8 = \frac{18}{10}$

$\displaystyle \left(\frac{18}{10}\right)^2 = \frac{324}{100} = 3.24$

Check: $\displaystyle \sqrt{3.24} = \frac{\sqrt{324}}{\sqrt{100}} = 1.8$