MCQ test, Set 9

Answer: (D) $c \times c$

$a^2 = a \times a$

So $c^2 = c \times c$ ($c$ multiplied by itself, not by $2$).

Answer: (D) $256$

$16^2 = 16 \times 16 = 256$

Answer: (A) $529$

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 $400$.

$23^2 = 729 - 200 = 529$

Answer: (C) $400$

$20^2 = 20 \times 20$

$2 \times 2 = 4$, then write two zeros: $400$

Answer: (C) $2 \times 3$

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

$2 \times 3 = 6$, so $25^2 = 625$

Answer: (B) $176$

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

$176^2 = 30976 \Rightarrow \sqrt{30976} = 176$

(The square root is not half the number.)

Answer: (D) $17$

$17^2 = 17 \times 17 = 289$

$\therefore \sqrt{289} = 17$

Answer: (A) $784$

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

$28^2 = 28 \times 28 = 784$

Answer: (A) $81$

Take one factor from each pair: $3 \times 3 \times 3 \times 3 = 81$

Answer: (B) $105$

$\sqrt{3 \times 3 \times 5 \times 5 \times 7 \times 7} = \sqrt{(3 \times 3) \times (5 \times 5) \times (7 \times 7)}$

$= 3 \times 5 \times 7 = 105$

Answer: (A) $37$

Pairs: $\overline{13}\;\overline{69}$.

$3 \times 3 = 9 \le 13$: first digit $3$, and $13 - 9 = 4$

Bring down $69$: we get $469$. New divisor: $3 \times 2 = 6$; $67 \times 7 = 469$ ($68 \times 8 = 544$ is too big); $469 - 469 = 0$

Remainder $0$. $\therefore \sqrt{1369} = 37$

Answer: (D) $8$

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

Then $85 \times 5 = 425$ ($86 \times 6 = 516$ is too big), so $\sqrt{2025} = 45$

Answer: (C) $348$

Pair the digits from the right: $\overline{12}\;\overline{11}\;\overline{04}$

$3 \times 3 = 9 \le 12$: first digit $3$, and $12 - 9 = 3$

Bring down $11$: we get $311$. New divisor: $3 \times 2 = 6$; $64 \times 4 = 256$ ($65 \times 5 = 325$ is too big); $311 - 256 = 55$

Bring down $04$: we get $5504$. New divisor: $34 \times 2 = 68$; $688 \times 8 = 5504$ ($689 \times 9 = 6201$ is too big); $5504 - 5504 = 0$

Remainder $0$. $\therefore \sqrt{121104} = 348$

Answer: (B) $30$

$\sqrt{10} \times \sqrt{90} = \sqrt{10 \times 90} = \sqrt{900} = 30$

Answer: (D) $43$

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

$\therefore (\sqrt{43})^2 = 43$

Answer: (B) $3$

$\displaystyle \frac{\sqrt{117}}{\sqrt{13}} = \sqrt{\frac{117}{13}} = \sqrt{9} = 3$

Answer: (C) $\sqrt{39}$

$\displaystyle \frac{39}{\sqrt{39}} = \frac{\sqrt{39} \times \sqrt{39}}{\sqrt{39}} = \sqrt{39}$

Answer: (A) $47$

$\sqrt{144} + \sqrt{1225} = 12 + 35 = 47$

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

Answer: (D) $\sqrt{(x + y)^2} = x + y$

$\sqrt{(x + y)^2} = x + y$ 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: (B) $61$

$\sqrt{11^2 + 60^2} = \sqrt{121 + 3600} = \sqrt{3721} = 61$

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

Answer: (A) $22$

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

$\sqrt{(9 + 13)^2} = \sqrt{22^2} = 22$

Answer: (B) $0.3$

$\displaystyle \sqrt{0.09} = \sqrt{\frac{9}{100}}$

$\displaystyle = \frac{\sqrt{9}}{\sqrt{100}}$

$\displaystyle = \frac{3}{10}$

$= 0.3$

Answer: (B) $0.008$

$\displaystyle \sqrt{0.000064} = \sqrt{\frac{64}{1000000}}$

$\displaystyle = \frac{\sqrt{64}}{\sqrt{1000000}}$

$\displaystyle = \frac{8}{1000}$

$= 0.008$

Answer: (C) $2.9$

$\displaystyle \sqrt{8.41} = \sqrt{\frac{841}{100}} = \frac{\sqrt{841}}{\sqrt{100}} = \frac{29}{10} = 2.9$

Answer: (C) $90.25$

$\displaystyle 9.5 = \frac{95}{10}$

$\displaystyle \left(\frac{95}{10}\right)^2 = \frac{9025}{100} = 90.25$

Check: $\displaystyle \sqrt{90.25} = \frac{\sqrt{9025}}{\sqrt{100}} = 9.5$