MCQ test, Set 10

Answer: (D) $b^{3}$

$b$ is multiplied by itself 3 times.

$b \times b \times b = b^{3}$

Answer: (C) $4096$

$2^{12} = 2^{11} \times 2$

$= 2048 \times 2 = 4096$

Answer: (A) $10{,}00{,}00{,}000$

$10^{8} = 1$ followed by 8 zeros

$= 10{,}00{,}00{,}000$

Answer: (D) $2$

Dividing repeatedly by $19$: $361, 19$, that is 2 times.

$361 = 19^{2}$

Answer: (D) $20^{20}$

Using $a^m \times a^n = a^{m+n}$

$20^{7} \times 20^{13} = 20^{7+13} = 20^{20}$

Answer: (A) $20^{4}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{20^{12}}{20^{8}} = 20^{12-8} = 20^{4}$

Answer: (C) $\displaystyle \left(\frac{3}{5}\right)^{6}$

Using $(a^m)^n = a^{m \times n}$

$\displaystyle (\left(\frac{3}{5}\right)^{2})^{3} = \left(\frac{3}{5}\right)^{2 \times 3} = \left(\frac{3}{5}\right)^{6}$

Answer: (B) $(70)^{2}$

Using $a^n \times b^n = (a \times b)^n$

$14^{2} \times 5^{2} = (14 \times 5)^{2} = (70)^{2}$

Answer: (B) $\displaystyle \frac{9}{4}$

Using $\displaystyle \frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$

$\displaystyle \frac{21^{2}}{14^{2}} = \left(\frac{21}{14}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4}$

Answer: (A) $4$

$2^{0} + 3^{0} = 1 + 1 = 2$

$2^{2} = 2 \times 2 = 4$

Answer: (B) $\displaystyle \frac{1}{100}$

Using $\displaystyle a^{-n} = \frac{1}{a^n}$

$\displaystyle 10^{-2} = \frac{1}{10^{2}} = \frac{1}{100}$

Answer: (C) $32$

Using $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$

$\displaystyle \left(\frac{1}{2}\right)^{-5} = 2^{5} = 32$

Answer: (C) $-1$

$333$ is odd and $(-1)^{\text{odd number}} = -1$.

$(-1)^{333} = -1$

Answer: (B) $900$

Using $(-a)^n = -a^n \text{ if } n \text{ is odd}, \; (-a)^n = a^n \text{ if } n \text{ is even}$

$(-30)^{2} = 30^{2} = 900$

Answer: (D) $5$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2x-1 = 9$

$\Rightarrow 2x = 9 + 1$

$\Rightarrow 2x = 10$

$\displaystyle \Rightarrow x = \frac{10}{2} = 5$

Answer: (A) $1$

Given, $5^{4y-1} = 125$

Dividing $125$ repeatedly by $5$: $125, 25, 5$, so $125 = 5^{3}$.

$\Rightarrow 5^{4y-1} = 5^{3}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 4y-1 = 3$

$\Rightarrow 4y = 3 + 1$

$\Rightarrow 4y = 4$

$\displaystyle \Rightarrow y = \frac{4}{4} = 1$

Answer: (C) $\displaystyle \left(\frac{4}{7}\right)^{-3}$

Using $\displaystyle \left(\frac{b}{a}\right)^n = \left(\frac{a}{b}\right)^{-n}$

$\displaystyle \left(\frac{7}{4}\right)^{3} = \left(\frac{4}{7}\right)^{-3}$

Answer: (D) $16n^{6}$

Using $(ab)^n = a^n \times b^n$ and $(a^m)^n = a^{m \times n}$

$(4n^{3})^{2} = 4^{2} \times (n^{3})^{2}$

$= 16 \times n^{6} = 16n^{6}$

Answer: (A) $11^{1}$

Using $a^m \times a^n = a^{m+n}$ and $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{11^{10} \times 11^{3}}{11^{12}} = \frac{11^{13}}{11^{12}}$

$= 11^{13-12} = 11^{1}$

Answer: (C) $\displaystyle \frac{1}{64}$

Using $\displaystyle a^{-n} = \frac{1}{a^n}$

$\displaystyle (-2)^{-6} = \frac{1}{(-2)^{6}}$

$\displaystyle = \frac{1}{64} = \frac{1}{64}$ ($6$ is even)

Answer: (A) $3$

Given, $(5^{3})^{x} \times 5^{2} = 5^{11}$

$\Rightarrow 5^{3x} \times 5^{2} = 5^{11}$ ($(a^m)^n = a^{m \times n}$)

$\Rightarrow 5^{3x+2} = 5^{11}$ ($a^m \times a^n = a^{m+n}$)

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 3x+2 = 11$

$\Rightarrow 3x = 11 - 2$

$\Rightarrow 3x = 9$

$\displaystyle \Rightarrow x = \frac{9}{3} = 3$

Answer: (D) $146$

Using $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$

$\displaystyle \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{7}\right)^{-2} + \left(\frac{1}{9}\right)^{-2} = (4)^{2} + (7)^{2} + (9)^{2}$

$= 16 + 49 + 81$

$= 146$

Answer: (B) $9$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle 3^{\frac{13}{4}} \div 3^{\frac{5}{4}} = 3^{\frac{13}{4} - \frac{5}{4}}$

$\displaystyle = 3^{\frac{8}{4}} = 3^{2}$

$= 9$

Answer: (D) $\displaystyle \frac{5}{12}$

$\displaystyle \left(4^{-1} - 6^{-1}\right) \div 5^{-1} = \left(\frac{1}{4} - \frac{1}{6}\right) \div \frac{1}{5}$

$\displaystyle = \frac{6 - 4}{24} \div \frac{1}{5} = \frac{1}{12} \div \frac{1}{5}$

$\displaystyle = \frac{1}{12} \times 5 = \frac{5}{12}$

Answer: (B) $\displaystyle \frac{1}{7}$

$\displaystyle \left(7^{1} \div 7^{3}\right)^{2} \times 7^{3} = \left(\frac{7^{1}}{7^{3}}\right)^{2} \times 7^{3}$

$= (7^{1-3})^{2} \times 7^{3} = (7^{-2})^{2} \times 7^{3}$

$= 7^{-4} \times 7^{3}$ ($(a^m)^n = a^{m \times n}$)

$= 7^{-4+3} = 7^{-1}$

$\displaystyle = \frac{1}{7^{1}} = \frac{1}{7}$