MCQ test, Set 7

Answer: (A) $\displaystyle \left(\frac{4}{9}\right)^{2}$

$\displaystyle \frac{4}{9}$ is multiplied by itself 2 times.

$\displaystyle \frac{4}{9} \times \frac{4}{9} = \left(\frac{4}{9}\right)^{2}$

Answer: (A) $2744$

$14^{3} = 14 \times 14 \times 14$

$= 2744$

Answer: (B) $4^{4}$

$4^{4} = 256$, $5^{3} = 125$, $3^{5} = 243$, $6^{3} = 216$

The largest is $4^{4}$.

Answer: (C) $18^{2}$

Dividing repeatedly by $18$: $324, 18$, that is 2 times.

$324 = 18^{2}$

Answer: (B) $y^{13}$

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

$y^{6} \times y^{7} = y^{6+7} = y^{13}$

Answer: (D) $t^{10}$

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

$\displaystyle \frac{t^{18}}{t^{8}} = t^{18-8} = t^{10}$

Answer: (C) $12^{12}$

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

$(12^{3})^{4} = 12^{3 \times 4} = 12^{12}$

Answer: (C) $(60)^{3}$

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

$5^{3} \times 12^{3} = (5 \times 12)^{3} = (60)^{3}$

Answer: (A) $\displaystyle \frac{49}{9}$

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

$\displaystyle \frac{14^{2}}{6^{2}} = \left(\frac{14}{6}\right)^{2} = \left(\frac{7}{3}\right)^{2} = \frac{49}{9}$

Answer: (B) $1$

$6^{0} = 1$

$6^{0} \times 6^{0} = 1 \times 1 = 1$

Answer: (C) $\displaystyle \frac{1}{16}$

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

$\displaystyle 2^{-4} = \frac{1}{2^{4}} = \frac{1}{16}$

Answer: (B) $7$

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

$\displaystyle \left(\frac{1}{7}\right)^{-1} = 7^{1} = 7$

Answer: (A) $2$

$30$ is even, so $(-1)^{30} = 1$; $50$ is even, so $(-1)^{50} = 1$.

$(-1)^{30} + (-1)^{50} = 1 + 1 = 2$

Answer: (B) $-8000$

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

$(-20)^{3} = -20^{3} = -8000$

Answer: (D) $2$

Given, $3^{3x+1} = 2187$

Dividing $2187$ repeatedly by $3$: $2187, 729, 243, 81, 27, 9, 3$, so $2187 = 3^{7}$.

$\Rightarrow 3^{3x+1} = 3^{7}$

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

$\Rightarrow 3x+1 = 7$

$\Rightarrow 3x = 7 - 1$

$\Rightarrow 3x = 6$

$\displaystyle \Rightarrow x = \frac{6}{3} = 2$

Answer: (A) $2$

Given, $2^{5x-4} = 64$

Dividing $64$ repeatedly by $2$: $64, 32, 16, 8, 4, 2$, so $64 = 2^{6}$.

$\Rightarrow 2^{5x-4} = 2^{6}$

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

$\Rightarrow 5x-4 = 6$

$\Rightarrow 5x = 6 + 4$

$\Rightarrow 5x = 10$

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

Answer: (B) $36 \times 6^{x}$

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

$6^{x+2} = 6^{x} \times 6^{2} = 36 \times 6^{x}$

Answer: (D) $49x^{6}$

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

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

$= 49 \times x^{6} = 49x^{6}$

Answer: (D) $a^{11}b^{9}$

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

$(a^{3}b^{2})^{2} \times (ab)^{5} = a^{6}b^{4} \times a^{5}b^{5}$

Using $a^m \times a^n = a^{m+n}$: $= a^{11}b^{9}$

Answer: (C) $\displaystyle \frac{1}{6}$

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

$2^{-1} \times 3^{-1} = (2 \times 3)^{-1} = 6^{-1}$

$\displaystyle = \frac{1}{6^{1}} = \frac{1}{6}$

Answer: (A) $6$

Given, $6^{x-1} \times 6^{3} = 6^{8}$

$\Rightarrow 6^{x-1+3} = 6^{8}$ ($a^m \times a^n = a^{m+n}$)

$\Rightarrow 6^{x+2} = 6^{8}$

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

$\Rightarrow x+2 = 8$

$\Rightarrow x = 8 - 2$

$\Rightarrow x = 6$

Answer: (B) $-919$

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

$(-10)^{3} + (-9)^{2} = -10^{3} + 9^{2}$

$= -1000 + 81$

$= -919$

Answer: (C) $\displaystyle \frac{1}{4}$

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

$\displaystyle 2^{\frac{5}{3}} \div 2^{\frac{11}{3}} = 2^{\frac{5}{3} - \frac{11}{3}}$

$\displaystyle = 2^{\frac{-6}{3}} = 2^{-2}$

$\displaystyle = \frac{1}{2^{2}} = \frac{1}{4}$

Answer: (D) $8$

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

$\displaystyle 125 \times \left(\frac{2}{5}\right)^{3} = 125 \times \frac{2^{3}}{5^{3}}$

$\displaystyle = 125 \times \frac{8}{125}$

$= 8$

Answer: (D) $1$

Using $a^0 = 1 \; (a \ne 0)$

The bracket is not zero, so its power 0 is $1$.