MCQ test, Set 6

Answer: (A) $q^{8}$

$q$ is multiplied by itself 8 times.

$q \times q \times q \times q \times q \times q \times q \times q = q^{8}$

Answer: (B) $1296$

$6^{4} = 6^{3} \times 6$

$= 216 \times 6 = 1296$

Answer: (B) $1{,}00{,}00{,}000$

$10^{7} = 1$ followed by 7 zeros

$= 1{,}00{,}00{,}000$

Answer: (C) $2$

Dividing repeatedly by $12$: $144, 12$, that is 2 times.

$144 = 12^{2}$

Answer: (A) $8^{17}$

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

$8^{12} \times 8^{5} = 8^{12+5} = 8^{17}$

Answer: (A) $17^{20}$

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

$17^{25} \div 17^{5} = 17^{25-5} = 17^{20}$

Answer: (C) $5^{18}$

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

$(5^{2})^{9} = 5^{2 \times 9} = 5^{18}$

Answer: (C) $(39)^{9}$

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

$3^{9} \times 13^{9} = (3 \times 13)^{9} = (39)^{9}$

Answer: (A) $\displaystyle \frac{9}{16}$

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

$\displaystyle \frac{6^{2}}{8^{2}} = \left(\frac{6}{8}\right)^{2} = \left(\frac{3}{4}\right)^{2} = \frac{9}{16}$

Answer: (B) undefined

The notes: $a^0 = 1$ only when $a \ne 0$.

$0^{0}$ is undefined.

Answer: (C) $12^{-3}$

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

$\displaystyle \frac{1}{12^{3}} = 12^{-3}$

Answer: (B) $\displaystyle \frac{27}{64}$

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

$\displaystyle \left(\frac{4}{3}\right)^{-3} = \left(\frac{3}{4}\right)^{3} = \frac{27}{64}$

Answer: (D) $-1$

$1001$ is odd and $(-1)^{\text{odd number}} = -1$.

$(-1)^{1001} = -1$

Answer: (C) $13^{20}$

$20$ is even, so $(-13)^{20} = 13^{20}$

Answer: (C) $1$

Given, $6^{2n} = 36$

Dividing $36$ repeatedly by $6$: $36, 6$, so $36 = 6^{2}$.

$\Rightarrow 6^{2n} = 6^{2}$

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

$\Rightarrow 2n = 2$

$\displaystyle \Rightarrow n = \frac{2}{2} = 1$

Answer: (D) $-4$

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

$\Rightarrow x+7 = 3$

$\Rightarrow x = 3 - 7$

$\Rightarrow x = -4$

Answer: (B) $(-2)^{7}$

$7$ is odd, so $-a^n = (-a)^n$: $-2^{7} = (-2)^{7}$

Answer: (D) $\displaystyle \frac{c^{3}}{1000}$

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

$\displaystyle \left(\frac{c}{10}\right)^{3} = \frac{c^{3}}{10^{3}}$

$\displaystyle = \frac{c^{3}}{1000}$

Answer: (A) $5^{2}$

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

$\displaystyle \frac{5^{7} \times 5^{4}}{5^{9}} = \frac{5^{11}}{5^{9}}$

$= 5^{11-9} = 5^{2}$

Answer: (D) $\displaystyle -\frac{1}{27}$

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

$\displaystyle (-3)^{-3} = \frac{1}{(-3)^{3}}$

$\displaystyle = \frac{1}{-27} = -\frac{1}{27}$ ($3$ is odd)

Answer: (B) $7$

Given, $5^{x} \div 5^{4} = 125$

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

$\Rightarrow 5^{x-4} = 5^{3}$ ($\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

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

$\Rightarrow x-4 = 3$

$\Rightarrow x = 3 + 4$

$\Rightarrow x = 7$

Answer: (A) $-3$

Using $(-1)^{\text{odd number}} = -1, \; (-1)^{\text{even number}} = 1$

$(-1)^{3} + (-1)^{5} + (-1)^{11} = (-1) + (-1) + (-1)$

$= -3$

Answer: (D) $6$

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

$\displaystyle \frac{6^{\frac{7}{2}} \times 6^{-\frac{3}{2}}}{6^{1}} = \frac{6^{\frac{7}{2} + (-\frac{3}{2})}}{6^{1}}$

$\displaystyle = \frac{6^{\frac{4}{2}}}{6^{1}} = \frac{6^{2}}{6^{1}}$

$= 6^{2-1} = 6^{1}$

$= 6$

Answer: (D) $9$

$\displaystyle \left(3^{9} \div 3^{7}\right)^{3} \times 3^{-4} = \left(\frac{3^{9}}{3^{7}}\right)^{3} \times 3^{-4}$

$= (3^{9-7})^{3} \times 3^{-4} = (3^{2})^{3} \times 3^{-4}$

$= 3^{6} \times 3^{-4}$ ($(a^m)^n = a^{m \times n}$)

$= 3^{6-4} = 3^{2}$

$= 9$

Answer: (B) $900b^{6}$

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

$(10b^{2})^{2} \times (3b)^{2} = 10^{2} \cdot (b^{2})^{2} \times 3^{2} \cdot b^{2}$

$= 100b^{4} \times 9b^{2}$

$= 900 \cdot b^{4+2} = 900b^{6}$