MCQ test, Set 8

Answer: (D) $p$ power $9$

$p^{9} = p \times p \times p \times p \times p \times p \times p \times p \times p$

It is read as $p$ power $9$.

Answer: (A) $625$

$5^{4} = 5^{3} \times 5$

$= 125 \times 5 = 625$

Answer: (C) $10^{4}$

$10{,}000$ is 1 followed by 4 zeros.

$10{,}000 = 10^{4}$

Answer: (D) $2$

Dividing repeatedly by $17$: $289, 17$, that is 2 times.

$289 = 17^{2}$

Answer: (C) $19^{7}$

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

$19^{3} \times 19^{4} = 19^{3+4} = 19^{7}$

Answer: (A) $4^{14}$

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

$\displaystyle \frac{4^{21}}{4^{7}} = 4^{21-7} = 4^{14}$

Answer: (D) $15^{12}$

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

$(15^{6})^{2} = 15^{6 \times 2} = 15^{12}$

Answer: (C) $(22)^{10}$

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

$11^{10} \times 2^{10} = (11 \times 2)^{10} = (22)^{10}$

Answer: (B) $125$

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

$\displaystyle \frac{50^{3}}{10^{3}} = \left(\frac{50}{10}\right)^{3} = 5^{3} = 125$

Answer: (D) $3$

$4^{0} = 3^{0} = 2^{0} = 1$

$4^{0} + 3^{0} + 2^{0} = 1 + 1 + 1 = 3$

Answer: (B) $\displaystyle \frac{1}{t^{9}}$

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

$\displaystyle t^{-9} = \frac{1}{t^{9}}$

Answer: (A) $\displaystyle \frac{100}{9}$

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

$\displaystyle \left(\frac{3}{10}\right)^{-2} = \left(\frac{10}{3}\right)^{2} = \frac{100}{9}$

Answer: (D) $1$

$64$ is even and $(-1)^{\text{even number}} = 1$.

$(-1)^{64} = 1$

Answer: (C) $-y^{11}$

$11$ is odd, so $(-y)^{11} = -y^{11}$

Answer: (B) $12$

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

$\Rightarrow x-6 = 6$

$\Rightarrow x = 6 + 6$

$\Rightarrow x = 12$

Answer: (B) $2$

Given, $8^{x+1} = 512$

Dividing $512$ repeatedly by $8$: $512, 64, 8$, so $512 = 8^{3}$.

$\Rightarrow 8^{x+1} = 8^{3}$

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

$\Rightarrow x+1 = 3$

$\Rightarrow x = 3 - 1$

$\Rightarrow x = 2$

Answer: (A) $\displaystyle \frac{2^{x}}{16}$

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

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

Answer: (A) $\displaystyle \frac{16}{q^{4}}$

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

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

$\displaystyle = \frac{16}{q^{4}}$

Answer: (B) $7^{3}$

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

$\displaystyle \frac{(7^{2})^{5} \times 7^{3}}{7^{10}} = \frac{7^{10} \times 7^{3}}{7^{10}} = \frac{7^{13}}{7^{10}}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$: $7^{13-10} = 7^{3}$

Answer: (A) $10^{3}$

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

$\displaystyle 10^{1} \div 10^{-2} = \frac{10^{1}}{10^{-2}} = 10^{1-(-2)} = 10^{1+2} = 10^{3}$

$= 1000$

Answer: (D) $-4$

Given, $\displaystyle 10^{x+1} = \frac{1}{1000}$

$\displaystyle \frac{1}{1000} = \frac{1}{10^{3}} = 10^{-3}$ (dividing $1000$ repeatedly by $10$: $1000, 100, 10$)

$\Rightarrow 10^{x+1} = 10^{-3}$

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

$\Rightarrow x+1 = -3$

$\Rightarrow x = -3 - 1$

$\Rightarrow x = -4$

Answer: (C) $8$

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

$= 2^{-4+9} \times 2^{-2} = 2^{5} \times 2^{-2}$

$= 2^{5-2} = 2^{3}$

$= 8$

Answer: (B) $121$

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

$\displaystyle 11^{\frac{3}{2}} \times 11^{\frac{1}{2}} = 11^{\frac{3}{2} + \frac{1}{2}}$

$\displaystyle = 11^{\frac{4}{2}} = 11^{2}$

$= 121$

Answer: (B) $1$

$\displaystyle \left(2^{-1} - 6^{-1}\right) \div 3^{-1} = \left(\frac{1}{2} - \frac{1}{6}\right) \div \frac{1}{3}$

$\displaystyle = \frac{6 - 2}{12} \div \frac{1}{3} = \frac{1}{3} \div \frac{1}{3}$

$\displaystyle = \frac{1}{3} \times 3 = 1$

Answer: (C) $\displaystyle \frac{21}{4}$

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

$\displaystyle \left(\frac{2}{7}\right)^{-4} \times \left(\frac{3}{2}\right)^{-2} \times \left(\frac{7}{3}\right)^{-3} = \left(\frac{7}{2}\right)^{4} \times \left(\frac{2}{3}\right)^{2} \times \left(\frac{3}{7}\right)^{3}$

$\displaystyle = \frac{7^{4}}{2^{4}} \times \frac{2^{2}}{3^{2}} \times \frac{3^{3}}{7^{3}}$

$= 2^{-2} \times 7^{1} \times 3^{1}$ ($a^m \times a^n = a^{m+n}$, $\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

$\displaystyle = \frac{1}{4} \times 7 \times 3$

$\displaystyle = \frac{21}{4}$