MCQ test, Set 5

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

$10$ is multiplied by itself 4 times.

$10 \times 10 \times 10 \times 10 = 10^{4}$

Answer: (D) $169$

$13^{2} = 13 \times 13$

$= 169$

Answer: (D) $2^{10}$

$2^{10} = 1024$, $10^{2} = 100$, $3^{6} = 729$, $6^{3} = 216$

The largest is $2^{10}$.

Answer: (C) $15^{2}$

Dividing repeatedly by $15$: $225, 15$, that is 2 times.

$225 = 15^{2}$

Answer: (C) $15^{11}$

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

$15^{2} \times 15^{9} = 15^{2+9} = 15^{11}$

Answer: (A) $9^{8}$

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

$\displaystyle \frac{9^{10}}{9^{2}} = 9^{10-2} = 9^{8}$

Answer: (B) $z^{24}$

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

$(z^{4})^{6} = z^{4 \times 6} = z^{24}$

Answer: (A) $(48)^{4}$

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

$6^{4} \times 8^{4} = (6 \times 8)^{4} = (48)^{4}$

Answer: (A) $\displaystyle \frac{125}{8}$

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

$\displaystyle \frac{10^{3}}{4^{3}} = \left(\frac{10}{4}\right)^{3} = \left(\frac{5}{2}\right)^{3} = \frac{125}{8}$

Answer: (D) $1$

Taking $a + 2b \ne 0$: Using $a^0 = 1 \; (a \ne 0)$

$(a + 2b)^{0} = 1$

Answer: (B) $\displaystyle \frac{1}{b^{4}}$

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

$\displaystyle b^{-4} = \frac{1}{b^{4}}$

Answer: (D) $\displaystyle \frac{81}{4}$

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

$\displaystyle \left(\frac{2}{9}\right)^{-2} = \left(\frac{9}{2}\right)^{2} = \frac{81}{4}$

Answer: (D) $1$

$13$ is odd, so $(-1)^{13} = -1$; $29$ is odd, so $(-1)^{29} = -1$.

$(-1)^{13} \times (-1)^{29} = -1 \times (-1) = 1$

Answer: (C) $121$

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

$(-11)^{2} = 11^{2} = 121$

Answer: (B) $7$

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

$\Rightarrow 2x = 14$

$\displaystyle \Rightarrow x = \frac{14}{2} = 7$

Answer: (D) $-2$

Given, $7^{x+5} = 343$

Dividing $343$ repeatedly by $7$: $343, 49, 7$, so $343 = 7^{3}$.

$\Rightarrow 7^{x+5} = 7^{3}$

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

$\Rightarrow x+5 = 3$

$\Rightarrow x = 3 - 5$

$\Rightarrow x = -2$

Answer: (A) $\displaystyle \left(\frac{8}{5}\right)^{-3}$

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

$\displaystyle \left(\frac{5}{8}\right)^{3} = \left(\frac{8}{5}\right)^{-3}$

Answer: (D) $125a^{6}$

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

$(5a^{2})^{3} = 5^{3} \times (a^{2})^{3}$

$= 125 \times a^{6} = 125a^{6}$

Answer: (A) $9^{15}$

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

$9^{5} \times 9^{2} \times 9^{8} = 9^{5+2+8} = 9^{15}$

Answer: (B) $10^{-4}$

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

$(10^{2})^{-2} = 10^{2 \times (-2)} = 10^{-4}$

$\displaystyle = \frac{1}{10^{4}} = \frac{1}{10{,}000}$

Answer: (C) $2$

Given, $(3^{2})^{x} = 81$

Dividing $81$ repeatedly by $3$: $81, 27, 9, 3$, so $81 = 3^{4}$.

$\Rightarrow 3^{2x} = 3^{4}$ ($(a^m)^n = a^{m \times n}$)

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

$\Rightarrow 2x = 4$

$\displaystyle \Rightarrow x = \frac{4}{2} = 2$

Answer: (B) $113$

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

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

$= 4 + 9 + 100$

$= 113$

Answer: (A) $27$

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

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

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

$= 27$

Answer: (C) $\displaystyle \frac{2}{3}$

$\displaystyle \left(3^{-1} - 6^{-1}\right) \div 4^{-1} = \left(\frac{1}{3} - \frac{1}{6}\right) \div \frac{1}{4}$

$\displaystyle = \frac{6 - 3}{18} \div \frac{1}{4} = \frac{1}{6} \div \frac{1}{4}$

$\displaystyle = \frac{1}{6} \times 4 = \frac{2}{3}$

Answer: (B) $\displaystyle \frac{1}{8}$

$\displaystyle \left(2^{3} \div 2^{7}\right)^{2} \times 2^{5} = \left(\frac{2^{3}}{2^{7}}\right)^{2} \times 2^{5}$

$= (2^{3-7})^{2} \times 2^{5} = (2^{-4})^{2} \times 2^{5}$

$= 2^{-8} \times 2^{5}$ ($(a^m)^n = a^{m \times n}$)

$= 2^{-8+5} = 2^{-3}$

$\displaystyle = \frac{1}{2^{3}} = \frac{1}{8}$