MCQ test, Set 2

Answer: (D) $z^{5}$

$z$ is multiplied by itself 5 times.

$z \times z \times z \times z \times z = z^{5}$

Answer: (A) $729$

$3^{6} = 3^{5} \times 3$

$= 243 \times 3 = 729$

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

$10^{6} = 1$ followed by 6 zeros

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

Answer: (C) $3$

Dividing repeatedly by $6$: $216, 36, 6$, that is 3 times.

$216 = 6^{3}$

Answer: (D) $3^{17}$

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

$3^{11} \times 3^{6} = 3^{11+6} = 3^{17}$

Answer: (A) $12^{16}$

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

$12^{20} \div 12^{4} = 12^{20-4} = 12^{16}$

Answer: (D) $3^{14}$

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

$(3^{7})^{2} = 3^{7 \times 2} = 3^{14}$

Answer: (C) $(44)^{6}$

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

$4^{6} \times 11^{6} = (4 \times 11)^{6} = (44)^{6}$

Answer: (A) $27$

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

$\displaystyle \frac{24^{3}}{8^{3}} = \left(\frac{24}{8}\right)^{3} = 3^{3} = 27$

Answer: (C) $1$

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

$\displaystyle \left(\frac{7}{12}\right)^{0} = 1$

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

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

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

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

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

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

Answer: (C) $0$

$45$ is odd, so $(-1)^{45} = -1$; $62$ is even, so $(-1)^{62} = 1$.

$(-1)^{45} + (-1)^{62} = -1 + 1 = 0$

Answer: (B) $-6^{15}$

$15$ is odd, so $(-6)^{15} = -6^{15}$

Answer: (A) $5$

Given, $3^{x-1} = 81$

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

$\Rightarrow 3^{x-1} = 3^{4}$

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

$\Rightarrow x-1 = 4$

$\Rightarrow x = 4 + 1$

$\Rightarrow x = 5$

Answer: (D) $1$

Given, $5^{2x+1} = 125$

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

$\Rightarrow 5^{2x+1} = 5^{3}$

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

$\Rightarrow 2x+1 = 3$

$\Rightarrow 2x = 3 - 1$

$\Rightarrow 2x = 2$

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

Answer: (A) $\displaystyle \frac{7^{x}}{49}$

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

$\displaystyle 7^{x-2} = \frac{7^{x}}{7^{2}} = \frac{7^{x}}{49}$

Answer: (C) $\displaystyle \frac{r^{6}}{64}$

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

$\displaystyle \left(\frac{r}{2}\right)^{6} = \frac{r^{6}}{2^{6}}$

$\displaystyle = \frac{r^{6}}{64}$

Answer: (A) $2^{3}$

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

$\displaystyle \frac{2^{6} \times 2^{5}}{2^{8}} = \frac{2^{11}}{2^{8}}$

$= 2^{11-8} = 2^{3}$

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

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

$7^{-9} \times 7^{12} = 7^{-9+12} = 7^{3}$

$= 343$

Answer: (B) $8$

Given, $2^{m} \times 2^{3} = 2048$

Dividing $2048$ repeatedly by $2$: $2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2$, so $2048 = 2^{11}$.

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

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

$\Rightarrow m+3 = 11$

$\Rightarrow m = 11 - 3$

$\Rightarrow m = 8$

Answer: (C) $125$

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

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

$= 5^{3} = 125$

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

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

$\displaystyle \frac{2^{\frac{15}{2}} \times 2^{-\frac{5}{2}}}{2^{7}} = \frac{2^{\frac{15}{2} + (-\frac{5}{2})}}{2^{7}}$

$\displaystyle = \frac{2^{\frac{10}{2}}}{2^{7}} = \frac{2^{5}}{2^{7}}$

$= 2^{5-7} = 2^{-2}$

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

Answer: (B) $343$

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

$\displaystyle 8 \times \left(\frac{7}{2}\right)^{3} = 8 \times \frac{7^{3}}{2^{3}}$

$\displaystyle = 8 \times \frac{343}{8}$

$= 343$

Answer: (D) $1$

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

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