MCQ test, Set 4

Answer: (C) $x$ power $3$ (or) $x$ cube

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

It is read as $x$ power $3$ (or) $x$ cube.

Answer: (B) $1024$

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

$= 256 \times 4 = 1024$

Answer: (D) $10^{5}$

$1{,}00{,}000$ is 1 followed by 5 zeros.

$1{,}00{,}000 = 10^{5}$

Answer: (B) $5$

Dividing repeatedly by $7$: $16{,}807, 2401, 343, 49, 7$, that is 5 times.

$16{,}807 = 7^{5}$

Answer: (B) $p^{11}$

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

$p^{8} \times p^{3} = p^{8+3} = p^{11}$

Answer: (A) $6^{7}$

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

$6^{14} \div 6^{7} = 6^{14-7} = 6^{7}$

Answer: (B) $2^{27}$

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

$(2^{9})^{3} = 2^{9 \times 3} = 2^{27}$

Answer: (B) $(30)^{7}$

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

$2^{7} \times 15^{7} = (2 \times 15)^{7} = (30)^{7}$

Answer: (C) $25$

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

$\displaystyle \frac{40^{2}}{8^{2}} = \left(\frac{40}{8}\right)^{2} = 5^{2} = 25$

Answer: (B) $2$

$9^{0} = 1$

$2 \times 9^{0} = 2 \times 1 = 2$

Answer: (D) $\displaystyle \frac{1}{25}$

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

$\displaystyle 5^{-2} = \frac{1}{5^{2}} = \frac{1}{25}$

Answer: (B) $\displaystyle \frac{4}{25}$

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

$\displaystyle \left(\frac{5}{2}\right)^{-2} = \left(\frac{2}{5}\right)^{2} = \frac{4}{25}$

Answer: (A) an odd number

The notes: $(-1)^{\text{odd number}} = -1$, e.g. $(-1)^{3} = -1$.

Answer: (C) $x^{8}$

$8$ is even, so $(-x)^{8} = x^{8}$

Answer: (A) $a \ne -1, 0, 1$

$1^{4} = 1^{7}$, $0^{2} = 0^{3}$, $(-1)^{2} = (-1)^{4}$ but the powers are not equal.

So the rule needs $a \ne -1, 0, 1$.

Answer: (D) $13$

Given, $2^{x-3} = 1024$

Dividing $1024$ repeatedly by $2$: $1024, 512, 256, 128, 64, 32, 16, 8, 4, 2$, so $1024 = 2^{10}$.

$\Rightarrow 2^{x-3} = 2^{10}$

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

$\Rightarrow x-3 = 10$

$\Rightarrow x = 10 + 3$

$\Rightarrow x = 13$

Answer: (A) $9^{-1}$

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

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

Answer: (C) $a^{3}b^{6}$

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

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

$= a^{3}b^{6}$

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

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

$\displaystyle \frac{(3^{3})^{2} \times 3^{4}}{3^{7}} = \frac{3^{6} \times 3^{4}}{3^{7}} = \frac{3^{10}}{3^{7}}$

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

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

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

$5^{-4} \times 5^{6} = 5^{-4+6} = 5^{2}$

$= 25$

Answer: (C) $-3$

Given, $\displaystyle 4^{x} = \frac{1}{64}$

$\displaystyle \frac{1}{64} = \frac{1}{4^{3}} = 4^{-3}$ (dividing $64$ repeatedly by $4$: $64, 16, 4$)

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

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

$\Rightarrow x = -3$

Answer: (C) $10$

$\displaystyle \left(2^{-1} \times 5^{-1}\right) \div 10^{-2} = \left(\frac{1}{2} \times \frac{1}{5}\right) \div \frac{1}{10^{2}}$

$\displaystyle = \frac{1}{10} \div \frac{1}{100}$

$\displaystyle = \frac{1}{10} \times 100 = 10$

Answer: (D) $4$

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

$\displaystyle 2^{\frac{5}{3}} \times 2^{-\frac{1}{3}} \times 2^{\frac{2}{3}} = 2^{\frac{5}{3} + (-\frac{1}{3}) + \frac{2}{3}}$

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

$= 4$

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

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

$(-2)^{-9} \div (-2)^{-5} = (-2)^{-9-(-5)} = (-2)^{-9+5} = (-2)^{-4}$

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

Answer: (A) $648t^{7}$

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

$(3t)^{4} \times (2t)^{3} = 3^{4} \cdot t^{4} \times 2^{3} \cdot t^{3}$

$= 81t^{4} \times 8t^{3}$

$= 648 \cdot t^{4+3} = 648t^{7}$