MCQ test, Set 1

Answer: (B) $4^{6}$

$4$ is multiplied by itself 6 times.

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

Answer: (D) $256$

$2^{8} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 256$

Answer: (A) $2^{8}$

$2^{8} = 256$, $8^{2} = 64$, $3^{4} = 81$, $5^{3} = 125$

The largest is $2^{8}$.

Answer: (C) $2^{6}$

Dividing repeatedly by $2$: $64, 32, 16, 8, 4, 2$, that is 6 times.

$64 = 2^{6}$

Answer: (A) $7^{14}$

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

$7^{5} \times 7^{9} = 7^{5+9} = 7^{14}$

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

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

$\displaystyle \frac{8^{11}}{8^{3}} = 8^{11-3} = 8^{8}$

Answer: (C) $6^{20}$

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

$(6^{5})^{4} = 6^{5 \times 4} = 6^{20}$

Answer: (D) $(21)^{8}$

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

$7^{8} \times 3^{8} = (7 \times 3)^{8} = (21)^{8}$

Answer: (B) $81$

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

$\displaystyle \frac{15^{4}}{5^{4}} = \left(\frac{15}{5}\right)^{4} = 3^{4} = 81$

Answer: (A) $1$

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

$31^{0} = 1$

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

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

$\displaystyle 3^{-3} = \frac{1}{3^{3}} = \frac{1}{27}$

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

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

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

Answer: (C) $-1$

$87$ is odd and $(-1)^{\text{odd number}} = -1$.

$(-1)^{87} = -1$

Answer: (C) $64$

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

$(-2)^{6} = 2^{6} = 64$

Answer: (B) $4$

Given, $2^{x+2} = 64$

Dividing $64$ repeatedly by $2$: $64, 32, 16, 8, 4, 2$, so $64 = 2^{6}$.

$\Rightarrow 2^{x+2} = 2^{6}$

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

$\Rightarrow x+2 = 6$

$\Rightarrow x = 6 - 2$

$\Rightarrow x = 4$

Answer: (C) $14$

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

$\Rightarrow x-5 = 9$

$\Rightarrow x = 9 + 5$

$\Rightarrow x = 14$

Answer: (B) $25 \times 5^{x}$

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

$5^{x+2} = 5^{x} \times 5^{2} = 25 \times 5^{x}$

Answer: (A) $32m^{5}$

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

$(2m)^{5} = 2^{5} \times m^{5}$

$= 32 \times m^{5} = 32m^{5}$

Answer: (D) $x^{8}y^{11}$

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

$(x^{2}y)^{3} \times (xy^{4})^{2} = x^{6}y^{3} \times x^{2}y^{8}$

Using $a^m \times a^n = a^{m+n}$: $= x^{8}y^{11}$

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

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

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

$= 32$

Answer: (B) $4$

Given, $3^{m} \div 3^{-2} = 729$

Dividing $729$ repeatedly by $3$: $729, 243, 81, 27, 9, 3$, so $729 = 3^{6}$.

$\Rightarrow 3^{m-(-2)} = 3^{6}$ ($\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

$\Rightarrow 3^{m+2} = 3^{6}$

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

$\Rightarrow m+2 = 6$

$\Rightarrow m = 6 - 2$

$\Rightarrow m = 4$

Answer: (B) $1000$

$\displaystyle \left(10^{9} \div 10^{4}\right) \times 10^{-2} = \frac{10^{9}}{10^{4}} \times 10^{-2}$

$= 10^{9-4} \times 10^{-2} = 10^{5} \times 10^{-2}$

$= 10^{5-2} = 10^{3}$

$= 1000$

Answer: (A) $36$

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

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

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

$= 36$

Answer: (B) $\displaystyle \frac{1}{100}$

$\displaystyle \left(10^{6} \div 10^{8}\right)^{3} \times 10^{4} = \left(\frac{10^{6}}{10^{8}}\right)^{3} \times 10^{4}$

$= (10^{6-8})^{3} \times 10^{4} = (10^{-2})^{3} \times 10^{4}$

$= 10^{-6} \times 10^{4}$ ($(a^m)^n = a^{m \times n}$)

$= 10^{-6+4} = 10^{-2}$

$\displaystyle = \frac{1}{10^{2}} = \frac{1}{100}$

Answer: (D) $128p^{7}$

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

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

$= 16p^{4} \times 8p^{3}$

$= 128 \cdot p^{4+3} = 128p^{7}$