MCQ test, Set 9

Answer: (D) $15^{3}$

$15$ is multiplied by itself 3 times.

$15 \times 15 \times 15 = 15^{3}$

Answer: (B) $256$

$16^{2} = 16 \times 16$

$= 256$

Answer: (B) $7^{2}$

$2^{7} = 128$, $7^{2} = 49$, $3^{4} = 81$, $5^{3} = 125$

The smallest is $7^{2}$.

Answer: (C) $11^{2}$

Dividing repeatedly by $11$: $121, 11$, that is 2 times.

$121 = 11^{2}$

Answer: (A) $5^{20}$

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

$5^{10} \times 5^{10} = 5^{10+10} = 5^{20}$

Answer: (B) $100^{6}$

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

$100^{9} \div 100^{3} = 100^{9-3} = 100^{6}$

Answer: (D) $m^{27}$

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

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

Answer: (C) $(16)^{5}$

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

$2^{5} \times 8^{5} = (2 \times 8)^{5} = (16)^{5}$

Answer: (C) $32$

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

$\displaystyle \frac{22^{5}}{11^{5}} = \left(\frac{22}{11}\right)^{5} = 2^{5} = 32$

Answer: (B) $1$

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

$-8 \ne 0$, so $(-8)^{0} = 1$

Answer: (D) $6^{-7}$

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

$\displaystyle \frac{1}{6^{7}} = 6^{-7}$

Answer: (D) $\displaystyle \frac{25}{36}$

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

$\displaystyle \left(\frac{6}{5}\right)^{-2} = \left(\frac{5}{6}\right)^{2} = \frac{25}{36}$

Answer: (A) an even number

The notes: $(-1)^{\text{even number}} = 1$, e.g. $(-1)^{4} = 1$.

Answer: (C) $81$

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

$(-9)^{2} = 9^{2} = 81$

Answer: (B) $a$ cannot be $1$

Every power of $1$ is $1$, so equal powers of 1 need not have equal exponents.

The rule holds only when $a \ne -1, 0, 1$.

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

Given, $11^{2x-1} = 121$

Dividing $121$ repeatedly by $11$: $121, 11$, so $121 = 11^{2}$.

$\Rightarrow 11^{2x-1} = 11^{2}$

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

$\Rightarrow 2x-1 = 2$

$\Rightarrow 2x = 2 + 1$

$\Rightarrow 2x = 3$

$\displaystyle \Rightarrow x = \frac{3}{2}$

Answer: (A) $32^{x}$

Using $a^{mn} = (a^m)^n$

$2^{5x} = (2^{5})^{x} = 32^{x}$

Answer: (A) $\displaystyle \frac{x^{8}}{9}$

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

$\displaystyle \left(\frac{x^{4}}{3}\right)^{2} = \frac{(x^{4})^{2}}{3^{2}}$

$\displaystyle = \frac{x^{8}}{9}$

Answer: (C) $13^{19}$

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

$13^{4} \times 13^{4} \times 13^{11} = 13^{4+4+11} = 13^{19}$

Answer: (A) $\displaystyle \frac{1}{30}$

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

$6^{-1} \times 5^{-1} = (6 \times 5)^{-1} = 30^{-1}$

$\displaystyle = \frac{1}{30^{1}} = \frac{1}{30}$

Answer: (D) $2$

Given, $2^{x} \div 2^{-3} = 32$

Dividing $32$ repeatedly by $2$: $32, 16, 8, 4, 2$, so $32 = 2^{5}$.

$\Rightarrow 2^{x-(-3)} = 2^{5}$ ($\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

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

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

$\Rightarrow x+3 = 5$

$\Rightarrow x = 5 - 3$

$\Rightarrow x = 2$

Answer: (A) $3$

Using $(-1)^{\text{odd number}} = -1, \; (-1)^{\text{even number}} = 1$

$(-1)^{2} + (-1)^{8} + (-1)^{20} = 1 + 1 + 1$

$= 3$

Answer: (B) $9$

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

$\displaystyle \frac{9^{\frac{11}{2}} \times 9^{-\frac{5}{2}}}{9^{2}} = \frac{9^{\frac{11}{2} + (-\frac{5}{2})}}{9^{2}}$

$\displaystyle = \frac{9^{\frac{6}{2}}}{9^{2}} = \frac{9^{3}}{9^{2}}$

$= 9^{3-2} = 9^{1}$

$= 9$

Answer: (B) $-1000$

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

$(-10)^{-6} \div (-10)^{-9} = (-10)^{-6-(-9)} = (-10)^{-6+9} = (-10)^{3}$

$= -1000$

Answer: (D) $100z^{12}$

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

$(2z^{5})^{2} \times (5z)^{2} = 2^{2} \cdot (z^{5})^{2} \times 5^{2} \cdot z^{2}$

$= 4z^{10} \times 25z^{2}$

$= 100 \cdot z^{10+2} = 100z^{12}$