Exponents (Level 1): MCQ test, Set 3

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

$\displaystyle \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} =$
A $\displaystyle \frac{3}{7^{3}}$ B $\displaystyle \left(\frac{3}{7}\right)^{3}$ C $\displaystyle \frac{3^{3}}{7}$ D $\displaystyle \frac{9}{7}$

Answer and solution

Answer: (B) $\displaystyle \left(\frac{3}{7}\right)^{3}$

$\displaystyle \frac{3}{7}$ is multiplied by itself 3 times.

$\displaystyle \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} = \left(\frac{3}{7}\right)^{3}$
$7^{3} =$
A $21$ B $2187$ C $49$ D $343$

Answer and solution

Answer: (D) $343$

$7^{3} = 7 \times 7 \times 7$

$= 343$
Which is the smallest of $2^{5}$ , $5^{2}$ , $3^{3}$ , $6^{2}$ ?
A $2^{5}$ B $3^{3}$ C $5^{2}$ D $6^{2}$

Answer and solution

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

$2^{5} = 32$ , $5^{2} = 25$ , $3^{3} = 27$ , $6^{2} = 36$

The smallest is $5^{2}$ .
$1024$ written as a power of $2$ is
A $2^{9}$ B $2^{11}$ C $2^{10}$ D $10^{2}$

Answer and solution

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

Dividing repeatedly by $2$ : $1024, 512, 256, 128, 64, 32, 16, 8, 4, 2$ , that is 10 times.

$1024 = 2^{10}$
$\displaystyle \left(\frac{1}{3}\right)^{5} \times \left(\frac{1}{3}\right)^{4} =$
A $\displaystyle \left(\frac{1}{9}\right)^{9}$ B $\displaystyle \left(\frac{1}{3}\right)^{1}$ C $\displaystyle \left(\frac{1}{3}\right)^{20}$ D $\displaystyle \left(\frac{1}{3}\right)^{9}$

Answer and solution

Answer: (D) $\displaystyle \left(\frac{1}{3}\right)^{9}$

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

$\displaystyle \left(\frac{1}{3}\right)^{5} \times \left(\frac{1}{3}\right)^{4} = \left(\frac{1}{3}\right)^{5+4} = \left(\frac{1}{3}\right)^{9}$
$\displaystyle \frac{x^{15}}{x^{6}} =$
A $x^{21}$ B $x^{9}$ C $1$ D $\displaystyle x^{\frac{5}{2}}$

Answer and solution

Answer: (B) $x^{9}$

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

$\displaystyle \frac{x^{15}}{x^{6}} = x^{15-6} = x^{9}$
$(10^{3})^{8} =$
A $30^{8}$ B $80^{3}$ C $10^{11}$ D $10^{24}$

Answer and solution

Answer: (D) $10^{24}$

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

$(10^{3})^{8} = 10^{3 \times 8} = 10^{24}$
$a^{5} \times b^{5} =$
A $(ab)^{5}$ B $(ab)^{10}$ C $ab^{5}$ D $(a+b)^{5}$

Answer and solution

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

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

$a^{5} \times b^{5} = (a \times b)^{5} = (ab)^{5}$
$\displaystyle \frac{28^{3}}{7^{3}} =$
A $4$ B $64$ C $12$ D $9261$

Answer and solution

Answer: (B) $64$

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

$\displaystyle \frac{28^{3}}{7^{3}} = \left(\frac{28}{7}\right)^{3} = 4^{3} = 64$
$5^{0} + 5^{1} =$
A $2$ B $10$ C $5$ D $6$

Answer and solution

Answer: (D) $6$

$5^{0} = 1$ , $5^{1} = 5$

$5^{0} + 5^{1} = 1 + 5 = 6$
$\displaystyle \frac{1}{8^{5}} =$
A $-8^{5}$ B $8^{5}$ C $8^{-5}$ D $5^{-8}$

Answer and solution

Answer: (C) $8^{-5}$

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

$\displaystyle \frac{1}{8^{5}} = 8^{-5}$
$\displaystyle \left(\frac{1}{4}\right)^{-3} =$
A $\displaystyle -\frac{1}{64}$ B $\displaystyle -\frac{3}{4}$ C $64$ D $\displaystyle \frac{1}{64}$

Answer and solution

Answer: (C) $64$

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

$\displaystyle \left(\frac{1}{4}\right)^{-3} = 4^{3} = 64$
$(-1)^{150} =$
A $1$ B $-150$ C $150$ D $-1$

Answer and solution

Answer: (A) $1$

$150$ is even and $(-1)^{\text{even number}} = 1$ .

$(-1)^{150} = 1$
$(-6)^{4} =$
A $24$ B $-1296$ C $1296$ D $-24$

Answer and solution

Answer: (C) $1296$

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

$(-6)^{4} = 6^{4} = 1296$
If $6^{x+8} = 6^{11}$ , then $x =$
A $3$ B $-3$ C $11$ D $19$

Answer and solution

Answer: (A) $3$

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

$\Rightarrow x+8 = 11$

$\Rightarrow x = 11 - 8$

$\Rightarrow x = 3$
If $4^{2m-3} = 64$ , then $m =$
A $6$ B $3$ C $\displaystyle \frac{67}{2}$ D $0$

Answer and solution

Answer: (B) $3$

Given, $4^{2m-3} = 64$

Dividing $64$ repeatedly by $4$ : $64, 16, 4$ , so $64 = 4^{3}$ .

$\Rightarrow 4^{2m-3} = 4^{3}$

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

$\Rightarrow 2m-3 = 3$

$\Rightarrow 2m = 3 + 3$

$\Rightarrow 2m = 6$

$\displaystyle \Rightarrow m = \frac{6}{2} = 3$
$3^{3x} =$
A $27^{x}$ B $3^{x+3}$ C $27x$ D $9^{x}$

Answer and solution

Answer: (A) $27^{x}$

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

$3^{3x} = (3^{3})^{x} = 27^{x}$
$(3y)^{4} =$
A $12y^{4}$ B $81y^{5}$ C $3y^{4}$ D $81y^{4}$

Answer and solution

Answer: (D) $81y^{4}$

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

$(3y)^{4} = 3^{4} \times y^{4}$

$= 81 \times y^{4} = 81y^{4}$
$6^{3} \times 6^{7} \times 6^{10} =$
A $6^{20}$ B $18^{20}$ C $6^{210}$ D $6^{10}$

Answer and solution

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

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

$6^{3} \times 6^{7} \times 6^{10} = 6^{3+7+10} = 6^{20}$
$(5^{-1})^{-3} =$
A $5^{4}$ B $5^{-4}$ C $5^{3}$ D $5^{-3}$

Answer and solution

Answer: (C) $5^{3}$

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

$(5^{-1})^{-3} = 5^{(-1) \times (-3)} = 5^{3}$

$= 125$
If $(-7)^{m-2} \times (-7)^{5} = (-7)^{10}$ , then $m =$
A $13$ B $-13$ C $7$ D $-7$

Answer and solution

Answer: (C) $7$

Given, $(-7)^{m-2} \times (-7)^{5} = (-7)^{10}$

$\Rightarrow (-7)^{m-2+5} = (-7)^{10}$ ( $a^m \times a^n = a^{m+n}$ )

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

$\Rightarrow m+3 = 10$

$\Rightarrow m = 10 - 3$

$\Rightarrow m = 7$
$\left(2^{0} + 6^{-1}\right) \times 6^{2} =$
A $42$ B $-180$ C $78$ D $6$

Answer and solution

Answer: (A) $42$

$\displaystyle \left(2^{0} + 6^{-1}\right) \times 6^{2} = \left(1 + \frac{1}{6}\right) \times 36$

$\displaystyle = \frac{7}{6} \times 36$

$= 42$
$\displaystyle 4^{\frac{7}{2}} \div 4^{\frac{3}{2}} =$
A $\displaystyle \frac{1}{16}$ B $16$ C $1024$ D $\displaystyle 4^{\frac{7}{3}}$

Answer and solution

Answer: (B) $16$

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

$\displaystyle 4^{\frac{7}{2}} \div 4^{\frac{3}{2}} = 4^{\frac{7}{2} - \frac{3}{2}}$

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

$= 16$
$\left(5^{-1} - 6^{-1}\right) \div 3^{-1} =$
A $\displaystyle \frac{11}{10}$ B $\displaystyle \frac{1}{90}$ C $4$ D $\displaystyle \frac{1}{10}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{10}$

$\displaystyle \left(5^{-1} - 6^{-1}\right) \div 3^{-1} = \left(\frac{1}{5} - \frac{1}{6}\right) \div \frac{1}{3}$

$\displaystyle = \frac{6 - 5}{30} \div \frac{1}{3} = \frac{1}{30} \div \frac{1}{3}$

$\displaystyle = \frac{1}{30} \times 3 = \frac{1}{10}$
$\displaystyle \left(\frac{5}{2}\right)^{-3} \times \left(\frac{3}{5}\right)^{-1} \times \left(\frac{2}{3}\right)^{-2} =$
A $\displaystyle \frac{3}{50}$ B $\displaystyle \frac{6}{25}$ C $\displaystyle \frac{25}{6}$ D $150$

Answer and solution

Answer: (B) $\displaystyle \frac{6}{25}$

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

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

$\displaystyle = \frac{2^{3}}{5^{3}} \times \frac{5^{1}}{3^{1}} \times \frac{3^{2}}{2^{2}}$

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

$\displaystyle = \frac{1}{25} \times 2 \times 3$

$\displaystyle = \frac{6}{25}$

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MCQ test, Set 3