MCQ test, Set 3

Answer: (C) $625$

$\displaystyle \sqrt[3]{5^{12}} = 5^{\frac{12}{3}} = 5^{4} = 625$

Answer: (C) $72$

$\displaystyle \left(\frac{1}{2}\right)^{-3} = 2^{3} = 8$

$\displaystyle \left(\frac{1}{3}\right)^{-2} = 3^{2} = 9$

$8 \times 9 = 72$

Answer: (A) $5$

$= (5^{4})^{2} \times 5^{-7}$

$= 5^{8} \times 5^{-7}$

$= 5^{8 - 7} = 5^{1}$

$= 5$

Answer: (C) $63$

$\displaystyle (7^{2})^{\frac{1}{2}} \times (3^{3})^{\frac{2}{3}}$

$= 7 \times 3^{2}$

$= 7 \times 9 = 63$

Answer: (A) $4$

$\displaystyle \left(\frac{1}{32}\right)^{-\frac{2}{5}} = \left[\left(\frac{1}{2}\right)^{5}\right]^{-\frac{2}{5}}$

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

$= 4$

Answer: (B) $\displaystyle -\frac{5}{2}$

$\displaystyle \left(-\frac{8}{125}\right)^{-\frac{1}{3}} = \left[\left(-\frac{2}{5}\right)^{3}\right]^{-\frac{1}{3}}$

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

$\displaystyle = -\frac{5}{2}$

Answer: (B) $36$

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

$\displaystyle = 2^{\frac{8}{3} - \frac{2}{3}} \times 3^{\frac{8}{3} - \frac{2}{3}}$

$= 2^{2} \times 3^{2}$

$= 36$

Answer: (D) $x^{a^{2}-c^{2}}$

$= x^{(a+b)(a-b)} \cdot x^{(b+c)(b-c)} = x^{a^{2}-b^{2}} \cdot x^{b^{2}-c^{2}}$

$= x^{a^{2}-b^{2}+b^{2}-c^{2}} = x^{a^{2}-c^{2}}$

Answer: (C) $-60$

$a + b + c = 4 + 1 - 5 = 0$

So $a^{3} + b^{3} + c^{3} = 3abc = 3 \times 4 \times 1 \times (-5) = -60$

Answer: (D) $-1$

$4^{3x} = 64 = 4^{3} \Rightarrow 3x = 3 \Rightarrow x = 1$

$10^{y} = 0.01 = 10^{-2} \Rightarrow y = -2$

$x + y = 1 + (-2) = -1$

Answer: (A) $2$

$\displaystyle \frac{2^{7x}}{2^{3x}} = 2^{\frac{24}{3}}$

$\Rightarrow 2^{7x - 3x} = 2^{8}$

$\Rightarrow 4x = 8 \Rightarrow x = 2$

Answer: (D) $1$

$25 = 5^{2} \times 3^{0}$

$\Rightarrow 5^{x + 1} \times 3^{x - 1} = 5^{2} \times 3^{0}$

$\Rightarrow x + 1 = 2$ and $x - 1 = 0$

$\Rightarrow x = 1$ and $x = 1$

$\therefore x = 1$

Answer: (B) $\displaystyle \frac{1}{6}$

$(10^{y})^{2} = 6^{2}$

$\Rightarrow 10^{y} = 6$

$\displaystyle 10^{-y} = \frac{1}{10^{y}} = \frac{1}{6}$

Answer: (B) $2$

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

$\displaystyle \Rightarrow \left(\frac{5}{4}\right)^{-x} \cdot \left(\frac{5}{4}\right)^{3x} = \left(\frac{5}{4}\right)^{4}$

$\displaystyle \Rightarrow \left(\frac{5}{4}\right)^{2x} = \left(\frac{5}{4}\right)^{4}$

$\Rightarrow 2x = 4 \Rightarrow x = 2$

Answer: (C) $2$

$(2^{3})^{x} + 2^{3x + 1} = 192$

$\Rightarrow 2^{3x} + 2^{3x + 1} = 192$

$\Rightarrow 2^{3x} + 2 \times 2^{3x} = 192$

$\Rightarrow 2^{3x}\left(1 + 2\right) = 192$

$\Rightarrow 2^{3x} \times 3 = 192$

$\displaystyle \Rightarrow 2^{3x} = 192 \times \frac{1}{3} = 64 = 2^{6}$

$\Rightarrow 3x = 6 \Rightarrow x = 2$

Answer: (B) $8$

$(5^{3})^{x} = (5^{2})^{x+2} \Rightarrow 5^{3x} = 5^{2x+4}$

$\Rightarrow 3x = 2x + 4 \Rightarrow x = 4$

$2^{4-1} = 2^{3} = 8$

Answer: (D) $1$

Write $\displaystyle x^{a-b} = \frac{x^{a}}{x^{b}}$ and so on; multiply the first fraction by $x^{-a}$ above and below

$\displaystyle = \frac{x^{-a}}{x^{-a} + x^{-b} + x^{-c}} + \frac{x^{-b}}{x^{-b} + x^{-c} + x^{-a}} + \frac{x^{-c}}{x^{-c} + x^{-a} + x^{-b}}$

$\displaystyle = \frac{x^{-a} + x^{-b} + x^{-c}}{x^{-a} + x^{-b} + x^{-c}} = 1$

Answer: (D) $1$

$= x^{a^{3}-b^{3}} \cdot x^{b^{3}-c^{3}} \cdot x^{c^{3}-a^{3}}$

$= x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}} = x^{0} = 1$

Answer: (A) $8$

$\displaystyle b = a^{\frac{x}{2}}$

$\displaystyle c^{2} = b^{y} = a^{\frac{xy}{2}} \Rightarrow c = a^{\frac{xy}{4}}$

$\displaystyle a^{2} = c^{z} = a^{\frac{xyz}{4}} \Rightarrow \frac{xyz}{4} = 2 \Rightarrow xyz = 8$

Answer: (C) $\displaystyle \frac{1}{x} + \frac{2}{y}$

Let each $= k$: $\displaystyle 3 = k^{\frac{1}{x}}$, $\displaystyle 2 = k^{\frac{1}{y}}$, $\displaystyle 12 = k^{\frac{1}{z}}$

$12 = 3 \times 2^{2}$, so $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{2}{y}$

Answer: (A) $0.2$

$\displaystyle 0.0016 = \frac{16}{10000} = \left(\frac{2}{10}\right)^{4} = (0.2)^{4}$

$\displaystyle (0.0016)^{\frac{1}{4}} = \left[(0.2)^{4}\right]^{\frac{1}{4}} = 0.2$

Answer: (A) $\displaystyle \frac{2}{9}$

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

$\displaystyle \left(\frac{3}{\sqrt{2}}\right)^{4} \div \left(\frac{3}{\sqrt{2}}\right)^{6} = \frac{1}{\left(\frac{3}{\sqrt{2}}\right)^{2}}$

$\displaystyle = \frac{2}{9}$

Answer: (D) $7$

$(a + b)(a^{2} + b^{2} - ab) = a^{3} + b^{3}$ with $\displaystyle a = 2^{\frac{1}{3}}$, $\displaystyle b = 5^{\frac{1}{3}}$

$= 2 + 5 = 7$

Answer: (C) $25$

$\displaystyle x^{2 - 1.5} = x^{0.5} = x^{\frac{1}{2}} = 5$

$x = 5^{2} = 25$

Answer: (B) $3$

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

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

$= 3^{1} = 3$