MCQ test, Set 4

Answer: (D) $3$

$\displaystyle 3^{\frac{2}{7}} \times 3^{\frac{5}{7}} = 3^{\frac{2}{7} + \frac{5}{7}} = 3^{1} = 3$

Answer: (B) $\displaystyle \frac{4}{9}$

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

$\displaystyle \left(\frac{2}{3}\right)^{5-3} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$

Answer: (D) $4$

$= (2^{7})^{2} \times (2^{2})^{-6}$

$= 2^{14} \times 2^{-12}$

$= 2^{14 - 12} = 2^{2}$

$= 4$

Answer: (B) $2000$

$\displaystyle (2^{6})^{\frac{2}{3}} \times (5^{2})^{\frac{3}{2}}$

$= 2^{4} \times 5^{3}$

$= 16 \times 125 = 2000$

Answer: (D) $\displaystyle \frac{8}{7}$

$\displaystyle \left(\frac{49}{64}\right)^{-\frac{1}{2}} = \left[\left(\frac{7}{8}\right)^{2}\right]^{-\frac{1}{2}}$

$\displaystyle = \left(\frac{7}{8}\right)^{-1} = \left(\frac{8}{7}\right)^{1}$

$\displaystyle = \frac{8}{7}$

Answer: (A) $\displaystyle \frac{27}{125}$

$\displaystyle \left(\frac{25}{9}\right)^{-\frac{3}{2}} = \left[\left(\frac{5}{3}\right)^{2}\right]^{-\frac{3}{2}}$

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

$\displaystyle = \frac{27}{125}$

Answer: (D) $15$

$\displaystyle = \frac{(3 \times 5)^{\frac{1}{5}} \times (5^{2})^{\frac{2}{5}}}{3^{-\frac{4}{5}}}$

$\displaystyle = 3^{\frac{1}{5} + \frac{4}{5}} \times 5^{\frac{1}{5} + \frac{4}{5}}$

$= 3^{1} \times 5^{1}$

$= 15$

Answer: (A) $1$

Powers: $(3 - 5)(8) + (5 - 6)(11) + (6 - 3)(9)$

$= -16 - 11 + 27 = 0$

Value $= 7^{0} = 1$

Answer: (B) $3$

$x + y + z = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3xyz$

$\displaystyle \frac{3xyz}{xyz} = 3$

Answer: (C) $3$

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

$\displaystyle (2^{3})^{x} = \left(2^{\frac{3}{2}}\right)^{6}$

$\Rightarrow 2^{3x} = 2^{9}$

$\Rightarrow 3x = 9 \Rightarrow x = 3$

Answer: (C) $1$

$\displaystyle \frac{5^{3x}}{5^{x}} \times (5^{2})^{3} \times 5^{3} = 5^{11}$

$\Rightarrow 5^{2x} \times 5^{6} \times 5^{3} = 5^{11}$

$\Rightarrow 5^{2x + 9} = 5^{11}$

$\Rightarrow 2x + 9 = 11 \Rightarrow 2x = 2 \Rightarrow x = 1$

Answer: (A) $2$

$40 = 2^{3} \times 5^{1}$

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

$\Rightarrow 2x - 1 = 3$ and $x - 1 = 1$

$\Rightarrow x = 2$ and $x = 2$

$\therefore x = 2$

Answer: (B) $2$

$\displaystyle 5^{x-3} = \frac{5^{x}}{5^{3}}$

$\displaystyle = \frac{250}{125} = 2$

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

$\displaystyle (2^{5})^{x} = \frac{1}{2^{2}}$

$\Rightarrow 2^{5x} = 2^{-2}$

$\Rightarrow 5x = -2$

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

Answer: (C) $1$

$(5^{2})^{x + 1} - 5^{2x + 1} = 500$

$\Rightarrow 5^{2x + 2} - 5^{2x + 1} = 500$

$\Rightarrow 25 \times 5^{2x} - 5 \times 5^{2x} = 500$

$\Rightarrow 5^{2x}\left(25 - 5\right) = 500$

$\Rightarrow 5^{2x} \times 20 = 500$

$\displaystyle \Rightarrow 5^{2x} = 500 \times \frac{1}{20} = 25 = 5^{2}$

$\Rightarrow 2x = 2 \Rightarrow x = 1$

Answer: (B) $16$

$\displaystyle 5^{\frac{n}{2}} = 625 = 5^{4} \Rightarrow \frac{n}{2} = 4 \Rightarrow n = 8$

$\displaystyle 2^{\frac{8}{2}} = 2^{4} = 16$

Answer: (D) $1$

$\displaystyle \frac{1}{1 + \frac{1}{9}} + \frac{1}{1 + 9} = \frac{9}{10} + \frac{1}{10} = 1$

Answer: (A) $x^{a^{3}+b^{3}}$

$= x^{(a+b)(a^{2}-ab+b^{2})}$

$= x^{a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}} = x^{a^{3}+b^{3}}$

Answer: (C) $2$

$2 = 3^{y} = (5^{x})^{y} = 5^{xy}$

$25 = 2^{z} = (5^{xy})^{z} = 5^{xyz}$

$5^{xyz} = 25 = 5^{2}$

$\Rightarrow xyz = 2$

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

Let each $= k$: $\displaystyle 5 = k^{\frac{1}{x}}$, $\displaystyle 3 = k^{\frac{1}{y}}$, $\displaystyle 45 = k^{\frac{1}{z}}$

$45 = 5 \times 3^{2}$, so $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{2}{y} = \frac{y + 2x}{xy}$, $\displaystyle z = \frac{xy}{y + 2x}$

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

$\displaystyle \sqrt[n]{a^{m}} = a^{\frac{m}{n}}$

$\displaystyle \sqrt[4]{81^{-3}} = 81^{-\frac{3}{4}} = (3^{4})^{-\frac{3}{4}}$

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

Answer: (C) $\displaystyle \frac{9}{5}$

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

$\displaystyle \left(\frac{\sqrt{5}}{3}\right)^{6} \div \left(\frac{\sqrt{5}}{3}\right)^{8} = \frac{1}{\left(\frac{\sqrt{5}}{3}\right)^{2}}$

$\displaystyle = \frac{9}{5}$

Answer: (C) $\displaystyle \frac{10}{3}$

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

$\displaystyle ab = x^{\frac{1}{3}} \cdot x^{-\frac{1}{3}} = x^{0} = 1$

So the value $\displaystyle = (x^{\frac{1}{3}})^{3} + (x^{-\frac{1}{3}})^{3} = x + \frac{1}{x}$

$\displaystyle = \frac{1}{3} + 3 = \frac{10}{3}$

Answer: (A) $8$

$= (4096)^{0.15 + 0.1} = (4096)^{0.25}$

$\displaystyle 0.25 = \frac{1}{4}$

$\displaystyle = (2^{12})^{\frac{1}{4}} = 8$

Answer: (A) $7$

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

$\displaystyle = 7^{(-1) \times 2 \times (-3) \times \frac{1}{6}}$

$= 7^{1}$

$= 7$