MCQ test, Set 10

Answer: (A) $\displaystyle \frac{1}{9}$

$\displaystyle \left(\sqrt{3}\right)^{-4} = \left(3^{\frac{1}{2}}\right)^{-4} = 3^{-2} = \frac{1}{9}$

Answer: (B) $0^{0}$

$a^{0} = 1$ only when $a \ne 0$; the notes say $0^{0}$ is undefined

Answer: (C) $36$

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

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

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

$= 36$

Answer: (D) $54$

$\displaystyle (6^{2})^{\frac{3}{2}} \div (2^{7})^{\frac{2}{7}}$

$= 6^{3} \div 2^{2}$

$= 216 \div 4 = 54$

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

$\displaystyle \left(-\frac{32}{243}\right)^{-\frac{1}{5}} = \left[\left(-\frac{2}{3}\right)^{5}\right]^{-\frac{1}{5}}$

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

$\displaystyle = -\frac{3}{2}$

Answer: (D) $1$

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

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

$3 - 2 = 1$

Answer: (B) $2$

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

$\displaystyle = 2^{\frac{8}{5} + \frac{3}{5} - \frac{6}{5}} \times 3^{\frac{3}{5} - \frac{3}{5}}$

$= 2^{1} \times 3^{0}$

$= 2$

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

$= x^{a^{2}-b^{2}} \div x^{a^{2}-c^{2}}$

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

Answer: (A) $90$

$a + b + c = -2 - 3 + 5 = 0$

So $a^{3} + b^{3} + c^{3} = 3abc = 3 \times (-2) \times (-3) \times 5 = 90$

Answer: (D) $-2$

$\displaystyle \frac{7^{4x}}{7^{x}} \times (7^{2})^{2} \times 7^{3} = 7^{1}$

$\Rightarrow 7^{3x} \times 7^{4} \times 7^{3} = 7^{1}$

$\Rightarrow 7^{3x + 7} = 7^{1}$

$\Rightarrow 3x + 7 = 1 \Rightarrow 3x = -6 \Rightarrow x = -2$

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

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

$\displaystyle (7^{2})^{x} = \left(7^{\frac{3}{2}}\right)^{2}$

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

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

Answer: (A) $5$

$52 = 2^{2} \times 13^{1}$

$\Rightarrow 2^{x - 3} \times 13^{x - 4} = 2^{2} \times 13^{1}$

$\Rightarrow x - 3 = 2$ and $x - 4 = 1$

$\Rightarrow x = 5$ and $x = 5$

$\therefore x = 5$

Answer: (A) $\displaystyle \frac{1}{5}$

$(7^{y})^{2} = 5^{2}$

$\Rightarrow 7^{y} = 5$

$\displaystyle 7^{-y} = \frac{1}{7^{y}} = \frac{1}{5}$

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

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

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

$\Rightarrow 2x = -3$

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

Answer: (B) $2$

$(2^{2})^{x} + 2^{2x + 1} = 48$

$\Rightarrow 2^{2x} + 2^{2x + 1} = 48$

$\Rightarrow 2^{2x} + 2 \times 2^{2x} = 48$

$\Rightarrow 2^{2x}\left(1 + 2\right) = 48$

$\Rightarrow 2^{2x} \times 3 = 48$

$\displaystyle \Rightarrow 2^{2x} = 48 \times \frac{1}{3} = 16 = 2^{4}$

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

Answer: (C) $125$

$(2^{2})^{x+1} = 2^{6} \Rightarrow 2x + 2 = 6 \Rightarrow x = 2$

$5^{2+1} = 5^{3} = 125$

Answer: (B) $1$

$\displaystyle 4^{\frac{1}{2}} = 2$

$\displaystyle \frac{1}{1 + 2} + \frac{1}{1 + \frac{1}{2}} = \frac{1}{3} + \frac{2}{3} = 1$

Answer: (B) $x^{p^{3}+2q^{3}+r^{3}}$

$= x^{(p+q)(p^{2}-pq+q^{2})} \cdot x^{(q+r)(q^{2}-qr+r^{2})}$

$= x^{p^{3}+q^{3}} \cdot x^{q^{3}+r^{3}} = x^{p^{3}+2q^{3}+r^{3}}$

Answer: (C) $-3$

$5 = 3^{q} = (2^{p})^{q} = 2^{pq}$

$7 = 5^{r} = (2^{pq})^{r} = 2^{pqr}$

$\displaystyle \frac{1}{8} = 7^{t} = (2^{pqr})^{t} = 2^{pqrt}$

$\displaystyle 2^{pqrt} = \frac{1}{8} = 2^{-3}$

$\Rightarrow pqrt = -3$

Answer: (A) $\displaystyle \frac{1}{x} + \frac{3}{y}$

Let each $= k$: $\displaystyle 7 = k^{\frac{1}{x}}$, $\displaystyle 2 = k^{\frac{1}{y}}$, $\displaystyle 56 = k^{\frac{1}{z}}$

$56 = 7 \times 2^{3}$, so $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{3}{y}$

Answer: (A) $0.008$

$\displaystyle 0.04 = \frac{4}{100} = \left(\frac{2}{10}\right)^{2} = (0.2)^{2}$

$\displaystyle (0.04)^{\frac{3}{2}} = \left[(0.2)^{2}\right]^{\frac{3}{2}} = (0.2)^{3} = 0.008$

Answer: (C) $\displaystyle \frac{9}{10}$

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

$\displaystyle \left(\frac{\sqrt{10}}{3}\right)^{4} \div \left(\frac{\sqrt{10}}{3}\right)^{6} = \frac{1}{\left(\frac{\sqrt{10}}{3}\right)^{2}}$

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

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

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

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

Answer: (A) $27$

$= (9)^{0.75 + 0.75} = (9)^{1.5}$

$\displaystyle 1.5 = \frac{3}{2}$

$\displaystyle = (3^{2})^{\frac{3}{2}} = 27$

Answer: (D) $1331$

$\displaystyle \frac{1}{11} = 11^{-1}$

$\displaystyle = 11^{(-1) \times 2 \times (-3) \times \frac{1}{2}}$

$= 11^{3}$

$= 1331$