MCQ test, Set 4

Answer: (B) $216$

$\sqrt[3]{216} = \sqrt[3]{6 \times 6 \times 6} = 6$

The other three are not perfect cubes, so their roots are surds.

Answer: (B) Every irrational number is a surd.

$\pi$ and $e$ are irrational numbers but not surds.

So 'every irrational number is a surd' is false.

Answer: (D) $5\sqrt[3]{29}$

$5\sqrt[3]{29} = 5 \cdot \sqrt[3]{29}$: a rational number times a surd.

$\sqrt[3]{44}$ is a pure surd; $9\sqrt{100} = 90$ and $\sqrt[4]{16} = 2$ are rational.

Answer: (C) $11 + \sqrt{21}$

Only the sign of the surd part changes: the conjugate of $11 - \sqrt{21}$ is $11 + \sqrt{21}$.

Answer: (D) $12$

$\sqrt{6} \times \sqrt{24} = \sqrt{144} = \sqrt{12 \times 12} = 12$

Answer: (C) $8$

$\displaystyle \frac{\sqrt{192}}{\sqrt{3}} = \sqrt{\frac{192}{3}}$

$= \sqrt{64} = \sqrt{8 \times 8} = 8$

Answer: (D) $\sqrt[6]{17}$

$\displaystyle a^{\frac{1}{n}} = \sqrt[n]{a}$, so $\displaystyle 17^{\frac{1}{6}} = \sqrt[6]{17}$

Answer: (B) $3$

$\displaystyle \frac{\sqrt[5]{729}}{\sqrt[5]{3}} = \sqrt[5]{\frac{729}{3}}$

$= \sqrt[5]{243} = \sqrt[5]{3^5} = 3$

Answer: (A) $28$

$(2\sqrt{7})^2 = 2^2 \times (\sqrt{7})^2$

$= 4 \times 7 = 28$

Answer: (C) $17 + 7\sqrt{5}$

$(4 + \sqrt{5})(3 + \sqrt{5}) = 12 + 4\sqrt{5} + 3\sqrt{5} + 5$

$= 17 + 7\sqrt{5}$

Answer: (A) $\sqrt{14}$

$5\sqrt{14} \times \sqrt{14} = 5 \times 14 = 70$, a rational number.

Answer: (C) $2$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{13}$, $y = \sqrt[3]{11}$.

Product $= (\sqrt[3]{13})^3 - (\sqrt[3]{11})^3 = 13 - 11 = 2$

Answer: (A) $10\sqrt{2}$

$\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}$

Answer: (B) $225$

$(\sqrt{15})^4 = [(\sqrt{15})^2]^2 = 15^2 = 225$

Answer: (A) $21 + 2\sqrt{110}$

$(\sqrt{11} + \sqrt{10})^2 = 11 + 10 + 2\sqrt{110} = 21 + 2\sqrt{110}$

Answer: (C) $5$

$(a + b)(a - b) = a^2 - b^2$

$= (\sqrt{19})^2 - (\sqrt{14})^2 = 19 - 14 = 5$

Answer: (C) $-166$

$(a + b)(a - b) = a^2 - b^2$

$= 3^2 - (5\sqrt{7})^2 = 9 - 175 = -166$

Answer: (B) $\displaystyle \frac{9 - \sqrt{11}}{70}$

Multiply above and below by $9 - \sqrt{11}$: $(9 + \sqrt{11})(9 - \sqrt{11}) = 81 - 11 = 70$

$\displaystyle = \frac{9 - \sqrt{11}}{70}$

Answer: (D) $5$

$\displaystyle \frac{6 + 3\sqrt{7}}{4 - \sqrt{7}} = \frac{(6 + 3\sqrt{7})(4 + \sqrt{7})}{(4 - \sqrt{7})(4 + \sqrt{7})}$

Numerator: $24 + 6\sqrt{7} + 12\sqrt{7} + 21 = 45 + 18\sqrt{7}$

Denominator: $4^2 - (\sqrt{7})^2 = 16 - 7 = 9$

$\displaystyle \frac{45 + 18\sqrt{7}}{9} = 5 + 2\sqrt{7}$

So $a = 5$, $b = 2$

Answer: (D) $\displaystyle \frac{24}{7}$

$\displaystyle = \frac{(\sqrt{19} + \sqrt{5})^2 + (\sqrt{19} - \sqrt{5})^2}{(\sqrt{19} + \sqrt{5})(\sqrt{19} - \sqrt{5})} = \frac{48}{14} = \frac{24}{7}$

The $\sqrt{95}$ terms cancel and $(\sqrt{19} + \sqrt{5})(\sqrt{19} - \sqrt{5}) = 19 - 5 = 14$.

Answer: (C) $\sqrt{7}$

$\displaystyle = \frac{(\sqrt{7} - \sqrt{5}) + (\sqrt{7} + \sqrt{5})}{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})} = \frac{2\sqrt{7}}{2} = \sqrt{7}$

Answer: (A) $30$

$x^3 = 15 - 4\sqrt{14}$, so $\displaystyle \frac{1}{x^3} = \frac{1}{15 - 4\sqrt{14}} = 15 + 4\sqrt{14}$

$\displaystyle x^3 + \frac{1}{x^3} = (15 - 4\sqrt{14}) + (15 + 4\sqrt{14}) = 30$

Answer: (C) $\sqrt{6} + \sqrt{3}$

$\sqrt{9 + 2\sqrt{18}} = \sqrt{6 + 3 + 2\sqrt{6 \times 3}}$

$= \sqrt{(\sqrt{6})^2 + (\sqrt{3})^2 + 2(\sqrt{6})(\sqrt{3})}$

$= \sqrt{(\sqrt{6} + \sqrt{3})^2} = \sqrt{6} + \sqrt{3}$

Answer: (B) $\sqrt{6} - \sqrt{5}$

$\sqrt{11 - 2\sqrt{30}} = \sqrt{6 + 5 - 2\sqrt{6 \times 5}}$

$= \sqrt{(\sqrt{6})^2 + (\sqrt{5})^2 - 2(\sqrt{6})(\sqrt{5})}$

$= \sqrt{(\sqrt{6} - \sqrt{5})^2} = \sqrt{6} - \sqrt{5}$

Answer: (D) $\displaystyle \frac{\sqrt{23} + 3}{\sqrt{2}}$

$\displaystyle \sqrt{16 + \sqrt{207}} = \sqrt{\frac{2(16 + \sqrt{207})}{2}} = \frac{\sqrt{32 + 2\sqrt{207}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{23 + 9 + 2\sqrt{23 \times 9}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23})^2 + (3)^2 + 2(\sqrt{23})(3)}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23} + 3)^2}}{\sqrt{2}} = \frac{\sqrt{23} + 3}{\sqrt{2}}$