MCQ test, Set 10

Answer: (B) $\sqrt{121}$

$\sqrt{121} = \sqrt{11 \times 11} = 11$, a rational number, so it is not a surd.

The other three are roots of numbers that are not perfect powers, so they are surds.

Answer: (D) $\sqrt[3]{46}$ only

$3^3 = 27$ and $4^3 = 64$, so $46$ is not a perfect cube. So $\sqrt[3]{46}$ is a surd.

$\pi$ is irrational but it is not a root of a rational number, so it is not a surd.

Answer: (A) $12$

$3^2 = 9$ and $4^2 = 16$, so $12$ is not a perfect square. So $6\sqrt{12}$ is a mixed surd.

For the others $\sqrt{n}$ is a whole number, so the product is rational.

Answer: (A) $-\sqrt{22} + 15$

Only the sign of the surd part changes: the conjugate of $\sqrt{22} + 15$ is $-\sqrt{22} + 15$.

Answer: (C) $\sqrt{10}$

$\sqrt{2} \times \sqrt{5} = \sqrt{2 \times 5} = \sqrt{10}$ (multiply the numbers under the roots).

Answer: (A) $3$

$\displaystyle \frac{\sqrt{135}}{\sqrt{15}} = \sqrt{\frac{135}{15}}$

$= \sqrt{9} = \sqrt{3 \times 3} = 3$

Answer: (B) $\displaystyle 15^{\frac{1}{4}}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$, so $\displaystyle \sqrt[4]{15} = 15^{\frac{1}{4}}$

Answer: (C) $10$

$\sqrt[3]{25} \times \sqrt[3]{40} = \sqrt[3]{25 \times 40}$

$= \sqrt[3]{1000} = \sqrt[3]{10^3} = 10$

Answer: (D) $6$

$\displaystyle \frac{\sqrt[3]{432}}{\sqrt[3]{2}} = \sqrt[3]{\frac{432}{2}}$

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

Answer: (B) $90$

$(3\sqrt{10})^2 = 3^2 \times (\sqrt{10})^2$

$= 9 \times 10 = 90$

Answer: (C) $5 - 2\sqrt{10}$

$(3 - \sqrt{10})(5 + \sqrt{10}) = 15 + 3\sqrt{10} - 5\sqrt{10} - 10$

$= 5 - 2\sqrt{10}$

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

$8\sqrt{7} \times \sqrt{7} = 8 \times 7 = 56$, a rational number.

Answer: (B) $19$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{10}$, $y = \sqrt[3]{9}$.

Product $= (\sqrt[3]{10})^3 + (\sqrt[3]{9})^3 = 10 + 9 = 19$

Answer: (D) $10\sqrt{5}$

$\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$

$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$

$6\sqrt{5} + 4\sqrt{5} = 10\sqrt{5}$ (add like surds)

Answer: (C) $9\sqrt{3}$

$(\sqrt{3})^{5} = [(\sqrt{3})^2]^{2} \times \sqrt{3} = 3^{2} \times \sqrt{3} = 9\sqrt{3}$

Answer: (B) $8 - 2\sqrt{15}$

$(\sqrt{5} - \sqrt{3})^2 = 5 + 3 - 2\sqrt{15} = 8 - 2\sqrt{15}$

Answer: (C) $3$

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

$= (\sqrt{10})^2 - (\sqrt{7})^2 = 10 - 7 = 3$

Answer: (A) $\displaystyle \frac{\sqrt{10} - 2}{6}$

Multiply above and below by $2 - \sqrt{10}$: $(2 + \sqrt{10})(2 - \sqrt{10}) = 4 - 10 = -6$

$\displaystyle = \frac{\sqrt{10} - 2}{6}$

Answer: (D) $13$

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

Numerator: $9 + 3\sqrt{6} + 9\sqrt{6} + 18 = 27 + 12\sqrt{6}$

Denominator: $3^2 - (\sqrt{6})^2 = 9 - 6 = 3$

$\displaystyle \frac{27 + 12\sqrt{6}}{3} = 9 + 4\sqrt{6}$

So $a = 9$, $b = 4$

Answer: (B) $3$

$\displaystyle = \frac{(5 + \sqrt{5})^2 + (5 - \sqrt{5})^2}{(5 + \sqrt{5})(5 - \sqrt{5})} = \frac{60}{20} = 3$

The $\sqrt{5}$ terms cancel and $(5 + \sqrt{5})(5 - \sqrt{5}) = 20$.

Answer: (D) $\displaystyle \frac{18}{71}$

$\displaystyle = \frac{(9 - \sqrt{10}) + (9 + \sqrt{10})}{(9 + \sqrt{10})(9 - \sqrt{10})} = \frac{18}{71}$

Answer: (A) $20$

$x = 10 + 3\sqrt{11}$ and $\displaystyle \frac{1}{x} = 10 - 3\sqrt{11}$

$\displaystyle x + \frac{1}{x} = 20$

Answer: (C) $\sqrt{13} + \sqrt{7}$

$\sqrt{20 + 2\sqrt{91}} = \sqrt{13 + 7 + 2\sqrt{13 \times 7}}$

$= \sqrt{(\sqrt{13})^2 + (\sqrt{7})^2 + 2(\sqrt{13})(\sqrt{7})}$

$= \sqrt{(\sqrt{13} + \sqrt{7})^2} = \sqrt{13} + \sqrt{7}$

Answer: (B) $\sqrt{14} - 2$

$\sqrt{224} = \sqrt{4 \times 56} = 2\sqrt{56}$

$\sqrt{18 - \sqrt{224}} = \sqrt{14 + 4 - 2\sqrt{14 \times 4}}$

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

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

Answer: (D) $\displaystyle \frac{\sqrt{17} - 1}{\sqrt{2}}$

$\displaystyle \sqrt{9 - \sqrt{17}} = \sqrt{\frac{2(9 - \sqrt{17})}{2}} = \frac{\sqrt{18 - 2\sqrt{17}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{17 + 1 - 2\sqrt{17 \times 1}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{17})^2 + (1)^2 - 2(\sqrt{17})(1)}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{17} - 1)^2}}{\sqrt{2}} = \frac{\sqrt{17} - 1}{\sqrt{2}}$