Rationalization and Surds: MCQ test, Set 1

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

Which of the following is a rational number?
A $\sqrt[3]{64}$ B $\sqrt{54}$ C $\sqrt{12}$ D $\sqrt[4]{17}$

Answer and solution

Answer: (A) $\sqrt[3]{64}$

$\sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4} = 4$ , 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.
$\sqrt[6]{15}$ and $e$ are both irrational. Which of them is a surd?
A $e$ only B $\sqrt[6]{15}$ only C Both D Neither

Answer and solution

Answer: (B) $\sqrt[6]{15}$ only

$1^6 = 1$ and $2^6 = 64$ , so $15$ is not a perfect sixth power. So $\sqrt[6]{15}$ is a surd.

$e$ is irrational but it is not a root of a rational number, so it is not a surd.
$9\sqrt[3]{n}$ is a mixed surd when $n$ is
A $64$ B $729$ C $512$ D $19$

Answer and solution

Answer: (D) $19$

$2^3 = 8$ and $3^3 = 27$ , so $19$ is not a perfect cube. So $9\sqrt[3]{19}$ is a mixed surd.

For the others $\sqrt[3]{n}$ is a whole number, so the product is rational.
The conjugate surd of $12 - \sqrt{30}$ is
A $-12 - \sqrt{30}$ B $12 + \sqrt{30}$ C $-12 + \sqrt{30}$ D $\displaystyle \frac{1}{12 - \sqrt{30}}$

Answer and solution

Answer: (B) $12 + \sqrt{30}$

Only the sign of the surd part changes: the conjugate of $12 - \sqrt{30}$ is $12 + \sqrt{30}$ .
$\sqrt{10} \times \sqrt{10} =$
A $2\sqrt{10}$ B $10$ C $\sqrt{10}$ D $100$

Answer and solution

Answer: (B) $10$

$\sqrt{10} \times \sqrt{10} = \sqrt{10^2} = 10$
$\displaystyle \frac{\sqrt{171}}{\sqrt{19}} =$
A $\sqrt{152}$ B $9$ C $3$ D $3\sqrt{19}$

Answer and solution

Answer: (C) $3$

$\displaystyle \frac{\sqrt{171}}{\sqrt{19}} = \sqrt{\frac{171}{19}}$

$= \sqrt{9} = \sqrt{3 \times 3} = 3$
$\displaystyle 64^{\frac{1}{6}} =$
A $4$ B $2$ C $\displaystyle \frac{64}{6}$ D $12$

Answer and solution

Answer: (B) $2$

$\displaystyle 64^{\frac{1}{6}} = \sqrt[6]{64} = \sqrt[6]{2^6} = 2$
$\displaystyle \frac{\sqrt[4]{80}}{\sqrt[4]{5}} =$
A $2$ B $\sqrt[4]{85}$ C $16$ D $\sqrt[4]{75}$

Answer and solution

Answer: (A) $2$

$\displaystyle \frac{\sqrt[4]{80}}{\sqrt[4]{5}} = \sqrt[4]{\frac{80}{5}}$

$= \sqrt[4]{16} = \sqrt[4]{2^4} = 2$
$(4\sqrt{2})^2 =$
A $32$ B $16\sqrt{2}$ C $64$ D $8$

Answer and solution

Answer: (A) $32$

$(4\sqrt{2})^2 = 4^2 \times (\sqrt{2})^2$

$= 16 \times 2 = 32$
$(5 - \sqrt{7})(7 + \sqrt{7}) =$
A $42 - 2\sqrt{7}$ B $28 - 2\sqrt{7}$ C $28$ D $28 + 2\sqrt{7}$

Answer and solution

Answer: (B) $28 - 2\sqrt{7}$

$(5 - \sqrt{7})(7 + \sqrt{7}) = 35 + 5\sqrt{7} - 7\sqrt{7} - 7$

$= 28 - 2\sqrt{7}$
A rationalising factor of $4\sqrt{10}$ is
A $\sqrt{10}$ B $4$ C $4 + \sqrt{10}$ D $10$

Answer and solution

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

$4\sqrt{10} \times \sqrt{10} = 4 \times 10 = 40$ , a rational number.
$(\sqrt[3]{11} - \sqrt[3]{4})\left((\sqrt[3]{11})^2 + \sqrt[3]{11} \cdot \sqrt[3]{4} + (\sqrt[3]{4})^2\right) =$
A $1267$ B $15$ C $7$ D $\sqrt[3]{7}$

Answer and solution

Answer: (C) $7$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{11}$ , $y = \sqrt[3]{4}$ .

Product $= (\sqrt[3]{11})^3 - (\sqrt[3]{4})^3 = 11 - 4 = 7$
$\sqrt{175} + \sqrt{252} =$
A $\sqrt{427}$ B $30\sqrt{7}$ C $11\sqrt{14}$ D $11\sqrt{7}$

Answer and solution

Answer: (D) $11\sqrt{7}$

$\sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7}$

$\sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7}$

$5\sqrt{7} + 6\sqrt{7} = 11\sqrt{7}$ (add like surds)
$\sqrt{28}$ in simplest form is
A $4\sqrt{7}$ B $7\sqrt{2}$ C $2\sqrt{7}$ D $2\sqrt{14}$

Answer and solution

Answer: (C) $2\sqrt{7}$

$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$
$(\sqrt{2})^{13} =$
A $64\sqrt{2}$ B $64$ C $32\sqrt{2}$ D $128\sqrt{2}$

Answer and solution

Answer: (A) $64\sqrt{2}$

$(\sqrt{2})^{13} = [(\sqrt{2})^2]^{6} \times \sqrt{2} = 2^{6} \times \sqrt{2} = 64\sqrt{2}$
$(\sqrt{10} - \sqrt{2})^2 =$
A $12 - 4\sqrt{5}$ B $8$ C $104 - 4\sqrt{5}$ D $12 + 4\sqrt{5}$

Answer and solution

Answer: (A) $12 - 4\sqrt{5}$

$(\sqrt{10} - \sqrt{2})^2 = 10 + 2 - 2\sqrt{20} = 12 - 4\sqrt{5}$
$(\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3}) =$
A $9$ B $27$ C $3$ D $\sqrt{3}$

Answer and solution

Answer: (C) $3$

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

$= (\sqrt{6})^2 - (\sqrt{3})^2 = 6 - 3 = 3$
$\displaystyle \frac{1}{\sqrt{17} - \sqrt{10}} =$
A $\displaystyle \frac{\sqrt{17} + \sqrt{10}}{27}$ B $\sqrt{17} + \sqrt{10}$ C $\displaystyle \frac{\sqrt{17} - \sqrt{10}}{7}$ D $\displaystyle \frac{\sqrt{17} + \sqrt{10}}{7}$

Answer and solution

Answer: (D) $\displaystyle \frac{\sqrt{17} + \sqrt{10}}{7}$

Multiply above and below by $\sqrt{17} + \sqrt{10}$ : $(\sqrt{17} - \sqrt{10})(\sqrt{17} + \sqrt{10}) = 17 - 10 = 7$

$\displaystyle = \frac{\sqrt{17} + \sqrt{10}}{7}$
$\displaystyle \frac{1}{7 + \sqrt{13}} =$
A $\displaystyle \frac{7 - \sqrt{13}}{62}$ B $\displaystyle \frac{\sqrt{13} - 7}{6}$ C $\displaystyle \frac{7 + \sqrt{13}}{36}$ D $\displaystyle \frac{7 - \sqrt{13}}{36}$

Answer and solution

Answer: (D) $\displaystyle \frac{7 - \sqrt{13}}{36}$

Multiply above and below by $7 - \sqrt{13}$ : $(7 + \sqrt{13})(7 - \sqrt{13}) = 49 - 13 = 36$

$\displaystyle = \frac{7 - \sqrt{13}}{36}$
If $\displaystyle \frac{3 + 5\sqrt{3}}{3 - \sqrt{3}} = a + b\sqrt{3}$ , then $b =$
A $12$ B $4$ C $3$ D $-3$

Answer and solution

Answer: (C) $3$

$\displaystyle \frac{3 + 5\sqrt{3}}{3 - \sqrt{3}} = \frac{(3 + 5\sqrt{3})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}$

Numerator: $9 + 3\sqrt{3} + 15\sqrt{3} + 15 = 24 + 18\sqrt{3}$

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

$\displaystyle \frac{24 + 18\sqrt{3}}{6} = 4 + 3\sqrt{3}$

So $a = 4$ , $b = 3$
$\displaystyle \frac{11 + \sqrt{17}}{11 - \sqrt{17}} + \frac{11 - \sqrt{17}}{11 + \sqrt{17}} =$
A $\displaystyle \frac{69}{52}$ B $\displaystyle -\frac{69}{26}$ C $\displaystyle \frac{11\sqrt{17}}{26}$ D $\displaystyle \frac{69}{26}$

Answer and solution

Answer: (D) $\displaystyle \frac{69}{26}$

$\displaystyle = \frac{(11 + \sqrt{17})^2 + (11 - \sqrt{17})^2}{(11 + \sqrt{17})(11 - \sqrt{17})} = \frac{276}{104} = \frac{69}{26}$

The $\sqrt{17}$ terms cancel and $(11 + \sqrt{17})(11 - \sqrt{17}) = 104$ .
$\displaystyle \frac{1}{8 + \sqrt{6}} + \frac{1}{8 - \sqrt{6}} =$
A $\displaystyle \frac{8}{29}$ B $\displaystyle \frac{\sqrt{6}}{29}$ C $0$ D $\displaystyle \frac{8}{35}$

Answer and solution

Answer: (A) $\displaystyle \frac{8}{29}$

$\displaystyle = \frac{(8 - \sqrt{6}) + (8 + \sqrt{6})}{(8 + \sqrt{6})(8 - \sqrt{6})} = \frac{16}{58} = \frac{8}{29}$
If $x = \sqrt[3]{13 + 2\sqrt{42}}$ , then $\displaystyle x^3 + \frac{1}{x^3} =$
A $4\sqrt{42}$ B $13$ C $0$ D $26$

Answer and solution

Answer: (D) $26$

$x^3 = 13 + 2\sqrt{42}$ , so $\displaystyle \frac{1}{x^3} = \frac{1}{13 + 2\sqrt{42}} = 13 - 2\sqrt{42}$

$\displaystyle x^3 + \frac{1}{x^3} = (13 + 2\sqrt{42}) + (13 - 2\sqrt{42}) = 26$
$\sqrt{15 + 10\sqrt{2}} =$
A $\sqrt{10} - \sqrt{5}$ B $\sqrt{50} + 1$ C $\sqrt{10} + \sqrt{5}$ D $\sqrt{5} - \sqrt{10}$

Answer and solution

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

$10\sqrt{2} = 2 \times 5\sqrt{2} = 2\sqrt{50}$

$\sqrt{15 + 10\sqrt{2}} = \sqrt{10 + 5 + 2\sqrt{10 \times 5}}$

$= \sqrt{(\sqrt{10})^2 + (\sqrt{5})^2 + 2(\sqrt{10})(\sqrt{5})}$

$= \sqrt{(\sqrt{10} + \sqrt{5})^2} = \sqrt{10} + \sqrt{5}$
$\sqrt{16 - \sqrt{87}} =$
A $\displaystyle \frac{\sqrt{29} - \sqrt{3}}{\sqrt{2}}$ B $\displaystyle \frac{\sqrt{29} - \sqrt{3}}{2}$ C $\sqrt{29} - \sqrt{3}$ D $\displaystyle \frac{\sqrt{29} + \sqrt{3}}{\sqrt{2}}$

Answer and solution

Answer: (A) $\displaystyle \frac{\sqrt{29} - \sqrt{3}}{\sqrt{2}}$

$\displaystyle \sqrt{16 - \sqrt{87}} = \sqrt{\frac{2(16 - \sqrt{87})}{2}} = \frac{\sqrt{32 - 2\sqrt{87}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{29 + 3 - 2\sqrt{29 \times 3}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{29})^2 + (\sqrt{3})^2 - 2(\sqrt{29})(\sqrt{3})}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{29} - \sqrt{3})^2}}{\sqrt{2}} = \frac{\sqrt{29} - \sqrt{3}}{\sqrt{2}}$

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MCQ test, Set 1