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Home > Practice > Arithmetic Aptitude > Surds and Indices > Miscellaneous

21. 9³ × 6² ÷ 3³ = ?

A. 948

B. 972

C. 984

D. 1012

Discuss

Solution:

\begin{aligned} \frac{9^{3} \times 6^{2}}{3^{3}} \\ = \frac{{(3^{2})}^{3} \times {(3 \times 2)}^{2}}{3^{3}} \\ = \frac{3^{(3 \times 2)} \times 3^{2} \times 2^{2}}{3^{3}} \\ = \frac{3^{(6 + 2)} \times 2^{2}}{3^{3}} \\ = 3^{(8 - 3)} \times 2^{2} \\ = 3^{5} \times 2^{2} \\ = 243 \times 4 \\ = 972 \end{aligned}

22. $(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1})$ Simplifies to :

A. 2 - $\sqrt{3}$

B. 2 + $\sqrt{3}$

C. 16 - $\sqrt{3}$

D. 40 - $\sqrt{3}$

Discuss

Solution:

(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1})

= {\frac{{(2 + \sqrt{3})}^{2} + {(2 - \sqrt{3})}^{2}}{(2 - \sqrt{3}) (2 + \sqrt{3})} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}}

= {\frac{4 + 3 + 4 \sqrt{3} + 4 + 3 - 4 \sqrt{3}}{4 - 3} + \frac{{(\sqrt{3} - 1)}^{2}}{3 - 1}}

\begin{aligned} = {14 + \frac{3 + 1 - 2 \sqrt{3}}{2}} \\ = 14 + \frac{2 (2 - \sqrt{3})}{2} \\ = 14 + 2 - \sqrt{3} \\ = 16 - \sqrt{3} \end{aligned}

23. The value of $\sqrt{\frac{(\sqrt{12} - \sqrt{8}) (\sqrt{3} + \sqrt{2})}{5 + \sqrt{24}}}$ is = ?

A. $\sqrt{6} - \sqrt{2}$

B. $\sqrt{6} + \sqrt{2}$

C. $\sqrt{6} - 2$

D. $2 - \sqrt{6}$

Discuss

Solution:

\begin{aligned} \sqrt{\frac{(\sqrt{12} - \sqrt{8}) (\sqrt{3} + \sqrt{2})}{5 + \sqrt{24}}} \\ = \sqrt{\frac{\sqrt{36} + \sqrt{24} - \sqrt{24} - \sqrt{16}}{5 + \sqrt{24}}} \\ = \sqrt{\frac{6 - 4}{5 + \sqrt{24}}} \\ = \sqrt{\frac{2}{5 + \sqrt{24}} \times \frac{5 - \sqrt{24}}{5 - \sqrt{24}}} \\ = \sqrt{\frac{2 (5 - \sqrt{24})}{25 - 24}} \\ = \sqrt{2 (5 - 2 \sqrt{6})} \\ = \sqrt{2 {{(\sqrt{3})}^{2} + {(\sqrt{2})}^{2} - 2 \sqrt{3} \times \sqrt{2}}} \\ = \sqrt{2 {(\sqrt{3} - \sqrt{2})}^{2}} \\ = \sqrt{2} (\sqrt{3} - \sqrt{2}) \\ = \sqrt{6} - 2 \end{aligned}

24. (19)¹² × (19)⁸ ÷ (19)⁴ = (19)^?

A. 6

B. 8

C. 12

D. 24

Discuss

Solution:

\begin{aligned} \frac{{(19)}^{12} \times {(19)}^{8}}{{(19)}^{4}} \\ = \frac{19^{(12 + 8)}}{{(19)}^{4}} \\ = \frac{{(19)}^{20}}{{(19)}^{4}} \\ = {(19)}^{(20 - 4)} \\ = {(19)}^{16} \end{aligned}

Hence the missing number = 16

25. (64)⁴ ÷ (8)⁵ = ?

A. (8)⁸

B. (8)²

C. (8)¹²

D. (8)⁴

Discuss

Solution:

\begin{aligned} {(64)}^{4} \div {(8)}^{5} \\ = {(8^{2})}^{4} \div {(8)}^{5} \\ = {(8)}^{(2 \times 4)} \div 8^{5} \\ = \frac{8^{8}}{8^{5}} \\ = 8^{(8 - 5)} \\ = 8^{3} \end{aligned}

26. The value of $\frac{1}{\sqrt{(12 - \sqrt{140})}}$ $- \frac{1}{\sqrt{(8 - \sqrt{60})}}$ $- \frac{2}{\sqrt{(10 + \sqrt{84})}}$ = is ?

A. 0

B. 1

C. 2

D. 3

Discuss

Solution:

\frac{1}{\sqrt{(12 - \sqrt{140})}} - \frac{1}{\sqrt{(8 - \sqrt{60})}} - \frac{2}{\sqrt{(10 + \sqrt{84})}}

= \frac{1}{\sqrt{(12 - \sqrt{4 \times 35})}} - \frac{1}{\sqrt{(8 - \sqrt{4 \times 15})}} - \frac{2}{\sqrt{(10 + \sqrt{4 \times 21})}}

= \frac{1}{\sqrt{(12 - 2 \sqrt{35})}} - \frac{1}{\sqrt{(8 - 2 \sqrt{15})}} - \frac{2}{\sqrt{(10 + 2 \sqrt{21})}}

= \frac{1}{\sqrt{{(\sqrt{7})}^{2} + {(\sqrt{5})}^{2} - 2. \sqrt{7} . \sqrt{5}}} - \frac{1}{\sqrt{{(\sqrt{5})}^{2} + {(\sqrt{3})}^{2} - 2. \sqrt{5} . \sqrt{3}}} - \frac{2}{\sqrt{{(\sqrt{7})}^{2} + {(\sqrt{3})}^{2} + 2. \sqrt{7} . \sqrt{3}}}

= \frac{1}{\sqrt{{(\sqrt{7} - \sqrt{5})}^{2}}} - \frac{1}{\sqrt{{(\sqrt{5} - \sqrt{3})}^{2}}} - \frac{2}{\sqrt{{(\sqrt{7} + \sqrt{3})}^{2}}}

= \frac{1}{\sqrt{7} - \sqrt{5}} - \frac{1}{\sqrt{5} - \sqrt{3}} - \frac{2}{\sqrt{7} + \sqrt{3}}

Rationalizing in above equation,

= \frac{1}{\sqrt{7} - \sqrt{5}} \times \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}} - \frac{1}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} - \frac{2}{\sqrt{7} + \sqrt{3}} \times \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} - \sqrt{3}}

= \frac{\sqrt{7} + \sqrt{5}}{2} - \frac{\sqrt{5} + \sqrt{3}}{2} - \frac{\sqrt{7} - \sqrt{3}}{2}

= \frac{\sqrt{7} + \sqrt{5} - \sqrt{5} - \sqrt{3} - \sqrt{7} + \sqrt{3}}{2}

= 0

27. (1000)¹² ÷ (10)³⁰ = ?

A. (1000)²

B. 10

C. 100

D. (100)¹²

Discuss

Solution:

\begin{aligned} \frac{{(1000)}^{12}}{{(10)}^{30}} \\ = \frac{{(10^{3})}^{12}}{{(10)}^{30}} \\ = \frac{{(10)}^{(3 \times 12)}}{{(10)}^{30}} \\ = \frac{{(10)}^{36}}{{(10)}^{30}} \\ = {(10)}^{(36 - 30)} \\ = 10^{6} \\ = {(10^{3})}^{2} \\ = {(1000)}^{2} \end{aligned}

28. Simplify : $\frac{{1.5}^{3} + {4.7}^{3} + {3.8}^{3} - 3 \times 1.5 \times 4.7 \times 3.8}{{1.5}^{2} + {4.7}^{2} + {3.8}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}$ = ?

A. 0

B. 1

C. 10

D. 30

Discuss

Solution:

\frac{{1.5}^{3} + {4.7}^{3} + {3.8}^{3} - 3 \times 1.5 \times 4.7 \times 3.8}{{1.5}^{2} + {4.7}^{2} + {3.8}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}

= \frac{(1.5 + 4.7 + 3.8) {{1.5}^{2} + {4.7}^{2} + {3.8}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}{{{1.5}^{2} + {4.7}^{2} + {3.8}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}

[∴ a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)]

\begin{aligned} = 1.5 + 4.7 + 3.8 \\ = 10.0 \\ = 10 \end{aligned}

29. Simplify : $\frac{0.41 \times 0.41 \times 0.41 + 0.69 \times 0.69 \times 0.69}{0.41 \times 0.41 - 0.41 \times 0.69 + 0.69 \times 0.69} = ?$

A. 0.28

B. 1.41

C. 1.1

D. 2.8

Discuss

Solution:

\begin{aligned} \frac{0.41 \times 0.41 \times 0.41 + 0.69 \times 0.69 \times 0.69}{0.41 \times 0.41 - 0.41 \times 0.69 + 0.69 \times 0.69} \\ = \frac{{(0.41)}^{3} + {(0.69)}^{3}}{{(0.41)}^{2} - 0.41 \times 0.69 + {(0.69)}^{2}} \\ . . . . . [∵ a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)] \\ = 0.41 + 0.69 \\ = 1.1 \end{aligned}

30. If 3¹⁰ × 27² = 9² × 3ⁿ then the value of n is = ?

A. 10

B. 12

C. 15

D. 30

Discuss

Solution:

\begin{aligned} 3^{10} \times 2 7^{2} = 9^{2} \times 3^{n} \\ \Rightarrow 3^{10} \times {(3^{3})}^{2} = {(3^{2})}^{2} \times 3^{n} \\ \Rightarrow 3^{10 + 6} = 3^{4 + n} \\ \Rightarrow 10 + 6 = 4 + n \\ \Rightarrow n = 12 \end{aligned}

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