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41. The value of ${(\frac{32}{243})}^{- \frac{4}{5}}$ is = ?

A. $\frac{4}{9}$

B. $\frac{9}{4}$

C. $\frac{16}{81}$

D. $\frac{81}{16}$

Discuss

Solution:

\begin{aligned} {(\frac{32}{243})}^{- \frac{4}{5}} \\ = {{(\frac{2}{3})}^{5}}^{- \frac{4}{5}} \\ = {(\frac{2}{3})}^{5 \times \frac{(- 4)}{5}} \\ = {(\frac{2}{3})}^{(- 4)} \\ = {(\frac{3}{2})}^{4} \\ = \frac{3^{4}}{2^{4}} \\ = \frac{81}{16} \end{aligned}

42. The value of $2 7^{- \frac{2}{3}}$ lies between = ?

A. 0 and 1

B. 1 and 2

C. 2 and 3

D. 3 and 4

Discuss

Solution:

\begin{aligned} 2 7^{- \frac{2}{3}} \\ = {(3^{3})}^{- \frac{2}{3}} \\ = 3^{[3 \times (- \frac{2}{3})]} \\ = 3^{- 2} \\ = \frac{1}{3^{2}} \\ = \frac{1}{9} \\ Clearly, 0 < \frac{1}{9} < 1 \end{aligned}

43. The value of $\frac{1}{\sqrt{3.25} + \sqrt{2.25}}$ $+ \frac{1}{\sqrt{4.25} + \sqrt{3.25}}$ $+ \frac{1}{\sqrt{5.25} + \sqrt{4.25}}$ $+ \frac{1}{\sqrt{6.25} + \sqrt{5.25}}$ is = ?

A. 1.00

B. 1.25

C. 1.50

D. 2.25

Discuss

Solution:

\begin{aligned} \frac{1}{\sqrt{3.25} + \sqrt{2.25}} \times \frac{\sqrt{3.25} - \sqrt{2.25}}{\sqrt{3.25} - \sqrt{2.25}} \\ = \frac{\sqrt{3.25} - \sqrt{2.25}}{3.25 - 2.25} \\ = \sqrt{3.25} - \sqrt{2.25} . . . . . . (i) \\ Similarly \\ \frac{1}{\sqrt{4.25} + \sqrt{3.25}} \\ = \sqrt{4.25} - \sqrt{3.25} . . . . . . . (i i) \\ \frac{1}{\sqrt{5.25} + \sqrt{4.25}} \\ = \sqrt{5.25} - \sqrt{4.25} . . . . . . . (i i i) \\ \frac{1}{\sqrt{6.25} + \sqrt{5.25}} \\ = \sqrt{6.25} - \sqrt{5.25} . . . . . . . (i v) \\ Now add all them \\ \sqrt{3.25} - \sqrt{2.25} + \sqrt{4.25} - \sqrt{3.25} + \sqrt{5.25} - \sqrt{4.25} + \sqrt{6.25} - \sqrt{5.25} \\ = \sqrt{6.25} - \sqrt{2.25} \\ = 2.5 - 1.5 \\ = 1 \end{aligned}

44. $\frac{3^{0} + 3^{- 1}}{3^{- 1} - 3^{0}}$ is simplified to = ?

A. -2

B. -1

C. 1

D. 2

Discuss

Solution:

\begin{aligned} \frac{3^{0} + 3^{- 1}}{3^{- 1} - 3^{0}} \\ = \frac{1 + \frac{1}{3}}{\frac{1}{3} - 1} \\ = \frac{\frac{4}{3}}{- \frac{2}{3}} \\ = - 2 \end{aligned}

45. The simplified form of $\frac{2}{\sqrt{7} + \sqrt{5}} +$ $\frac{7}{\sqrt{12} - \sqrt{5}} -$ $\frac{5}{\sqrt{12} - \sqrt{7}}$ is = ?

A. 5

B. 2

C. 1

D. 0

Discuss

Solution:

\frac{2}{\sqrt{7} + \sqrt{5}} + \frac{7}{\sqrt{12} - \sqrt{5}} - \frac{5}{\sqrt{12} - \sqrt{7}}

= \frac{2}{\sqrt{7} + \sqrt{5}} \times

\frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}} +

\frac{7}{\sqrt{12} - \sqrt{5}} \times

\frac{\sqrt{12} + \sqrt{5}}{\sqrt{12} + \sqrt{5}} -

(\frac{5}{\sqrt{12} - \sqrt{7}} \times \frac{\sqrt{12} + \sqrt{7}}{\sqrt{12} + \sqrt{7}})

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

\frac{7 (\sqrt{12} + \sqrt{5})}{7} -

\frac{5 (\sqrt{12} + \sqrt{7})}{5}

\begin{aligned} = \sqrt{7} - \sqrt{5} + \sqrt{12} + \sqrt{5} - \sqrt{12} - \sqrt{7} \\ = 0 \end{aligned}

46. The value of $\sqrt[3]{2^{4} \sqrt{2^{- 5} \sqrt{2^{6}}}}$ is = ?

A. 1

B. 2

C. $2^{\frac{5}{3}}$

D. 2⁵

Discuss

Solution:

\begin{aligned} \sqrt[3]{2^{4} \sqrt{2^{- 5} \sqrt{2^{6}}}} \\ = \sqrt[3]{2^{4} \sqrt{2^{- 5} {(2^{6})}^{\frac{1}{2}}}} \\ = \sqrt[3]{2^{4} \sqrt{2^{- 5} {(2)}^{(6 \times \frac{1}{2})}}} \\ = \sqrt[3]{2^{4} \sqrt{2^{- 5} {.2}^{3}}} \\ = \sqrt[3]{2^{4} \sqrt{2^{(- 5 + 3)}}} \\ = \sqrt[3]{2^{4} \sqrt{2^{- 2}}} \\ = \sqrt[3]{2^{4} . {(2^{2})}^{\frac{1}{2}}} \\ = \sqrt[3]{2^{4} {.2}^{(- 2 \times \frac{1}{2})}} \\ = \sqrt[3]{2^{4} {.2}^{(- 1)}} \\ = \sqrt[3]{2^{(4 - 1)}} \\ = \sqrt[3]{2^{3}} \\ = {(2^{3})}^{\frac{1}{3}} \\ = 2^{(3 \times \frac{1}{3})} \\ = 2 \end{aligned}

47. The value of $\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}} = ?$

A. $2^{\frac{29}{31}}$

B. $2^{\frac{31}{32}}$

C. $2^{\frac{9}{2}}$

D. $2^{\frac{11}{2}}$

Discuss

Solution:

\begin{aligned} \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{{2.2}^{\frac{1}{2}}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2^{(1 + \frac{1}{2})}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2^{\frac{3}{2}}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2. {(2^{\frac{3}{2}})}^{\frac{1}{2}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{{2.2}^{(\frac{3}{2} \times \frac{1}{2})}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{{2.2}^{\frac{3}{4}}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2^{(1 + \frac{3}{4})}}}} \\ = \sqrt{2 \sqrt{2 \sqrt{2^{\frac{7}{4}}}}} \\ = \sqrt{2 \sqrt{2 {(2^{\frac{7}{4}})}^{\frac{1}{2}}}} \\ = \sqrt{2 \sqrt{{2.2}^{(\frac{7}{4} \times \frac{1}{2})}}} \\ = \sqrt{2 \sqrt{{2.2}^{\frac{7}{8}}}} \\ = \sqrt{2 \sqrt{2^{(1 + \frac{7}{8})}}} \\ = \sqrt{2 \sqrt{2^{\frac{15}{8}}}} \\ = \sqrt{2 {(2^{\frac{15}{8}})}^{\frac{1}{2}}} \\ = \sqrt{{2.2}^{\frac{15}{16}}} \\ = \sqrt{2^{(1 + \frac{15}{16})}} \\ = \sqrt{2^{\frac{31}{16}}} \\ = {(2^{\frac{31}{16}})}^{\frac{1}{2}} \\ = 2^{(\frac{31}{16} \times \frac{1}{2})} \\ = 2^{\frac{31}{32}} \end{aligned}

48. The value of ${(0.03125)}^{- \frac{2}{5}}$ is = ?

A. 4

B. 9

C. 12

D. 31.25

Discuss

Solution:

\begin{aligned} {(0.03125)}^{- \frac{2}{5}} \\ = {[{(0.5)}^{5}]}^{- \frac{2}{5}} \\ = {0.5}^{[5 \times (- \frac{2}{5})]} \\ = {(0.5)}^{- 2} \\ = \frac{1}{{(0.5)}^{2}} \\ = \frac{1}{0.25} \\ = 4 \end{aligned}

49. Simplify : $\frac{1}{\sqrt{3} + \sqrt{4}} +$ $\frac{1}{\sqrt{4} + \sqrt{5}} +$ $\frac{1}{\sqrt{5} + \sqrt{6}} +$ $\frac{1}{\sqrt{6} + \sqrt{7}} +$ $\frac{1}{\sqrt{7} + \sqrt{8}} +$ $\frac{1}{\sqrt{8} + \sqrt{9}} = ?$

A. $\sqrt{3}$

B. $3 \sqrt{3}$

C. $3 - \sqrt{3}$

D. $5 - \sqrt{3}$

Discuss

Solution:

\begin{aligned} \frac{1}{\sqrt{3} + \sqrt{4}} \\ = \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} \\ = \frac{\sqrt{4} - \sqrt{3}}{1} \\ = \sqrt{4} - \sqrt{3} \\ Similarly \\ \frac{1}{\sqrt{4} + \sqrt{5}} = \sqrt{5} - \sqrt{4} \\ \Rightarrow \frac{1}{\sqrt{5} + \sqrt{6}} = \sqrt{6} - \sqrt{5} \\ \Rightarrow \frac{1}{\sqrt{6} + \sqrt{7}} = \sqrt{7} - \sqrt{6} \\ \Rightarrow \frac{1}{\sqrt{7} + \sqrt{8}} = \sqrt{8} - \sqrt{7} \\ \Rightarrow \frac{1}{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8} \\ Now put values \\ \Rightarrow \sqrt{4} - \sqrt{3} + \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8} \\ \Rightarrow \sqrt{9} - \sqrt{3} \\ \Rightarrow 3 - \sqrt{3} \end{aligned}

50. $[\frac{{(0.73)}^{3} + {(0.27)}^{3}}{{(0.73)}^{2} + {(0.27)}^{2} - (0.73) \times (0.27)}]$ simplifies to ?

A. 1

B. 0.4087

C. 0.73

D. 0.27

Discuss

Solution:

\begin{aligned} let \\ a = 0.73 \\ b = 0.27 \\ = \frac{a^{3} + b^{3}}{a^{2} + b^{2} - a b} \\ = \frac{(a + b) (a^{2} + b^{2} - a b)}{(a^{2} + b^{2} - a b)} \\ = (a + b) \\ = (0.73 + 0.27) \\ = 1 \end{aligned}

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