Question Detail
Find the value of,
\begin{aligned}
\frac{1}{216^{-\frac{2}{3}}}+\frac{1}{256^{-\frac{3}{4}}}+\frac{1}{32^{-\frac{1}{5}}}
\end{aligned}
- 100
- 101
- 102
- 103
Answer: Option C
1. Evaluate \begin{aligned} 256^{0.16} \times (256)^{0.09} \end{aligned}
- 2
- 4
- 8
- 16
Answer: Option B
Explanation:
\begin{aligned}
= 256^{0.16+0.09} = 256^{0.25} = 256^{\frac{25}{100}}
\end{aligned}
\begin{aligned}
= 256^{\frac{1}{4}}= (4^4)^{\frac{1}{4}}
\end{aligned}
\begin{aligned}
=(4)^{4 \times \frac{1}{4}} = 4
\end{aligned}
2. \begin{aligned}
x = 3 + 2\sqrt{2}, \text{ then the value of }\\
(\sqrt{x} - \frac{1}{\sqrt{x}})
\end{aligned}
- 1
- 2
- 3
- 4
Answer: Option B
Explanation:
Clue:
\begin{aligned}
(\sqrt{x} - \frac{1}{\sqrt{x}})^2 = x + \frac{1}{x} - 2 \
\end{aligned}
Now put the value of x to calculate the answer :)
3. \begin{aligned}
\frac{1}{1+a^{(n-m)}} + \frac{1}{1+a^{(m-n)}} = ?
\end{aligned}
- 1
- 2
- 3
- 4
Answer: Option A
Explanation:
\begin{aligned}
= \frac{1}{\left( 1 + \frac{a^n}{a^m} \right)} +
\frac{1}{\left( 1 + \frac{a^m}{a^n} \right)} \\
= \frac{a^m}{(a^m+a^n)} + \frac{a^n}{(a^m+a^n)} \\
= \frac{(a^m+a^n)}{(a^m+a^n)} = 1
\end{aligned}
4. \begin{aligned}
\left(25 \right)^{7.5} \times \left(5 \right)^{2.5} \div \left(125 \right)^{1.5} = 5^?
\end{aligned}
- 9.7
- 11.5
- 12
- 13
Answer: Option D
Explanation:
Lets assume,
\begin{aligned}
\left(25 \right)^{7.5} \times \left(5 \right)^{2.5} \div \left(125 \right)^{1.5} = 5^x \\
\text{then, } \frac{ \left( 5^2 \right)^{7.5} \times \left(5 \right)^{2.5} }{\left(5^3 \right)^{1.5} } = 5^x \\
=> \text{then, } \frac{ \left( 5^{15} \right) \times \left(5^{2.5} \right) }{\left(5^{4.5} \right) } = 5^x \\
=> 5^x = 5^{15 + 2.5 - 4.5} \\
=> 5^x = 5^{13} \\
\text{Hence, } x = 13
\end{aligned}
5. \begin{aligned}
\text{If }x = \left(8 + 3\sqrt{7}\right),\text{ what is the value of }\\\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)?
\end{aligned}
- \begin{aligned} \sqrt{13} \end{aligned}
- \begin{aligned} \sqrt{14} \end{aligned}
- \begin{aligned} \sqrt{15} \end{aligned}
- \begin{aligned} \sqrt{16} \end{aligned}
Answer: Option B
Explanation:
\begin{align}&\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2\\\\
&= x - 2 + \dfrac{1}{x}\\\\
&= x + \dfrac{1}{x} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{1}{\left(8 + 3\sqrt{7}\right)} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{\left(8 + 3\sqrt{7}\right)\left(8 - 3\sqrt{7}\right)} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{8^2 - \left(3\sqrt{7}\right)^2} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{64 - 63} - 2 \\\\
&= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{1} - 2 \\\\
&= 8 + 3\sqrt{7} + 8 - 3\sqrt{7} - 2 \\\\
&= 14 \\\\
&\text{as }\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2 = 14\\\\
&\text{so ,}\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right) = \sqrt{14}\end{align}