Free Practice Questions for Fraction in Maths

$3^{-2} \times 81^{\frac{3}{4}} \div(729)^{-1 / 3}$
$\Rightarrow \frac{1}{9} \times\left(3^{4}\right)^{\frac{3}{4}} \div\left[(9)^{3}\right]^{-1 / 3}$
$\Rightarrow \frac{1}{9} \times 27 \div(9)^{-1}$
$\Rightarrow \frac{1}{9} \times 27 \times 9 \Rightarrow 27$

$\left(\frac{343}{1024 \times 8 \times 4}\right)^{1 / 3} \times(256)^{1 / 2}$
$\Rightarrow\left[\frac{7^{3}}{32^{3}}\right]^{1 / 3} \times 16$
$\Rightarrow \frac{7}{32} \times 16 \Rightarrow \frac{7}{2}$

Make the power equal and compare the denominators.
$
\begin{aligned}
&2^{1 / 2}=64^{1 / 12} \\
&3^{1 / 3}=81^{1 / 12} \\
&4^{1 / 4}=64^{1 / 12} \\
&6^{1 / 6}=36^{1 / 12}
\end{aligned}
$
$81^{1 / 12}$ is the greatest $=3^{1 / 3}$ is the greatest.

$\frac{7}{8}$ of $60 \%$ of 80
$=\frac{7}{8} \times \frac{60}{100} \times 80$
$=42$

We may solve by percentage method
$
\begin{aligned}
&\frac{2}{4} \times 100=50 \% \\
&0.6=\frac{6}{10} \times 100 \\
&=\frac{6}{10} \times 100=60 \%
\end{aligned}
$
The number should be between $50 \%$ and $60 \%$
$
\frac{11}{25} \times 100=44 \%
$
So this option is wrong
$
\frac{21}{40} \times 100=52.5 \%
$
So right option may be this
$
\frac{3}{4} \times 100=75 \%
$
So this option is wrong
$
\frac{11}{4} \times 100=275 \%
$
It is clearly that only one option is between $50 \%$ and $60 \%$ that is option $\mathrm{b}$
Hence right answer is option b

$6 \frac{1}{4} \%=\frac{25}{400}=1 / 16$

$\frac{x - 4}{x}$                           (step 1)

$\frac{x - 4 - 2}{x + 1}$               (step 2) 

A.T.Q

8 ( x - 6 ) = ( x + 1 ) 

8x - 48 = x + 1

7x = 49 

x = 7 , Now put the value of x in step 1, we got 

$\frac{3}{7}$