Free Practice Questions for Fraction in Maths

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

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

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

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

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

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

$\frac{x-4}{x}$                           (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 

$\frac{3}{7}$