A's income is 20 % less than that of B's. How much per cent is B's income more than A's ? Solve it by two methods. First by 100. Second by taking "x"?

Suppose B's income is x.
Now, A's income is 20 % less than that of B's.
So, A's income = $x-x\times \frac{20}{100} = x-\frac{x}{5} = \frac{4x}{5}$
Then, difference between A's income and B's income = $x-\frac{4x}{5} = \frac{x}{5}$
So, B's income is $\frac{x}{5}$ more than A's income.
Therefore required percentage = $\frac{{\displaystyle \frac{x}{5}}\times 100}{{\displaystyle \frac{4x}{5}}} = \frac{x}{5}\times \frac{5}{4x}\times 100 = 25\%$
Therefore B's income is 25% more than A's income.

Second method:
Suppose B's income is 100.
Now, A's income is 20 % less than that of B's.
So, A's income = $100-100\times \frac{20}{100} = 100-20= 80$
Then, difference between A's income and B's income = $100-80 = 20$
So, B's income is 20 more than A's income.
Therefore required percentage = $\frac{20\times 100}{80} = 25\%$
Therefore B's income is 25% more than A's income.