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"?

*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.

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