If A has got 20% more marks than B, then by what percent marks of B are less than the marks of A?
Let B's marks = 100
=> A's marks = $$100+(\frac{20}{100}\times100)=120$$
$$\therefore$$ Percent by which B's marks are less than A's = $$\frac{(120-100)}{120}\times100$$
= $$\frac{1}{6}\times100=16.67\%$$
=> Ans - (A)
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