Question 83

If A has got 20% more marks than B, then by what percent marks of B are less than the marks of A?

Solution

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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