In an examination, B obtained 20% more marks than those obtained by A, and A obtained 10% less marks than those obtained by C. D obtained 20% more marks than those obtained by C. By what percentage are the marks obtained by D more than those obtained by A?

(a) 33\displaystyle\frac{1}{3}\%

(b) 13\displaystyle\frac{1}{3}\%

(c) 43\displaystyle\frac{1}{3}\%

(d) 23\displaystyle\frac{1}{3}\%

{Previously asked in SSC CGL Mains 2020}

**Ans**: (a) 33\displaystyle\frac{1}{3}\%

**Explanation**:

We would try to solve by using ratio approach

First of all write the ratio data.

We have got **two A** and **two C**. First we will make equal the** A**s by taking LCM(5, 9)

Now, we will make equal the **C**s by taking their LCM(10×5, 5)=LCM(50, 5) = 50

\displaystyle\frac{\text{A}}{\text{D}} = \displaystyle\frac{5\times 9}{6\times 10} = \displaystyle\frac{3}{4}

∴ D is % more than A = \displaystyle\frac{1}{3}\times 100 = 33\displaystyle\frac{1}{3}\%

**Aliter**

Given that, B = 120% of A

\qquad\qquadA = 90% of C

\qquad\qquadD = 120% of C

∴ \displaystyle\frac{\text{A}}{\text{D}} = \displaystyle\frac{90\% \ \text{of C}}{120\% \ \text{of C}} = \displaystyle\frac{3}{4}

Hence, D is % more than A = \displaystyle\frac{1}{3}\times 100 = 33\displaystyle\frac{1}{3}\%

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