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Markup percentage calculation

CAT 1997 Slot 1 · Quantitative Ability · Medium · Arithmetic

This is a medium Quantitative Ability question from the CAT 1997 Slot 1 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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After allowing a discount of 11.11%11.11\%, a trader still makes a gain of 14.28%14.28\%. At how many percentage above the cost price does he mark on his goods?

  1. A.

    28.56%

  2. B.

    35%

  3. C.

    22.22%

  4. D.

    None of these

Answer

A

Explanation

Note that 11.11%=1911.11\% = \frac{1}{9} and 14.28%=1714.28\% = \frac{1}{7}.

Let Cost Price (CP) = 1. Gain = 17\frac{1}{7}, so Selling Price (SP) = 1+17=871 + \frac{1}{7} = \frac{8}{7}.

Discount = 11.11%=1911.11\% = \frac{1}{9} of Marked Price (MP). So, SP=MP×(1−19)=89MP\text{SP} = \text{MP} \times \left(1 - \frac{1}{9}\right) = \frac{8}{9} \text{MP}.

Equating SP: 89MP=87  ⟹  MP=97\frac{8}{9} \text{MP} = \frac{8}{7} \implies \text{MP} = \frac{9}{7}

Markup above CP = MP−CP=97−1=27\text{MP} - \text{CP} = \frac{9}{7} - 1 = \frac{2}{7}.

Percentage markup = 27×100%=2×14.28%=28.56%\frac{2}{7} \times 100\% = 2 \times 14.28\% = 28.56\%.

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