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Profit percentage adjustment

CAT 2020 Slot 3 · Quantitative Ability · Medium · Profit and loss

This is a medium Quantitative Ability question from the CAT 2020 Slot 3 paper. It tests Profit and loss. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A man buys 35 kg35\text{ kg} of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg5\text{ kg} at this price, and 15 kg15\text{ kg} at a 10% discount. Accidentally, 3 kg3\text{ kg} of sugar is wasted. He sells the remaining sugar by raising the marked price by pp percent so as to make an overall profit of 15%. Then pp is nearest to

  1. A.

    35

  2. B.

    31

  3. C.

    22

  4. D.

    25

Answer

D

Explanation

Let CP per kg = 100. Total CP = 35×100=350035 \times 100 = 3500. Target overall profit = 15%, so total SP = 3500×1.15=40253500 \times 1.15 = 4025. Marked price per kg MP = 100×1.20=120100 \times 1.20 = 120.

  1. 5 kg5\text{ kg} sold at MP = 5×120=6005 \times 120 = 600.
  2. 15 kg15\text{ kg} sold at 10% discount on MP (120×0.9=108120 \times 0.9 = 108) = 15×108=162015 \times 108 = 1620.
  3. 3 kg3\text{ kg} wasted (revenue = 0). Remaining sugar = 35−5−15−3=12 kg35 - 5 - 15 - 3 = 12\text{ kg}.

Total revenue so far = 600+1620=2220600 + 1620 = 2220. Required revenue from remaining 12 kg=4025−2220=180512\text{ kg} = 4025 - 2220 = 1805. Selling price per kg for remaining sugar = 180512≈150.417\frac{1805}{12} \approx 150.417.

New MP = 120×(1+p/100)=150.417  ⟹  1+p/100=150.417120≈1.2534120 \times (1 + p/100) = 150.417 \implies 1 + p/100 = \frac{150.417}{120} \approx 1.2534. So p≈25.34%p \approx 25.34\%, which is nearest to 25.

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