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Tea mixture pricing for profit

CAT 2021 Slot 2 · Quantitative Ability · Easy · Ratios and Proportions

This is an easy Quantitative Ability question from the CAT 2021 Slot 2 paper. It tests Ratios and Proportions. 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 person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2:3:52 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

  1. A.

    692

  2. B.

    688

  3. C.

    653

  4. D.

    675

Answer

B

Explanation

Let the amounts of tea bought be 2k,3k,5k2k, 3k, 5k kg. Total weight = 10k10k kg. Total Cost Price (CP) = 2k(800)+3k(500)+5k(300)=1600k+1500k+1500k=4600k2k(800) + 3k(500) + 5k(300) = 1600k + 1500k + 1500k = 4600k. Average CP per kg = 4600k10k=₹460\frac{4600k}{10k} = ₹ 460.

Target Selling Price (SP) for 50% profit: Total Target SP = 1.5×4600k=6900k1.5 \times 4600k = 6900k.

Selling 16\frac{1}{6} of the mixture (=10k6=\frac{10k}{6} kg) at ₹ 700 per kg: Revenue1=10k6×700=7000k6=3500k3\text{Revenue}_1 = \frac{10k}{6} \times 700 = \frac{7000k}{6} = \frac{3500k}{3}

Remaining weight = 10k−10k6=50k6=25k310k - \frac{10k}{6} = \frac{50k}{6} = \frac{25k}{3} kg. Let PP be the selling price per kg of remaining tea. Revenue2=25k3×P\text{Revenue}_2 = \frac{25k}{3} \times P

Total Revenue = 3500k3+25k3P=6900k\frac{3500k}{3} + \frac{25k}{3} P = 6900k 3500+25P=3×6900=207003500 + 25P = 3 \times 6900 = 20700 25P=1720025P = 17200 P=1720025=688P = \frac{17200}{25} = 688.

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