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Sum of prices of three tea cup sizes

CAT 2021 Slot 3 · Quantitative Ability · Easy · Linear Equations

This is an easy Quantitative Ability question from the CAT 2021 Slot 3 paper. It tests Linear Equations. 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 tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Answer

34

Explanation

Let prices be S,M,LS, M, L. Given S⋅M⋅L=800S \cdot M \cdot L = 800. S:M=2:5  ⟹  S=2x,M=5xS : M = 2 : 5 \implies S = 2x, M = 5x.

New prices: (2x+6),(5x+6),L(2x + 6), (5x + 6), L. (2x+6)(5x+6)L=3200(2x + 6)(5x + 6)L = 3200

Dividing new product by old product: (2x+6)(5x+6)L(2x)(5x)L=3200800=4\frac{(2x + 6)(5x + 6)L}{(2x)(5x)L} = \frac{3200}{800} = 4 10x2+42x+3610x2=4  ⟹  30x2−42x−36=0  ⟹  5x2−7x−6=0\frac{10x^2 + 42x + 36}{10x^2} = 4 \implies 30x^2 - 42x - 36 = 0 \implies 5x^2 - 7x - 6 = 0 (5x+3)(x−2)=0  ⟹  x=2(5x + 3)(x - 2) = 0 \implies x = 2

Thus, S=4,M=10S = 4, M = 10. L=80040=20L = \frac{800}{40} = 20. Sum of original prices = 4+10+20=344 + 10 + 20 = 34.

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