Question

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

Accepted Answer

34. Let prices be $S, M, L$. Given $S \cdot M \cdot L = 800$. $S : M = 2 : 5 \implies S = 2x, M = 5x$. New prices: $(2x + 6), (5x + 6), L$. $$(2x + 6)(5x + 6)L = 3200$$ Dividing new product by old product: $$\frac{(2x + 6)(5x + 6)L}{(2x)(5x)L} = \frac{3200}{800} = 4$$ $$\frac{10x^2 + 42x + 36}{10x^2} = 4 \implies 30x^2 - 42x - 36 = 0 \implies 5x^2 - 7x - 6 = 0$$ $$(5x + 3)(x - 2) = 0 \implies x = 2$$ Thus, $S = 4, M = 10$. $L = \frac{800}{40} = 20$. Sum of original prices = $4 + 10 + 20 = 34$.

Sum of prices of three tea cup sizes

Explanation

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