Question:medium

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

Updated On: Jan 15, 2026
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Solution and Explanation

Let the prices of the small, medium, and large cups be \(2x\), \(5x\), and \(y\) respectively.

Step 1: Establish the equations 

From the given information: 
\((2x)(5x)(y) = 800 \quad \text{(1)}\)

When the prices of the small and medium cups are increased by 6, the product becomes: 
\((2x + 6)(5x + 6)(y) = 3200 \quad \text{(2)}\)

Step 2: Isolate \(y\)

Divide Equation (2) by Equation (1): 
\(\frac{(2x + 6)(5x + 6)(y)}{(2x)(5x)(y)} = \frac{3200}{800}\)

This simplifies to: 
\(\frac{(2x+6)(5x+6)}{10x^2} = 4\)

Step 3: Solve the equation

Multiply both sides by \(10x^2\): 
\((2x+6)(5x+6) = 40x^2\)

Expand the left side: 
\(10x^2 + 30x + 12x + 36 = 10x^2 + 42x + 36\)

Rearrange the terms: 
\(10x^2 + 42x + 36 = 40x^2\) 
\(30x^2 - 42x - 36 = 0\)

Divide by 2: 
\(15x^2 - 21x - 18 = 0\)

Step 4: Factor the quadratic equation

\((5x + 6)(3x - 2) = 0\) 
The possible values for \(x\) are \(x = -\frac{6}{5}\) or \(x = 2\).

Since a price cannot be negative, we select \(x = 2\).

Step 5: Determine the value of \(y\)

Substitute \(x = 2\) into Equation (1): 
\((2 \cdot 2)(5 \cdot 2)(y) = 800\) 
\(4 \cdot 10 \cdot y = 800 \Rightarrow 40y = 800 \Rightarrow y = 20\)

Final Result:

The sum of the prices is \(2x + 5x + y = 4 + 10 + 20 = \boxed{34}\).

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