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

Answer 1

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

Step 1: Form the equations 

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

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

Step 2: Eliminate \(y\)

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

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-hand side: 
\(10x^2 + 30x + 12x + 36 = 10x^2 + 42x + 36\)

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

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

Step 4: Factorize

\((5x + 6)(3x - 2) = 0\) 
So, \(x = -\frac{6}{5}\) or \(x = 2\)

Since price can't be negative, take \(x = 2\)

Step 5: Find \(y\)

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

Final Answer:

\(2x + 5x + y = 4 + 10 + 20 = \boxed{34}\)