Question

A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is:

• A. \(2.5\)
• B. \(5\)
• C. \(4\)
• D. \(6\)

Hint:

When mixtures of the same two ingredients are given with different compositions and costs, set up linear equations using the percentage of each ingredient and solve for the individual prices. Then use those prices to find the composition of any new mixture.

Answer 1

The Correct Option is B

Let the cost of 1 kg of coffee be \(C_f\) and the cost of 1 kg of cocoa be \(C_c\). Step 1: Use the two given mixtures to find \(C_f\) and \(C_c\). \underline{Mixture 1:} \(16%\) coffee, \(84%\) cocoa, cost Rs 240/kg: \[ 0.16 C_f + 0.84 C_c = 240 \tag{1} \] \underline{Mixture 2:} \(36%\) coffee, \(64%\) cocoa, cost Rs 320/kg: \[ 0.36 C_f + 0.64 C_c = 320 \tag{2} \] Subtract (1) from (2): \[ (0.36 C_f + 0.64 C_c) - (0.16 C_f + 0.84 C_c) = 320 - 240 \] \[ 0.20 C_f - 0.20 C_c = 80 \] Divide by \(0.20\): \[ C_f - C_c = 400 \quad\Rightarrow\quad C_f = C_c + 400. \] Substitute into (1): \[ 0.16 (C_c + 400) + 0.84 C_c = 240 \] \[ 0.16 C_c + 64 + 0.84 C_c = 240 \] \[ 1.00 C_c = 240 - 64 = 176. \] So \[ C_c = 176,\quad C_f = 176 + 400 = 576. \]
Step 2: Find the composition of the new mixture. Let the fraction of coffee in the new mixture be \(x\). Then the fraction of cocoa is \(1 - x\). Given its cost is Rs 376/kg: \[ x C_f + (1-x) C_c = 376 \] \[ x \cdot 576 + (1-x)\cdot 176 = 376 \] \[ 576x + 176 - 176x = 376 \] \[ 400x + 176 = 376 \Rightarrow 400x = 200 \Rightarrow x = \frac{200}{400} = \frac{1}{2}. \] So the new mixture is \(50%\) coffee.
Step 3: Coffee content in 10 kg of the new mixture. \[ \text{Coffee} = \frac{1}{2} \times 10 = 5\ \text{kg}. \] Therefore, the quantity of coffee in 10 kg of the new mixture is \(5\) kg.