CAT 2025

Slot 2

QA

Question & SolutionQues & Sol

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

1. Concept Used

• Topic: Mixtures and Alligations — Weighted Average Price of Blends
• Formula: $\text{Price of mixture} = p \cdot C + (1 - p) \cdot K$

where $C$ = price of coffee (Rs/kg), $K$ = price of cocoa (Rs/kg), and $p$ = fraction of coffee in the mixture.

2. Calculation

Let the price of pure coffee be $C$ Rs/kg and pure cocoa be $K$ Rs/kg.

From the two given mixtures, we set up the following system of equations:

$0.16C + 0.84K = 240 \quad \text{...(1)}$

$0.36C + 0.64K = 320 \quad \text{...(2)}$

Multiplying both equations by 100 to eliminate decimals:

$16C + 84K = 24000 \quad \text{...(1')}$

$36C + 64K = 32000 \quad \text{...(2')}$

Subtracting equation (1') from equation (2'):

$(36C + 64K) - (16C + 84K) = 32000 - 24000$

$20C - 20K = 8000$

$C - K = 400 \quad \text{...(3)}$

Substituting $C = K + 400$ into equation (1'):

$16(K + 400) + 84K = 24000$

$16K + 6400 + 84K = 24000$

$100K = 17600$

$K = 176 \text{ Rs/kg}$

Therefore: $C = 176 + 400 = 576 \text{ Rs/kg}$

Now, for the new mixture priced at Rs 376/kg, let $p$ be the fraction of coffee:

$p \cdot 576 + (1 - p) \cdot 176 = 376$

$576p + 176 - 176p = 376$

$400p = 200$

$p = \frac{1}{2} = 50%$

So, coffee constitutes 50% of the new mixture.

In 10 kg of the new mixture, quantity of coffee:

$= 10 \times \frac{1}{2} = 5 \text{ kg}$

3. Solution

Answer = Option 1 ✅

The quantity of coffee in 10 kg of the new mixture is 5 kg.