CAT 2025Slot 2QAQuestion & Solution
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
Options
5
4
2.5
6
Solution
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.