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

Let coffee price = C Rs/kg and cocoa price = K Rs/kg.

From the two given mixtures:

$$0.16C + 0.84K = 240$$

$$0.36C + 0.64K = 320$$

Multiply both equations by 100 to remove decimals:

$$16C + 84K = 24000$$ 

$$36C + 64K = 32000$$

Subtract the first from the second:

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

$$20C - 20K = 8000 \implies C - K = 400$$

Put $$C=K+400$$ into $$16C+84K=24000$$:

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

$$16K+6400+84K=24000 $$

$$100K = 17600 \implies K = 176$$

So $$C = 176 + 400 = 576$$ (Rs/kg).

For the new mixture priced at Rs 376/kg, let the coffee fraction be p. Then

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

Upon solving $$p = \tfrac{1}{2}$$

Thus, coffee is (50%) of the new mixture. In 10 kg of this mixture, coffee = $$10\times 0.5 = 5\text{kg}$$