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 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:
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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.
Updated On: Jul 4, 2026
- \(2.5\)
- \(5\)
- \(4\)
- \(6\)
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The Correct Option is B
Approach Solution - 1
Approach: Each mixture's price is a weighted average of the (unknown) prices of pure coffee and pure cocoa. Two mixtures give two equations to pin those prices, then the third mixture's price tells us its coffee fraction.
Step 1: Let pure coffee cost \(C_f\) and pure cocoa cost \(C_c\) (Rs/kg). Mixture 1 (16% coffee): \[ 0.16C_f+0.84C_c=240. \tag{1} \] Mixture 2 (36% coffee): \[ 0.36C_f+0.64C_c=320. \tag{2} \]
Step 2: Subtract (1) from (2): \(0.20C_f-0.20C_c=80\Rightarrow C_f-C_c=400\).
Step 3: Put \(C_f=C_c+400\) into (1): \(0.16(C_c+400)+0.84C_c=240\Rightarrow C_c+64=240\Rightarrow C_c=176\), hence \(C_f=576\).
Step 4: Let the new mixture be a fraction \(x\) coffee, \(1-x\) cocoa, costing 376: \[ 576x+176(1-x)=376\Rightarrow 400x=200\Rightarrow x=\tfrac12. \] So it is 50% coffee.
Step 5: Coffee in 10 kg \(=\tfrac12\times 10=5\) kg.
Answer: 5 kg (option 2).
Step 1: Let pure coffee cost \(C_f\) and pure cocoa cost \(C_c\) (Rs/kg). Mixture 1 (16% coffee): \[ 0.16C_f+0.84C_c=240. \tag{1} \] Mixture 2 (36% coffee): \[ 0.36C_f+0.64C_c=320. \tag{2} \]
Step 2: Subtract (1) from (2): \(0.20C_f-0.20C_c=80\Rightarrow C_f-C_c=400\).
Step 3: Put \(C_f=C_c+400\) into (1): \(0.16(C_c+400)+0.84C_c=240\Rightarrow C_c+64=240\Rightarrow C_c=176\), hence \(C_f=576\).
Step 4: Let the new mixture be a fraction \(x\) coffee, \(1-x\) cocoa, costing 376: \[ 576x+176(1-x)=376\Rightarrow 400x=200\Rightarrow x=\tfrac12. \] So it is 50% coffee.
Step 5: Coffee in 10 kg \(=\tfrac12\times 10=5\) kg.
Answer: 5 kg (option 2).
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Approach Solution -2
Alternate approach: alligation by external division.
Picture the new mixture (Rs 376/kg) as a blend of mixture 1 (16% coffee, Rs 240) and mixture 2 (36% coffee, Rs 320). Since Rs 376 is higher than both given prices, no ordinary blend of the two can reach it — the blend has to be external, i.e. one component effectively carries a negative share.
By alligation, the (external) ratio in which mixture 1 and mixture 2 combine to price 376 is
\[ \frac{\text{mixture 1}}{\text{mixture 2}} = \frac{320-376}{376-240} = \frac{-56}{136} = -\frac{7}{17}. \]
Applying this same ratio to the coffee percentages of the two mixtures gives the new mixture's coffee share:
\[ p = \frac{(-7)(16) + (17)(36)}{-7+17} = \frac{-112+612}{10} = 50\%. \]
So 50% of the new mixture is coffee, and in 10 kg that is \(0.50 \times 10 = 5\) kg.
Answer: 5 kg (option 2).
Picture the new mixture (Rs 376/kg) as a blend of mixture 1 (16% coffee, Rs 240) and mixture 2 (36% coffee, Rs 320). Since Rs 376 is higher than both given prices, no ordinary blend of the two can reach it — the blend has to be external, i.e. one component effectively carries a negative share.
By alligation, the (external) ratio in which mixture 1 and mixture 2 combine to price 376 is
\[ \frac{\text{mixture 1}}{\text{mixture 2}} = \frac{320-376}{376-240} = \frac{-56}{136} = -\frac{7}{17}. \]
Applying this same ratio to the coffee percentages of the two mixtures gives the new mixture's coffee share:
\[ p = \frac{(-7)(16) + (17)(36)}{-7+17} = \frac{-112+612}{10} = 50\%. \]
So 50% of the new mixture is coffee, and in 10 kg that is \(0.50 \times 10 = 5\) kg.
Answer: 5 kg (option 2).
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