Arithmetic

Mixture & Alligation

Medium

A trader sells $10$ litres of a mixture of paints $A$ and $B$ , where the amount of $B$ in the mixture does not exceed that of $A$ . The cost of paint A per litre is Rs. $8$ more than that of paint B. If the trader sells the entire mixture for Rs. $264$ and makes a profit of $10 \%$ , then the highest possible cost of paint B, in Rs. per litre, is

$20$

$22$

$16$

$26$

✅ Correct Option: 1

Given Information:

Total mixture = 10 litres (paints A and B combined)
Amount of B ≤ Amount of A (key constraint!)
Cost of A = Cost of B + Rs. 8 per litre
Selling price = Rs. 264
Profit = 10%

Since the trader makes a 10% profit:

Selling Price = Cost Price + 10% of Cost Price

264 = Cost Price × (1 + 0.10)

264 = Cost Price × 1.10

Cost Price = 264 ÷ 1.10 = Rs. 240

Therefore, average cost per litre = 240 ÷ 10 = Rs. 24

Let us define:

Cost of paint B = Rs. x per litre
Cost of paint A = Rs. (x + 8) per litre (since A costs Rs. 8 more than B)
Quantity of paint A = a litres
Quantity of paint B = b litres

From the problem:

Total quantity: a + b = 10

Amount constraint: b ≤ a (amount of B doesn't exceed amount of A)

Cost equation: a(x + 8) + bx = 240

From the cost equation:

a(x + 8) + bx = 240

ax + 8a + bx = 240

x(a + b) + 8a = 240

Since a + b = 10:

x(10) + 8a = 240

10x + 8a = 240

x = $\dfrac{240 - 8a}{10}$ = 24 - 0.8a

To maximize x (cost of B), we need to minimize a (quantity of A).

From the constraint b ≤ a and a + b = 10:

b ≤ a

Since b = 10 - a, we get: 10 - a ≤ a

10 ≤ 2a

a ≥ 5

So the minimum value of a is 5, which means b = 5.

When a = 5:

x = 24 - 0.8(5) = 24 - 4 = Rs. 20

Let us check if this works:

Cost of A = Rs. 28 per litre, Cost of B = Rs. 20 per litre

Total cost = 5(28) + 5(20) = 140 + 100 = Rs. 240

Constraint satisfied: 5 litres of B ≤ 5 litres of A

Therefore, the highest possible cost of paint B is Rs. 20 per litre.

Key Insight: The maximum cost of B occurs when we use the minimum allowed quantity of A (and maximum allowed quantity of B), which happens when both paints are used in equal amounts - the boundary condition of our constraint.

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