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CAT 2018 Slot 1 — QA Question 13

MixturesEasy

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

Answer & solution

Correct answer: 20

  • 20

  • B

    26

  • C

    16

  • D

    22

Solution

Easy

Translate the cost relation and the profit into one linear equation in the per-litre cost of B and the volume of A. The cost of B is largest when the volume of A is at its allowed minimum.

1

Set up variables. Let the mixture hold YY litres of A and (10−Y)(10-Y) litres of B. Let B cost $X$\$X\$ per litre, so A costs $(X+8)$\$(X+8)\$ per litre. Since B does not exceed A:

10−Y≤Y⇒ Y≥5(B does not exceed A)\begin{aligned} &10-Y \le Y\\ &\Rightarrow\ Y \ge 5 \quad\text{(B does not exceed A)} \end{aligned}
2

Recover the cost price from the profit. A 10%10\% profit gives selling price 264264:

CP×1.1=264⇒ CP=2641.1=240(remove the 10% markup)\begin{aligned} &\text{CP}\times 1.1 = 264\\ &\Rightarrow\ \text{CP} = \frac{264}{1.1} = 240 \quad\text{(remove the }10\%\text{ markup)} \end{aligned}
3

Cost equation. Total cost of the two paints equals 240240:

(X+8)Y+(10−Y)X=240⇒ 10X+8Y=240(expand and simplify)⇒ X=24−0.8Y\begin{aligned} &(X+8)Y + (10-Y)X = 240\\ &\Rightarrow\ 10X + 8Y = 240 \quad\text{(expand and simplify)}\\ &\Rightarrow\ X = 24 - 0.8Y \end{aligned}
4

Maximise XX. XX decreases as YY grows, so take the smallest allowed YY, namely Y=5Y=5 from step 1:

X=24−0.8×5⇒ X=24−4=20\begin{aligned} &X = 24 - 0.8\times 5\\ &\Rightarrow\ X = 24 - 4 = 20 \end{aligned}
Highest cost of B=20 Rs. per litre\text{Highest cost of B} = \mathbf{20}\ \text{Rs. per litre}

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CAT 2018 Slot 1 QA Q13: A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed — Solution | TheCATExam