TheCATExam
Sign in

CAT 2020 Slot 1 — QA Question 20

MixturesEasy

An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg, of the metal C is

Answer & solution

  • A

    70

  • 84

  • C

    96

  • D

    48

Solution

Easy

Pick a convenient block of the alloy using the volume proportion, multiply each volume by its per-unit weight to get component weights, find the total weight of that block, then scale up to 130130 kg and read off C's share.

1

Choose a sample block. Take volumes A:B:C=3:4:7A:B:C = 3:4:7 (litres), with unit weights 5,2,65,2,6 per litre.

WA=3×5=15WB=4×2=8WC=7×6=42\begin{aligned} &W_A = 3\times 5 = 15\\ &W_B = 4\times 2 = 8\\ &W_C = 7\times 6 = 42 \end{aligned}
2

Total weight of the block.

W=15+8+42=65 kg⇒ C is 42 kg out of every 65 kg of alloy\begin{aligned} &W = 15+8+42 = 65\ \text{kg}\\ &\Rightarrow\ \text{C is } 42 \text{ kg out of every } 65 \text{ kg of alloy} \end{aligned}
3

Scale to 130130 kg. Since 130=2×65130 = 2\times 65.

WC=4265×130=2×42=84 kg\begin{aligned} &W_C = \frac{42}{65}\times 130 = 2\times 42 = 84\ \text{kg} \end{aligned}
weight of C=84 kg— option (b)\text{weight of C} = 84\ \text{kg}\quad\text{— option (b)}
CAT 2020 Slot 1 QA Q20: An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the s — Solution | TheCATExam