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CAT 2021 Slot 3 — QA Question 15

AlligationEasy

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

Answer & solution

  • A

    2.5

  • B

    3.5

  • 3

  • D

    4

Solution

Easy

Let the initial alloy weigh xx kg with silver fraction p%p\%. Write the silver balance for each mixing (with pure silver, and with the 90%90\% alloy). Two equations in xx and pp solve cleanly.

1

Mix with 33 kg pure silver →90%\to 90\%. Silver in == silver out.

p100x+3=0.90 (x+3)(silver balance)⇒ p100x+3=0.9x+2.7⇒ p100x=0.9x−0.3...(1)\begin{aligned} &\frac{p}{100}x + 3 = 0.90\,(x+3) \quad\text{(silver balance)}\\ &\Rightarrow\ \frac{p}{100}x + 3 = 0.9x + 2.7\\ &\Rightarrow\ \frac{p}{100}x = 0.9x - 0.3 \quad\text{...(1)} \end{aligned}
2

Mix with 22 kg of 90%90\% alloy →84%\to 84\%. Again balance silver.

p100x+0.9⋅2=0.84 (x+2)⇒ p100x+1.8=0.84x+1.68⇒ p100x=0.84x−0.12...(2)\begin{aligned} &\frac{p}{100}x + 0.9\cdot 2 = 0.84\,(x+2)\\ &\Rightarrow\ \frac{p}{100}x + 1.8 = 0.84x + 1.68\\ &\Rightarrow\ \frac{p}{100}x = 0.84x - 0.12 \quad\text{...(2)} \end{aligned}
3

Subtract to find xx. The left sides of (1) and (2) are identical.

0.9x−0.3=0.84x−0.12[(1)=(2)]⇒ 0.06x=0.18⇒ x=3\begin{aligned} &0.9x - 0.3 = 0.84x - 0.12 \quad\text{[(1)=(2)]}\\ &\Rightarrow\ 0.06x = 0.18\\ &\Rightarrow\ x = 3 \end{aligned}

(Back-substituting gives p=80%p = 80\%, a valid silver content, confirming the answer.)

x=3 kgx = 3\ \text{kg}
CAT 2021 Slot 3 QA Q15: If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy wi — Solution | TheCATExam