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CAT 2020 Slot 3 — QA Question 22

MixturesEasy

Two alcohol solutions, A and B, are mixed in the proportion 1 : 3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

Answer & solution

  • 92%

  • B

    90%

  • C

    89%

  • D

    94%

Solution

Medium

Choose convenient volumes in the 1:31:3 ratio, then double the total by adding solution A. The alcohol in the final mixture equals the alcohol from A plus the alcohol from B; balance this to find B's strength.

1

Initial volumes. Mix A and B in 1:31:3; take 1010 L of A and 3030 L of B (total 4040 L).

A=10 L,B=30 L,total=40 L\begin{aligned} &\text{A}=10\ \text{L},\qquad \text{B}=30\ \text{L},\qquad \text{total}=40\ \text{L} \end{aligned}
2

Double the volume with A. Add 4040 L of A so the total becomes 8080 L.

A=10+40=50 L,B=30 L,total=80 L\begin{aligned} &\text{A}=10+40=50\ \text{L},\qquad \text{B}=30\ \text{L},\qquad \text{total}=80\ \text{L} \end{aligned}
3

Alcohol balance. A is 60%60\% alcohol, B is b%b\%, and the final 8080 L is 72%72\% alcohol.

80×72%=50×60%+30×b%⇒ 57.6=30+0.30 b⇒ 0.30 b=27.6⇒ b=92\begin{aligned} &80\times 72\%=50\times 60\%+30\times b\%\\ &\Rightarrow\ 57.6=30+0.30\,b\\ &\Rightarrow\ 0.30\,b=27.6\\ &\Rightarrow\ b=92 \end{aligned}
Alcohol in B=92%\text{Alcohol in B}=92\%
CAT 2020 Slot 3 QA Q22: Two alcohol solutions, A and B, are mixed in the proportion 1 : 3 by volume. The volume of the mixture is then — Solution | TheCATExam