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CAT 2018 Slot 2 — QA Question 14

MixturesEasy

A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1 : 3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

Answer & solution

Correct answer: 50%

  • A

    52%

  • 50%

  • C

    48%

  • D

    55%

Solution

Easy

Track the total ethanol against the total volume. Pick convenient volumes for the 1:31:3 mix, then add an equal volume of the 20%20\% solution and force the final concentration to 31.25%31.25\%.

1

Set up the first blend. Mix the 20%20\% solution with SS (concentration x%x\%) in ratio 1:31:3. Take 1010 L of the 20%20\% and 3030 L of SS, giving 4040 L of intermediate mix.

ethanol so far=20%⋅10+x⋅30=200+30x(in litre-percent units)\begin{aligned} &\text{ethanol so far}=20\%\cdot 10+x\cdot 30=200+30x \quad\text{(in litre-percent units)} \end{aligned}
2

Add an equal volume of 20%20\%. "Equal volume" means another 4040 L of the 20%20\% solution. Total volume =40+40=80=40+40=80 L; that extra 4040 L carries 20%⋅40=80020\%\cdot 40=800 units of ethanol.

total ethanol=(200+30x)+800=1000+30xtotal volume=80 L\begin{aligned} &\text{total ethanol}=(200+30x)+800=1000+30x\\ &\text{total volume}=80\text{ L} \end{aligned}
3

Apply the final concentration. The result is 31.25%31.25\%.

1000+30x80=31.25⇒ 1000+30x=2500(80×31.25=2500)⇒ 30x=1500⇒ x=50\begin{aligned} &\frac{1000+30x}{80}=31.25\\ &\Rightarrow\ 1000+30x=2500 \quad\text{(}80\times31.25=2500\text{)}\\ &\Rightarrow\ 30x=1500\\ &\Rightarrow\ x=50 \end{aligned}
S=50%S=50\%

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CAT 2018 Slot 2 QA Q14: A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proporti — Solution | TheCATExam