TheCATExam
Sign in

CAT 2019 Slot 1 — QA Question 5

AlligationEasy

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

Answer & solution

Correct answer: 80

  • A

    70

  • B

    85

  • 80

  • D

    75

Solution

Easy

Convert all weights to densities in g/L. The mixture's density is a volume-weighted average of the two, so apply alligation between the two densities and the mixture density to get the volume ratio, then convert to a percentage.

1

Find the densities. Liquid 1: 11 kg/L =1000=1000 g/L. Liquid 2: 800800 g/L.

d1=1000 g/L,d2=800 g/L\begin{aligned} &d_1 = 1000\ \text{g/L},\qquad d_2 = 800\ \text{g/L} \end{aligned}
2

Density of the mixture. Half a litre weighs 480480 g, so one litre weighs 960960 g.

dm=480 g0.5 L=960 g/L\begin{aligned} &d_m = \frac{480\ \text{g}}{0.5\ \text{L}} = 960\ \text{g/L} \end{aligned}
3

Alligation for the volume ratio. The ratio of volumes is the inverse ratio of the distances from the mean.

V1V2=dm−d2d1−dm=960−8001000−960=16040=41\begin{aligned} &\frac{V_1}{V_2} = \frac{d_m-d_2}{d_1-d_m} = \frac{960-800}{1000-960} = \frac{160}{40} = \frac{4}{1} \end{aligned}
4

Convert to a percentage. Liquid 1 is 44 parts out of 4+1=54+1=5.

V1V1+V2×100=45×100=80%\begin{aligned} &\frac{V_1}{V_1+V_2}\times 100 = \frac{4}{5}\times 100 = 80\% \end{aligned}
Liquid 1=80%\text{Liquid 1} = 80\%

Related Alligation questions

See all Mixture, Alligation, Removal & Replacement questions →
CAT 2019 Slot 1 QA Q5: A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm — Solution | TheCATExam