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CAT 2024 Slot 3 — QA Question 12

Removal & ReplacementEasy

A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was

Answer & solution

Answer: 30

Solution

Hard

Let ww be the water initially poured into the 300 L container (rest is milk). Remove 2w2w litres of the mixture (which is proportionally milk and water), then top up with water. Track the milk only: milk is never added, so set final milk =72%=72\% of 300300.

1

Initial milk. Water =w=w, so milk =300−w=300-w.

milk0=300−w\begin{aligned} &\text{milk}_0=300-w \end{aligned}
2

Remove 2w2w litres of solution. The fraction of milk in the mix is 300−w300\dfrac{300-w}{300}. Removing 2w2w litres removes that fraction of milk; adding pure water adds no milk.

milk1=(300−w)−2w⋅300−w300⇒ milk1=(300−w) ⁣(1−2w300)(factor)⇒ milk1=(300−w)⋅300−2w300\begin{aligned} &\text{milk}_1=(300-w)-2w\cdot\frac{300-w}{300}\\ &\Rightarrow\ \text{milk}_1=(300-w)\!\left(1-\frac{2w}{300}\right)\quad\text{(factor)}\\ &\Rightarrow\ \text{milk}_1=(300-w)\cdot\frac{300-2w}{300} \end{aligned}
3

Final milk is 72% of 300. After refilling, total is again 300300 L with 72%72\% milk =216=216 L.

(300−w)⋅300−2w300=216⇒ (300−w)(300−2w)=64800(×300)⇒ 2w2−900w+90000=64800(expand)⇒ 2w2−900w+25200=0⇒ w2−450w+12600=0\begin{aligned} &(300-w)\cdot\frac{300-2w}{300}=216\\ &\Rightarrow\ (300-w)(300-2w)=64800\quad\text{(}\times300\text{)}\\ &\Rightarrow\ 2w^2-900w+90000=64800\quad\text{(expand)}\\ &\Rightarrow\ 2w^2-900w+25200=0\\ &\Rightarrow\ w^2-450w+12600=0 \end{aligned}
4

Solve and pick the valid root. We need 2w≤3002w\le 300, i.e. w≤150w\le 150.

w=450±4502−4(12600)2⇒ w=450±202500−504002=450±1521002⇒ w=450±3902(152100=390)⇒ w=30 or 420⇒ w=30(420 exceeds 150, rejected)\begin{aligned} &w=\frac{450\pm\sqrt{450^2-4(12600)}}{2}\\ &\Rightarrow\ w=\frac{450\pm\sqrt{202500-50400}}{2}=\frac{450\pm\sqrt{152100}}{2}\\ &\Rightarrow\ w=\frac{450\pm 390}{2}\quad\text{(}\sqrt{152100}=390\text{)}\\ &\Rightarrow\ w=30\ \text{or}\ 420\\ &\Rightarrow\ w=30\quad\text{(}420\text{ exceeds 150, rejected)} \end{aligned}
30 litres30\text{ litres}
CAT 2024 Slot 3 QA Q12: A certain amount of water was poured into a 300 litre container and the remaining portion of the container was — Solution | TheCATExam