CSAT Solved Papers/ 2022/Q69

2022 CSAT — Q69

Quant Arithmetic & numeracy 2.5 marks Hard

There are two containers $X$ and $Y$ . $X$ contains $100$ ml of milk and $Y$ contains $100$ ml of water. $20$ ml of milk from $X$ is transferred to $Y$ . After mixing well, $20$ ml of the mixture in $Y$ is transferred back to $X$ . If $m$ denotes the proportion of milk in $X$ and $n$ denotes the proportion of water in $Y$ , then which one of the following is correct?

A m = n Answer
B m > n
C m < n
D Cannot be determined due to insufficient data

Worked rationale

Step 1: Move $20$ ml milk $X \to Y$ . Now $X$ has $80$ ml milk; $Y$ has $20$ ml milk $+ 100$ ml water $= 120$ ml, with milk fraction $\tfrac{20}{120} = \tfrac16$ .

Step 2: Move $20$ ml of $Y$ ‘s mixture back to $X$ . That $20$ ml carries milk $= 20\cdot\tfrac16 = \tfrac{10}{3}$ ml and water $= 20\cdot\tfrac{5}{6} = \tfrac{50}{3}$ ml.

Final $X$ ( $100$ ml): milk $= 80 + \tfrac{10}{3} = \tfrac{250}{3}$ , water $= \tfrac{50}{3}$ . So $m = \dfrac{250/3}{100} = \dfrac{5}{6}$ .

Final $Y$ ( $100$ ml): water $= 100 - \tfrac{50}{3} = \tfrac{250}{3}$ , milk $= \tfrac{50}{3}$ . So $n = \dfrac{250/3}{100} = \dfrac{5}{6}$ .

Hence $m = n$ .

Answer: (a) m = n.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2022 CSAT Q69 — Arithmetic & numeracy — Tap the drawing to open it full size for the fine detail.

Why the other options miss

B

broke the symmetry: assumes $X$ “keeps more of its own” milk than the water displaced into $Y$ , breaking the conservation symmetry.
C

the mirror error: over-crediting the dilution of $X$ .
D

gave up too early: thinks the mixing fraction leaves the result open, missing the exact equality.

Specialist insight

This is the classic displacement-symmetry result: after equal-volume transfers between two equal containers, the amount of milk that ends up in $Y$ equals the amount of water that ends up in $X$ (both volumes return to $100$ ml, and what leaves one as milk is replaced by exactly that much water). So the milk fraction of $X$ equals the water fraction of $Y$ — always $m = n$ , no arithmetic needed. The full computation merely confirms the invariant.

The trap, in one line

Equal-volume swap between equal containers $\Rightarrow$ milk in $Y$ $=$ water in $X$ , so $m = n = \tfrac56$ $=$ (a).

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