CAT 2023 Slot 3 · Quant · Question 12

Let the three people weigh S kg in total, so the current average is \frac{S}{3}.

If D joins, there are four people weighing S+D, and the average falls by x.

So \frac{S+D}{4}=\frac{S}{3}-x.

If E joins instead, there are again four people, and the average rises by 2x.

So \frac{S+E}{4}=\frac{S}{3}+2x.

Subtracting the first equation from the second, the \frac{S}{3} terms cancel.

That leaves \frac{E-D}{4}=3x.

So E-D=12x.

But E is stated to be 12 kg heavier than D, so E-D=12.

Therefore 12x=12, giving x=1 — the fourth option.

Shortcut. Only the difference between the two scenarios matters, and subtracting them removes S entirely. The headcount is 4 either way, so the weight gap E-D equals 4 times the average gap, which is x+2x=3x — one line to E-D=12x.