CAT 2023 Slot 1 · Quant · Question 13

Let the common selling price in the first situation be S.

A 20% profit on A means S=1.2\times{\text{CP}}_{A}, so {\text{CP}}_{A}=\frac{S}{1.2}.

A 10% loss on B means S=0.9\times{\text{CP}}_{B}, so {\text{CP}}_{B}=\frac{S}{0.9}.

In the new situation both are sold at a common price S', with a 10% profit on B.

So S'=1.1\times{\text{CP}}_{B}=1.1\times \frac{S}{0.9}=\frac{11S}{9}.

The profit percentage on A is \frac{S'-{\text{CP}}_{A}}{{\text{CP}}_{A}}=\frac{S'}{{\text{CP}}_{A}}-1.

Substituting, \frac{S'}{{\text{CP}}_{A}}=\frac{11S}{9}\times \frac{1.2}{S}=\frac{13.2}{9}.

So the profit fraction is \frac{13.2}{9}-1=\frac{4.2}{9}=0.4667.

That is about 46.67%, nearest to 47% — the second option.

Shortcut. Only the ratio of the two selling prices matters, and it is \frac{1.1}{0.9}=\frac{11}{9}. So A's profit factor rises from 1.2 to 1.2\times \frac{11}{9}=\frac{22}{15}\approx 1.467 in one multiplication, and no cost price is ever needed.