CAT 2023 Slot 2 · Quant · Question 17

The three figures ABP, APQ and AQCD together tile the whole rectangle, whose area is 9\times 6=54 sq cm.

Let their areas be the geometric progression t, tr and t{r}^{2}.

The last is stated to be four times the first, so t{r}^{2}=4t, giving {r}^{2}=4 and r=2 since areas are positive.

So the three areas are t, 2t and 4t, and they sum to the rectangle: 7t=54.

Now relate the areas to the lengths along BC.

Triangles ABP and APQ both have AB as their height from the line BC, and AB is 9 cm.

So the area of ABP is \frac{1}{2}\times BP\times 9 and the area of APQ is \frac{1}{2}\times PQ\times 9.

Since these areas are t and 2t, the bases are in the same ratio: BP:PQ=1:2.

For QC, note that the area of AQCD is 54-\frac{1}{2}\times BQ\times 9, where BQ=BP+PQ.

Writing BP=p, we have PQ=2p and BQ=3p, so the area of AQCD is 54-\frac{27p}{2}.

Setting this equal to 4t=4\times \frac{9p}{2}=18p gives 54-\frac{27p}{2}=18p, so 54=\frac{63p}{2} and p=\frac{12}{7}.

Then PQ=\frac{24}{7} and QC=6-3p=6-\frac{36}{7}=\frac{6}{7}.

The ratio BP:PQ:QC is \frac{12}{7}:\frac{24}{7}:\frac{6}{7}, which is 12:24:6, that is 2:4:1 — the third option.

Shortcut. The first two figures are triangles on the same height, so their areas and bases share the ratio 1:2 — and only option 3 and option 2 begin that way. Since BP, PQ and QC must add to 6 while BQ stays under 6, option 2 would need the parts to total 1+2+4=7 units with QC the largest, which contradicts AQCD being only four times ABP.