IPMAT Indore 2026 · Quantitative Ability · Question 24

Step 1 — identify the ratio. Each leg is 80\% of the previous one, so r=0.8 (check: 1000\to 800\to 640\to 512 ✓).

Step 2 — group the legs by direction. Directions repeat every 4 moves, so each direction forms a GP with ratio {r}^{4}.

East (moves 1, 5, 9, …): \frac{1000}{1-{r}^{4}}

West (moves 3, 7, …): \frac{1000{r}^{2}}{1-{r}^{4}}

\text{Net east}=\frac{1000\left( 1-{r}^{2}\right)}{1-{r}^{4}}=\frac{1000}{1+{r}^{2}}=\frac{1000}{1.64}\approx 609.76

using 1-{r}^{4}=\left( 1-{r}^{2}\right)\left( 1+{r}^{2}\right).

Step 3 — same for north/south.

\text{Net north}=\frac{1000r}{1+{r}^{2}}=\frac{800}{1.64}\approx 487.80

Step 4 — combine the two perpendicular displacements.

\sqrt{{609.76}^{2}+{487.80}^{2}}=\sqrt{371,808+237,949}\approx \sqrt{609,757}\approx 781

Answer: Option 3 — 781

Shortcut. The factorisation 1-{r}^{4}=\left( 1-{r}^{2}\right)\left( 1+{r}^{2}\right) turns four infinite sums into the single tidy result \frac{\text{first leg}}{1+{r}^{2}} for each axis. Note the net north is exactly 0.8\times the net east, so the displacement vector has the same 5:4 shape as the first two legs.