IPMAT Indore 2022 · Quantitative Ability · Question 40

Step 1 — set up. Let A be Ayesha’s position atop the tower, B the foot of the tower, and C, D the car’s positions at 11:00 and 11:02. The tower height is AB=200 m.

Step 2 — use the 45° observation. The angle of depression equals the angle of elevation from the car, so in triangle ABC:

\tan{45}^{∘}=1=\frac{AB}{BC}\Rightarrow BC=200 m

Step 3 — use the 30° observation.

\tan{30}^{∘}=\frac{1}{\sqrt{3}}=\frac{AB}{BD}\Rightarrow BD=200\sqrt{3}m

Step 4 — find the distance travelled.

CD=BD-BC=200\sqrt{3}-200=200\left( \sqrt{3}-1 \right)\approx 200\left( 0.732 \right)=146.4 m

Step 5 — convert to speed. That distance was covered in 2 minutes:

v=\frac{0.1464\text{km}}{2/60 h}=0.1464\times 30=4.392\text{km/h}

Answer: Option 4 — approximately 4.39 km per hour

Shortcut. The 45° reading hands you the first distance free: a depression of 45° means the horizontal distance equals the tower height, 200 m. No trigonometry needed for that step.

Then the gap is 200\left( \sqrt{3}-1 \right)\approx 146.4 m in 2 minutes, and multiplying by 30 converts m-per-2-minutes into km/h in one step:

0.1464\times 30=4.39\text{km/h}

Sanity-check the magnitude before computing precisely. Roughly 150 m in 2 minutes is about 4.5 km/h — a brisk walk. Only one option is anywhere near that, so the question can be answered on estimation alone.