A hare pursued by a hound is 60 of her own leaps before him. When the hare takes 4 leaps, the hound takes 3. In one leap, the hare goes $1\frac{3}{4}$ metres and the hound $2\frac{3}{4}$ metres. In how many leaps will the hound overtake the hare?

### Concept & Logic This is a classic relative distance problem. We need to find the effective distance the hound gains on the hare in a specific time interval (measured in leaps). ### Step-by-Step Solution 1. **Initial Distance:** The hare is 60 of her leaps ahead. Distance = $60 \times 1\frac{3}{4} = 60 \times \frac{7}{4} = 15 \times 7 = 105 \text{ metres}$. 2. **Interval Analysis:** Consider an interval where the hare takes 4 leaps and the hound takes 3 leaps. Distance covered by hare in this interval = $4 \times 1\frac{3}{4} = 4 \times \frac{7}{4} = 7 \text{ metres}$. Distance covered by hound in this interval = $3 \times 2\frac{3}{4} = 3 \times \frac{11}{4} = \frac{33}{4} = 8.25 \text{ metres}$. 3. **Relative Distance Gained:** In one such interval, the distance the hound gains on the hare is: Gain = $8.25 - 7 = 1.25 \text{ metres} = \frac{5}{4} \text{ metres}$. 4. **Total Intervals Needed:** Number of intervals required to cover the initial 105 metres gap: Number of intervals = $\frac{105}{1.25} = 105 \times \frac{4}{5} = 21 \times 4 = 84 \text{ intervals}$. 5. **Total Hound Leaps:** In each interval, the hound takes 3 leaps. Total leaps = $84 \times 3 = 252 \text{ leaps}$. ### Exam Strategy & Shortcut Convert everything to improper fractions immediately to avoid calculation errors with mixed numbers. Track "intervals" instead of time to bypass complex rate equations. ### Common Pitfall A common mistake is confusing the hound's leaps with the hare's leaps when calculating the final answer. The question asks for the hound's leaps. ### Final Answer Therefore, the correct answer is **252**.