Question : A man makes four trips of equal distances. His speed on the first trip was 720 km/hr and in each subsequent trip, his speed was half of the previous trip. What is the average speed of the man in these four trips?

Option 1: 104 km/hr

Option 2: 156 km/hr

Option 3: 192 km/hr

Option 4: 288 km/hr

Correct Answer: 192 km/hr

Solution : A man makes four trips of equal distances.
His speed on the first trip was 720 km/hr.
In each subsequent trip, his speed was half of the previous trip.
Let's denote the distance of each trip as d.
For the first trip, his speed was 720 km/hr, so the time taken was = $\frac{\text{d}}{720}$
For the second trip, his speed was $\frac{720}{2}$ = 360 km/hr
So the time taken was = $\frac{\text{d}}{360}$
For the third trip, his speed was $\frac{360}{2}$ = 180 km/hr
So the time taken was = $\frac{\text{d}}{180}$
For the fourth trip, his speed was $\frac{180}{2}$ = 90 km/hr
So the time taken was = $\frac{\text{d}}{90}$
So, the total distance travelled was 4d, and the total time taken was = $\frac{\text{d}}{720}$ + $\frac{\text{d}}{360}$ + $\frac{\text{d}}{180}$ + $\frac{\text{d}}{180}$
Average speed = $\frac{\text{Total distance}}{\text{Total time}}$
= $\frac{4\text{d}}{\frac{\text{d}}{720} + \frac{\text{d}}{360} + \frac{\text{d}}{180} + \frac{\text{d}}{180}}$
= $\frac{4}{ (\frac{1}{720} + \frac{2}{720} + \frac{4}{720} + \frac{8}{720})}$
= $\frac{4} {\frac{15}{720}}$
= $4 \times {\frac{720}{15}}$
= 192 km/hr
Hence, the correct answer is 192 km/hr.