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
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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.
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