Subject: General Aptitude · Chapter: Time and Distance · Exam: 2010 · Marks: · Difficulty:
A man drives 150 km to the seashore in 3 hours 20 min. He returns from the shore to the starting point in 4 hours 10 min. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$, in kilometres per hour, by
(a)2
(b)4
(c)$4 \frac{1}{2}$
(d)5
Answer
Answer (as printed): D
Explanation
Time taken to cover 150 km in going trip $=3 \mathrm{hr} 20 \mathrm{min}=3 \frac{20}{60} \mathrm{hr}=3 \frac{1}{3} \mathrm{hr}=\frac{10}{3} \mathrm{hr}$. Speed in going trip $=\left(150 \times \frac{3}{10}\right) \mathrm{km} / \mathrm{hr}=45 \mathrm{km} / \mathrm{hr}$. Time taken to cover 150 km in return trip = 4 hr 10 min. $$=4 \frac{1}{6} \mathrm{hr}=\frac{25}{6} \mathrm{hr} .$$ Speed in return trip $=\left(150 \times \frac{6}{25}\right) \mathrm{km} / \mathrm{hr}=36 \mathrm{km} / \mathrm{hr}$. ∴ Average speed $=\left(\frac{2 \times 45 \times 36}{45+36}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{2 \times 45 \times 36}{81}\right) \mathrm{km} / \mathrm{hr}=40 \mathrm{km} / \mathrm{hr}$. Required difference $=(45-40) \mathrm{km} / \mathrm{hr}=5 \mathrm{km} / \mathrm{hr}$.