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Numerical Ability
Time Distance and Speed
A man drives 150 km from to in 3 hours and 20 minutes and returns to the place in 4 hours and 10 minutes. If is the average speed of the entire trip, then the average speed of the trip from to , exceeds by:
A. 5.0 km/hour
B. 4.5 km/hour
C. 4.0 km/hour
D. 2.5 km/hour
E. 3.5 km/hour
Read Solution (Total 3)
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- Suppose Man drives From A to B (i.e 150 Km) in 3 hours 20 minutes (i.e 3 + 1/3 = 10/3 hours) then Speed is 150/(10/3) = 45km/hr. Now again he drives from B to A (same 150 km) in 4 hours 10 minutes (i.e 4+1/6 = 25/6 hr) then speed = 150/(25/6) = 36km/hr. Now Average Speed is (45+36)/2 = 40.5 km/hr. Question is to find avergae speed exceeds Speed from to by (i.e from A to B by) which is 45 - 40.5 = 4.5km/hr.
- 7 years agoHelpfull: Yes(2) No(3)
- 3 hrs 20 min = 3 + (20/60) = 10/3
4 hrs 10 min = 4 + (10/60) = 25/6
S1 = 150/(10/3) = (150 * 3)/10 = 45 k/h
S2 = 150/(25/6) = (150 * 6)/25 = 36 k/h
Average Speed = (2 * 45 * 36)/(45 + 36) = 40
By seeing we can say that difference b/w average speed and S1 is 5 k/h
- 7 years agoHelpfull: Yes(2) No(0)
- 4.5 is the correct answer
- 5 years agoHelpfull: Yes(0) No(0)
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