# Accenture Sample Problems On Speed

Below are three problems dealing with speed and time calculations.

Question 1
Two buses leaving from two stations 20 km away from each other travel with constant speed of 45km/hr towards each other.Find the time taken to cross each other if the length of each bus is 100m.
a)10sec b)4sec c)15sec d)20sec
Solution :
The time taken to cross each other = (a + b) / (u + v) sec.
Here, a = b = length of the two buses = 100m = 1/10 km.
and u = v = speed of the two buses = 45km/hr.
Time taken to cross each other = (a + b) / (u + v) sec = (1/10 + 1/10) / (45 + 45) hours
1 / (10 x 90) = 1/900 hours.
since 1 hour = 3600 sec, 1/900 hour = 3600/900 = 4 sec
Hence the answer is 4 sec.

Question 2
A bus starts from A at 7 a.m and reaches the destination B at 7.30 a.m while a cyclist starts from B at 7 a.m and reaches A at 8.30 a.m. At what time the bus and the cyclist will cross each other?
a)7.34 a.m b)7.49 a.m c)7.23 a.m d)8.01 a.m
Solution :
Let the distance between A and B is X km
Time taken by the bus to cover X km = 1/2 hour .(i.e., 7 a.m to 7.30 a.m)
Time taken by the cyclist to cover X km = 3/2 hour (i.e., 7 a.m to 8.30 a.m)
Speed = distance / time
Speed of the bus = X / 1/2 = 2X km/hr
And the speed of the cyclist = X / 3/2 = 2X / 3 km/hr.
Let they meet at point C at Y hours after 7 a.m.
Distance covered by the bus in Y hours = Speed X Time = 2XY km = AC
And the distance covered by the cyclist in Y hours = Speed x Time = 2XY/3 km = BC
Since C is the point where Bus and Cyclist cross each other, AC + BC = AB
In other words, 2XY + (2/3)XY = X
Y(2 + 2/3) = 1
Y = 3/8 hours.
Expressing in minutes, Y = 3/8 x 60 minutes = 22.5 minutes
Y = 23 minutes (approximately)
Therefore they meet at 7 a.m + 23 minutes = 7.23 a.m
Hence the answer is 7.23 a.m.

Question 3
A man rides a bike with a constant speed of 20 km/hr from A at 4 p.m and travels towards a destination B. Another rider with a speed of 25 km/hr starts from B at 5 p.m and travels towards A. If the straight distance between A and B is 110km then at what time they will meet?
a)7 p.m b)8 p.m c)7.30 p.m d)6 p.m
Solution :
Let the two riders meet at X hours after 4 p.m.
Speed of the first rider is 20km/hr and the distance covered by him in X hours = 20X km
Speed of the 2nd rider is 25 Km/hr and he starts at 5 p.m.
Distance covered by him in (X - 1) hours = 25(X - 1) km
Distance between A and B is 110km.
Therefore 20X + 25(X - 1) = 110
45X = 135
X = 135/45 = 3
So they meet at 3 hours after 4 p.m.
Hence the answer is 7 p.m.