# Accenture Sample Problems On Speed

Below are three problems dealing with speed and time calculations.

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.

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.

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.

**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

**Answer :**b)4secSolution :

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

**Answer :**c)7.23a.mSolution :

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

**Answer :**a)7 p.mSolution :

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.

Accenture Sample Problems On Speed
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