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.

## No comments:

## Post a Comment

Your Comment Here!.....