### IBM Sample Problem Using Speed

**Question 1**

A policeman starts chasing a thief 30 minutes after the thief had run from a spot. With an average speed of 20km per hour, he takes 2 hours to catch the thief. What is the average speed of the thief?

a)16km/hr b)25km/hr c)24km/hr d)18km/hr

**Answer :**a)16km/hr

Solution:

As given, the average speed of the policeman = 20km/hr.

He takes 2 hours to catch the thief, so from formula, "distance = speed x time" we have

The total distance covered by the police to catch the thief = 20 x 2 = 40 km (This value is also equal to the distance run by thief before being caught by Police.)

Policeman had started late by 30 minutes and took 2 hours to catch the running thief.

Above means that the thief takes (30minutes + 2 hours =) 5/2 hours to reach 40km.

So the speed of the thief = 40/(5/2) = 40 x 2 / 5 = 16 km/hr.

Hence the answer is 16km/hr.

**Question 2**

From a particular spot, Tom started to chase Jerry which had left the spot before 30 minutes. Tom ran across a highway and three streets. After travelling 1 hour Tom met Jerry at a distance if 120 meters. Find the ratio of the speed of Tom to that of Jerry.

a)X=2,Y=1 b)X=3,Y=2 c)X=5,Y=2 d)X=4,Y=3

**Answer :**b)X=3,Y=2

Solution :

**Part :1**To find the speed of Tom

As given in the question, the total time taken by Tom = 1 hour and

The total distance = 120 meters = 0.12 km

Now, the speed of Tom = distance / time = 0.12 / 1 = 0.12km/hr

**Part :2**To find the speed of Jerry

The total time taken by Jerry = 30 minutes + 1 hour = 3/2 hour

Distance = 0.12km

Then, the speed of Jerry = 0.12/(3/2) = 0.08km/hr.

**Part :3**To find the ratio of the speed of Tom to that of Jerry

Based on part 1 and 2, ratio of the speed of Tom and Jerry = 0.12:0.08 = 3:2

**Question 3**

A call taxi started from certain place A at 7.00am. The taxi reached its destination, waited there for 30 minutes and returned back to A. The destination is 100 meters from A. The speed of the taxi is 50 meters per hour. In return journey, taxi travels with 20% greater speed (as compared to onward journey speed). At what time it will return to A?

a)11.00am b)11.10am c)11.20am d)12.10pm

**Answer :**b)11.10am

Solution:

From the question, the distance between A and the destination is 100 meters and speed of the taxi is 50mph.

Then the time taken to reach the destination = Time = distance / speed = 100/50 hr = 2 hours.

Since the taxi started at 7.00 am, it should had reached the destination at 9.00am.

It is given that the taxi stayed at destination for 30 minutes. Therefore, it should have started from destination at 9.30am.

On return journey its speed had increased by 20%. This means, its new speed = Original speed + 20% of Original speed = 50 + (20/100)50 = 60mph.

Then the time taken to reach A = 100/60 = 1 + 2/3 hours = 1 hour and 40 minutes.

So it should had reached A at 11.10am.

Hence the answer is 11.10am.

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