# Time, Speed Distance and Work CAT Problems

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1998

Time Speed Distance CAT Problems consists of important Time and Work CAT Questions with Solutions. This Time and Speed test is very important for Quantitative aptitude for CAT. Taking a free mock test for CAT and solving CAT previous papers will help you more on this topic.

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Instructions

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

Question 1

At what time do Ram and Shyam first meet each other?

Solution:

Let the time at which they meet be t minutes past 10.

So, distance run by Ram + distance run by Shyam = 10 km

=> (60+t)5/60 [t+60 because he would have traveled for 9 am to 10 am and t minutes more before meeting Shyam]+ (15+t)*10/60 [15+t because he would have traveled from 9:45 to 10:00 and t minutes more]= 10

=> 300+5t+150+10t = 600 => t = 10

So, they meet at 10.10 am

Question 2

At what time does Shyam overtake Ram?

Solution:

Let the time at which Shyam overtakes Ram be t minutes past 10.
So, distance run by both of them is the same till that moment.
(60+t)5 = (15+t)10 => 300 + 5t = 150 + 10t => 5t = 150 => t = 30 min.
So, at 10.30 am, Shyam overtakes Ram

Question 3

Prof. Mandal walks to the market and comes back in an auto. It takes him 90 minutes to make the round trip. If he takes an auto both ways it takes him 30 minutes. On Sunday, he decides to walk both ways. How long would it take him?

Solution:

If he takes auto both ways it takes him 30 minutes, so one way trip by auto takes 15 min. So, one way trip by walk takes 90 - 15 = 75 minutes. So, a round trip by walk takes 150 min.

Question 4

If Ram drives at 60 km/hr, he reaches the office 15 minutes early. If he drives at 30 km/hr, he reaches the office 25 minutes late. Find the distance between his office and home?

Solution:

Let D be the distance and T the time required to reach office on time. Hence, D/60 = T - 15/60 and D/30 = T + 25/60. 60T-D=15 and 60T-2D=-25 So, D=40km.

Alternate solution:

The difference in time taken arises out of the difference in speed.
If he drives at 60 kmph, he reaches 15 minutes early. If he drives at 30 kmph, he reaches 25 minutes late. There is a net time difference of 40 minutes.
Therefore, d/30 - d/60 = 40/60
d/60 = 2/3
=> d = 40 km.

Question 5

On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?

Solution:

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5