# Linear Races Questions for CAT Exam

Sunday, August 9th, 2020 1. In a race of 1000m, A beats B by 50m or 5 seconds. Find:

• B’s speed
• A’s speed
• Time is taken by A to complete the race

Sol: Since A beats B by 50m. It means by the time A reaches the winning point, B is 50 m away and as A beats B by 5 seconds, it means B takes another 5 seconds to reach the winning point. This means B covers 50m in 5 seconds i.e., B’s speed is 50/5 = 10m/s. Since A wins by 50m, in the time A covers 1000m, B covers 950m in 95 seconds (950/10). Hence A will take 95 seconds to cover 1000m.

Therefore, A’s speed is 1000/95= 10 10/19 m/sec.

In this question, we firstly find the speed of B by analyzing the distance it cover within the given time, then within the same time A had covered 1000m. By this we get the speed of A too.

2. Rakesh runs 4/3 times as fast as Mukesh. In a race, if Rakesh gives a lead of 60m to Mukesh, find the distance from the starting point where both of them will meet.

Sol: Since Rakesh runs 1 1/3 times as fast as Mukesh, in the time Mukesh runs 3 meters, Rakesh will run 4 meters i.e., Rakesh gains 1 m for every 4 meters he runs.

Since he has given a lead of 60m, he will gain this distance by covering 4 x 60= 240m.

Hence they will meet at a point 240m from the starting point.

In this question, we understand that if speeds of A & B are in the ratio m:n (m>n) then A will gain (m-n)units of distance. Now if he had given a headstart of let say h meters then to cover this headstart he had to cover {hxm/(m-n)} units.

3. In a 1500m race, A beats B by 150m and in the same race B beats C by 75m. By what distance does A beats C?

Sol: Let’s write the data given as below
A                B                         C

1500          1350m                   ?

1500m            1425m

This can be worked this way :

In the time, B runs 1500m, C runs 1425m. In the time B runs 1350m, how much distance C run?

This will be equal to (1350*1425)/1500 =1282.5

But B running 1350m is the same as A running 1500m.

Hence, A beats C by 217.5m.

This is a simple question involving the concept of unitary method. We are given distance comparison between A & B and B & C. We have to find the difference between A & C by using B.

4. In a 500m race, the ratio of speeds of two runners P and Q is 3:5. P has a start of 200m. Who wins the race and what is the distance between P and Q at the finish of the race?

Sol: Since the ratio of speeds of P and Q is 3:5, in the time P runs 300m, Q runs 500m.

Since P has a start of 200m, at the Q starts at the starting point, P has already covered 200m and he has another 300m to cover. In the time P covers this 300m, Q can cover 500m, thus reaching the finish point exactly at the same time as P.

Therefore, Both P and Q reach the finishing point at the same time.

5. Three runners A, B and C run a race with A finishing 20m ahead of B and 34m ahead of C, while B finishes 21m ahead of C. Each runner travels the entire distance at a constant speed, what was the length of the race (in m)?

Sol: When A finished the race, B is 14m ahead of C ad B has to travel 20m to finish the race. When B finished the race, he is ahead of C by 21m.

∴ , when B covered 20m, then the difference between the distance covered by B and C is 7m.

∴ to have the difference of 21m between the distances covered by B and C, B has to cover 60m.

∴ the length of the race is 60m.

In simpler way, there are 2 cases. In case 1,
Difference between B & C is 14m.

In Case 2:

After B covering 20m, difference is 21m
It means B have to run 20 m to gain 7m difference. Therefore for getting 21m difference, he have to run 60m.

6. In a 1000m race, P beats Q by 100m. Q beats R by 50m in a 500m race If P beats R by 30 seconds in 2000m race, how long would car Q take to travel a distance of 3000m?

Sol: Ratio of speeds of P & Q : 1000/900= 10/9

Ratio of speeds of Q & R : 500/450= 10/9

Ratio of speeds of P & R : (10/9) (x) (10/9) = 100/81
Since, ratio of speeds of P & R is 100/81, when P covers 2000m, R covers (81*2000)/100=1620m

⇒ R covered (2000 -1620)m in 30 seconds
⇒ Speed of R = 380/30 m/s
⇒ Speed of Q = (380/30) (x) (10/9) =380/27 m/s
⇒ Q takes 3000 ÷ 380/27= 4050/19seconds.

This question involves ratios more. We find the ratio of each car and then compared the distance travelled by P and R by which we get speed of R and then in turn speed of Q.

7. In a 200m race, if A gives B a start of 25 metres, then A wins the race by 10 seconds. Alternatively, if A gives B a start of 45 metersthe race ends in a dead heat. How long does A take to run 200m?

Sol: A gives B a start of 25 metersand still wins the race by 10 seconds.

Alternatively, if A gives B a start of 45 metres, then the race ends in a dead heat.

Therefore, the additional 20 metersstart given to B compensates for the 10 seconds.

i.e., B runs 20 metersin 10 seconds.

Hence, B will take 100 seconds to run 200 metres.

We know that A gives B a start of 45 metres. B will take 22.5 seconds to run this 45 metersas B runs 20 metersin 10 seconds or at the speed of 2 m/s.

Hence, A will take 22.5 seconds lesser than B or 100 – 22.5 = 77.5 seconds to complete the race.

In this question, we firstly find the speed of B by analyzing the difference between headstart and time.
By this we get how much time B take and then how much time will take by analyzing the headstart condition.

8. Karan and Arjun run a 100 m race, where Karan beats Arjun by 10 m. To do a favour to Arjun, Karan starts 10 m behind the starting line in a second 100 race. They both run at their earlier speeds. Which of the following is true in connection with the race?

a) Karan and Arjun reach at the same time
b) Arjun beats Karan by 1 m.
c) Arjun beats Karan by 11 m.
d) Karan beats Arjun by 1 m.

Sol: Initially it seems that option A is the correct answer. But before considering it, let’s delve into the question:

Case 1: In whatever time Karan covers a distance of 100 m, Arjun covers 90 m in the same time.

Case 2: Now Karan is 10 m behind the starting point. Once again to cover 100 m from this new point Karan will be taking the same time as before. In this time Arjun will be covering 90 meters only.

This means that now both of them will be at the same point, which will be 10 meters away from the finish point.
Since both of them are required to cover the same distance of 10 m now and Karan has a higher speed, he will beat Arjun.

No need for calculations as option (4) is the only such option.

Alternatively:

In the 1st race when Karan runs 100 m , Arjun runs only 90 m.
Hence, the ratio of speeds of Arjun and Karan is 90:100 = 9:10.
In 2nd race, Karan has run 110 m. when he finishes the race, Arjun would have run (9/10)*110 = 99 m.

9. Two cars P and Q are moving at uniform speeds, 50 km/hr and 25 km/hr respectively, on two straight roads intersecting at right angle to each other. P passes the intersecting point of the roads when Q has still to travel 50 km to reach it. What is the shortest distance between the cars during the journey?

Sol: Let P’s distance be Y and Q’s be X.

Then, Y/50 = (X-50)/25
Y=2X-100
say Y=100, X=100

Distance at any time = d
d^2 = (100-50t)^2 + (100-25t)^2

Now, for minimum value differentiate this equation w.r.t t, therefore,
(100-50t)*2 +(100-25t) = 0
t=12/5
At t=12/5, d becomes

20^2 + 40^2=d^2
d= 20√5

This question is slightly different. It also involves the concept of maxima and minima.

Circular Races Questions for CAT Exam