Quant (Concepts): Time,Speed& Distance

Quant (Concepts): Time,Speed& Distance

She Runs Faster than He/Him

Concept of Races:

Race is a competition in which two or more participants aim to reach a specified point before other competitors. There can be many kinds of races. The concepts of time, speed and distance are used in solving questions based on races, which may be linear or circular. In this article, you will learn concepts related to Linear Races.

Imagine, A and B are two Contestants in a Race,

1. The statements “A gives B a start of x meters in a 1 km race” or, “A beats B by x meters in a 1 km race”, implies that in a race of 1000 meters, when A covers 1000 meters B only covers  (1000-x) meters. e.g. In a 100m race, the statements “A can give B 20 m start” or “A beats B by 20 m”, mean that in the time A runs 100 m, B runs 80 m. In this case, 20 meters can be called as beat distance.
2. Similarly, the statements “A can give B 20 seconds start” or, “A beats B by 20 seconds”, means that if the given distance is covered by A in x seconds, then B will take (x – 20) seconds.
3. “A beats B by x meters or t seconds” implies that, B runs x meters in t seconds. Hence, the speed of B is x/t m/s.
4. Loser’s time = Winner’s time + (start time + beat time). (Remember that the winner’s time is less than loser’s time).
5. A dead heat means the contestants reached the end point at same time. So in this case, beat distance = 0 and beat time = 0.

You will understand the concepts better with the help of the following Examples:

Example 1:

A can give B a 20 m start and C a 69 m start in a race of 1000 m. By how many meters could B beat C in the same race of 1000 m?

Solution:

A runs 1000m, while B runs (1000-20) = 980m.
A runs 1000m, while C runs (1000-40) = 931 m.

Therefore, B runs 980m, while C runs 931 m.

applying Unitary Method,

You can find that when B runs 1000 m, C will run 1000 × 931/980 = 950 m.

Hence, B can give C (1000- 950) = 50 m start.

Example 2:

In a 1 km race, A beats B by 20 m or 5 seconds. Find A’s and B’s time over the course?

Solution:

Here, B runs 20 m in 5 seconds.
Therefore, the speed of B is 20/5 = 4 m/s.

Time taken by B = Distance/Speed = 1000/4
= 250 sec.

Hence, A’s time over the course = (250-5)
= 245 sec.

Example 3:

In a 400m race, if A gives B a start of 50 m, then A wins the race by 20 seconds. Alternatively, if A gives B a start of 90 m, then the race ends in a dead heat. How long does A take to run 400m?

Solution:

Case (i)

     Distance(m)  Time Taken(s)
A        400                 t
B        350                 t+20

Case (ii)

     Distance(m)  Time Taken(s)
A        400                 t
B        310                 t

Therefore, if you compare B in the two cases we can conclude that B covers the additional 40 min 20 seconds or at the speed of 2 m/s.

Time taken by B to cover 310 m= 310/2 = 155 s.

As can be seen from case (ii) the time taken by A to cover 400m = Time taken by B to cover 310 m. Hence, the required answer is 155 s.

Example 4:

A runs 2 3/8 times as fast as B. A gives B a start of 110 meters. How far should the winning post be so that the race may end in a dead heat?

Solution:

This implies that in a race of 19 m,

B will only cover 8 m in the same time in which A will cover 19 m.

So, A can give B a start of 11m out of a total race of 19 m.

Applying unitary method, you can find that to give a start of 110 m the race must be of 110 × 19/11 = 190 m.