In this article, we will discuss an important concept of Time, Speed and Distance i.e. Races and Games.

A race is a competition in which the competitors have to cover a given distance in the least time. A race may involve running, riding, driving, sailing, rowing etc. The ground or path on which a race takes place is called a racecourse. The point from where a race starts is called the starting point, and the winning point or goal is the point where a race finishes.

Since Races and Games is a sub-topic of Time, Speed and Distance, all the formulas of time, speed and distance are applicable to Races and Games questions.

**'A gives B a start of x meters’:**This statement implies that, while A starts the race from starting point, whereas, B starts 10 meters ahead of A. To cover a race of 100 meters in this case, A will have to cover 100 meters while B will have to cover only (100 - x).**'A beats B by x m':**This statement implies that in the same time, while A reached the winning point, whereas, B is behind A by x m. To cover a race of 100 meters in this case, A has covered 100 meters while B has covered only (100 - x )**A can give B a start of t minutes:**This statement implies that A will start t minutes after B starts from the starting point. Both A and B will reach the finishing point at the same time.**A gives B x meters and t minutes:**This statement implies that A and B start from the starting point at the same instant, but while A reaches the finishing point, B is behind by x meters, and, B takes t minutes compared to A to complete the race. So, B covers remaining x meters in extra t minutes. This gives the speed of B as x/t.**Dead Heat:**A dead heat situation is when all participants reach the finishing point at the same instant of time.

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Let us solve some questions based on the above discussed concepts.

**Example 1:** In a 50 m race, A can give a start of 5 m to B and a start of 14 m to C. In the same race, how much start can B give to C?

**Solution:** In the same time, A covers 50 m, B covers (50 – 5) = 45 m, and C covers (50 – 14) = 36 m. If the race is between B and C only, applying unitary method, when B runs 45 m, C runs 36 m. Hence, when B runs 50 m, C runs (36/45 × 50 = 40) This means B gives C a 10 m start.

**Example 2: **In a 500 m race, A reaches the final point in 28 s and B reaches in 35 s. By how much distance does A beat B?

**Solution:** In this question, we have to find the distance covered by B, after A has finished the race i.e. distance covered by B in (35 – 28 = 7s).

Distance = 500/35 × 7 = 100 m

A beats B by 100 m.

**Example 3:** A runs 4 times as fast as B. If A gives B a start of 60 m, how far must the goal on the race course be so that A and B reach it at the same time?

**Solution:** Ratio of the speeds of A and B = 4:1.

In a race of 4 m, A gains 3 m over B.60 meters will be gained by A in race of (4/2 × 60 = 80 m). Thus, Winning post must be at a distance of 80 m from the starting point.

This topic is a frequently asked topic in CAT. Let us take a look at some previous year CAT questions.

**Question 1: **In a 10 km race. A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C? **(CAT 2017)**

*Answer:* 1900

*Explanation:* A beats B by 1 km, means A travels 10 km in the same time that B travels 9 km. The ratio of speeds of A and B is 10 : 9. Similarly, the ratio of speeds of B and C is 10 : 9.

A : B = 10 : 9

B : C = 10 : 9

⇒ The ratio of speeds of A : B : C = 100 : 90 : 81

In the same time that A travels 100 m, C travels 81 metres ⇒ In the same time that A travels 10000 m, C would travel 8100 m or A would beat C by 1900 m.

**Question 2: **In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? One mile is 1,600 m.**(CAT 2016)**

*Answer:* Chinmay, 1/16 mile

*Explanation:* In a mile race, Akshay can be given a start of 128 m by Bhairav. This means that Bhairav can afford to start after Akshay has travelled 128 m and still complete one mile with him. In other words, Bhairav can travel one mile, i.e. 1,600 m in the same time as Akshay can travel (1600 – 128) = 1,472 m. Hence, the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time = 1600/1472 = 25 : 23.

Bhairav can give Chinmay a start of 4 miles. This means that in the time Bhairav runs 100 m, Chinmay only runs 96 m. So the ratio of the speeds of Bhairav and Chinmay = 96 : 100 = 25 : 24.

Hence, we have B : A = 25 : 23 and B : C = 25 : 24. So A : B : C = 23 : 25 : 24. This means that in the time Chinmay covers 24 m, Akshay only covers 23 m. In other words, Chinmay is faster than Akshay. So if they race for 1.5 miles = 2,400 m, Chinmay will complete the race first and by this time Akshay would only complete 2,300 m. In other words, Chinmay would beat Akshay by 100 m = 1/16 mile.