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.
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)
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.