Races & Games | Applied Mathematics for Class 12 - Commerce PDF Download

Introduction

Some Terminologies used in Races

• '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.

Solved Examples

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?

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?

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.