**Quantitative Aptitude Races and Games Study Material**

**Quantitative Aptitude Races and Games ( Download PDF )**

**Race** : A contest of speed over a specified distance is called race. The contest may be in running, riding, driving, sailing, rowing etc. The participants in the contest may be persons, animals or vehicles. The place, ground, field, path or track on which the contest is held is called Race Course. The point from where a race begins is called Starting Point. The point set to bound a race is called Winning post or Goal. The participant who first reaches the goal is declared Winner; When all the participants contesting a race reach the goal exactly at the same time, the race is called a Dead-heat race.

Suppose A and B are two participants in a race. If at the start of the race, A is at the starting point and B is ahead of A at a distance x meters, then A is said to have given B a start of x meters or a lead of x meters. For instance, in a race of 100m, if at the time of start, A is at the starting point and B is 5 m ahead of A, then A is said to have given B a start or lead of 5 m. In this case it means that A will have to cover a distance of 100 m whereas B will have to cover (100-5) = 95 m only. .

**Game** : When we say that It is a game of 100, it means that the participant who scores 100 points first is the winner. If A scores 100 points while scores only 70 points, we say that “A can give B (100 – 70) = 30 points” in a game of 100.

**Ex. 1. In a race of 5 kms Ram beats Shyam by 200 m or 50 seconds. Find their time and speed over the course.**

**Sol**. Shyam is 200 m behind Ram when Ram reaches the winning post.

Also Shyam takes 50 seconds to cover this distance of 200 m because he reaches the winning post 50 seconds after Ram. ‘

Shyam’s speeds m = 4m / s 50

**Ex. 2. A’s speed is 1(1/2) times of B’s. In a race A gives B a start of 300 m. Flow long should the race course be so that both reach the winning post simultaneously?**

**Sol**. A’s speed : B’s speed = 1(1/2): 1 = 3/2 : 1 = 3:2

It means that in a race of 3 m, A gains (3 – 2) = lm over B

1 m is gained by A in a race of 3 m

.’. 300 m is gained by A in a race of

= 3/1 x 300 = 900m

Therefore, to cover up the lead of 300 m, the race course must be 900 m long.

**Ex. 3. A is n times as fast as B. A gives B a start of x m. How long should the race course be so that both reach the winning post at the same time?**

**Ex 4. A can run 2 km. race in 5 minutes and B in 5 minutes 20 seconds. By what distance A beats B ?**

**Sol**. The distance by which A beats B is equal to the distance covered by B in the last 20 seconds

(=5 min 20 sec – 5 min) 2 km = 2000 m

5 min and 20 secs. = (5 x 60) + 20 = 320 seconds.

In 320 seconds B covers 2000 m

In 20 seconds B covers 2000/320*20 = 125m

Hence, A beats B by 125 m

**Ex. 5. A can run 500 m in 30 seconds and B in 35 seconds. How many metres start can A give to B in a km race so that the race may end in a dead-heat?**

Sol. Time taken by A to run 1 km = 30 x 2 = 60 sec.

Time taken by B to run 1km = 35 x 2 = 70 sec.

.’. A can give B a start of (70 – 60) = 10 sec.

In 35 sec. B runs 500 m.

. T m D 500 1000 6

.’. In 10 sec. B runs = 500/30×10 = 1000/7 = 142(6/7) m

So, A can give B a start of 142 6/7 metres in a km race Ans.

**Ex. 6. A beats B by 29 m and C by 10 m in a 200 m race. By how many metres will C beat B in 500m race?**

**Ex. 7. In a km race A beats B by 30 seconds and B beats C by 15 seconds. If A beats C by 180 m, find the time taken by each to run a km.**

**Sol**. A beats B by 30 sec. and B beats C by 15 sec.

Therefore, A beats Cby (30 + 15) = 45 sec.

Also, A beats Cby 180 m.

.•. Time taken by C to run 1 km = (45/180 x 1000) =250 sec

Time taken by B to run 1 km = 250 – 15 = 235 sec.

Time taken by A to run 1 km = 250 – 45 = 205 sec

**Ex. 8. In a race of 1000 m A can beat B by 100 m. In a race of400 m B beats C by 40 m. By how many metres will A beat C in a race of 500 m?**

**Sol**. When A runs 1000 m, Bruns (1000-100) = 900 m.

When A runs 500 m, B runs = 900*500/1000 = 450m

When B runs 400 m, C runs (400-40) = 360 m.

When B runs 450 m, C runs = 360/400 x 450 = 405 m

So, A beats C by (500 – 405) = 95 m

**Quantitative Aptitude Races and Games ( Download PDF )**