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Races and Games

Theory and Formula


Question


Q1. A can run 100 m in 15 sec and B in 20 sec , A will beat by ___________ m .

Solution.
Distance covered by B in 5 sec =
100/25
x 5 = 25 m

So, A beats B by 25 m .
Solution

Q2. In a 100m race . A beats B by 10m and C by 13m . In a race of 180m . B beats C by ___________m .

A:B:C = 100:10:100 - 13

A:B:C = 100:90:87

∴
B/C
=
90/87
=
180/174


So, B beats C by 180 - 174 = 6m .
Solution

Q3. A is
3/2
times as fast as B . If a gives B a start of a 60 m , how long should the race course B so that both of them reach at the same time .

Solution.
Quicker Method

A's speed : B's speed =
3/ 2
: 1 = 3 : 2

Course of race =
Lead/1 -
B's speed /A's speed


=
60 x 1/1 -
2/3


= 60 x 3 = 180
Solution

Q4. A runs 1
2/3
times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Solution.
Ratio of the speeds of A and B =
5/3
: 1 = 5 : 3.

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of
5/2
x 80 m = 200 m.

∴ Winning post is 200 m away from the starting point.

Solution

Q5. In a 500m race , the ratio of speeds of two runners A and B is 3:4 . A has a start of 140m . Then A wins by .

Solution.
To reach the winning points A covers 500 - 140 = 360m .

∴ B covers 360 (
4/3
) = 480m when A reaches the winning points .

So, A reaches the winning point while B remains 20m behind .

∴ A wins by 20m .
Solution