Races and games
Race :-
When a contestant participate in any contest , then a contest of speed in running, driving, riding sailing or rowing over a specified distance is called race.
Race Course :-
The ground or path on which contests are organised in a systematic way , is called a race course.
Starting Point :-
The exact point or place from where a race begins is called the starting point.
For example :- If A and B are the contestants for a 100 m race and A has to cover 100 m , while B has to cover ( 100 - 20 ) = 80 m
Winning Post (or Goal) :-
The point where the race finishes is called the winning post or finishing point or goal.
For example :- In a 800 m race , A gives some start to B and this makes the length of race for B 725 m .What start does B get from A ?
Sol :- Start given by A to B = ( 800 - 725 )m = 75 m
Dead-heat Race :-
If all the persons contesting a race reach the finishing point exactly at the same time, then the race is called a dead-heat race.
Winner :-
The person who first reaches the finishing point is the winner.
For example :- If in game of 100 points means that the contestant who scores 100 points first , is declared the winner .While who scores 85 points , is second .
Some Facts about Race :-
1. Let two contestants are A and B .If A beats B by d m , then
Distance covered by B who is loser = ( L - d ) m
Example 1. In a race of 100 m , A gives B a start of 25 m .What distance will be covered by B ?
Sol :- Here , Distance covered by A = 100 m and d = 25 m
Using the above formula , we have
Distance covered by B = 100 - 25 = 75 m
2. Two contestants are A and B .If B begins from d m ahead of A , then
A starts at the starting point , B starts d m ahead from the starting point at the same time .
Example 2. In a race of 200 m , If B begins from 50 m ahead of A at same time .What distance will be covered by B ?
Sol :- Here , L = 200 m and d = 50 m
Using the above formula , we have
Distance covered by B = 200 - 50 = 150 m
3. A and B are two contestants .If A beats B by T seconds , then
A and B both starts from starting point .
Example 3. In 1 km race , A beats B by 18 m or 9 s .Find the A' s time over the course ?
Sol :- Given , B covers 18 m or 9 s .
B' s time over the course = ( 9 / 8 ) x 1000 = 500 s
Using the above formula ,
A's time over the course = 500 - 9 = 491 s
4. Let A and B are two contestants .A game in which the participant scoring 100 points first is the winner .
5. If both of the contestants A and B get at finishing point at the same time , then
, Difference in distance of defeat = 0
Some Useful Shortcut Method :-
1. If A is n times as fast as B and A gives B a start of d metres .If both A and B reach the winning post at the same time, then the length of the race course must be
d | metres | |||
( n - 1 ) |
Example 4. A is 3 times as fast as B. If A gives B a start of 80 m. How long should the race course be so that both of them reach at the same time?
Sol :- Here , n = 3 times and d = 80 m
Using the above given formula , we have
Length of the race course = 80 x | meters | |
( 3 - 1 ) |
= 80 x | = 40 x 3 = 120 metres. | |
2 |
2. If A can run d m race in T_{1} seconds and B in T_{2} seconds ,then A beats B by a distance is
x ( T_{2} - T_{1} ) meters | ||||
T_{2} |
Example 5. A can run 200 m in 20 seconds and B in 25 seconds. By what distance A beats B?
Sol :- Here , T_{1} = 20 seconds , T_{2} = 25 seconds and d = 200 m
Using the above given formula ,
A beats b by a distance = 80 x | x ( 25 - 20 ) = 8 x 5 = 40 metres | |
25 |
3. Let A and B are two contestants.If A is winner and B is loser , then the following relation will be
= | ||
Distance covered by B | Distance covered by A |
⇒ | = | ||
Distance covered by B | ( Difference of winning distance + initial distance ) |
Example 6. In a 1000 m race , P beats Q by 72 metres in 12 seconds .Find the P's time over the course .
Sol :- Here , L = 1000 m and d = 72 metres
Difference of winning time = 12 seconds
Difference of winning distance = 72 metres
Using the above given formula , we get
= | |||
( 1000 - 72 ) | 72 |
= | |||
928 | 72 |
Time taken by A = 928 × | = | ||
72 | 3 |
= 154 | meters | |
3 |
4. If three contestants A , B and C are participants in a race of L m , A contestant beats B contestant and C contestant by distances of d_{12} and d_{13} metres respectively and B contestant beats the C contestant by a distance of d_{23} , then we get the following relation between them
Example 7. A , B and C are three contestants in a 500 m race .If A can give B a start of 20 m and A can give C a start of 32 m. How many metres start can B give to C ?
Sol :- Here , d_{12} = 20 m , d_{13} = 32 m , d_{23} = ?
and L = 500 m
Using the above given formula ,
( 500 - 20 )d_{23} = 500 ( 32 - 20 ) = ( 25 - 20 )
⇒ 480d_{23} = 500 x 12
d_{23} = | ||
480 |
d_{23} = | = 12.5 metres | |
4 |
5. A , B and C are any three contestants .If in a race of L_{1} m , A contestant beats B contestant by a distance of d_{12} , in a race of L_{2} m B contestant beats C contestant by a distance of d_{23} and in a race of L_{3} m , A contestant beats C contestant by a distance of d_{13} , then for a race of L m
A_{12} = | × L , A_{23} = | × L and A_{13} = | × L | |||
L_{1} | L_{2} | L_{3} |
Example 8. In a race of 1000 m , A can beat by 80 m and in a race of 640 m , B can beat C by 160 m .By how many metres will A beat C in a race of 900 m ?
Sol :- Here , L_{1} = 1000 m , d_{12} = 80 m , d_{23} = 160 m , L_{1} = 640 m , L = 900 m and A_{13} = ?
Using the above given formula ,
A_{12} = | × 900 = 8 x 9 = 72 m | |
1000 |
A_{23} = | × 900 = | = 250 m | ||
640 | 4 |
Now , ( 900 - 72 ) x 250 = 900 ( A_{13} - 72 )
⇒ 828 x 250 = 900A_{13} - 900 x 72 ⇒ 207000 = 900A_{13} - 64800
⇒ 900A_{13} = 207000 + 64800
⇒ A_{13} = | = 302 m | |
900 |
Example 9. Arun and Bhaskar start from place P at 6 : 00 am and 7 : 30 am, respectively and run in the same direction. Arun and Bhaskar run at 8 km/h and 12 km/h. respectively. Bhaskar overtakes
Arun at ?
Sol :- Distance between Arun and Bhaskar at 7 : 30 am = 8 x | = 12 km | |
2 |
Time taken by Bhaskar in covering a distance of 12 km = | = 3 h | |
( 12 - 8 ) |
Example 10. A 10 km race is organised at 800 m circular race course. P and Q are the contestants of the race. If the ratio of the speed of P and Q is 5 : 4, how many times will the winner overtake the loser ?
Sol :- Speed of P : Speed of Q = 5 : 4
Time taken by P to cover 5 rounds = Time taken by Q to cover 4 rounds
Distance covered by P in 5 rounds = 5 x | = 4 km | |
1000 |
Distance covered by Q in 4 rounds = 4 x | = | Km | ||
1000 | 5 |
∴ After covering 10 km P will overtake Q = | x 10 = | times | ||
4 | 2 |
Example 11. In a race of 600 m, A can beat B by 60 m and in a race of 500 m, B can beat C by 50 m. By how many meters will A beat C in a race of 400 m?
Sol :- Clearly, if A runs 600 meters, B runs = 540 m.
∴ If A runs 400 m, B runs = 540 × | = 540 × | = 90 × 4 = 360 m | ||
600 | 6 |
∴ When B runs 360 m, C runs = 450 × | = 45 × | = 9 × 36 = 324 m | ||
500 | 5 |