Aptitude - Races and Games - Discussion
Discussion Forum : Races and Games - General Questions (Q.No. 12)
12.
A runs 1
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?
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?
Answer: Option
Explanation:
| Ratio of the speeds of A and B = | 5 | : 1 = 5 : 3. |
| 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 | x 80 |  m |
= 200 m. |
| 2 |
Winning post is 200 m away from the starting point.
Discussion:
15 comments Page 2 of 2.
Vinay billa said:
1 decade ago
An alternate solution which is more general.
Let the speed of B = x.
Hence speed of A = 5/3 x.
Let the distance from B to the finish point is D.
The distance to be covered by A = 80 + D as A is 80 m behind B.
The time taken by both will be the same.
So, D/x = (80 + D)/(5x/3) => D = 120 m.
Hence the total distance from A is = 120+80 = 200m.
Let the speed of B = x.
Hence speed of A = 5/3 x.
Let the distance from B to the finish point is D.
The distance to be covered by A = 80 + D as A is 80 m behind B.
The time taken by both will be the same.
So, D/x = (80 + D)/(5x/3) => D = 120 m.
Hence the total distance from A is = 120+80 = 200m.
Niraj said:
1 decade ago
After every 5 m, A takes a lead of 2 m.
And B already 80 m ahead of A.
So A has to cover 80 m.
A will be take 40min to cover 80 m.
40 m*5 = 200.
And B already 80 m ahead of A.
So A has to cover 80 m.
A will be take 40min to cover 80 m.
40 m*5 = 200.
Kani said:
1 decade ago
I can't understand please explain by shortcuts.
Atul said:
1 decade ago
After every 5m A takaes a lead 2m
so as A moves forward another 5m i.e 5+5=10
the lead increase by 2.& i.e 2+2=4
there foe 15 it is 6, for 20 it is 8.
to compensate B's lead of 80,
and reach the point both a&b at the same time .
it must 200 i.e 80m lead and both will reach at the same time.
so as A moves forward another 5m i.e 5+5=10
the lead increase by 2.& i.e 2+2=4
there foe 15 it is 6, for 20 it is 8.
to compensate B's lead of 80,
and reach the point both a&b at the same time .
it must 200 i.e 80m lead and both will reach at the same time.
Gowthami said:
1 decade ago
Can you explain briefly any one?
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