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 1 of 2.
Gowthami said:
1 decade ago
Can you explain briefly any one?
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.
Kani said:
1 decade ago
I can't understand please explain by shortcuts.
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.
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.
Vinay sv said:
1 decade ago
x is B speed then A speed is 5x/3 let pole is at distance y.
Then y-80/x=t for b y/5x/3=t for a equate both you get y as 200,
Why to equate? because both should come at same time.
Then y-80/x=t for b y/5x/3=t for a equate both you get y as 200,
Why to equate? because both should come at same time.
Shivakrishna said:
1 decade ago
The difference between A and B is 80m.
Let x be the total distance to cover in race.
In the problem he given A and B to reach at same time. i.e.4.
Time taken by A = time taken by B.
x/5 = (x-80)/3.
3x = 5x - 400.
2x = 400.
x = 200.
Let x be the total distance to cover in race.
In the problem he given A and B to reach at same time. i.e.4.
Time taken by A = time taken by B.
x/5 = (x-80)/3.
3x = 5x - 400.
2x = 400.
x = 200.
(4)
Anjali said:
1 decade ago
Hello @Shivakrishna A and B to reach at same time i.e., 4 has not mentioned in the problem see the problem clearly write.
Anonmyous said:
10 years ago
Given a is 1(2/3) as fast as b.
Ratio of their speeds a:b = 5:3.
Speed is directly proportional to distance traveled.
d1:d2 = 5:3.
Assume a travels 5 m in t time then b travels 3 m in the same time.
That's it and continue as explained in the answer.
Ratio of their speeds a:b = 5:3.
Speed is directly proportional to distance traveled.
d1:d2 = 5:3.
Assume a travels 5 m in t time then b travels 3 m in the same time.
That's it and continue as explained in the answer.
Pranky said:
10 years ago
Speed of A:B = 5:3.
Time A = Time B.
Distance A/Speed A = Distance B/Speed B.
Distance A/Distance B = 5/3.
A/A+80 = 5/3.
3A = 5A+400.
2A = 400 (Neglecting negative sign as its distance).
A = 200 m.
Time A = Time B.
Distance A/Speed A = Distance B/Speed B.
Distance A/Distance B = 5/3.
A/A+80 = 5/3.
3A = 5A+400.
2A = 400 (Neglecting negative sign as its distance).
A = 200 m.
(2)
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