Aptitude - Races and Games - Discussion
Discussion Forum : Races and Games - General Questions (Q.No. 3)
3.
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
Answer: Option
Explanation:
To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.
While A covers 3 m, B covers 4 m.
| While A covers 360 m, B covers |  |
4 | x 360 |  m |
= 480 m. |
| 3 |
Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
A wins by 20 m.
Discussion:
14 comments Page 2 of 2.
Rajesh said:
1 decade ago
Can be solved by simple calculation.
A covers 360m (ie) 500-140.
Ratio of speeds of A:B = 3:4.
Let us consider A covers 3m/sec & B covers 4m/sec.
Time taken by A to cover 360m=(360/3)=120seconds.
So,Distance covered by B in 120s =120*4=480m.
Therefore,A wins by 20 metres which is (500-480).
A covers 360m (ie) 500-140.
Ratio of speeds of A:B = 3:4.
Let us consider A covers 3m/sec & B covers 4m/sec.
Time taken by A to cover 360m=(360/3)=120seconds.
So,Distance covered by B in 120s =120*4=480m.
Therefore,A wins by 20 metres which is (500-480).
Srinu said:
1 decade ago
Time = dis\speed.
T = 360\3 = 120sec.
Now dis = sxT = 4x120 = 480m.
So 500-480 = 20m.
This is the answer.
T = 360\3 = 120sec.
Now dis = sxT = 4x120 = 480m.
So 500-480 = 20m.
This is the answer.
Damini.arora said:
1 decade ago
Here, because A has a start of 140, he has already covered 140 m when B starts. So when A covers rest of 360, B covers 480.
Hence B loses race by 20 m.
Hence B loses race by 20 m.
Yadu said:
3 years ago
How come we take 3 as 3m and 4 as 4m If time is not a constant here?
Anyone, explain to me.
Anyone, explain to me.
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