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.
Navneet said:
9 years ago
In the question, its is given that A runs 1 2/3 times as fast as B.
They why the speed of A : B = 5 : 3 is taken in solution.
They why the speed of A : B = 5 : 3 is taken in solution.
Anu said:
8 years ago
A runs 1 2/3 times faster than B so it can be written as (1*3+2)/3 : 1 => 5/3 :1.On taking LCM we have 5 : 3
Salman said:
8 years ago
If A runs 1 times as fast as B then how can be the ratio is 5:3?
It should be 2:1.
It should be 2:1.
Mainak Mukherjee said:
8 years ago
Let total distance is "y" mtr.
As per the question A gives B a start of 80 mtr.
So we can say that A runs "y" mtr while B covers (y-80) mtr.
Now B's speed is X m/sec (say)
And A's speed is 5X/3 m/sec
Now as they both have to reach the finish point at the same time then we can say that time taken by both the players to reach the same spot is equal.
Time taken by A = y/(5X/3) secs --> (i)
and B = (y-80)/X secs --> (ii)
Now eqating the equation i and ii we get ,
y/5X/3= (y-80)/X
3y= 5y-400
2y= 400
y= 200 m (ans.)
As per the question A gives B a start of 80 mtr.
So we can say that A runs "y" mtr while B covers (y-80) mtr.
Now B's speed is X m/sec (say)
And A's speed is 5X/3 m/sec
Now as they both have to reach the finish point at the same time then we can say that time taken by both the players to reach the same spot is equal.
Time taken by A = y/(5X/3) secs --> (i)
and B = (y-80)/X secs --> (ii)
Now eqating the equation i and ii we get ,
y/5X/3= (y-80)/X
3y= 5y-400
2y= 400
y= 200 m (ans.)
Md. Iqbal Hasan said:
4 years ago
Awesome explanation. Thanks @Shivakrishna.
(1)
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