in a race, amit gave rahul a head start of 'x' m and took 20 sec to ge

chetan2u

I have applied the same reasoning but I concluded that Amit traveled (x+y) m.And my final answer was 5/12.

<-----Amit----><------Rahul------>
---------------------------------------
<-------y------>/<-----------x------->

Can you please explain why Amit traveled 2x+y ??

Hi..
Rahul is x ahead of Amit, but then in 20 secs, goes ahead of Rahul by x..
So if Rahul travelled y...
Position at start..
A.....<--x--->.....R
In 20 secs..
|<--x-->|<--y-->|R|<--x-->A
So R traveled y, while A traveled x+y+x

Thank you.The wording was little confusing.Could this type of question appear on gmat?

let A be the speed of Amit and R be the speed of Raul
the speed of R is 1/60 (fraction of race/sec)
there for the speed of A is 1/48
Amit took 10sec to catch up to Raul and 10 more seconds to beat him by x meters
therefore the total distance the Raul travelled would be: 10/48 = 5/24

chetan2u

I think I answered my own question. My brain is fried tonight for some reason. So I might as well double check.

In the question, Rahul is running those first X meters before Amit even starts.

“Head-start” in this sense means he started running for some time before Amit even left. It doesn’t mean that Rahul started at a line that was X meters ahead of where Amir started.

In other words, they both started running from the same starting line. Rahul just left earlier.

I don’t know why the word “head start” was causing my brain to close up.


Posted from my mobile device

I didn’t follow this approach, but I do know that the answer you got answers the question of how much A covered when he was X meters ahead of R......NOT when A passed R and they were even.


Posted from my mobile device

Since the question asks for a Ratio answer, we can use some values in line with the given information in order to make the work easier.


Let speed of R = 4 m/s

Speed of A = 5 m/s


R gets a head start for some time and then A leaves after R is X meters in front.

They then both run for 20 seconds.

After this time, A is in front of R by X meters

Using our chosen numbers:

In 20 seconds:

A covered distance = (5) (20) = 100 m
R covered distance = (4) (20) = 80 m

This means R manages to run for 10 m before A started and then fell behind another 10 m (X and X).


Over the constant distance of the entire race, the ratio of the speeds will be inversely proportional to the time taken.

Speed of A to Speed of R = 5 : 4

Time A to finish race : Time R to finish race = 4x : 5x

Since R takes 60 seconds ——-> A would take 48 seconds to finish the race.

At our assume speed of R = 5 m/s ———> this would make the race distance = (5) (48) = 240


Lastly, we need to know how much distance A covered when he passed R.

From above, based on our chosen numbers, R starts off with a 10 m head start. Then A leaves.

Same direction Relative Speed = 5 - 4 = 1 meter per second

Every 1 second, A catches up 1 meter.

Thus, it will take A 10 seconds to catch up the 10 meters.

For this travel time, running at a speed of 5 m/s———> A will have traveled a distance = (5) (10) = 50 m when he passes R


From above, total distance of race was 240 m

Fraction of race A covers when he passes R = 50/240 = 5/24

Posted from my mobile device

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