Rates & Work

Let s be the number of seconds each runner runs before they meet.
Distance covered by Cristina = 5*s
Distance covered by Nicky = 3*s
Distance covered by Cristina = Distance covered by Nicky + 36 meters
5*s = 3*s + 36
s = 18 seconds

gmattesttaker2 wrote:Hello,

Can you please help with this problem here:

Nicky and Cristina are running a race. Since Christina is faster than Nicky, she gives him a 36 meter head start. If Cristina runs at a pace of 5 meters per second and Nicky runs at a pace of only 3 meters per second, how many seconds will Nicky have run before Cristina catches up to him?

Ans: 18 sec

I was thinking that it should be [spoiler]18 + 12 = 30 seconds[/spoiler]


Thanks for your help.

Best Regards,
Sri

The CATCH-UP rate = faster rate - slower rate.
Since Cristina travels 5 meters per second and Nicky travels 3 meters per second, every second Cristina catches up by 5-3 = 2 meters.
Since Cristina is 36 meters behind Nicky, the time for Cristina to catch up = distance/(catch-up rate) = 36/2 = 18 seconds.

thank u mitch !!

These tidbits helps us in saving time on the real GMAT.

Hello,

I was trying to approach this problem as follows:

Rate x Time = Distance
N: 3 x t = d
C: 5 x t = d + 36

So, 3t = d
and 5t = d + 36

=> 5t = 3t + 36
=> t = 18 sec.

However, Since Cristina gave Nick a 36 meter head start, Nick has already ran for 36 meters. So we need to calculate the time taken to run these 36 meters as well:

Rate x Time = Distance
N: 3 x t = 36

So, t = 12 sec.

Hence, Nick would have run 12 sec. (for this 36 meter headstart) + 18 sec. (when Cristina starts running and catches up with him) = 30 sec.

However, the book answer is 18 sec.

Can you please clarify? Thanks a lot for your help.

Best Regards,
Sri