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by gmattesttaker2 » Wed Nov 06, 2013 9:34 pm
Hello,

Can you please help with this problem here. This is from MGMAT

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?

OA: 18 sec

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

Thanks for your help.

Best Regards,
Sri
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by Brent@GMATPrepNow » Wed Nov 06, 2013 9:51 pm
gmattesttaker2 wrote: 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?
In 1 second, Nicky runs 3 meters, and Christina runs 5 meters.
So, in ONE SECOND the gap between Nicky and Christina decreases by 2 meters.

At the beginning, the gap between them is 36 meters. So, since (18)(2) = 36, we can see that it will take 18 seconds for the 36-meter gap to reduce to zero meters.

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by theCodeToGMAT » Wed Nov 06, 2013 10:06 pm
This question tests the logic of Catch & meet

Catch (when bodies are moving in same direction) = (Distance GAP)/(Difference of Speeds)

Meet (When bodies are moving towards each other, opposite direction) = (Distance GAP)/(Sum of Speeds)

Now, according to question:

It's the case of "CATCH"

= (36)/(5-3) = 36/2 = 18 Sec
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by Patrick_GMATFix » Wed Nov 06, 2013 11:17 pm
An alternative approach uses the Distance = Rate x Time formula. If we call t the span of time spent running, Chris' distance will the product of her rate and time, or 5t (note that since 5 is in meters per second, t must be in seconds). Likewise, Nicky's distance will be the product of her rate and time, or 3t.

We know that between the time Christina starts running and the time she catches Nick, the distance she will need to cover is 36 meters longer. In other words, over this span of t seconds, Chris' distance is 36 longer than Nicky's distance. This translates to 5t = 3t + 36. Solving yields that t=18 seconds.

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by gmattesttaker2 » Thu Nov 07, 2013 12:00 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: 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?
In 1 second, Nicky runs 3 meters, and Christina runs 5 meters.
So, in ONE SECOND the gap between Nicky and Christina decreases by 2 meters.

At the beginning, the gap between them is 36 meters. So, since (18)(2) = 36, we can see that it will take 18 seconds for the 36-meter gap to reduce to zero meters.

Cheers,
Brent

Hello Brent,

Thanks a lot for your explanation. I was trying to solve this problem as follows: Since Cristina gives Nicky a 36 meter head start, Nicky first runs 36 meters and then Cristina starts running (while Nicky continues to run). Hence in the time "t" that it takes Cristina to catch up to Nicky, Nicky has covered distance "d" miles while Cristina has covered 36 + d miles. Hence, I have the following:

R x T = D
N: 3 x t = d
C: 5 x t = 36 + d

Solving, 5t = 36 + (3t)
=> 2t = 36
=> t = 18 sec.

So Cristina runs 18 seconds when she catches up to Nicky but then we also need to find out how long it took Nicky to run the initial 36 meters i.e.

3 x t = 36
=> t = 12 sec.

Hence, Nicky ran for 12 sec (to cover the initial 36 meters) + 18 secs (which is also the time ran by Cristina to catchup with Nicky) = 30 sec.

I was wondering if this reasoning is correct? Thanks a lot for your valuable time and help.

Best Regards,
Sri
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by Patrick_GMATFix » Thu Nov 07, 2013 12:52 am
Original Q wrote:Since Christina is faster than Nicky, she gives him a 36 meter head start.
Since Cristina gives Nicky a 36 meter head start, Nicky first runs 36 meters and then Cristina starts running
The question is a bit ambiguous. when I first read it, I had the same interpretation as you Sri. Under that interpretation, your logic and answer are correct. However, a "36 meter head start" in this case simply means that Nicky starts 36 meters ahead and they start the race at the same time (ie they did not begin from the same starting line).
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by amirhakimi » Thu Nov 07, 2013 2:10 am
This figure might help when dealing with relative rates.
I've redraw it from MGMAT book :D
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by theCodeToGMAT » Thu Nov 07, 2013 3:21 am
amirhakimi wrote:This figure might help when dealing with relative rates.
I've redraw it from MGMAT book :D
Image
nice pictorial representation ;)
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by GMATGuruNY » Thu Nov 07, 2013 4:01 am
gmattesttaker2 wrote: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?

OA: 18 sec
Time to catch up = (distance behind)/(catch-up rate).

The CATCH-UP rate is the DIFFERENCE between the two rates:
5-3 = 2 meters per second.
Here is the reasoning:
Every second Christina travels 5 meters, while Nicky travels 3 meters.
Result:
Christina travels 2 MORE METERS than Nicky, allowing Christina to CATCH UP by 2 meters every second.

Since Christina is 36 meters behind, we get:
t = d/r = 36/2 = 18 seconds.

An alternate approach is to WRITE IT OUT.

Every second, Christina travels 5 more meters, while Nicky travels 3 more meters.
Thus, every 5 seconds, Christina travels 25 more meters, while Nicky travels 15 more meters.
Calculate the distances at every 5-second mark until the distances are almost equal.
Then calculate the distances at every 1-second mark.

Start: C = 0 meters, N = 36 meters
After 5 seconds: C = 0+25 = 25 meters, N = 36+15 = 51 meters
After 10 seconds: C = 25+25 = 50 meters, N = 51+15 = 66 meters
After 15 seconds: C = 50+25 = 75 meters, N = 66+15 = 81 meters
After 16 seconds: C = 75+5 = 80 meters, N = 81+3 = 84 meters
After 17 seconds: C = 80+5 = 85 meters, N = 84+3 = 87 meters
After 18 seconds: C = 85+5 = 90 meters, N = 87+3 = 90 meters

Since the distances are equal after 18 seconds, the time for Christina to catch up to Nicky = 18 seconds.
Last edited by GMATGuruNY on Thu Nov 07, 2013 9:49 pm, edited 1 time in total.
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by Digvijay01 » Thu Nov 07, 2013 12:53 pm
@gmattesttaker2-- There are a lot of good approaches above.

I just want to clarify the ambiguity you are facing.. A headstart of 36m means..Nicky is 36m ahead of Christina at the start of the race..and they are starting simultaneously..

If it would have been said..that a headstart of some x seconds is given.. then you would have to factor in the distance covered in those seconds by Christina before Nicky starts running.

Hope this clarifies... :)
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by gmatclubmember » Mon Nov 11, 2013 9:45 am
gmattesttaker2 wrote:
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: 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?
In 1 second, Nicky runs 3 meters, and Christina runs 5 meters.
So, in ONE SECOND the gap between Nicky and Christina decreases by 2 meters.

At the beginning, the gap between them is 36 meters. So, since (18)(2) = 36, we can see that it will take 18 seconds for the 36-meter gap to reduce to zero meters.

Cheers,
Brent

Hello Brent,

Thanks a lot for your explanation. I was trying to solve this problem as follows: Since Cristina gives Nicky a 36 meter head start, Nicky first runs 36 meters and then Cristina starts running (while Nicky continues to run). Hence in the time "t" that it takes Cristina to catch up to Nicky, Nicky has covered distance "d" miles while Cristina has covered 36 + d miles. Hence, I have the following:

R x T = D
N: 3 x t = d
C: 5 x t = 36 + d

Solving, 5t = 36 + (3t)
=> 2t = 36
=> t = 18 sec.

So Cristina runs 18 seconds when she catches up to Nicky but then we also need to find out how long it took Nicky to run the initial 36 meters i.e.

3 x t = 36
=> t = 12 sec.

Hence, Nicky ran for 12 sec (to cover the initial 36 meters) + 18 secs (which is also the time ran by Cristina to catchup with Nicky) = 30 sec.

I was wondering if this reasoning is correct? Thanks a lot for your valuable time and help.

Best Regards,
Sri
Hello Gmattesttaker:
Please check what is the 't' time that you have assumed. This is the time to catch up with Nicky.
This is what you need to find out and this is what you yourself calculated as 18 seconds. That IS the answer.
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by Mathsbuddy » Sun Nov 17, 2013 11:46 am
gmattesttaker2 wrote:Hello,

Can you please help with this problem here. This is from MGMAT

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?

OA: 18 sec

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

Thanks for your help.

Best Regards,
Sri
A fast method is this:

distance = speed * time + advantage distance

We want to find t when

N's distance = C's distance:

5t = 3t + 36

2t = 36

t = 18
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