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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Cristina gave Nicky a head start of 36 meters.

Cristina's speed = 5 m/s
Nicky's speed = 3 m/s

Since both are running in the same direction, the relative speed will be 5-3 = 2 m/s.

how many seconds will Nicky have run before Cristina catches up to him = time taken by Cristina to cross Nicky
time taken by Cristina to cross Nicky = distance they are apart/ relative speed

36/2 = 18 seconds
So Nicky will run for 18 seconds by the time Cristina catches him.

Ans : B
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
My answer D.

My explanation as follows


[........................[.....................]
A B C
36 mts x mts


Suppose Christina and Niky meet at point C ,which is x meters away from X.

Christina starts the race exactly when Nicky is at point B, which is 36 meters away from point A.


X/3 = 36+X /5 => X= 54. (So, the distance from point A to point C= 36+54= 90 mts) Christina and Nicky meet a point C and from point C Christina leads the race.


To cover the 90mts, Niky has to run for 90/ 3 seconds= 30 seconds (Option D)


Bunuel wrote:
Nicky and Cristina are running a race. Since Cristina 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?

(A) 15 seconds
(B) 18 seconds
(C) 25 seconds
(D) 30 seconds
(E) 45 seconds

Kudos for a correct solution.
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Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Bunuel wrote:
Nicky and Cristina are running a race. Since Cristina 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?

(A) 15 seconds
(B) 18 seconds
(C) 25 seconds
(D) 30 seconds
(E) 45 seconds

Kudos for a correct solution.


Relative Distance = Head Start = 36 meters
Relative Speed = C-N = 5-3 = 3 meters per second

Time = Relative Distance / Relative Speed

i.e. Time = 36/2 = 18 Second Taken by Christina to catch up Nicky

i.e. Time taken by Nicky to have run till then = (36/3) Time taken to make headstart + 18 (After Christina starts running to catch up with Nicky)



Answer: option

Originally posted by GMATinsight on 23 Jun 2015, 08:52.
Last edited by GMATinsight on 01 Aug 2020, 08:48, edited 1 time in total.
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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carcass wrote:
The distace between the will have decreased by 36 meters, this is our distance.

This distance is decreasing between the two at a rate of 5-3=2 meters per second

R*T=D ----> 2*T=36 ------> T=18 seconds

B should be the answer.


Hi carcass

Everything is fine but you may have read the last line of question wrong. :wink:

Question is : how many seconds will Nicky have run before Cristina catches up to him?

So the Total Time taken = Time taken to make Headstart of 36 meters + time taken by christina to catch up Nicky = (36/3) + 18 = 30 Second

What does "headstart" mean? For me a headstart is when both starting at the same time but from different locations, i.E. Nicky starts 36 meters in front of Cristina, starting both at the same time. No? :(
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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used pluging in method
say t is the time for cristina to catch up with Nicky, the equation will be as under:
for Nicky=N=3*t+36
for Cristina=C=5*t

@t=25, N=111 & C=125 X
@t=18, N=90 & C=90 right answer
ans:B
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Cristina-----------------Nicky--------------------------------Catching Point
...............36 meters.....................x meters

At catching point Time will be same for Cristina and Nicky

Time taken for Cristina to travel 36 + x meters = 36+x/5 ---------> T1
Time taken for Nicky to travel x meters = x/3 ---------> T2

Equating T1 = T2

36+x/5 = x/3 ....so x = 54 meters

So total distance traveled by Nicky including the head start (this is a trap) = 36 + 54 = 90 Meters

Hence, time taken by Nicky to travel 90 meters = 90/3 = 30 Seconds

Ans D
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Bunuel wrote:
Nicky and Cristina are running a race. Since Cristina 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?

(A) 15 seconds
(B) 18 seconds
(C) 25 seconds
(D) 30 seconds
(E) 45 seconds

Kudos for a correct solution.


Relative Speed is 5 - 3 = 2 meters per second
Distance to run: 36m
Time to catch up: 36/2 = 18 seconds

Answer B
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
GMATinsight wrote:
carcass wrote:
The distace between the will have decreased by 36 meters, this is our distance.

This distance is decreasing between the two at a rate of 5-3=2 meters per second

R*T=D ----> 2*T=36 ------> T=18 seconds

B should be the answer.


Hi carcass

Everything is fine but you may have read the last line of question wrong. :wink:

Question is : how many seconds will Nicky have run before Cristina catches up to him?

So the Total Time taken = Time taken to make Headstart of 36 meters + time taken by christina to catch up Nicky = (36/3) + 18 = 30 Second

What does "headstart" mean? For me a headstart is when both starting at the same time but from different locations, i.E. Nicky starts 36 meters in front of Cristina, starting both at the same time. No? :(


Hi GMATinsight

Lets assume it to be a 100m race
(start)0m(Cristina)--------------36m(Nicky)---------------------------------------------------100m(Finish)

The above positions are at the start of the race. The race ideally starts when both the contestants start the run at the same time.
Given a headstart(before the race starts) of 36 meters meaning, both the contestants start to run at the same time but Nicky starts from the 36m mark and Cristina from 0m mark.

So we should not count the 12 sec that Nicky took to reach 36m mark.

Correct if I am missing anything.

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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
Bunuel wrote:
Nicky and Cristina are running a race. Since Cristina 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?

(A) 15 seconds
(B) 18 seconds
(C) 25 seconds
(D) 30 seconds
(E) 45 seconds


Kudos for a correct solution.


Since both are running in same direction, Relative speed between N and C = C's Speed - N's Speed = 5-3 = 2 meters per second.

Now the relative distance they C has to cover is 36 meters.

Hence Option B should be correct.

Hence C will take 18 seconds to cover 36 meter with speed of 2 meters per second.
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Head start = 36meters;Nicky runs at a pace of only 3 meters per second-> 36/3 = 12sec head start
now let x be the seconds Nicky will have run before Cristina catches up to him!
now catches up means same at same distance from the start ; eq = 5x=3(12+x)
only 18 satisfies the equation hence x=18
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Bunuel wrote:
Nicky and Cristina are running a race. Since Cristina 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?

(A) 15 seconds
(B) 18 seconds
(C) 25 seconds
(D) 30 seconds
(E) 45 seconds

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Save time on this problem by dealing with the rate at which the distance between Cristina and Nicky changes. Nicky is originally 36 meters ahead of Cristina. If Nicky runs at a rate of 3 meters per second and Cristina runs at a rate of 5 meters per second, then the distance between the two runners is shrinking at a rate of 5 - 3 = 2 meters per second.

You can now figure out how long it will take for Cristina to catch Nicky using a single RT = D equation. The rate at which the distance between the two runners is shrinking is 2 meters per second, and the distance is 36 meters (because thats how far apart Nicky and Cristina are):

2t = 36;
t = 18.

Answer: B.

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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
Hello,

I like the relative rate idea, but for some reason very rarely think of using it... Anyway, this is what I did:
5T = 36 + 3T --> 2T = 36 --> T = 18.

So, 5T is the distance for Christina, which should be the same as the distance for Nicky, plus the extra 36 meters Nicky ran. So, we add 3T to 36, as we don't know how many minutes extra Nicky ran.
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
For those of you getting the Ans as 30 (aka Option D)- I had got that too intially- but I realized I had misunderstood the question. Headstart means that Nicky and Christina started the race at THE SAME TIME but Nicky started from d point which was 36mt ahead of the point Christina started the race from.
The answer would probably have been 30 if the question read : "Christina let Nicky run for 36 mt before she started running the race"
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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Dont overcomplicate things here.

Headstart means that total distance covered by Cristina will be 36 m more than what Nicky would cover. The only time that we need to worry about is when Cristina started the race and took to cover D+36 when D = T*3 m
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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laveen_g wrote:
Nicky and Cristina are running a race. Since Cristina 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 run before Cristina catches up to him?
a. 15 seconds
b. 18 seconds
c. 25 seconds
d. 30 seconds
e. 45 seconds


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.

Answer: B

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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
as given
speed of Cristina =5 metres per second
Speed of Nicky =3 metres per second
head start =36 meters (Relative distance)
relative speed =2 metres per second
time taken = 36/2=18 seconds
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Re: Nicky and Cristina are running a race. Since Cristina is faster than N [#permalink]
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This type of word problems often trap between its wordings.
Just look at the fist line of Q - "Nicky and Cristina are running a race. Since Cristina is faster than Nicky, she gives him a 36 meter head start"
Headstart is actually a trap. The answer should include time taken by Nicky for this headstart also.
Answer should be 30 seconds option D

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