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# Car A is 20 miles behind car B, which is traveling in the

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19 Feb 2011, 06:51
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Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour.How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5

I tried two formulas to calculate catch up time
1) catch up time when both r moving in same direction=distance between the two/Speed of Car A- Speed of Car B
=> 20/58-50 = 5/2=2.5
to drive 8 miles ahead of Car B by Car A
RT=D
58T=8
T=8/58

Total time= 2.5+8/58 = 2.63 ~=3.0
2) As Car A is 20 miles behind it must have started late than Car B have
so time taken by Car B will be t+x nd
time taken by Car A will be t

As both are moving in the same direction ,both will travel the same distance till catch up time --d

using this information i created two equations and try to solve but
i am not getting correct answer
[Reveal] Spoiler: OA

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19 Feb 2011, 07:08
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GMATD11 wrote:
Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

a)1.5
b)2.0
c)2.5
d)3.0
e)3.5

I tried two formulas to calculate catch up time
1) catch up time when both r moving in same direction=distance between the two/Speed of Car A- Speed of Car B
=> 20/58-50 = 5/2=2.5
to drive 8 miles ahead of Car B by Car A
RT=D
58T=8
T=8/58

Total time= 2.5+8/58 = 2.63 ~=3.0

2) As Car A is 20 miles behind it must have started late than Car B have
so time taken by Car B will be t+x nd
time taken by Car A will be t

As both are moving in the same direction ,both will travel the same distance till catch up time --d

using this information i created two equations and try to solve but
i am not getting correct answer

Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it'll need 28/8=3.5 hours.

P.S. The way you are doing it should be 20/8+8/8=2.5+1=3.5 hours.
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19 Feb 2011, 09:35
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Question says that B gets a head start of 20 miles from A.

Say, B traveled x miles when A was already 8 miles ahead of B.

Thus,
Total distance traveled by B = x miles @ speed 50m/h
Total distance traveled by A = 20+x+8 miles @ speed 58m/h ("A" covered the 20m lag; covered the distance x that B covered and got a lead of 8 miles)

It is given that they both traveled these distances in the same time/duration;
Time spent by B = Time spent by A

Time = Distance/Speed

x/50 = (20+x+8)/58
58x = 1000+50x+400
8x = 1400
x = 175 miles

We know the value of x.

We need to find out the time taken by A to travel the distance = 20+x+8 = 20+175+8 = 203miles

A traveled the distance of 203 @ 58m/h in 203/58 h = 3.5h.

Ans: "E"
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12 Mar 2011, 09:02
Hey everyone, I am having problems with this exercise:

OG 12 PS 206

Car A is 20 miles behind Car B, which is travelling in the same direction along the same route as Car A. Car A is traveling at a constant speed of 58 miles per hour and Car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

a) 1.5 b) 2.0 c) 2.5 d)3.0 and e)3.5

My approach to this problem was that BOTH cars are moving at the same time, therefore:

1st see how much time it takes to overtake (be on the same place)
distance for Car B = 20 + x
distance for Car A = x

time = distance/rate

times are equal hence:

(20+x)/58 = x/50

x = 25/2

Car B reaches Car A 37,5 miles ahead.

Time for this to happen = (37,50/58)

Now for Car B to overtake and drive 8 miles ahead:

distance Car A = z + 8
distance Car B= z

Assuming it takes them the same time
(z+8)/58 = z/50

z = 5

Car B takes 13/58 time to do this.

Thereore the total time should be 37,50/58 + 13/58 which is < 1

however the answer on OG states the following explanation:

Understand that Car A first as to travel 20 miles behind car B to catch up to car B and then has to travel an additional 8 miles ahead of Car B, a total of 28 extra miles to travel relative to Car B. It can be stated that Car A is traveling 58 - 50 = 8 miles per hour faster than Car B.

By substitution of the distance/rate = time formula we have : 28 miles/8 miles per hour = 3.5 hours

THE ANSWER WOULD BE E) 3.5 HOURS.

I think this appoach only valid if we DO NOT TAKE in consideration that while Car A is moving Car B is also moving.

My question is, where is my mistake on reading the problem?

what do you think?
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12 Mar 2011, 09:39
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Using relative speed concept-
-----------------------------
Relative speed = the distance will shrink between the two cars effectively @ 8 miles per hour

Distance to be covered by A = Difference + Lead = 20 + 8 = 28 miles

Time = Distance / Relative Speed = 28 / 8 = 7/2 = 3.5 hrs.

using time distance concept -
----------------------------
Let t hrs be the time to overtake

Distance traveled by A = 58t miles = 20 + Distance traveled by B = (20 + 50t)

=> 58t = 50t + 20

t = 2.5 hrs

Let k hrs be the time to cover extra 8 miles by A

Distance traveled by A = 58k miles = 8 + Distance traveled by B in k hrs = (8 + 50k) miles

=>8 + 50k = 58k
k = 1.0 hrs

Total time = t + k = 2.5 + 1.0 = 3.5 hrs
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12 Mar 2011, 10:50
Thanks a lot! Kudos!

I still dont understand what is wrong with my approach though...
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Re: Catch up time and time required to travel ahead of other [#permalink]

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12 Mar 2011, 12:49
so silly...thanks a lot for the help guys kudos for everyone!
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19 Mar 2013, 12:28
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Just making some formulas :

1. If Objects move in samedirection : The relative speed can be Added / Substracted .
ie. If Car A is moving at X km/hr and Car B is moving at Y km/hr . X > Y
imagine car B to be stopped at car A moving at X-Y km/hr

2. If Objects moving in oppdirection (towards OR away from each other)
if If Car A is moving at X km/hr from Left to right And Car B @ Y km/hr from R2L
imagine car B stationary and A moving toward it at X+Y kms/hr

-Eski
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20 Mar 2013, 09:01
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car a 58mph...car b 50mph..so in 1hour..car goes 8mile ahead..to cover 20mile car a will take 2.5hr..hence an additional 1hour to go 8mile ahead..hence 3.5 in total
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Re: Catch up time and time required to travel ahead of other [#permalink]

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16 Apr 2013, 11:26
Bunuel wrote:
GMATD11 wrote:
Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

a)1.5
b)2.0
c)2.5
d)3.0
e)3.5

I tried two formulas to calculate catch up time
1) catch up time when both r moving in same direction=distance between the two/Speed of Car A- Speed of Car B
=> 20/58-50 = 5/2=2.5
to drive 8 miles ahead of Car B by Car A
RT=D
58T=8
T=8/58

Total time= 2.5+8/58 = 2.63 ~=3.0

2) As Car A is 20 miles behind it must have started late than Car B have
so time taken by Car B will be t+x nd
time taken by Car A will be t

As both are moving in the same direction ,both will travel the same distance till catch up time --d

using this information i created two equations and try to solve but
i am not getting correct answer

Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it'll need 28/8=3.5 hours.

P.S. The way you are doing it should be 20/8+8/8=2.5+1=3.5 hours.

Car A need to drive 28 mile, so time for A= 28/58
But as car B is moving too 28/58-8/50=3
Can you help me, what i am missing?
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Re: Catch up time and time required to travel ahead of other [#permalink]

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16 Apr 2013, 11:42
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Marchikn wrote:
Car A need to drive 28 mile, so time for A= 28/58
But as car B is moving too 28/58-8/50=3
Can you help me, what i am missing?

Hi Marchikn, you are missing the fact that while a "catches up" B is still driving forward.

Here is my solution: $$speedA = 58, speedB=50$$ this means that every hour A gains $$58-50=8$$ miles on B.
In how many hours will A reach B?
space=time*speed, space = distance between cars, speed is the rate at which A is catching up (difference of the speeds A-B)
$$t=\frac{20}{8}=2.5h$$ A will take 2.5 h to reach B.
But the question asks "when A will be 8 miles ahead", so we need to add to this time the time it will take A to do so.
$$8=t*8$$, $$t=1$$, 1h to get 8 miles ahead + 2.5 h to reach B = $$3.5h$$

Let me know if it's clear.
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16 Apr 2013, 23:29
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Marchikn wrote:
Bunuel wrote:
GMATD11 wrote:
Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

a)1.5
b)2.0
c)2.5
d)3.0
e)3.5

I tried two formulas to calculate catch up time
1) catch up time when both r moving in same direction=distance between the two/Speed of Car A- Speed of Car B
=> 20/58-50 = 5/2=2.5
to drive 8 miles ahead of Car B by Car A
RT=D
58T=8
T=8/58

Total time= 2.5+8/58 = 2.63 ~=3.0

2) As Car A is 20 miles behind it must have started late than Car B have
so time taken by Car B will be t+x nd
time taken by Car A will be t

As both are moving in the same direction ,both will travel the same distance till catch up time --d

using this information i created two equations and try to solve but
i am not getting correct answer

Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it'll need 28/8=3.5 hours.

P.S. The way you are doing it should be 20/8+8/8=2.5+1=3.5 hours.

Car A need to drive 28 mile, so time for A= 28/58
But as car B is moving too 28/58-8/50=3
Can you help me, what i am missing?

The concept being used here is relative speed. Check out my post:
http://www.veritasprep.com/blog/2012/07 ... elatively/

Time taken by A = 28/(58 - 50) = 3.5 hrs
We subtract 50 because of what you said - B is moving too and B's speed is 50. The concept will be clear once you through the post.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 Jan 2013 Posts: 35 Location: Spain Concentration: Finance, Other Schools: Wharton '17 (S) GMAT 1: 740 Q49 V42 Followers: 1 Kudos [?]: 64 [1] , given: 12 Re: Car A is 20 miles behind car B, which is traveling in the [#permalink] ### Show Tags 17 Apr 2013, 04:18 1 This post received KUDOS Let t be time car A needs to overtake car B and drive 8 miles ahead. Then: 58t - 20 = 50t + 8 t=28/8=3.5 hours Hope this helps Posted from my mobile device GMAT Club Legend Joined: 09 Sep 2013 Posts: 13537 Followers: 577 Kudos [?]: 163 [0], given: 0 Re: Car A is 20 miles behind car B, which is traveling in the [#permalink] ### Show Tags 26 Apr 2014, 19:17 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 05 May 2014 Posts: 5 Followers: 0 Kudos [?]: 0 [0], given: 10 Re: Car A is 20 miles behind car B, which is traveling in the [#permalink] ### Show Tags 07 Jun 2014, 03:59 Hi Karishma, I read your document at below: veritasprep/blog/2012/07/quarter-wit-quarter-wisdom-speeding-relatively/ Could you please share the next week doc which is mentioned there for some tougher relative speed questions? "Next week, we will look at some tougher relative speed questions" Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7130 Location: Pune, India Followers: 2140 Kudos [?]: 13709 [2] , given: 222 Re: Car A is 20 miles behind car B, which is traveling in the [#permalink] ### Show Tags 08 Jun 2014, 21:59 2 This post received KUDOS Expert's post 2 This post was BOOKMARKED Nishant1234567 wrote: Hi Karishma, I read your document at below: veritasprep/blog/2012/07/quarter-wit-quarter-wisdom-speeding-relatively/ Could you please share the next week doc which is mentioned there for some tougher relative speed questions? "Next week, we will look at some tougher relative speed questions" Here are two posts discussing relative speed questions: http://www.veritasprep.com/blog/2012/08 ... -speeding/ http://www.veritasprep.com/blog/2012/08 ... -concepts/ P.S. - In case you are looking for posts on particular topics, just use the search feature e.g. type "relative speed quarter wit" in the Search field of our blog. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]

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17 Jun 2014, 01:14
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To solve this problem we need to know just the difference in Kilometers between A and B. The difference = 20 Miles + 8 Miles = 28
8*T=28 --> T=28/8 = 3,5 E
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Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]

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12 Aug 2014, 18:19
Bunuel wrote:
GMATD11 wrote:
Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

a)1.5
b)2.0
c)2.5
d)3.0
e)3.5

I tried two formulas to calculate catch up time
1) catch up time when both r moving in same direction=distance between the two/Speed of Car A- Speed of Car B
=> 20/58-50 = 5/2=2.5
to drive 8 miles ahead of Car B by Car A
RT=D
58T=8
T=8/58

Total time= 2.5+8/58 = 2.63 ~=3.0

2) As Car A is 20 miles behind it must have started late than Car B have
so time taken by Car B will be t+x nd
time taken by Car A will be t

As both are moving in the same direction ,both will travel the same distance till catch up time --d

using this information i created two equations and try to solve but
i am not getting correct answer

Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it'll need 28/8=3.5 hours.

P.S. The way you are doing it should be 20/8+8/8=2.5+1=3.5 hours.

Hi Bunuel,

Can you please guide me to the location of additional relative rate problems?

Thanks!
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12 Aug 2014, 20:40
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I am sure Bunuel will give you a dozen links for practice questions but meanwhile, here are some of my posts with relative speed practice questions:

http://www.veritasprep.com/blog/2012/07 ... elatively/
http://www.veritasprep.com/blog/2012/08 ... -speeding/
http://www.veritasprep.com/blog/2012/08 ... -concepts/
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Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]

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25 Oct 2014, 09:16
I've done these problems numerous times but got stuck into a very easy problem in the main exam. I feel that its useful to have a uniform strategy for these two types of these distance-time problems
1. A overtakes B ( using relative speed concept)
2. A is travelling toward B. ( e.g the NY-Dallas problem).

Thanks to experts' posts.
Re: Car A is 20 miles behind car B, which is traveling in the   [#permalink] 25 Oct 2014, 09:16

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