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Car A is 20 miles behind car B, which is traveling in the [#permalink]
19 Feb 2011, 06:51

10

This post was BOOKMARKED

00:00

A

B

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E

Difficulty:

5% (low)

Question Stats:

78% (02:01) correct
22% (01:36) wrong based on 631 sessions

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 But answer is different 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

Re: Catch up time and time required to travel ahead of other [#permalink]
19 Feb 2011, 07:08

2

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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 But answer is different 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.

Answer: E.

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

Re: Catch up time and time required to travel ahead of other [#permalink]
19 Feb 2011, 09:35

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This post received KUDOS

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.

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?

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

Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]
19 Mar 2013, 12:28

1

This post received KUDOS

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

Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]
20 Mar 2013, 09:01

1

This post was BOOKMARKED

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

Re: Catch up time and time required to travel ahead of other [#permalink]
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 But answer is different 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.

Answer: E.

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?

Re: Catch up time and time required to travel ahead of other [#permalink]
16 Apr 2013, 11:42

1

This post received KUDOS

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. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Catch up time and time required to travel ahead of other [#permalink]
16 Apr 2013, 23:29

1

This post received KUDOS

Expert's post

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 But answer is different 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.

Answer: E.

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?

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. _________________

Re: Car A is 20 miles behind car B, which is traveling in the [#permalink]
26 Apr 2014, 19:17

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

Answer: E.

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?

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

gmatclubot

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

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