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

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24 Jul 2016, 00:17

In first hour ( Assume A is at 0 and B is at 20 miles on a scale:)

1st hour A will cover 58 miles ( A covers 58 miles per hour) and B will have covered 20+50 miles ( 20 as B took lead before and the speed is 50 mile per hour)

2nd hour A= 58*2 = 116 miles B in same time = 70 + 50 miles=120 miles

3rd hour A = 116+58= 174 miles B = 120+50 = 170 miles

So you know A has taken lead in 3 hours , you need to check for a lead of 8 miles.

3.5 hours -> since a takes 58 miles an hour, in .5 hours it will cover 29 miles (half) A will cover = 174 + 29 = 203 miles B will cover = 170 + 25 miles ( half of 50 miles an hour) = 195 miles (203-195= lead of 8 miles) Here you get to see that A is ahead of B by 8 miles in the 3.5 hour course.

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

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28 Nov 2016, 20:17

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

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

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09 Dec 2016, 05:32

I am awful at complex/advanced distance/rate problems. Anyways, with this one I used Magoosh's approach: since the cars are going the same direction you subtract their speed and think of that speed as the speed at which the gap between the two cars is shrinking. Hence, it will take car A 3.5 hours to close the gap and be 8 miles ahead of car B.

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

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20 Apr 2017, 15:45

For each catch up or meeting questions; I've found that relative speed concept is very useful!

1. When catching up, the relative speed is the difference between rate A vs rate B = which is 58 mph - 50 mph = 8 mph.

2. First, we want to to know when the A catch up the B : - Rate : 8 mph - Distance : 20 miles (since 20 miles is the difference) - So, we can calculate Time = \(\frac{Distance}{Rate}\) = \(\frac{20}{8}\) = 2,5 hrs.

3. Second, since we must calculate when A is 8 miles ahead, so we must calculate how much time A will set this difference. - Rate : 8 mph - Distance : 8 miles (we want to set 8 miles difference) - Time = \(\frac{Distance}{Rate}\) =\(\frac{8}{8}\)= 1 hour

4. The sum of number (2) and (3) is the answer = 3,5 hrs. Hence E.
_________________

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

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31 May 2017, 22:43

the distance between car A and car B is 20 miles so this 20 miles could be completed in 20/58-50= in 2.5 hours and for the next 8 miles car A and car B have to travel another 1 hour so total time to cross car B and to travel 8 miles ahead will be for car A is 3.5 hours .

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

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11 Jun 2017, 20:38

Cars traveling in same direction and with constant speed. Best way to solve is relative speed

Relative speed of Car A with respect to Car B = (58 - 50) => 8 mph So after rephrasing the question => with relative speed of 8mph Car A will take how much time to cover (20 + 8 ) miles, which is => (28 / 8) => 3.5 hr

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

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13 Aug 2017, 10:07

Relative distance = 20 miles relative speed = 8 Mph, or let us assume B is stationary and A's speed is 8 ( 58-50)Mph time to meet = 20/8 hrs = 2.5 hrs

After this instant car A will start overtaking B. Since we assumed B is stopped, with speed of 8Mph 8 miles will be covered in 1 hr. total time will be 2.5 hr to meet + overtaking by 8 mile time i.e. 1 hr = 2.5+1 = 3.5 hrs

E is the answer _________________

Give Kudos for correct answer and/or if you like the solution.

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

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13 Aug 2017, 12:25

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?

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

I think a very simple way of justifying the answer for this question is simply

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

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14 Sep 2017, 21:16

[quote="GMATD11"]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

My answer ---

Using simple formula --- Time(difference) = distance(difference)/speed(difference) t= 20/58-50=> 2.5hrs

2.5hrs —> 20 miles x hrs —> 8 miles => solving for x => 2.5*8/20 => 1hr