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

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

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19 Feb 2011, 06:51
6
46
00:00

Difficulty:

15% (low)

Question Stats:

78% (01:32) correct 22% (02:02) wrong based on 2032 sessions

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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

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

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19 Feb 2011, 07:08
12
12
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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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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

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

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

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19 Mar 2013, 12:28
5
2
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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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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20 Mar 2013, 09:01
2
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: Car A is 20 miles behind car B, which is traveling in the  [#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: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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16 Apr 2013, 11:42
2
1
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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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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16 Apr 2013, 23:29
2
4
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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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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17 Apr 2013, 04:18
2
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
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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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07 Jun 2014, 03:59
Hi Karishma,

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

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08 Jun 2014, 21:59
2
2
Nishant1234567 wrote:
Hi Karishma,

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.
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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
1
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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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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12 Aug 2014, 20:40
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.
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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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11 Nov 2015, 18:41
1
I find it easier to understand by looking at the distances:

lets say distance of car A is (Rate x Time) ---> 58xTime
car B is ---> 50xTime

So at some Time car A with its 58 miles per hour rate: will manage to cover 20 miles to break even + 8 miles to get ahead

Distance of A = Distance of B +20 +8
58xTime = 50xTime + 28

8Time = 28
Time = 3.5
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Re: Car A is 20 miles behind car B, which is traveling in the  [#permalink]

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27 May 2016, 22:37
For every hour car b travels 58 miles and car a travels 50 miles. Therefore, for every 1 mile of car a, car b travels 8 miles every hour.
In first hour - car a travel 50 miles, car b travel 58 miles - distance left 20 - 8 =12
In Second hour - car a travel 100 miles, car b travel 116 miles - distance left 20 - 16 =4
In third hour - car a travel 150 miles, car b travel 174 miles - distance left 20 - 24 = -4 (4 km ahead- car b has overtaken car a)
In fourth hour - car a travel 200 miles, car b travel 232 miles - distance left 20 - 32 = -12 ( 12 km ahead)

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

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13 Jul 2016, 06:44
1
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

We can classify this problem as a “catch up and pass” rate question. This means that one car is catching up to the other car and passing it by some distance.

Since there is a change in rate as well as a change in distance between the two cars, we can use the formula:

time = (change in distance)/(change in rate)

We are given that car A is 20 miles behind car B and we need to determine the time when car A is 8 miles ahead of car B. Thus, we can say that the change in distance is 20 + 8 = 28 miles.

We are also given that car A travels at a constant speed of 58 mph and car B travels a constant speed of 50 miles per hour. Thus, we can say that the change in rate is 58 – 50 = 8 mph.

We can plug this information into our equation:

time = (change in distance)/(change in rate)

time = 28/8 = 7/2 = 3.5 hours

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Re: Car A is 20 miles behind car B, which is traveling in the   [#permalink] 13 Jul 2016, 06:44

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