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Car A and Car B are each traveling along the same stretch of highway

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Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 17 Jul 2016, 08:46
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Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?

A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm
[Reveal] Spoiler: OA

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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 17 Jul 2016, 09:16
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Bunuel wrote:
Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?

A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm


Backsolved the question.


Let's check option C.
At 4:30, car B is travelling for 6 hrs. and hence will travel a distance of 6*50= 300 miles

At 4:30 car A is travelling for 6.5 hrs, and will travel a distance of 6.5*40= 260 miles

Hence at 4:30 car A will be 300-260=40 miles ahead of car B

C is the answer
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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 17 Jul 2016, 09:21
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Bunuel wrote:
Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?

A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm



When Car B Starts at 10:30, Car A has already covered 20 Miles(Speed is 40 Miles per hour).

=> The gap between the two is 20 miles.

Thus, Car A and Car B will meet after time = Gap/Difference in speeds ( Difference is taken as both are travelling in the same direction)

=> Time = 20/50-40=20/10=2 Hours.

Thus, both the cars will meet after 2 hours. Now , in order to make a gap of another 40 miles, another period of time required = 40/50-10=4 hours.

Thus , a total of 2+4=6 hours is required for making Car B 40 miles ahead of Car A.

Thus, 10:30 a.m. + 6 hours = 4:30 p.m.

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Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 17 Jul 2016, 09:26
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Relative speed = 10 mph

At 10.30 car A would have traveled 20 miles

If car B has to be 40 miles ahead of car A, then it has cover the above 20 miles and be ahead by 40 miles. i.e. Net additional distance car B has to cover = 60 miles

Time taken to complete 60 miles = 60/10 = 6 hours

Car B will be ahead of car A by 40 miles at 10.30 am + 6hrs = 4.30 pm

Answer: C

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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 23 Jul 2016, 03:38
Let X be the total distance traveled by CAR A in (T+0.5) - Starting Half Hour earlier.
= X=40(T+0.5)
Total distance traveled by CAR B in time T,
= X+ 40 = 50 T

Put Value of X from eq 1 in eq 2
= 40 (T+0.5) + 40 = 50 T
deriving
=T= 6 Hours

Starting at 10:30 AM, At 16:30 PM (6 Hours)

Hope it helps.

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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 23 Jul 2016, 07:47
A starts at 10:00 and by the time B starts A is 20 miles ahead
now B has to go 40 mile ahead, and b has to cover 20 miles also
so total 60 miles
50-40 = 10 miles/ hour relative seed ....
6 hours 10:30+6 = 4:30

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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 29 Aug 2017, 15:37
Bunuel wrote:
Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?

A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm


Bunuel

Could we simply do

40 +20/ (50-40) = 6

10:30 + 6 hours?

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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 05 Sep 2017, 18:10
Bunuel wrote:
Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?

A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm


Car A’s speed is 40 mph and it starts at 10:00 a.m., so when Car B starts at 10:30 a.m., Car B will be 40 x 1/2 = 20 miles behind Car A. To determine what time will Car B be 40 miles ahead of Car A, we can use the following formula:

change in distance/change in rate = time

The change in distance is 60 miles, since Car B needs to catch up to Car A starting from 20 miles behind and needing to be 40 miles ahead. The change in rate is 50 - 40 = 10 mph. Thus:

60/10 = 6 hours

So, Car B needs 6 hours to be 40 miles ahead of Car A. Since Car B starts at 10:30 a.m., it will be 40 miles ahead of Car A at 10:30 a.m. + 6 hours = 4:30 p.m.

Answer: C
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Re: Car A and Car B are each traveling along the same stretch of highway [#permalink]

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New post 12 Sep 2017, 17:44
When car B starts its trip, it's already 20 miles behind. For it to be 40 miles ahead of A, 60 miles have to be "recovered".

Since car B is 10 mph faster than A, car B will take 6 hours to "recover" the 60 miles, which will be achieved at 4:30 pm.

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Re: Car A and Car B are each traveling along the same stretch of highway   [#permalink] 12 Sep 2017, 17:44
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