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A car going at 40 miles per hour set out on an 80-mile trip at 9:00

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A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 01 Dec 2015, 04:27
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A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60

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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 01 Dec 2015, 05:54
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Let Car A = car that starts at 9 AM
Car B = car that starts at 9:10 AM
Time for which car A travels at speed of 40 m per hour = 1.5 hours
Distance travelled by Car A = 40 *1.5 = 60 miles

Since Car B catches up Car A at 10:30 , time = 80 mins = 4/3 hour
Speed of Car B = 60/(4/3) = 45 miles per hour

Answer A
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 01 Dec 2015, 12:04
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Bunuel wrote:
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. ....... if it caught up with the first car at 10:30 A.M.?
Distance covered by the car in 1.5 hours ( 10:30 A.M. - 9:00 A.M.) is 1.5 * 40 = 60 miles
Bunuel wrote:
a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?
Now car A has covered 60 miles by 10:30 A.M , so car B must also cover the same distance in ( 10:30 A.M - 9:10 A.M.) 80/60 hr

So, speed of car B must be = 60 * 6/8 i.e. 45 miles per hour

So, Answer must be (A)
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 02 Dec 2015, 04:17
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Distnace covered in 1.5 hr (90 minutes i.e. between 09:00AM - 10:30AM ) = 60 miles (40 * 1.5)

Second covered 60 miles in 80 minutes (i.e. 10 minutes less than the first car because second car started at 09:10AM).

Speed of the second car = 60miles / (80/60) hour = 45miles/hour

Option A is correct.
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 02 Dec 2015, 08:29
Bunuel wrote:
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60


Distance travelled by first car in 10 mins = 40/6 = 20/3 miles

Relative Distance between two cars at the time second car starts = 20/3 miles

Relative speed of two cars = (s-14) where s is the speed of second car

Time taken by second car to catch first car = (9:10 to 10:30) = 1 hour 20 mins = 4/3 hours

Relative speed = Relative Distance / time

(s-40) = (20/3)/(4/3) = 5
i.e. s = 5+40 = 45 miles per hour

Answer: Option A
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 07 May 2016, 10:02
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Here is one other way to solve the problem.
Car 1 took 90 minutes to cover some distance, whereas the second car took 80 minutes to do the same. We can set up following equation-
Ratio of time = inverse of the ratio of speed
80/90 = 40/x => 2/90=1/x=> x = 45m.

the problem can also be solved by the relative speed equation
distance to be caught up by the second car is equal to the distance that the first car did in first 10 minutes
40/60*10
This is to be caught up in 80 minutes or 8/6 hours
40/60*10/(x-40) = 8/6. Solve this to get x = 45.
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A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 01 Jun 2017, 10:10
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Bunuel wrote:
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60


We are given that a car left at 9 a.m. traveling at a rate of 40 mph and another car left 10 minutes later. Since the second car caught up to the first car at 10:30 a.m., the time of the first car is 1.5 hours and the distance is 1.5 x 40 = 60 miles.

If we let r = the rate of the second car, the distance of the second car is:

r x (1 hour 20 minutes)

r x (1 1/3 hours)

r x (4/3) = 4r/3

Since the second car catches up to the first car we can set the distances equal and determine r:

4r/3 = 60

4r = 180

r = 45

Answer: A
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A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 25 Jun 2017, 16:29
Very little calculation here.
The second car travelled for 80 minutes and covered 60 miles. The question asks the speed of the second car in miles per hour. OR
How many miles did the second car travel in 60mins?

80mins------60miles
60mins------?
\(\frac{(60*60)}{80}\) = \(\frac{(3600)}{80}\) = slightly over 40 but less than 50

Answer is A
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00  [#permalink]

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New post 17 May 2019, 18:47
Bunuel wrote:
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60


Hi GMATPrepNow

COuld you please help me with this solution . I got the realtive distance of the gap as 20/3 miles. I am not sure how to proceed next
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Re: A car going at 40 miles per hour set out on an 80-mile trip at 9:00   [#permalink] 17 May 2019, 18:47
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