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Both car A and car B set out from their original locations at exactly

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Both car A and car B set out from their original locations at exactly  [#permalink]

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New post 14 Mar 2017, 04:06
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Question Stats:

73% (03:00) correct 27% (03:20) wrong based on 131 sessions

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Both car A and car B set out from their original locations at exactly the same time and on exactly the same route. Car A drives from Morse to Houston at an average speed of 65 miles per hour. Car B drives from Houston to Morse at 50 miles per hour, and then immediately returns to Houston at the same speed and on the same route. If car B arrives in Houston 2 hours after car A, how many hours did it take car A to make its trip?

(A) 0.50

(B) 1.00

(C) 1.25

(D) 1.33

(E) 2.00
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Re: Both car A and car B set out from their original locations at exactly  [#permalink]

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New post 14 Mar 2017, 08:39
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Quote:
Both car A and car B set out from their original locations at exactly the same time and on exactly the same route. Car A drives from Morse to Houston at an average speed of 65 miles per hour. Car B drives from Houston to Morse at 50 miles per hour, and then immediately returns to Houston at the same speed and on the same route. If car B arrives in Houston 2 hours after car A, how many hours did it take car A to make its trip?


Car A: 65m/h
Car B: 50m/h

Houston to morse = m = distance
\(speed=\frac{dist}{time}\)

Car A:
\(65m/h=\frac{m}{t}\)
\(m=65*t\) ....equation 1

Car B:
\(50m/h=\frac{2m}{(t+2)}\)
\(2m=50*t+50*2\)
\(m=25*t+50\) ....equation 2

Combining equation 1 and 2
\(65*t=25*t+50\)
\(40*t=50\)
\(t=5/4\)
\(t=1,25\)
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Both car A and car B set out from their original locations at exactly  [#permalink]

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New post 14 Mar 2017, 13:45
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vikasp99 wrote:
Both car A and car B set out from their original locations at exactly the same time and on exactly the same route. Car A drives from Morse to Houston at an average speed of 65 miles per hour. Car B drives from Houston to Morse at 50 miles per hour, and then immediately returns to Houston at the same speed and on the same route. If car B arrives in Houston 2 hours after car A, how many hours did it take car A to make its trip?

(A) 0.50

(B) 1.00

(C) 1.25

(D) 1.33

(E) 2.00


65t=d
50(t+2)=2d
130t=50t+100
t=5/4=1.25 hours
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Re: Both car A and car B set out from their original locations at exactly  [#permalink]

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New post 05 Jul 2018, 16:53
vikasp99 wrote:
Both car A and car B set out from their original locations at exactly the same time and on exactly the same route. Car A drives from Morse to Houston at an average speed of 65 miles per hour. Car B drives from Houston to Morse at 50 miles per hour, and then immediately returns to Houston at the same speed and on the same route. If car B arrives in Houston 2 hours after car A, how many hours did it take car A to make its trip?

(A) 0.50

(B) 1.00

(C) 1.25

(D) 1.33

(E) 2.00



We can let the distance between Morse and Houston = d miles. So the time it takes car A to drive from Morse to Houston is d/65. Since car B arrives in Houston 2 hours after car A and it also drives double the distance, we can create the following equation:

2d/50 = d/65 + 2

Multiplying both sides by 650, we have:

26d = 10d + 1300

16d = 1300

d = 81.25

Therefore, it takes car A 81.25/65 = 1.25 hours to make the trip from Morse to Houston.

Alternate Solution:

Let the time car A takes to drive from Morse to Houston be t. Since car A drives at 65 mph, the distance between Morse and Houston, in terms of t, is 65t.

We are given that it takes t + 2 hours for car B to drive twice the distance between Morse and Houston; thus:

65t = [50(t + 2)]/2

130t = 50t + 100

80t = 100

t = 100/80 = 5/4 = 1.25 hours

Answer: C
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Re: Both car A and car B set out from their original locations at exactly &nbs [#permalink] 05 Jul 2018, 16:53
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