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

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

by BTGmoderatorDC » Fri Oct 12, 2018 7:51 pm

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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

OA C

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by nonplus2 » Fri Oct 12, 2018 11:47 pm
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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by Scott@TargetTestPrep » Sun Oct 28, 2018 5:19 pm
BTGmoderatorDC 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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by [email protected] » Sun Oct 28, 2018 6:57 pm
Hi All,

We're told that 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 and 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, arriving in Houston 2 hours after car A. We're asked how many hours it took car A to make its trip. This question can be solved in a couple of different ways, including by TESTing THE ANSWERS. Let's TEST Answer B first...

Answer B: 1 hour
If Car A travels for 1 hour, then it travels (1)(65) = 65 miles, meaning that the distance between the two cities is 65 miles.
Car B would have to travel 65+65 = 130 miles at 50 miles/hour.
130 miles = (50 mph)(T hours)
130/50 = T
2.6 hours = T
This is 2.6 - 1 = 1.6 hours after Car A arrived, but this is NOT a match for what we were told (it's relatively close, but it's supposed to be a 2 hour difference). Thus, we need the distance between the cities to be LARGER (but not too much larger).

Answer C: 1.25 hours
If Car A travels for 1.25 hours = 5/4 hours, then it travels (5/4)(65) = 325/4 = 81.25 miles, meaning that the distance between the two cities is 81.25 miles.
Car B would have to travel 81.25 + 81.25 = 162.5 miles at 50 miles/hour.
162.5 miles = (50 mph)(T hours)
162.5/50 = T
3.25 hours = T
The difference is 3.25 - 1.25 = 2 hours. This is an exact match for what we were told, so this must be the answer.

Final Answer: C

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Rich
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