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

Question

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

(A) 0.50

(B) 1.00

(C) 1.25

(D) 1.33

(E) 2.00

carlo.damore1@gmail.com · Accepted Answer

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

gracie · Answer

65t=d
50(t+2)=2d
130t=50t+100
t=5/4=1.25 hours
C

ScottTargetTestPrep · Answer

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 =  /2

130t = 50t + 100

80t = 100

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

Answer: C

EMPOWERgmatRichC · Answer

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

GMAT assassins aren't born, they're made, 
Rich

Saiganesh999 · Answer

@Buneul I need your help in this.

I tried to solve conceptually, but stuck between B, C and D. Here's my approach.
Before car B finishes its first trip from Houston to Morse, car A would have already finished its trip, as speed of car A > speed of car B, and by this time, car B has got little distance to finish its first trip. Hence, '2' hours of car B given in the later part of the statement includes
1. The time required for car B to complete remaining trip from Houston to Morse, after car A has already reached its destination.
AND
2. Time required for car B to travel from Morse to Houston back

From the '2 hour' commute for car B, I can assume that time required for car B to travel from Morse to Houston is greater than 1 (something like 1.5 may be), since it takes lesser time to complete its first trip than to make a full trip from Morse to Houston.

Now since I'm sure that car B with a speed of 50mph takes approx 1.5 hours, car A with 65mph on the same distance takes lesser than 1.5 hours.

So that eliminates E. Choice A is too far fetched, for its value 0.5. Eliminate A too!
So now, between B, C and D, can you help me how to eliminate (possibly) wrong answers or may be to directly get to the right one, just by using conceptual understanding and not by messy algebraic approach?

Thanks in advance! I appreciate your help.

kashyap89 · Answer

Above question is having incomplete literature. It should have been one trip instead of trip. 
One more query if 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.

It implies distance covered Houston- Morse - 2hrs after car A.then why distance should be considered twice in calculation.

hero_with_1000_faces · Answer

In case of A
as D=RT, therefore Distance =65T
Now B travels back and forth so distance travelled by B is 2(65T).

Now D = RT for B is,

130T = 50(T+2)

You get 1.25. Simple

Rainman91 · Answer

Can someone explain to me why we set T + 2 for B's rate? B drove for 2 hours longer. So if we were to set the equation, shouldn't we add the 2 to A's time?

TA + 2 = TB

D/65 + 2 = 2D/50

Solve for D to get total distance. 
Divide by 65.

bumpbot · Answer

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