If a car traveled from Townsend to Smallville at an average

[mastevano]

Hello All,

I'm struggling with the following question:

If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?

(1) The return trip took 50% longer than the trip there.
(2) The distance from Townsend to Smallville is 165 miles.

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

So, I understand how to display the data in a chart, and I understand that I need to find the total time and total distance to find the average speed for the entire trip.

Here's the table I came up with for (1)

Going Returning Total
40mph ?? ??
t 1.5t 2.5t
d d 2d

In order to find the average rate I know I need to find 2d/2.5t, however I was not given t or d! How can the answer be A when either of these variables are not given??

Thanks for your help!
Monica

[Guest]

In order to find the average rate I know I need to find 2d/2.5t, however I was not given t or d!

You know d/t = 40. That's the initial rate.
So 2d/2.5t = 2*40/2.5.
Sufficient.

[GMATPaduan]

If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?

(1) The return trip took 50% longer than the trip there.
(2) The distance from Townsend to Smallville is 165 miles.

As you noted: Average Speed = Total Distance / Total Time

Statement 1 tells us:
Townsend to Smallville Smallville to Townsend
R 40 mph 20mph (took twice as long for the same distance, then must have gone 1/2 as slow)
T t .5t
D d d

Since RT = Distance; Total Distance = 40t + 20(.5t)
Total Time = t + .5t = 1.5t

40t + 10t/1.5t = 50t/1.5t = 50 / 1.5; therefore, this equation can be solved to get the average speed -- Sufficient


(2) No information on the speed of the return trip -- not sufficient


Hope that helps....

[StaceyKoprince]

Yep, we can figure it out. Just watch out GMATPaduan - you said it took twice as long on the way back, but that's not what the statement says. It says it took 50% longer than the trip there. That's technically 1.5x as long, not 2x as long.

Remember d=rt
On the way there, we have d=40(T1)
On the way back, we have d=(R2)(T2)
Note the distance is the same but the rates and times are not.

We are given a way to equate the times though: 1.5T1=T2.
So, on the way back, we have d=(R2)(1.5T1)
The d's are equal, so I can set the two equations equal: 40(T1) = (R2)(1.5T1)
Divide by 1.5T1 to lose the T1 and get: 40/1.5 = R2. I can now solve for R2, which means I can solve for T1 and T2... which means I'm going to be able to calculate the average speed.

[mastevano]

It makes sense now :D