ETS Paper Test Rates Problem. Need help with the Algebra

[tonebeeze]

Hey All,

This question is not difficult. I got the question correct rather quickly by using the Plug in method. I want to make sure I understand how to quickly solve this problem using algebra. Can someone please walk me through this. Thanks!

"On the first day of her vacation, Louisa traveled 216 miles. On the second day, traveling at the same average speed, she traveled 378 miles. If the 216-mile trip took 3 hours less than the 378-mile trip, what was the average speed, in miles per hour?"

1. 31
b. 38
c. 50
d. 54
3. 56

[beat_gmat_09]

Let S = average speed for both the trips.
t = time taken for travelling 378 miles.

216/(t-3) = S ........ (1) ....... for 216 miles
378/t = S...............(2) .........for 378 miles.
equate (1) and (2) as speed is same
216/(t-3) = 378/t

= 72/(t-3) = 126/t
4/(t-3) = 7/t
4t = 7t - 21
3t = 21
t = 7

S = 378/7
S = 54 miles/hr

[N:Dure]

s= 216/t

s= 378 /t+3


216/t = 378/ t+3

72/t = 126 / t+3

72 t + 216 = 126 t

t= 4

s = 216/4 = 54

[Scott@TargetTestPrep]

On the first day of her vacation, Louisa traveled 216 miles. On the second day, traveling at the same average speed, she traveled 378 miles. If the 216-mile trip took 3 hours less than the 378-mile trip, what was the average speed, in miles per hour?"

1. 31
b. 38
c. 50
d. 54
3. 56

We can let the time to travel the 216 miles = t and the time to travel the 378 miles = t + 3. Since the rates are the same and rate = distance/time, we have:

216/t = 378/(t + 3)

216(t + 3) = 378t

216t + 648 = 378t

648 = 162t

4 = t

Thus, the average speed was 216/4 = 54 mph.

Answer: D