Question 52337: Hi There! I am studying rate questions for the GMAT, and feel pretty comfortable on the basic forms such questions can take. I am completely stuck, however, on rate questions that pose hypothetical questions, like "If the train went 4 mph faster, it would have arrived in 1/2 the time." The question below is taken from an online course, and the answer is also provided. I just don't see how they figured it out.
The Question:
A train traveled 960 miles from Town A to Town B. If the train had traveled from Town A to Town B at an average speed of 12 miles per hour less, it would have taken 4 more hours.
What was the train's average speed during that journey?
The Author's Answer:
We need an average speed to have sufficiency. Let's call the train's average speed s miles per hour and let's call the time that it took the train to travel the 960 miles t hours. The distance formula says that Distance = Rate * Time. So here, st = 960.
We learn that (s 12)(t + 4) = 960. Including the equation st = 960 from the question, we now have two equations for two variables, and we could solve. In the solution, we can substitute (960/s) for t in the equation (s 12)(t + 4) = 960. Then (s 12)((960/s) + 4) = 960. When this equation is solved for s, there will be two solutions, one positive and one negative. (The solution s = 60 corresponds to t = 16; the other case s = 48 which would correspond to t = 20 but would be discarded because its negative.) So there is just one answer to the question, which is that the average speed was 60 miles per hour.
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Whoa! "When this equation is solved for s..?" How about showing how to solve for it? I follow everything up until this point. After that, I'm lost. Two different solutions? What, is this a quadratic equation? Help!
Thank you so much for your help!
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website!
The Question:
A train traveled 960 miles from Town A to Town B.
If the train had traveled from Town A to Town B
at an average speed of 12 miles per hour less,
it would have taken 4 more hours.
What was the train's average speed during that
journey?
The explanations in some of those online courses are
not so good. Here's the way I would solve it:
Make this chart:
DISTANCE RATE TIME
ACTUAL TRIP
HYPOTHETICAL TRIP
Let the speed on the actual trip be x. Fill that in
for the RATE of the ACTUAL TRIP.
DISTANCE RATE TIME
ACTUAL TRIP x
HYPOTHETICAL TRIP
>>...If the train had traveled from Town A to Town B
at an average speed of 12 miles per hour less...<<
So we subtract 12 from x, getting x-12 and fill that
in for the RATE for the HYPOTHETICAL TRIP.
DISTANCE RATE TIME
ACTUAL TRIP x
HYPOTHETICAL TRIP x-12
We are told that the DISTANCE for both the ACTUAL
and the HYPOTHETICAL trip is 960 miles, so we fill
that in:
DISTANCE RATE TIME
ACTUAL TRIP 960 x
HYPOTHETICAL TRIP 960 x-12
Now we fill in the TIMEs using TIME = DISTANCE/RATE
DISTANCE RATE TIME
ACTUAL TRIP 960 x 960/x
HYPOTHETICAL TRIP 960 x-12 960/(x-12)
Now we look for what we haven't used:
>>...it [THE HYPOTHETICAL TRIP] would have taken 4
more hours...<<
So
HYPOTHETICAL TRIP TIME = ACTUAL TRIP TIME + 4 HOURS
or
960/(x-12) = 960/x + 4
Divide thru by 4
240/(x-12) = 240/x + 1
Multiply through by LCD = x(x-12)
240x = 240(x-12) + x(x-12)
240x = 240x - 2880 + xē - 12x
0 = xē - 12x - 2880
Solve this by the quadratic formula
xē - 12x - 2880 = 0
The quadratic formula is:
______
-b ą Öbē-4ac
x =
2a
where a = 1; b = -12; c = -2880
__________________
-(-12) ą Ö(-12)ē-4(1)(-2880)
x =
2(1)
_________
12 ą Ö144+11520
x =
2
_____
12 ą Ö11664
x =
2
12 ą 108
x =
2
Using the +
12 + 108
x =
2
120
x = = 60 mph
2
Using the -
12 - 108
x =
2
-96
x = = -48 mph
2
so we discard that and the only answer is 60 mph.
Edwin
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