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frikkin easy rate problem driving me mad

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frikkin easy rate problem driving me mad

by jzw » Sun Mar 04, 2012 12:40 pm
Josh drives from his house to work and back. Same distance each way. On his way to work he drives @ an average of 60 mph. Oh his way back, he drives @ an average of 80 mph. What was his average speed?

(a) 60.0
(b) 68.6
(c) 70.0
(d) 71.4
(e) 80.0

Ok, so I see that he is spending "more time" at 60 mph rate and that since the distance is the same it has to be less than 70 mph but not too much less, so clearly the answer is (b). But how does one set up the math for this question?
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by neelgandham » Sun Mar 04, 2012 1:37 pm
Josh drives from his house to work and back. Same distance each way. On his way to work he drives @ an average of 60 mph. Oh his way back, he drives @ an average of 80 mph. What was his average speed?

(a) 60.0
(b) 68.6
(c) 70.0
(d) 71.4
(e) 80.0

Alright, Here it is

Let the distance from House to Work be x.
Time taken to drive to work = x/60 hours
Time taken to drive back home = x/80 hour

Average speed = Total Distance/ Total time taken
Average Speed = 2x/((x/60)+(x/80)) = ~68.6 mph
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by krusta80 » Sun Mar 04, 2012 2:05 pm
jzw wrote:Josh drives from his house to work and back. Same distance each way. On his way to work he drives @ an average of 60 mph. Oh his way back, he drives @ an average of 80 mph. What was his average speed?

(a) 60.0
(b) 68.6
(c) 70.0
(d) 71.4
(e) 80.0

Ok, so I see that he is spending "more time" at 60 mph rate and that since the distance is the same it has to be less than 70 mph but not too much less, so clearly the answer is (b). But how does one set up the math for this question?
First off, you're approach to answering the question is by far the most efficient...nicely done.

That said, I remeber seeing a question like this years ago on a practice SAT. Without being able to prove it but having an intuitive understanding of why it worked, I came up with the following formula:

average speed = 2*speed1*speed2/(speed1+speed2) = 2*60*80/140 = 960/14 = 68 and 4/7 mph

Now onto deriving my formula, which I did one long and boring evening in college. lol

Let t1 represent the time going and t2 the time returning
Let s1 represent the speed going and s2 the speed returning
Let d represent the one-way distance for the trip

d = s1*t1 = s2*t2
s1 = s2*t2/t1

average speed = 2d/(t1+t2)
Now let's plug in for t1 and t2...

t2 = d/s2
t1 = d/s1

2d/(t1+t2) = 2*d/(d/s1+d/s2) = 2*d/[(s2*d+s1*d)/s1*s2]
The d's cancel out...

average speed = 2*s1*s2/(s1+s2)
Last edited by krusta80 on Sun Mar 04, 2012 2:07 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Sun Mar 04, 2012 2:06 pm
jzw wrote:Josh drives from his house to work and back. Same distance each way. On his way to work he drives @ an average of 60 mph. Oh his way back, he drives @ an average of 80 mph. What was his average speed?

(a) 60.0
(b) 68.6
(c) 70.0
(d) 71.4
(e) 80.0
Another option is to choose a "nice" distance between Josh's house and work (a distance that works well with 60mph and 80mph)

Let's say that it's 240 miles between Josh's house and work.
Total distance = 240 + 240 = 480 miles

Time at 60mph = distance/rate = 240/60 = 4 hrs
Time at 80mph = distance/rate = 240/80 = 3 hrs
Total time = 7 hours

Average speed = (total distance)/(total time) = 480/7
Rather than calculate/simplify this fraction, we can see that 490/7 = 70mph
So, 480/7 will be a little bit less than 70mph

Choose B

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Brent
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