I need many ways for this one too!

[winniethepooh]

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180 - x) /2
B. (x + 60) /4
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

[Ashley@VeritasPrep]

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180 - x) /2
B. (x + 60) /4
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

An important thing to realize here in terms of making the problem easier is that it doesn't actually matter at all how long the trip was, since how these speeds will weigh against each other is determined exclusively by their weight relative to each other. So we may as well say that the trip was 100 miles long, just to make it easy since there are percents here. In that case, we've got d=rt --> r = d/t, and here, we break the distance up into two parts, d1 and d2 (d1 representing the portion traveled at 40mph and d2 representing the portion traveled at 60mph. Overall avg rate = (total distance)/(total time). We'll break the time up in the same fashion into t1 and t2. So, again supposing that the trip is 100 miles long,

d1 = x miles t1 = d1/r1 = x/40
d2 = 100-x miles t2 = d2/r2 = (100-x)/60
d"total" = 100 t"total" = t1 + t2 = x/40 + (100-x)/60 = 3x/120 + (200-2x)/120 = [3x+(200-2x)]/120 = (x+200)/120

avg. rate = (total distance)/(total time) = 100/[(x+200)/120] = 100 * [120/(x+200)] = 12000/(x+200).

[MM_Ed]

It's C. Here's a quick-and-dirty way of doing this:

Assume x=50. Now her trip is 50% @ 40 mph and 50% @ 60 mph, so the average speed is midway at 50 mph. See which option gives you 50 mph if x=50. C works, so C it is.

[winniethepooh]

Well, I would have done what you want (If you found one of my answers useful, amusing, or poetic, hit the shiny Thanks button and make my day); but you're wrong!

[Ian Stewart]

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180 - x) /2
B. (x + 60) /4
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

I'd probably just solve this algebraically, but since you asked for alternative approaches:

* when you plug in x=0, the answer needs to be 60

* when you plug in x=100, the answer needs to be 40

* now, that gets you down to C or E. Average speed is a weighted average, but it's weighted by the *time* spent at each speed (and not by the *distances*). If you travel equal distances at 40mph and 60mph, you will spend longer traveling at 40mph, so your average speed will be closer to 40 than to 60. So when you plug in x=50, you must get an answer which is strictly between 40 and 50. So C cannot be right, and E must be the answer.

[GMATGuruNY]

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180 - x) /2
B. (x + 60) /4
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Let x=50.
Let total distance = 240.
Since x=50, half the total distance will be traveled at 40 miles per hour, the other half at 60 miles per hour.
Time for first half = 120/40 = 3 hours.
Time for second half = 120/60 = 2 hours.
Total time = 3+2 = 5 hours.
Average speed for the whole trip = 240/5 = 48 miles per hour. This is our target.

Now we plug x=50 into the answers to see which yields our target of 48.

Only answer choice E works:
12,000 / ( x + 200) = 12,000/(50+200) = 48.

The correct answer is E.

[blackjack]

@ Mitch, I have 2 questions:
1. Do you think this is the fastest approach?
2. In general, when a question talks about percentages, do you think assuming x=50% is a fair and safe choice?

[GMATGuruNY]

@ Mitch, I have 2 questions:
1. Do you think this is the fastest approach?

What's fastest depends on the test-taker. Plugging in numbers certainly is an efficient way to solve this -- and many other -- problems.

2. In general, when a question talks about percentages, do you think assuming x=50% is a fair and safe choice?

Choose numbers that make the math easy.
50% makes this problem easy because then half the distance is traveled at one speed and half the distance is traveled at the other speed.
x = 10% and a total distance of 400 miles also would work well: 40 miles would be traveled at 40 miles per hour, 360 miles would be traveled at 60 miles per hour, resulting in an average speed for the whole trip of 400/7 miles per hour.