Average speed

[netcaesar]

Francine travel x percent of the distance of a trip at a constant speed of 40 mph and the rest of the trip at a constant speed of 60 mph. What is the average speed for the trip in terms of x?
a) (x+180)/2
b) 2400/(x+100)
c) 12000/(x+200)
d) 200/(x+180)
e) (x+240)/180

[Kevin]

A cheater could reason that if x=0, the answer must be 60 and get C as the answer

Otherwise, given a distance d, xd/100 miles at 40 mph and (100d-xd)/100 miles at 60 mph

Total time = xd/4000 + (100d-xd)/6000 = d(x+200)/12000

Average speed= d/(d(x+200)/12000)= 12000/(x+200)

[netcaesar]

Thanks Kevin.

With your explanation now it is a silly question....

A cheater could reason that if x=0, the answer must be 60 and get C as the answer

Otherwise, given a distance d, xd/100 miles at 40 mph and (100d-xd)/100 miles at 60 mph

Total time = xd/4000 + (100d-xd)/6000 = d(x+200)/12000

Average speed= d/(d(x+200)/12000)= 12000/(x+200)

[Shaji]

This "Cheater" has the management aptitude that top management schools look for.

[RonPurewal]

A cheater could reason that if x=0, the answer must be 60 and get C as the answer

that isn't cheating at all. that's an astute application of "plugging in numbers" / "VIC method", a perfectly legitimate problem-solving method.
in fact, any competent GMAT taker should include such methods as automatic backups in case they don't come up with "textbook" solutions to problems like these right away. these methods enable you to "solve" problems that you don't know how to solve the regular way.

if you actually regard this as "cheating" and aren't just being facetious, then your attitude about test-taking, and about the mental flexibility that characterizes good test-taking, is completely the opposite of the attitude required for success.

[Guest]

You're right, Ron, I was being facetious. Such experiments are often the FIRST thing you should try. In this case, we save over one minute! I would say that more than 70% of test takers would LOVE an extra minute in the quant section.

[StaceyKoprince]

Totally agree - I'll take an extra minute any day. :)

[gmatwork]

One way to solve this question is to use smart numbers; as per the strategy guide we should not use smart numbers when we have known quantities in a problem, although this problem has known quantities (speed) still the problem works well with smart numbers...why is that?

Please clarify when not to use smart numbers and when are we fine using them.

[RonPurewal]

One way to solve this question is to use smart numbers; as per the strategy guide we should not use smart numbers when we have known quantities in a problem, although this problem has known quantities (speed) still the problem works well with smart numbers...why is that?

Please clarify when not to use smart numbers and when are we fine using them.

the guide shouldn't say that. are you sure it says that?

the basic idea is this: if you don't find out the value of a quantity, then you can pick your own value for it.
if the problem contains OTHER quantities with known values, then that is immaterial. you should only avoid subbing in your own numbers if the actual quantity you're trying to pick has a determinable value.

[OmarQ]

Hey guys - thanks for the explanations thus far. I am trying and failing to do this question as a weighted averages question. I'm either a) making an error in my calculations or b) applying an incorrect strategy, and I'm not sure what I'm doing wrong.

My logic is this:

First, here's how I understand weighted averages. If 2/3 of the class weighs 60 pounds and 1/3 weigh 90, then the average for the class is
2/3 x 60 + 1/3 x 90 = 40 + 30 = 70 pounds.

In this case, Francine is going 40 mph x percent of the time and 60 mph 100-x percent of the time. So I get the following equation (disregarding distance completely):

40(x/100) + 60(1 - x/100)

40x/100 + 60 - 60x/100
60 - 20x/100

and at this point it seems like I'm off the rails. Sorry I'm not able to be more specific - but my guess is that my weighted average approach is fundamentally flawed and I can't see why. Any thoughts on if this is the case?

[RonPurewal]

In this case, Francine is going 40 mph x percent of the time and 60 mph 100-x percent of the time.

if those red things were actually 'time' in the problem, then this approach would be flawless.

...but they aren't. the problem is stated in terms of percentages of distance, not percentages of time. so, essentially, you're solving a problem that's not actually there.

[RonPurewal]

So I get the following equation (disregarding distance completely):

well, you can't do that ^^ because the problem is stated in terms of distance!

perhaps you were assuming that 'percentages of time' and 'percentages of distance' work out to be the same.
whenever you think you've invented a 'shortcut', though, you need to TEST IT WITH EASY CASES.

here, it's easy to prove to yourself that the above hypothesis is wrong (i.e., that the percentages of time DO NOT correspond to the percentages of distance).
just think about any easy example. as long as your example isn't '100 percent at one speed and 0 percent at the other one', there will be a difference.

in fact, you don't even need to work out specific numbers to see that this is so.
consider:
* i drive 100 miles (at freeway speed).
* my car breaks down, and i have to walk 100 miles back home.
here, i'm driving 50 percent of the distance, and walking the other 50 percent of the distance. on the other hand—just use your common sense—i'm definitely going to be walking for way, way, WAY more than 50 percent of the time.

so...nope. can't use that 'shortcut', because it doesn't work.