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31 Oct 2016, 06:12
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WilDThiNg
Yes, you can plug in values like that. If multiple options give you the correct answer, try some other values on the shortlisted options only.
For example, next you should try x = 0 (all distance travelled at 60 only and hence avg is 60). You get options (C) and (E).
Next try x = 50
Avg Speed = 2*40*60/(40+60) = 48
Out of options (C) and (E), only option (E) gives 48.
16 Jul 2017, 16:37
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Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?
Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.
I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?
Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.
I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
17 Jul 2017, 05:07
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suntorytea
How do you get 12000/(x + 2)?
You don't need to assume a value for d.
Average Speed = Total Distance / Total Time
Total Distance = d
Time 1 \(= \frac{D1}{S1} = \frac{(\frac{x}{100})*d}{40} = \frac{xd}{4000}\)
Time 2 \(= \frac{D2}{S2} = \frac{(1 - x/100)*d}{60} = \frac{(100 - x)d}{6000}\)
Average Speed \(= \frac{d}{\frac{xd}{4000} + \frac{(100-x)d}{6000}}\)
We take d common from the denominator and it cancels out with the d in the numerator
Average Speed \(= \frac{d}{d * (\frac{x}{4000} + \frac{(100-x)}{6000})}\)
Average Speed \(= \frac{1}{(\frac{x}{4000} + \frac{(100-x)}{6000})}\)
Average Speed \(= \frac{12000}{x + 200}\)
