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Mischa drove from A to B, a distance of 100 miles

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monsama
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Mischa drove from A to B, a distance of 100 miles [#permalink] New post   Updated on: 25 Aug 2014, 03:55
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Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour, and then back from B to A along the same route at an average speed of m miles per hour. What is the value of m?

(1) Mischa’s average speed for the entire round trip, excluding any time he spent at point B, was \(\frac{m+50}{2}\) miles per hour.

(2) If Mischa’s average speed from B to A had been 20% slower, the total time for the entire round trip, excluding any time Mischa spent at point B, would have been 12.5% greater.
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D

Originally posted by monsama on 25 Aug 2014, 03:51.
Last edited by Gnpth on 25 Aug 2014, 03:55, edited 1 time in total.
formatted the Topic.
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink] New post   25 Aug 2014, 13:53
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You don't have to do any calculation to solve this, because both statements present an equation with only one variable. (Yes, the first one ends up being a quadratic, but we know m can't be negative.)

Here's how I thought about it (without doing the algebra):

(1) This will result in an equation that has only one unknown (m) on both sides. It looks like I can solve for m. SUFFICIENT.

(2) This relates two different versions of the same formula (total distance). There are no new variables--just numbers--so again I can solve for m. SUFFICIENT.

Just for fun (we don't need to do this to solve), there's also a quick way to tell that m=50 from Statement 1. It's telling us that the average speed for the whole trip is simply the average of the rates coming and going. When can you average rates? Only when the times are the same. If the distance (100 miles) and the time (2 hours) are the same for both parts, then so are the rates. m=50. We can be extra sure of this, because if m were greater than 50, it would make the time shorter, and if m were less than 50, it would make the time longer. This is contradicted by the formula in Statement 1, in which the bigger m is, the bigger the time is. Further confirmation that m must be 50!
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink] New post   25 Aug 2014, 05:15
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Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour. SO time taken to go from A to B = 100/50 hrs = 2hrs

Average speed for going to B and coming back from B to A = 2uv/(u+v)
where u = speed from A to B = 50
v = speed from B to A = m
Average Speed = 100m/(50+m)
Total time taken = total distance/ avg speed = 200/(100m/(50+m)) = (100+2m)/m hrs
STAT1
Average speed = (50+m)/2
= 100m/(50+m)
=> 2500 + m^2 + 100m = 200m
=> m^2 - 100m + 2500 = 0
=> m = 50
So, SUFFICIENT

STAT2
If Mischa's avg speed from B to A was 20% slower => New Avg speed = 2*50*(0.8m)/(50+0.8m)
= 80m/(50+0.8m) = 400m/(250+4m)

Total time would be 12.5% greater => new total time = old total time *1.125 = 1.125 *(100+2m)/m ...(1)

New total time = total distance / new avg speed = 200/ (400m/250+4m))
= (250+4m)/2m = 1.125*(100+2m)/m (from ...(1))
=> (250+4m) = 2.25*(100+2m)
=> 25 = 0.5m
=> m = 50
So, SUFFICIENT

So, Answer will be D

Hope it helps!


MonSama wrote:
Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour, and then back from B to A along the same route at an average speed of m miles per hour. What is the value of m?

(1) Mischa’s average speed for the entire round trip, excluding any time he spent at point B, was \(\frac{m+50}{2}\) miles per hour.

(2) If Mischa’s average speed from B to A had been 20% slower, the total time for the entire round trip, excluding any time Mischa spent at point B, would have been 12.5% greater.
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink] New post   25 Aug 2014, 07:53
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MonSama wrote:
Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour, and then back from B to A along the same route at an average speed of m miles per hour. What is the value of m?

(1) Mischa’s average speed for the entire round trip, excluding any time he spent at point B, was \(\frac{m+50}{2}\) miles per hour.

(2) If Mischa’s average speed from B to A had been 20% slower, the total time for the entire round trip, excluding any time Mischa spent at point B, would have been 12.5% greater.


You actually don't need to get to the absolute value of m (as this is not PS :p )

Average speed = Total Distance/Total time.

(1) Let time take to travel from B to A be t hours.
=> t = 100/m -- (A)

by FS 1 - 200/(t+2) = (m+50)/2

we have already derived a relation between t and m in (A)
so m can easily be worked out (i did not solve the quadratic)

(2) if the avg speed from B to A would be have been 20% lesser i.e. avg speed shall be 0.8m
so it took 12.5% extra time for the entire round trip because of this change in speed

Let original time taken for the round tripe be T.
T = 200/(t+2)
now when m becomes 0.8m, T becomes 1.125T.
in this case also we have a relation between T and t, t and m. so m can be worked out.

Answer, therefore, should be D.
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink] New post   10 Jun 2017, 15:33
Here's what I think:

(1) To average out speeds, you divide Total Distance by Total Time. This equation would give the correct answer only if m=50, therefore sufficient.
(2) There is a definite co-relation between the 20% and 12.5% and my gut says this can be solved. But plugging in m=50, 20% slower would be m=40 at which speed it'll take 2.5 hours instead of the 2 hours for the return journey increasing the total time from 4 hours to 4.5 hours. 0.5 / 4 = 12.5%. Sufficient.

Disclaimer: I do a lot of my quant on number crunching and gut, this solution might not resonate with everyone nor would it be applicable in other similar questions, but please tell me if specifically for this question, the approach is wrong. My GMAT's up very soon :)
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink] New post   30 Apr 2023, 01:20
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Re: Mischa drove from A to B, a distance of 100 miles [#permalink]
30 Apr 2023, 01:20
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