Difficult PS Q?

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briology: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Sun Sep 18, 2011 1:14 am

by briology » Fri Sep 23, 2011 10:55 pm

Hi All--

I got this problem during a practice test today.

Train M and Train N are traveling in the same direction on parallel sets of tracks. Train M is traveling at 2^m miles per hour, and Train N is traveling at 2^n miles per hour, with N>M. If Train M is initially 2^m+1 miles ahead of Train N, how many hours will it take Train N to catch up with Train M?

Two Questions:

1. I'm curious how you guys would rank this problem, difficulty-wise. To me, it seemed very tough. Is this something I'd actually see on the GMAT? And if so, do you think it'd be considered a 700lvl question, or am I just failing at math?

2. I tried to solve the question algebraically, but it was such a long calculation. I ended up sinking about 5 minutes into the problem, and I still got it wrong in the end. Is there a better way to solve it?

Thanks for the help!

Last edited by briology on Fri Sep 23, 2011 11:42 pm, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Fri Sep 23, 2011 10:55 pm

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Fri Sep 23, 2011 11:00 pm

Just miles ahead? or is there a number?

briology wrote:Hi All--

I got this problem during a practice test today.

Train M and Train N are traveling in the same direction on parallel sets of tracks. Train M is traveling at 2^m miles per hour, and Train N is traveling at 2^n miles per hour, with N>M. If Train M is initially miles ahead of Train N, how many hours will it take Train N to catch up with Train M?

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briology: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Sun Sep 18, 2011 1:14 am

by briology » Fri Sep 23, 2011 11:42 pm

Sorry! Train M is initially 2^m+1 miles ahead

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gmatclubmember: Master | Next Rank: 500 Posts; Posts: 158; Joined: Sat Sep 03, 2011 10:31 amFollowed by: 2 members; Thanked: 29 times

by gmatclubmember » Sat Sep 24, 2011 12:31 am

The faster train is gaining distance over the slower train at the speed of 2^n - 2^m.
Total distance to be covered is 2^(m+1)
Time required to catch up with the slower train is:

2^(m+1)/(2^n-2^m)

Is this the OA?

Cheers
Ami/-

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sat Sep 24, 2011 12:40 am

The faster train needs to cover a distance of 2^(m+1) to catch up the slower train at the rate of 2^n - 2^m (Relative speed, given n>m)

Distance = Speed * Time

Time = 2^(m+1)/2^(n-m)

= 2^(2m-n+1)

Its just complicated in its wordings I guess, and given answer choices, you could always substitute values of m and n and see which one satisfies, guess that would make it a lot more easier

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gmatclubmember: Master | Next Rank: 500 Posts; Posts: 158; Joined: Sat Sep 03, 2011 10:31 amFollowed by: 2 members; Thanked: 29 times

by gmatclubmember » Sat Sep 24, 2011 12:46 am

shankar.ashwin wrote:The faster train needs to cover a distance of 2^(m+1) to catch up the slower train at the rate of 2^n - 2^m (Relative speed, given n>m)

Distance = Speed * Time

Time = 2^(m+1)/2^(n-m)

= 2^(2m-n+1)

Its just complicated in its wordings I guess, and given answer choices, you could always substitute values of m and n and see which one satisfies, guess that would make it a lot more easier

Ashwin
The calculation 2^(n-m) is incorrect in your answer. 2^n-2^m cannot be 2^(n-m).

Cheers
Ami/-

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briology: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Sun Sep 18, 2011 1:14 am

by briology » Sat Sep 24, 2011 1:50 am

OA = 2/(2^n-m - 1)

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sat Sep 24, 2011 2:22 am

Oops my bad... Just didnt realize. Thanks BTW

gmatclubmember wrote:
shankar.ashwin wrote:The faster train needs to cover a distance of 2^(m+1) to catch up the slower train at the rate of 2^n - 2^m (Relative speed, given n>m)

Distance = Speed * Time

Time = 2^(m+1)/2^(n-m)

= 2^(2m-n+1)

Its just complicated in its wordings I guess, and given answer choices, you could always substitute values of m and n and see which one satisfies, guess that would make it a lot more easier

Ashwin
The calculation 2^(n-m) is incorrect in your answer. 2^n-2^m cannot be 2^(n-m).

Cheers
Ami/-

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gmatclubmember: Master | Next Rank: 500 Posts; Posts: 158; Joined: Sat Sep 03, 2011 10:31 amFollowed by: 2 members; Thanked: 29 times

by gmatclubmember » Sat Sep 24, 2011 2:35 am

briology wrote:OA = 2/(2^n-m - 1)

You will get this answer when you divide 2^(m+1)/(2^n-2^m) by 2^m in numerator and denominator.

Cheers
Ami/-

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