BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Difficult PS Q?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 22
Joined: Sun Sep 18, 2011 1:14 am

Difficult PS Q?

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.
Top
Source: — Problem Solving |
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

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?
Top
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
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Sat Sep 03, 2011 10:31 am
Thanked: 29 times
Followed by:2 members

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/-
Top
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

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
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Sat Sep 03, 2011 10:31 am
Thanked: 29 times
Followed by:2 members

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/-
Top
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)
Top
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

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/-
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Sat Sep 03, 2011 10:31 am
Thanked: 29 times
Followed by:2 members

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/-
Top
Post new topic Post Reply
• Page 1 of 1