Questions about the world of GMAT Math from other sources and general math related questions.
PudGe
Students
 
Posts: 12
Joined: Mon Sep 05, 2011 4:24 pm
 

P1, P2, P3, …. Pn, Pn+1 = Pn/3

by PudGe Tue May 29, 2012 3:37 am

Q) A series of positive numbers P1, P2, P3, .... Pn, Pn+1 = Pn/3 for all values of n ≥ 1. what is the value of 2P4?
(1) P2 = 1/3
(2) P2 - P3 = 2/9

OA is D
need some help in going about this type of problem.
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by jnelson0612 Sat Jun 02, 2012 10:54 pm

Hey PudGe, again, according to forum guidelines, we need an original source for each question. Please provide and we're happy to help. :-)
Jamie Nelson
ManhattanGMAT Instructor
navneesh
Students
 
Posts: 1
Joined: Wed Dec 31, 1969 8:00 pm
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by navneesh Tue Jun 26, 2012 2:29 pm

Hello Nelson,

This question is from the Mock test of Jamboree classes in India. This is from their free test, so It shouldn't be an issue to solve it.

This is a challenging question though...:p

Thanks,
krishnan.anju1987
Forum Guests
 
Posts: 125
Joined: Mon May 07, 2012 6:13 pm
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by krishnan.anju1987 Wed Jun 27, 2012 9:32 am

Hi,

I believe the OA is D. Please correct me if I am wrong.
krishnan.anju1987
Forum Guests
 
Posts: 125
Joined: Mon May 07, 2012 6:13 pm
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by krishnan.anju1987 Fri Jun 29, 2012 6:42 pm

Hi,

The question does mention that Pn+1 = Pn/3 which is the equivalent of your statement. So I am unable to find anything wrong with the question. Could you please share what is wrong with the question. We do need the statements (each statement would suffice by itself) to provide us the first value of the expression.
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by jnelson0612 Sun Jul 01, 2012 10:57 pm

krishnan.anju1987 Wrote:Hi,

The question does mention that Pn+1 = Pn/3 which is the equivalent of your statement. So I am unable to find anything wrong with the question. Could you please share what is wrong with the question. We do need the statements (each statement would suffice by itself) to provide us the first value of the expression.


Thanks krishnan! I get it now . . . any value in this sequence is the previous value divided by 3. This is an unusual way to express this and that is what threw me. Usually the testwriters would express this as Pn= P(n-1)/3.
Jamie Nelson
ManhattanGMAT Instructor
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3

by jnelson0612 Sun Jul 01, 2012 11:06 pm

PudGe Wrote:Q) A series of positive numbers P1, P2, P3, .... Pn, Pn+1 = Pn/3 for all values of n ≥ 1. what is the value of 2P4?
(1) P2 = 1/3
(2) P2 - P3 = 2/9

OA is D
need some help in going about this type of problem.


Okay, our first step is to define the formula for this series. As mentioned in my previous post, any value is obtained simply by taking the previous value and dividing that value by 3. That is what the P(n+1) = P(n)/3 means.

Okay, so if I want 2 * P(4), then I just need to know P(4) to be able to solve the question. Really, if I know any value in the sequence I could obtain P(4).

Statement 1 says that P(2) = 1/3. Sufficient. P(2) is 1/3, so P(3) is 1/9 (1/3 divided by 3). P(4) will be 1/9 divided by 3, or 1/27.

Statement 2 is only a little trickier:
Let's call P(2) = x
Because P(3) is P(2) divided by 3, P(3) = x/3.

Statement 2 says that P(2) - P(3) = 2/9. Thus x - x/3 = 2/9. 2/3x=2/9.
(3/2)2/3x = 2/9(3/2)
x=1/3
So we know that P(2) is 1/3. Sufficient.
Please note that we did not have to do much of this work for this statement. Once we had "x - x/3 = 2/9" we knew we could solve for x, which is standing in for P(2), and we should immediately have just said sufficient and moved on.
Jamie Nelson
ManhattanGMAT Instructor