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# M15 #3

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Intern
Joined: 19 Jun 2008
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22 Oct 2008, 13:23
1
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Series A(n) is such that $$A(n) = \frac{A(n-1)}{n}$$ . How many elements of the series are bigger than 1/2?

1. $$A(2) = 5$$
2. $$A(1) - A(2) = 5$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

1) Explanation 1:

Together with the information from the stem S1 completely defines the series. Thus, S1 is sufficient to answer the question.

S2: $$A(1) - A(2) = 5 = A(1) - \frac{A(1)}{2}$$ from where $$A(1) = 10$$ . S2 is also sufficient to answer the question.

2) Alternative Explanation from GMAT Club member laxieqv:

Statement 1: $$A(2)= 5 --> A(1) = 10$$ , $$A(3) = \frac{5}{3}$$ , $$A(4)= \frac{5}{12}$$ ( all are according to the provided formula) . Notice that the bigger n is, the smaller A(n) is. For n>=4, $$A(n) <= \frac{5}{12} < \frac{1}{2}$$ ---> there are only A(1), A(2), A(3) which are bigger than 1/2 ---> this statement is sufficient.

Statement 2: $$A(1)- A(2)= 5$$ , together with $$A(2)= \frac{A(1)}{2}$$ ---> $$A (1) =10$$ and $$A(2)= 5$$ ---> come back to statement 1 ---> also sufficient.

That's why it's D.
The correct answer is D.

I don't understand in dealing with Statement 2 why A(2) is not set equal to 0. 1-1 is 0 so A(1) should be equal to 0 while A(2) is equal to A1/2.
CIO
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Re: M15 #3 [#permalink]

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23 Oct 2008, 06:23
You might be confused with the series markup. $$A(n)$$ signifies a series member. Thus, $$A(1)$$ is the first term of the series and so forth. We need to know at least one one term of this series to be able to find the rest. Do you think the question stem would look better like this:

Series A(n) is such that $$A_n = \frac{A_{n-1}}{n}$$ . How many elements of the series are greater than $$\frac{1}{2}$$?

Tell me what you think.
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Intern
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Re: M15 #3 [#permalink]

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26 Jan 2010, 11:01
option is D
in Ds, the questions have yes/no solution, we dont have to really find the exact answer just make sure whether we can get the answer or no.
by solving both the statement individually we will e able to find the elements of the series. once the series elements are known the question can be easily answered.
hence option D
Intern
Joined: 12 Jul 2008
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Re: M15 #3 [#permalink]

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26 Jan 2010, 11:55
Wouldn' this question make more sense if they give the range values for 'n' say 'n' is a natural no. that varies from 1 to infinity.

I found some inconsistencies while solving for lower values of 'n' such as 0 and 1/2
Eg:
if A(1) = 10 then A(1) = A(0) / 1 and hence A(0) is also 10 .
But solving for A(0) :
A(0) = A(-1) / 0 which is not a valid number and hence A(0) cannot be 10 .
You get what I am saying . If this a wrong way to deal with series questions or should the question be more precise.
Manager
Joined: 23 Nov 2009
Posts: 52
Re: M15 #3 [#permalink]

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26 Jan 2010, 18:29
Question stem could be revamped as suggested. It looks very confusing.
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A kudos would greatly help

Tuhin

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Re: M15 #3 [#permalink]

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27 Jan 2010, 07:14
1
KUDOS
given

A(n)=A(n-1)/n

a) A(2)=5

so A(2)=A(1)/2=5 ==> A(1)=10 similarly we can identify

A(3)=5/3
A(2)=5
A(1)=10
A(0)=10
A(-1)=0
....

so we know only for 3 elements the value is greater than 1/2

b)A1-A2=5
A1=A2+5

by using the formula A(n)=A(n-1)/n
A2=A1/2
so A1=A1/5+5 ==> A1=(A1+10)/2 ==> 2A1=A1+10 ==> A1=10

so applying the same logic as of choice 1 we know that only for 3 elements the value is greater than 1/2

son my ans choice is D
Intern
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Re: M15 #3 [#permalink]

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25 Jan 2011, 06:22
If A(n) = A(n-1)/n, how many elements of the series are larger than 1/2?

For part 1. A(2) =5

How do you input A(2)=5 into the formula above to get the ((A(1) =10, A(3)= 5/3, A(4)= 5/12)?
Intern
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Re: M15 #3 [#permalink]

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28 Jan 2011, 08:43
not sure how you implement A(3) to get 5/3 and A(4) to get 5/12. can someone please walk me through it. thanks in advance
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Re: M15 #3 [#permalink]

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01 Feb 2011, 02:34
Can somebody will explain the theory behind this question? I didn't understand the question. What exactly it wants to ask??
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Re: M15 #3 [#permalink]

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02 Feb 2012, 06:23
Never-mind... I looked over silasaaa2's post in more detail. Small mistake in the post was confusing me but after going over the logic I caught it and it makes sense now. THANK YOU!!!

------------
Sorry I am still not understanding how S2 gives us 10 for A(1). When I do the math I get A(0)/1 and this is confusing me. Can someone walk through the steps with greater detail?

Of course knowing what we learned in S1 I see how it is "correct" and how S2 supports what we have already seen in S1, but how we get there independently is where my issue lies.

Thank you

Last edited by Reighngold on 02 Feb 2012, 06:48, edited 1 time in total.
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Re: M15 #3 [#permalink]

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02 Feb 2012, 06:45
not sure how you implement A(3) to get 5/3 and A(4) to get 5/12. can someone please walk me through it. thanks in advance

This threw me for a loop as well when I first saw it here is how I figured it out.

In both S1 and S2 we have already established that A(2) = 5

So when we substitute n with 3 we get:
A(3)=A(3-1)/3
A(3)=A(2)/3

Since we know that A(2) = 5 we can substitute 5 and we get:
A(3)=5/3

Following the same pattern you then get:
A(4)=A(4-1)/4
A(4)=A(3)/4
A(4)=5/3/4 <-- again substituting the pre derived value for A(3)
A(4)=5/3*1/4
A(4)=5/12

I hope this helps.
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Re: M15 #3 [#permalink]

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02 Feb 2012, 13:54
2
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Expert's post
rajeshaaidu wrote:
Can somebody will explain the theory behind this question? I didn't understand the question. What exactly it wants to ask??

Series A(n) is such that $$A(n) = \frac{A(n-1)}{n}$$ . How many elements of the series are bigger than 1/2?

Probably te question should have been worded in a different manner:

The sequence $$A_1$$, $$A_2$$, ... is defined such that $$A_{n+1}=\frac{A_{n}}{n+1}$$ for all n>1. How many terms of the sequence is greater than 1/2? (Note that it's the same exact formula just written in different manner)

Basically we have a sequence of numbers which is defined with some formula. For example: $$A_{2}=\frac{A_{1}}{1+1}$$, $$A_{3}=\frac{A_{2}}{2+1}$$, $$A_{4}=\frac{A_{3}}{3+1}$$, ... The question asks: how many numbers from the sequence is greater than 1/2. Notice that if we knew ANY term of the sequence we would be able to get all the other terms/numbers and thus answer the question.

(1) $$A_2=5$$. As discussed above this statement is sufficient as we can write down all the terms. For example: $$A_{2}=\frac{A_{1}}{1+1}=5$$ --> $$A_1=10$$. $$A_{3}=\frac{A_{2}}{2+1}=\frac{5}{3}$$, and so on.

(2) $$A_1-A_2=5$$ --> $$A_1-\frac{A_{1}}{1+1}=5$$ --> we can solve for $$A_1$$ and thus will have the same case of knowing one term. Sufficient.

Hope it's clear.
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Re: M15 #3 [#permalink]

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04 Feb 2012, 18:37
Question statement 2 was real tricky. I knew that statement 1 is sufficient once I found out A1=10

but for statement 2, I decided that A(1) is = to 0 since the numerator is being multiplied by 0.

Thus I got 0-A(2)=5 therefore -A(2)=5 which got me A=-10. I thus picked A but after reading this thread and seeing how everyone got for statement 2, A=10, I just thought it was tricky the way the other people solved it.

Where did I go wrong in my thought process for statement 2?
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Re: M15 #3 [#permalink]

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04 Feb 2012, 18:44
AzWildcat1 wrote:
Question statement 2 was real tricky. I knew that statement 1 is sufficient once I found out A1=10

but for statement 2, I decided that A(1) is = to 0 since the numerator is being multiplied by 0.

Thus I got 0-A(2)=5 therefore -A(2)=5 which got me A=-10. I thus picked A but after reading this thread and seeing how everyone got for statement 2, A=10, I just thought it was tricky the way the other people solved it.

Where did I go wrong in my thought process for statement 2?

I think you multiplied by n-1 when you got 0 for A1, but n-1 is a subscript not a multiple. Refer to my post above for proper version of this question with a solution.

Hope it helps.
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Re: M15 #3 [#permalink]

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04 Feb 2012, 22:52
Bunuel wrote:
AzWildcat1 wrote:
Question statement 2 was real tricky. I knew that statement 1 is sufficient once I found out A1=10

but for statement 2, I decided that A(1) is = to 0 since the numerator is being multiplied by 0.

Thus I got 0-A(2)=5 therefore -A(2)=5 which got me A=-10. I thus picked A but after reading this thread and seeing how everyone got for statement 2, A=10, I just thought it was tricky the way the other people solved it.

Where did I go wrong in my thought process for statement 2?

I think you multiplied by n-1 when you got 0 for A1, but n-1 is a subscript not a multiple. Refer to my post above for proper version of this question with a solution.

Hope it helps.

Ahhh I see now. Thanks. It does sorta look like a multiple with (n-1) being multiplied by A. I now see that it is a subscript. Thanks
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Re: M15 #3 [#permalink]

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05 Feb 2012, 07:10
I am sorry but i did not get this question! in the first post how does A(2)=5 imply A(1)=10. can anyone explian?
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Re: M15 #3 [#permalink]

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05 Feb 2012, 07:13
akhilright wrote:
I am sorry but i did not get this question! in the first post how does A(2)=5 imply A(1)=10. can anyone explian?

Please refer to this post: m15-71960.html#p1038539, for proper version of this question with detailed solution.

Hope it helps.
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Re: M15 #3 [#permalink]

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05 Feb 2012, 07:19
Thanks Bunuel,
I understood that particular reply, but fundamentaly i did not get how has the reply reworded A (n+1)=An/n+1, from the real equation given in the question
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Re: M15 #3 [#permalink]

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05 Feb 2012, 07:29
akhilright wrote:
Thanks Bunuel,
I understood that particular reply, but fundamentaly i did not get how has the reply reworded A (n+1)=An/n+1, from the real equation given in the question

First of all, if you look at the formula given in the initial post and the one given in mine, you'll notice that they're basically the same: just replaced n with n+1. Next, I didn't reworded the question, I changed it so that it would be more GMAT type.
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Re: M15 #3 [#permalink]

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07 Jan 2013, 06:13
Bunuel.
Since its given that n>1, then how can we even deal with A1.
This is why I got this one wrong.
Chose A, which is wrong.
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Re: M15 #3   [#permalink] 07 Jan 2013, 06:13

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