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A certain established organization has exactly 4096 members.

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A certain established organization has exactly 4096 members. [#permalink] New post 03 May 2005, 07:49
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A certain established organization has exactly 4096 members. A certain new organization has exactly 4 members. Every 5 months the membership of the established organization increases by 100 percent. Every 10 months the membership of the new organization increases by 700 percent. New members join the organizations only on the last day of each 5- or 10-month period. Assuming that no member leaves the organizations, after how many months will the two groups have exactly the same number of members?

(A) 20
(B) 40
(C) 50
(D) 80
(E) 100
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 [#permalink] New post 03 May 2005, 08:49
4096.2^(X/5)=4.8^(X/10)

so 1024.2^(X/5)=2^(3X/10)

id est 1024 = 2^10 = 2^(3X/10-X/5)

X=100
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 [#permalink] New post 03 May 2005, 10:50
Hey twixt could you explain your answer to all is non-math genius's??? :-D
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Oraganisation soln. [#permalink] New post 04 May 2005, 00:32
greenandwise wrote:
Hey twixt could you explain your answer to all is non-math genius's??? :-D


Consider n to be the no. of 10 month period for the existing org to be levelled to the new org's employee strength.

for every n 10 month periods we have 2n 5 month periods.

new org increases it strengh 8(2^3) fold every 5 month period.
old org increases it strengh 2 fold every 10 month.

So we have
new org strengh at the end of (2n) 5 month periods = old org strenght at the end of n 10 month period.

i.e

4*8^2n = 4096*2^n

2^(6n+2) = 2^(n + 12)
OR N=2 i.e 2 10 month periods i.e 20 months in all.


month/ new/ old
5/32/4096
10/256/8192
15/2048/8192
20/16384/16384

Hope this helps.
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 [#permalink] New post 04 May 2005, 03:39
HMTG,

your reasonning is perfect but you mixed the organisations and their time scale of growth.
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 [#permalink] New post 04 May 2005, 05:15
twixt wrote:
HMTG,

your reasonning is perfect but you mixed the organisations and their time scale of growth.


twixt,
Thanks for pointing the flaw.You are absolutely right.
I messed that up.

So I guess the revised soln. would look like

4*8^n = 4096*2^2n
3n+2 = 12 + 2n
n=10
or 100 months.

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 [#permalink] New post 04 May 2005, 05:46
Hint:

Since the membership of the new group increases by 700% every 10 months, after the first 10-month period the new group will have (8)(4) members (remember that to increase by 700% is to increase eightfold). After the second 10-month period, it will have (8)(8)(4) members. After the third 10-month period, it will have (8)(8)(8)(4) members. We can now see a pattern: the number of members in the new group can be expressed as , where x is the number of 10-month periods that have elapsed.

Since the membership of the established group doubles every 5 months (remember, to increase by 100% is to double), it will have (2)(4096) after the first 5-month period. After another 5 months, it will have (2)((2)(4096)) members. After another 5 months, it will have (2)((2)(2)(4096)). We can now see a pattern: the number of members in the established group will be equal to , where y represents the number of 5-month periods that have elapsed.

The question asks us after how many months the two groups will have the same number of members. Essentially, then, we need to know when . Since y represents the number of 5-month periods and x represents the number of 10-month periods, we know that y = 2x. We now need to solve for x, which represents the number of 10-month periods that elapse before the two groups have the same number of members.

I think most of you can do the math from here, otherwise I will post it longhand upon request...
  [#permalink] 04 May 2005, 05:46
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