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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]

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New post Updated on: 21 Apr 2014, 06:35
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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

Originally posted by JCLEONES on 08 Jan 2008, 11:23.
Last edited by Bunuel on 21 Apr 2014, 06:35, edited 1 time in total.
Added the OA.
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 21 Apr 2014, 06:56
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virajgawade wrote:
can someone pls explain how does 100 months come???


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

Every 5 months the membership of the established organization increases by 100 percent, means that every 5 months the membership of the established organization doubles.

Every 10 months the membership of the new organization increases by 700 percent, means that every 10 months the membership of the new organization increases 8 times.

Say these organizations will have the same number of members in x months.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).

Equate: \(2^{12}*2^{\frac{x}{5}}=2^2*2^{\frac{3x}{10}}\) --> reduce by 2^2: \(2^{10 +\frac{x}{5}}=2^{\frac{3x}{10}}\) --> \(10 +\frac{x}{5}=\frac{3x}{10}\) --> \(x=100\).

Answer: E.

Hope it's clear.
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 11:39
3
E

4096*2^[m/5]=4*8^[m/10]
where, m - the number of months.
2^12*2^[m/5]=2^2*2^(3*[m/10])
2^(12+[m/5])=2^(2+3*[m/10])
12+[m/5]=2+3*[m/10]
10+[m/5]=3*[m/10]

10+2x=3x (x=[m/10])
x=10 m=100.

100/5=20, 100/10=10. are integers.
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 11:42
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1
4096 = 2^12 members for old group at start
4 = 2^2 members for new group at start

2^2x where x = number of 10 month periods
2^3x where x = number of 10 month periods

(2^12)(2^2x) = (2^2)(2^3x)
(2^10)(2^2x) = (2^3x)
(2^10) = (2^x)
x = 10

Since x = number of 10 month periods (10*10) = 100

Answer E
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 16:20
eschn3am wrote:
4096 = 2^12 members for old group at start
4 = 2^2 members for new group at start

2^2x where x = number of 10 month periods
2^3x where x = number of 10 month periods

(2^12)(2^2x) = (2^2)(2^3x)
(2^10)(2^2x) = (2^3x)
(2^10) = (2^x)
x = 10

Since x = number of 10 month periods (10*10) = 100

Answer E


I tried setting up an equation such as this but couldn't get it... why do you have 2^2x and 2^3x???
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 18:47
JCLEONES wrote:
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


For the established group the membership is doubling every 5 months. If you set x = to # of 5 month periods you could say that the membership is equal to original membership*2^x. However, since the other group is expanding every 10 months, and the answers were all in increments of 10, I set x equal to 10 months. Of course 10 months = two 5 month periods so the equation is 2^2x instead of 2^x. The 2^2 = 4 comes from the population quadrupling every 10 months. You could also have the equation set up as 4^x where x is the number of 10 month periods. You get the same results, I just wanted to keep everything in powers of 2.

The new organization is 8x bigger after 10 months (growing by 700 percent = x+7x = 8x). 2^3 = 8 so 2^3(x) = membership after x number of 10 month periods.

Try plugging in numbers and testing them on a calculator to see for yourself. Start with a membership size of 1 person and play around with different time periods to see how it works.
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 20:10
Answer E

Explanation..

Est organisation has 4096 members = 2 ^ 12
New organisation has 4 members = 2 ^ 2

After 5 months.. Est Org members = 2 ^ 12 + 2 ^ 12 = 2 ^ 13 = 2 ^ ( 12 + 1 )
After 10 months.. Est org members = 2 ^ 14 = 2 ^ ( 12 + 2 )

After 10 months.. New org members = 2 ^ 2 + 7 * 2 ^ 2 = 2 ^ 2 ( 1 + 7 )

Let X be the months after which both orgs have same strenthg..

Let Y be X / 10

then 2 ^ ( 12 + 2 * Y ) = ( 2 ^ 2 ) * ( (1 + 7) ^ Y )

this gives Y = 10, that means X = 100
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Re: PS Membership  [#permalink]

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New post 08 Jan 2008, 20:30
GMATBLACKBELT wrote:
eschn3am wrote:
4096 = 2^12 members for old group at start
4 = 2^2 members for new group at start

2^2x where x = number of 10 month periods
2^3x where x = number of 10 month periods

(2^12)(2^2x) = (2^2)(2^3x)
(2^10)(2^2x) = (2^3x)
(2^10) = (2^x)
x = 10

Since x = number of 10 month periods (10*10) = 100

Answer E


I tried setting up an equation such as this but couldn't get it... why do you have 2^2x and 2^3x???


2^2x because an increase of 100% essentially means that you are doubling your membership

2^3x because an increase of 700% means that your increasing your membership 8-fold, i.e. 2^3
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 21 Apr 2014, 06:24
can someone pls explain how does 100 months come???
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 21 Apr 2014, 20:41
1
Looks like compound interest problem, that is why complicated for backsolving

1. First we shold fit 5 months in established group to 10 months in new one. So we get 2*2=4 times

2. Create equation and solve it: 4096*4^x=4*8^x => 1024=2^x, x=10 times and 10*10=100 months

Answer is E
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 03 Jun 2015, 03:24
1
Bunuel wrote:
virajgawade wrote:
can someone pls explain how does 100 months come???


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

Every 5 months the membership of the established organization increases by 100 percent, means that every 5 months the membership of the established organization doubles.

Every 10 months the membership of the new organization increases by 700 percent, means that every 10 months the membership of the new organization increases 8 times.

Say these organizations will have the same number of members in x months.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).

Equate: \(2^{12}*2^{\frac{x}{5}}=2^2*2^{\frac{3x}{10}}\) --> reduce by 2^2: \(2^{10 +\frac{x}{5}}=2^{\frac{3x}{10}}\) --> \(10 +\frac{x}{5}=\frac{3x}{10}\) --> \(x=100\).

Answer: E.

Hope it's clear.




Hi Bunnel

Can the above problem be solved using the concept of geometric progression ?

I tried but couldn't make it.
Formula used : nth term of GP = ar^(n-1)
4096 (2)^n-1=4(8)^m-1
n: number of 5 month periods
m: number of 10 month periods

so, n=2m

not getting where am i going wrong

can you please help ?
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 03 Jun 2015, 03:41
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JCLEONES wrote:
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


Established Organization growth = 100% i.e. x becomes x+(100/100)x = 2x in 5 months
New Organization growth = 700% i.e. x becomes x+(700/100)x = 8x in 10 months

Which means the population of established organization become 4 times of previous population every 10 months
and
the population of New organization become 8 times of previous population every 10 months


Let's say after a cycles of 10 months the population of both organization becomes equal

Population of New Organization after 'y' cycles of 10 months = 4 x (8^y) = (2^2) x [2^(3y)] = [2^(2+3y)]

Population of Established Organization after 'y' cycles of 10 months = 4096 x (4^y) = (2^12) x [2^(2y)] = [2^(12+2y)]

i.e. [2^(2+3y)] = [2^(12+2y)]

i.e. 2+3y = 12+2y

i.e. y = 10

i.e. 10 cycles of 10 months i.e. 10x10 = 100 months

Answer: Option
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A certain established organization has exactly 4096 members.  [#permalink]

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New post 03 Jun 2015, 04:14
Radhika11 wrote:
Bunuel wrote:
virajgawade wrote:
can someone pls explain how does 100 months come???


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

Every 5 months the membership of the established organization increases by 100 percent, means that every 5 months the membership of the established organization doubles.

Every 10 months the membership of the new organization increases by 700 percent, means that every 10 months the membership of the new organization increases 8 times.

Say these organizations will have the same number of members in x months.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).

Equate: \(2^{12}*2^{\frac{x}{5}}=2^2*2^{\frac{3x}{10}}\) --> reduce by 2^2: \(2^{10 +\frac{x}{5}}=2^{\frac{3x}{10}}\) --> \(10 +\frac{x}{5}=\frac{3x}{10}\) --> \(x=100\).

Answer: E.

Hope it's clear.




Hi Bunnel

Can the above problem be solved using the concept of geometric progression ?

I tried but couldn't make it.
Formula used : nth term of GP = ar^(n-1)
4096 (2)^n-1=4(8)^m-1
n: number of 5 month periods
m: number of 10 month periods

so, n=2m

not getting where am i going wrong

can you please help ?


Hi Radhika,

Please check the below solution as you want it to be



You need to understand here that the the terms are increasing by common ratio of 2 and 8 for Established and New organization respectively the period of increase are different so you need to avoid taking (-1) in the power because (n-1) will lead to a different number of 5 months period than no. of (m-1) periods of 10 months

After n periods of 5 months the Established organization will have population = 4096 (2)^n
and
After m periods of 10 months the New organization will have population = 4 (8)^m

4096 (2)^n=4(8)^m
n: number of 5 month periods
m: number of 10 month periods

so, n=2m

4096 (2)^(2m)=4(8)^(m)

i.e. (2^12)[2^(2m)] = (2^2)[2^(3m)] ------ [Please note that 8 = 2^3 therefore 8^(m) = 2^3(m) = 2^(3m)]

i.e. [2^(12+2m)] = [2^(2+3m)]

i.e. (12+2m) = (2+3m)

i.e. 12+2m = 3m+2

i.e. m=10

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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 03 Apr 2016, 10:07
Hi Bhoopendra (@GMATinsight)

Could you please elaborate on why we cant solve this using G.P and why we shouldn't take n-1 in the power ? Here is how i solved the prob :

Old organization = 2^12 * 2^0 (initial value); r = 2^2
New organization = 2^2*2^0 (initial value) ; r = 2^3

as per G.P, equating the two GPs to get n

2^12*[2^(2n-2)]=2^2[2^(3n-3)]
I get n = 11. Not sure what the issue is in this method.

TIA!
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 23 Apr 2016, 05:35
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Keysersoze10 wrote:
Hi Bhoopendra (@GMATinsight)

Could you please elaborate on why we cant solve this using G.P and why we shouldn't take n-1 in the power ? Here is how i solved the prob :

Old organization = 2^12 * 2^0 (initial value); r = 2^2
New organization = 2^2*2^0 (initial value) ; r = 2^3

as per G.P, equating the two GPs to get n

2^12*[2^(2n-2)]=2^2[2^(3n-3)]
I get n = 11. Not sure what the issue is in this method.

TIA!



As per your solution

Formula used : nth term of GP = ar^(n-1)

a= 4096 for Established organization and r=2^2

@n=1, The Total Employees = 4096
@n=2, The Total Employees = 4096*2^2(1) [After FIRST cycle of 10 months]
@n=3, The Total Employees = 4096*2^2(2) [After SECOND cycle of 10 months]
@n=4, The Total Employees = 4096*2^2(3) [After THIRD cycle of 10 months]
---
---
---
@n=11, The Total Employees = 4096*2^2(n-1) [After TENTH cycle of 10 months] i.e.Total Month passes = 10*10 = 100


So n=11 tells that 11th term equates the two numbers (no. of employees for two organization) but 11th term comes after 10 cycles of 10 months hence 100 months


I hope this helps!!!
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 16 May 2016, 08:59
dvdv wrote:
Answer E

Explanation..

Est organisation has 4096 members = 2 ^ 12
New organisation has 4 members = 2 ^ 2

After 5 months.. Est Org members = 2 ^ 12 + 2 ^ 12 = 2 ^ 13 = 2 ^ ( 12 + 1 )
After 10 months.. Est org members = 2 ^ 14 = 2 ^ ( 12 + 2 )

After 10 months.. New org members = 2 ^ 2 + 7 * 2 ^ 2 = 2 ^ 2 ( 1 + 7 )

Let X be the months after which both orgs have same strenthg..

Let Y be X / 10

then 2 ^ ( 12 + 2 * Y ) = ( 2 ^ 2 ) * ( (1 + 7) ^ Y )

this gives Y = 10, that means X = 100


Can you explain as to why have you considered y be x/10?
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 17 May 2016, 03:45
Bunuel wrote:
virajgawade wrote:
can someone pls explain how does 100 months come???


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

Every 5 months the membership of the established organization increases by 100 percent, means that every 5 months the membership of the established organization doubles.

Every 10 months the membership of the new organization increases by 700 percent, means that every 10 months the membership of the new organization increases 8 times.

Say these organizations will have the same number of members in x months.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).

Equate: \(2^{12}*2^{\frac{x}{5}}=2^2*2^{\frac{3x}{10}}\) --> reduce by 2^2: \(2^{10 +\frac{x}{5}}=2^{\frac{3x}{10}}\) --> \(10 +\frac{x}{5}=\frac{3x}{10}\) --> \(x=100\).

Answer: E.

Hope it's clear.



Bunuel I have a query re the question above and a few similar questions where you seem to have used the formula for GP without subtracting 1 from the power of r, ar^n (instead ar^n-1).

Can you please explain the reasoning behind this.?

I solved this as follows -

ar^n-1
Established company => 4096 * 2^2n-1 = 2^12 * 2^2n-1
New company => 4*2^n-1 = 2^2 * 8^n-1

Now as these are both equal,
2^12*2^2n-1 = 2^2 * 2^3(n-1)

Hence we get n=12
Therefore the number of months = 12*10 = 120.

Where am i going wrong? Can you please explain?

Thank you so much!
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 30 May 2016, 00:07
eschn3am wrote:
JCLEONES wrote:
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


For the established group the membership is doubling every 5 months. If you set x = to # of 5 month periods you could say that the membership is equal to original membership*2^x. However, since the other group is expanding every 10 months, and the answers were all in increments of 10, I set x equal to 10 months. Of course 10 months = two 5 month periods so the equation is 2^2x instead of 2^x. The 2^2 = 4 comes from the population quadrupling every 10 months. You could also have the equation set up as 4^x where x is the number of 10 month periods. You get the same results, I just wanted to keep everything in powers of 2.

The new organization is 8x bigger after 10 months (growing by 700 percent = x+7x = 8x). 2^3 = 8 so 2^3(x) = membership after x number of 10 month periods.

Try plugging in numbers and testing them on a calculator to see for yourself. Start with a membership size of 1 person and play around with different time periods to see how it works.


I fail to understand how your are considering x = Time period in months and then raising it as a power(Can you do that?). I understood the following with your explanation.

>> Upon prime factorisation we get 2^12. Then as it doubles every 5 months, we can say population after 5 months is 2^12 x 2. Hence after 10 months is 2^12 x 2 x 2 = 2^12 x 2^2. (We are considering 2^12 as the months are mentioned in 10 month period for the 2nd organisation)

>>2nd Organization - Population after 10 months increases by 700%, therefore total increase is 100% + 700% = 800% Therefore 2^2 x 2^3.

I have no idea how to take it ahead from here. Could you please guide me?

Thanks a ton
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A certain established organization has exactly 4096 members.  [#permalink]

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New post 19 Aug 2017, 19:47
Bunuel wrote:
virajgawade wrote:
can someone pls explain how does 100 months come???


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

Every 5 months the membership of the established organization increases by 100 percent, means that every 5 months the membership of the established organization doubles.

Every 10 months the membership of the new organization increases by 700 percent, means that every 10 months the membership of the new organization increases 8 times.

Say these organizations will have the same number of members in x months.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).

Equate: \(2^{12}*2^{\frac{x}{5}}=2^2*2^{\frac{3x}{10}}\) --> reduce by 2^2: \(2^{10 +\frac{x}{5}}=2^{\frac{3x}{10}}\) --> \(10 +\frac{x}{5}=\frac{3x}{10}\) --> \(x=100\).

Answer: E.

Hope it's clear.


Hi Bunuel,
In above solution, i couldn't understand below red mark. Could you please let me know how and why did you take x/5 & x/10.

The number of members in the established organization in x months will double x/5 times (since it doubles every 5 months), and become \(4096*2^{\frac{x}{5}}=2^{12}*2^{\frac{x}{5}}\);

The number of members in the new organization in x months will increase 8 times x/10 times (since it increases 8 times every 10 months), and become \(4*8^{\frac{x}{10}}=2^2*2^{\frac{3x}{10}}\).
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Re: A certain established organization has exactly 4096 members.  [#permalink]

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New post 19 Oct 2018, 23:44
JCLEONES wrote:
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



Here is my approach (I find it simpler :-D :-D )

Lets make the months in equal unit say..Lets take 5 month period..Now every 5 month

when MEMBERSHIP becomes 2 times its orginal..So after 10 months again it will be become 2 times its orginal..this implies aftre every 10 months the value becomes 2^2 or 4 times previou..Now let n denote the number of UNITs/INTERVAL of 10//..So (4^n).4096 = 8^n. 4 This implies 2^12 * 2^(2n) = (2^2) *( 2^(3n)) implies n=10 .Now since n is the units of 10 monhs so total number of months = 100
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Re: A certain established organization has exactly 4096 members. &nbs [#permalink] 19 Oct 2018, 23:44
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