A certain established organization has exactly 4096 members.

Question

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

(A) 20
(B) 40
(C) 50
(D) 80
(E) 100

Bunuel · Accepted Answer

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

(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.

GMATinsight · Answer

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   =

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

i.e.   =

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

i.e. y = 10

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

Answer: Option  E

walker · Answer

E

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

10+2x=3x (x= )
x=10 m=100.

100/5=20, 100/10=10. are integers.

eschn3am · Answer

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

eschn3am · Answer

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.

dvdv · Answer

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

virajgawade · Answer

can someone pls explain how does 100 months come???

Temurkhon · Answer

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

Radhika11 · Answer

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 ?

GMATinsight · Answer

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)  = (2^2)  ------

i.e.   =

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

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

i.e. m=10

Keysersoze10 · Answer

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* =2^2 
I get n = 11. Not sure what the issue is in this method.

TIA!

GMATinsight · Answer

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)  
@n=3, The Total Employees = 4096*2^2(2)  
@n=4, The Total Employees = 4096*2^2(3)  
---
---
---
@n=11, The Total Employees =  4096*2^2(n-1)    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!!!

KarishmaB · Answer

Old org - No of members = 4096 = 2^12 - Every 5 months, increase of 100% i.e. it becomes 2 times its previous value. 
New org - No of members = 4 = 2^2 - Every 10 months, increase of 700% i.e. it becomes 8 times its previous value.

There is a difference of 10 multiplications of 2 between them.

Note what happens every 10 months - the old org grows to 4 times but the new org grows to 8 times. So the new org gets 1 extra multiplication by 2 in 10 months. To get 10 extra multiplications by 2, the new org will need 10 * 10 = 100 months.

Answer (E)

Think of it in terms of relative speed concept.

Fdambro294 · Answer

In a weird way, you can think of this question as almost a Relative Speed Question in which the new organization is behind the established organization and is trying to “catch up and meet.”

4,096 = (2)^12

Every 5 months —— increase by 100% = 2 times

So every 10 months ——- increase by *(2)^2

“Speed” = 2 powers every 10 months

New organization:

4 = (2)^2

Every 10 months ——— increase by 700% is equivalent to a multiplier of 8 or (2)^3

“Speed” of new organization = 3 powers every 10 months

The starting “gap distance” that the new organization must overcome is a Power of 2 Difference = (12) - (2) = 10

Every 10 months: the “relative speed in the same direction” is equal to the (3) - (2) = 1 power of 2 at which the new organization catches up

Time in 10-month-periods = (Gap Distance of 10 powers) / (relative speed of 1) =

10 of the 10-month periods are needed until the two organizations “meet”, I.e., they have the same number of members

(10) x (10) = 100 months

E

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imaityaroy · Answer

Established organization increases by 100% every 5 month, this means the group becomes 2 times every 5 months. This means the group becomes 4 times every 10 months.
  
  New organization increases by 700% every 10 month, this means it the group becomes 8 times every 10 months

Time Periods   
   Established group # of members   
   New group # of members   
 
    1 ten month period ie 10 months   
   4096*4 = 4096*4  ^1   
   4*8 = 4*8  ^1   
 
    2 ten month period ie 20 months   
   4096*4*4 = 4096*4  ^2   
   4*8*8 = 4*8  ^2   
 
    3 ten month period ie 30 months   
   4096*4*4*4 = 4096*4  ^3   
   4*8*8*8 = 4*8  ^3

We see a pattern here 
 If it takes X ten months period for the number of members of established group be equal to the number of members of new group then we can make the equation

4096*4^X = 4*8 ^X  
 X = 10  
 therefore it will take 10 ten month periods ie 10*10 = 100 months for the # of members to be equal

A certain established organization has exactly 4096 members.

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Time Periods	Established group # of members	New group # of members
1 ten month period ie 10 months	40964 = 40964^1	48 = 48^1
2 ten month period ie 20 months	409644 = 4096*4^2	488 = 4*8^2
3 ten month period ie 30 months	4096444 = 40964^3	4888 = 48^3

GMAT Problem Solving (PS) Questions

A certain established organization has exactly 4096 members.

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