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Number Systems -Sequence and Series

Expert replies
by sukhman » Thu Oct 17, 2013 10:18 am
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 neworganization 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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Source: — Problem Solving |

by Mike@Magoosh » Thu Oct 17, 2013 3:30 pm
sukhman 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 neworganization 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
Dear sukhman,
I'm happy to help. :-)

It would seem that this question hinges on, among other things, the recognition that 4096 = 2^12. That's something I see right away, because I'm a math nerd, but the GMAT certainly doesn't expect typically test takers to see this, and if you don't see it, this problem becomes considerably more difficult.

Club #1. Start at 4096 = 2^12. Every five months, they double ---- (2^12)*2 = 2^13. In ten months, they double and double again, multiplied by 4 = 2^2. In x ten-month periods, they would be multiplied by 4 = 2^2 a total of x times, 4^x = 2^(2x). Thus, the total, after x ten-month periods, would be (2^12)*(2^(2x)) = 2^(12+2x)

Club #2. Start with 4= 2^2 member. Every ten month, they increase by 700%, which means they are multiplied by 8. In x ten-month periods, they would be multiplied by 8 = 2^3 a total of x times, 8^x = 2^(3x). Thus, the total, after x ten-month periods, would be (2^2)*(2^(3x)) = 2^(2+3x)

Equate these expressions. Because the bases are equal, we simply can equate the exponents.

12 + 2x = 2 + 3x
10 = x

So it would take 10 ten-week periods, or 100 weeks, answer = [spoiler](E)[/spoiler].

Incidentally, after 100 weeks, each club would have
2^32 = 4,294,967,296 members (!!)
As with the question I answered earlier, this is more than all the humans alive now and all the humans who ever lived. Once again, the story in the question is a joke, a shallow ill-conceived plot to convey a question about number properties. This question also falls very short of the standard held by the GMAT. It appears to be a question written someone who knows a great deal about math, but not much about the standards to which GMAT questions conform. It is dubious whether practicing such questions will prepare you adequately for the GMAT.

Does all this make sense?
Mike :-)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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by ganeshrkamath » Thu Oct 17, 2013 7:47 pm
sukhman 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 neworganization 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 => E
New organization => N

Every 5 months: E' = 2E
Every 10 months: E'' = 2E' = 4E

Every 10 months: N' = 8N

E = 4096, N = 4
4096*4^m = 4*8^m
(2^12) * (2^(2m)) = 2^2 * (2^(3m))
2^10 = 2^m
m = 10
10 * 10 months = 100 months

Choose E

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Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
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