A certain club has exactly 5 new members at the end of its

Hello Bunuel. This is geometric progression with a formula b12=b1*q^11, which is y=5*x^11 for our example. However I couldn't figure out the answer, thank you for the explanation.

Shouldn't the question read that 'y' is the number of members brought into the club at the beginning of the twelfth week instead of during?
Cause during the week is not the 'stock' it is the new members brought in.

But how does 5x^11 = 3^11 x 5^12, 3^11 and 5^12 have unlike bases.

\(5x^{11}=3^{11} * 5^{12}\) --> reduce by 5: \(x^{11}=3^{11} * 5^{11}=(3*5)^{11}\) --> \(x=3*5=15\).

Hope it's clear now!

in 2nd week=(5x).x
in 12th week:(5 x^11)=y
now value of 5 will be 1 power more than 3.
D answer.

How you consider that the number of new memebers added each week is 15

please explain
where 15 comes from?

lets say each week new member brings 1 new member.
can u calculate it for me please?

\(5x^{11}=3^{11} * 5^{12}\);

\(x^{11}=3^{11} * 5^{11}\);

\(x^{11}=15^{11}\);

x=15.

I have 3 questions
1) why we have to consider 15 new member?
2) can we consider that each member brigs 1 new member?
3) this kind of quastion are common on gmat?

Hello All,

Need some quick help.When I read the question,my approach was

1st week end : 5 members
2nd week end : 5+x members
3rd week end : 5 + x + x members (x members have bought x new members).
.
.
12 th week end : 5 + 11 x = y.

Can anyone please explain where I am missing ? In the heat of the exam I might end up with the same approach.

Let me ask you a question: if a club has 5 members and EACH member will bring 2 new members, how many NEW members you'll have 5 + 2 = 7 or 5*2 = 10?

1. Number of new members in the second week is 5*(x) and in the third week 5*(x^2) and so on and in the 12th week 5*(x^11)
2. So we see in the choices that 5 has to be raised at least to the power of 11 and so is 3 which means x should be a multiple of 5 and 3 i.e, a multiple of 15.
3. We can eliminate choice A and rewrite choice E as 60* (60 ^11) or 5* (12 *(60^11) ). We can eliminate this also as this does not fit the format of x^11
4. We get only D with x =15.

It is a very tricky question. Though it seems easy in the first instance but when we start solving it, the question gets tougher because of the unknown x. The questions asks which of the options could be y. That means we will have to somehow eliminate the options apart from correct one.

So. Lets start solving
The number of members added at the end of 1st week = 5
Total members = m+5 ( Let me be the members during the start of 1st week)
New members = 5

There on after at the end of each week, no. of members added = x times the new members last week.
2nd week : total members = m + 5 + 5x ; New members = 5x

3rd week : total members = m+ 5+5x + 5x^2 ; New members = 5X^2

..
..

12th week : Total members = m + 5+ 5x+ 5x^2 +..... +5x^11 ; New members = 5x^11.


Now as explained above we will have to check option for integral value of x (As x is an integer.)

Option 1 : 5^1/12 = 5 X^11 => x^11 = 5^(1/12 -1)= 5^(-11/12) (x is not an integer)
Option 2 : 3^11 * 5^11 = 5 X^11 => 3^11 *5^10 = x^11 (x is not an integer)
Option 3 : 3^12 * 5^12 = 5 X^11 => 3^12 * 5^11 = x^11 (x is not an integer)
Option 4 : 3^11 * 5^12 = 5 x^11 => 3^11 * 5^11 = x^11 => x = 15 (x is an integer)
Option 5 : 60^12 = 5 X^11 => 12 * 60^12 = x^11 (x is not an integer)


So Option D gives us an integral value of X

Answer D

Bunuel