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

Question

A certain club has exactly 5 new members at the end of its first week. Every subsequent week, each of the previous week's new members (and only these members) brings exactly x new members into the club. If y is the number of new members brought into the club during the twelfth week, which of the following could be y?

(A) 5^1/12
(B) 3^11 * 5^11
(C) 3^12 * 5^12
(D) 3^11 * 5^12
(E) 60^12

I found it difficult and tricky. 
Started with previous week= m, 
first week= m+5, 
second week = m+5+5x, then got stuck.

Bunuel · Accepted Answer

At the end of the first week, there are 5 new members;
At the end of the second week, there are 5x new members (since each 5 new members from the previous week brings x new members);
At the end of the third week, there are 5x^2 new members (since each 5x new members from the previous week brings x new members);
...
At the end of the twelfth week, there are 5x^11 new members (since each 5x^10 new members from the previous week brings x new members).

We are given that 5x^11=y. Out of the answers only D yields integer value for x: 5x^11=3^11 * 5^12 --> x=3*5=15.

Answer: D.

Marcab · Answer

At the end of 2nd week, the number of new members are 5x.
At the end of third week, the number of new members are (5x)*x.
Therefore at the end of 12 th week, the number of new members are 5x^11.

Now here lies the trick. Consider the number of new members added each week be 15. Therefore x=15.
Put this in above relation. It will come out to be 5*(15)^11 or 5*(3*5)^11 or 5^12 * 3^11.

D is the answer.

fukirua · Answer

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.

jlgdr · Answer

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.

sagnik242 · Answer

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

Bunuel · Answer

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!

ssriva2 · Answer

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.

Vins7 · Answer

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

azamaka · Answer

please explain
where 15 comes from?

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

Bunuel · Answer

5x^{11}=3^{11} * 5^{12} ;

x^{11}=3^{11} * 5^{11} ;

x^{11}=15^{11} ;

x=15.

azamaka · Answer

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?

GouthamNandu · Answer

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.

Bunuel · Answer

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?

SVaidyaraman · Answer

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.

shashankism · Answer

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

deep1624 · Answer

Bunuel

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