Every member of a certain club volunteers to contribute equa

Every member of a certain club volunteers to contribute equally to the purchase of a $60 gift certificate. How many members does the club have?

Let the # of members be \(n\).

(1) Each member's contribution is to be $4 --> \(n=\frac{60}{4}=15\). Sufficient.

(2) If 5 club members fail to contribute, the share of each contributing member will increase by $2 --> share of each contributing member is \(\frac{60}{n}\), if 5 club members fail to contribute, the share of each contributing member will be \(\frac{60}{n-5}\) and we are told that this values is $2 more than the previous one --> \(\frac{60}{n}=\frac{60}{n-5}-2\) --> \(n=15\). Sufficient.

Answer: D.
_________________

Equations such as these: \(\frac{60}{n}=\frac{60}{n-5}-2\)
are solved using hit and trial.

Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator.
Say, if n = 10, you get 60/10 = 60/5 - 2 which is not true.\(\frac{60}{n}\)and \(\frac{60}{n-5}\) need to be much closer to each other so that the difference between them is 2.
Next put n = 15. It satisfies.
Notice that this equation gives us a quadratic so be careful while working on DS questions. A little manipulation shows us that the product of the roots of this quadratic will be -ve. So one root is positive i.e. 15, hence the other must be -ve and we can ignore it.

Notice that when you see such an equation in PS questions, the options can give you ideas on values on n.
Here, you anyway got that n = 15 from the first statement so you know which value you must try.
_________________

how can you get the n=15.

you have to solve the equation.

and this is time consuming?

Hi thangvietnam,

Every member of a certain club volunteers to contribute equally to the purchase of a $60 gift certificate. How many members does the club have?

(1) Each member's contribution is to be $4.
(2) If 5 club members fail to contribute, the share of each contributing member will increase by $2.

The problem states that each member contributes equally to the gift certificate. Let the number of people be n and the amount contributed by each member be x. Now, from the question we know that total amount contributed=$60.

Hence, n*x=60

1) This statement tells us that contribution of each member=$4. i.e x=4
Hence, n=60/4=15
SUFFICIENT

2) This tells us that if 5 members fail to contribute every person's contribution share increases by $2.
We know n=60/x
old n=n
new n=n-5
Old contribution=60/n
new contribution=60/(n-5)

Now, new contribution=old contribution+2
i.e 60/(n-5)=2+ (60/n)

60/(n-5)=(2n+60)/n

Cross multiplying
60n=(2n+60)(n-5)
60n = 2n^2+60n-10n-300

Bolded terms get cancelled

What remains is 2n^2-10n-300.
Take out 2
n^2-5n-150=0

Factorize 150, you get 5*3*2*5
15-10=5
Hence, n^2-15n+10n-150=0
Take out common factors
n(n-15)+10(n-15)=0
n=-15 or 10. Since, the number of people cannot be negative it has to be 15.

SUFFICIENT
Let me know if I can clarify something else.

I took the quadradic path as i usually do, but dint solve the equation to see if stmt 2 was sufficient, here's how
Given : nx=60 (n: number of ppl, x: contribution of each)
Stmt 1 : gives x, can solve for n : sufficient
Stmt 2 : (n-5)(x+2)=60
=>(n-5)(60/n+2)=60
Simple rearrangement
=>2n^2-10n-300=0
I stopped here knowing this equation will have one positive and one negative root (or solution)
here's how : ax^2+bx+c=0
b=sum of roots (or solutions) (or -600
c=product of roots
=> 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is
-ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.

Could someone cpls verify if this method has any flaw?

It's fine. Since the product is -ve, one root will be positive and one will be negative.
Check out this video that explains Quadratics: https://youtu.be/QOSVZ7JLuH0
_________________

Thanks Karishma. I chanced upon that article and then got deep into the pool of other articles you've published. Really good, would have been great if i wasnt 1 month away from my gmat...better late than after GMAT :D

Easiest way to tackle the problem is as follows
This problem can be solved by factorization
factors of 60 or 60 can be evenly divided by following numbers = 2,3,4,5,6,10,12,15,20,30,60
Statement 1 is simple as we can see from our list, only 15 can be the ans.
Statement 2: if 5 club members fail = as per our list, only two combinations are possible (5,10) and (10,15) ...plug values,,,, the ans is (10,15) set...
Both the statements are sufficient.
Ans is D

Statement 2:
Let,
The total members of club be \(T\)
\(\frac{60}{T}+2=\frac{60}{T-5}\)
This is the quadratic equation (it'll give the 2 values of \(T\)).
But, we should not consider the negative value of \(T\), because the member of the club CAN'T be negative. The equation will give you a \(definite\) value. So, we should not make any calculation here.
-->Sufficient.
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________