Every member of a certain club volunteers to contribute equa

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.

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