To rent an office, each member of a club must pay n dollars. If two mo

Question

To rent an office, each member of a club must pay n dollars. If two more members join the club, the per-member payment would be reduced by two dollars. Which of the following could be the number of members currently in the club?

I. 16
II. 17
III. 18

A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III

I. 16
II. 17
III. 18

A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III

Bunuel · Accepted Answer

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chetan2u · Answer

Hi,

info from Q 
There are no retrictions on either number of members or amount of payment..
Increase of 2 members results in contribution reducing by 2 dollars..

As member and contribution can be changed, all integers should satisfy the member number..
so E..

BUT lets see by ALGEBRIC way..

let the number of member =x and the contribution be n per member..
so total cost= xn..

in the secon scenario, x increases by 2 and n decreases by 2..
so cost= (x+2)(n-2)..
 
cost is he same both times, so:-
xn=(x+2)(n-2)..
xn=xn+2n-2x-4..
or 2x=2n-4..
 x=n-2...

hence x can take any values, as n changes..
E

generis · Answer

I went off sheer intuition here and tested numbers for Option I, then saw a pattern. The answer I got: all three are possible. Please correct me if I am mistaken.

I. 16

If there are 16 members, add 2 (=18), and multiply 16 * 18.

Do not find LCM; we need the product of 16 and 18 so that when we divide the product (total $) by (16 + 2), we will get $16.

16 members * $18 ea = $288
18 members would pay $288/18 = $16 each

16 members pay $18 each
18 members pay $16 each
2 more members, $2 less per person

II. 17 
If there are 17 members, add 2 and multiply 17 * $19 = $323
19 members will pay $323/19 = $17 each

17 members pay $19 each
19 members pay $17 each
2 more members, $2 less per person

III. 18
Add 2. Multiply 18 * $20 ea = $360
20 members would pay $360/20 = $18 each

18 members pay $20 each
20 members pay $18 each
2 more members, $2 less per person

All three are possible.

Answer E

* I started algebraically, but its implications befuddled me, so I switched to testing numbers.

A * n = S, where S must remain the same 
A = amount paid per person 
n = number of members
x = $ per person (i.e., is A)

Original: 
x * n = xn = SUM

New
(x - 2)(n + 2) = same SUM

xn = (x - 2)( n + 2)
xn = xn + 2x - 2n - 4
4 = 2x - 2n
4 = 2(x - n)
2 = x - n
n + 2 = x

Don't laugh too hard if I'm over the cliff. I  think  this means the same as what I did instinctively: add 2 to the original number of members to get a dollar amount for what each pays originally.

That dollar amount turns out to be the new number of total members.

ScottTargetTestPrep · Answer

If we let m = the number of members in the club, then the total money they will pay for the rental is mn. When two more members join, the number of members is now (m + 2), and now each person pays two dollars less (n - 2). Thus,we can create the equation:

mn = (m + 2)(n - 2)

mn = mn + 2n - 2m - 4

2m = 2n - 4

m = n - 2

We see that the number of members in the club is 2 fewer than the number of dollars contributed per person. Since there is no restriction on the value of n (as long as n is greater than 2), we see that m can be any of the three given values in the Roman numerals.

Answer: E

Fdambro294 · Answer

Let the $Total Cost split evenly among the members = C

I. 16 members

if there are 16 members originally, each member would pay:

N = C / 16

in which case ----> C = 16N

and if 2 more members show up, then the Cost per Member drops by -$2

N - $2 = 16N / 18

18N - 36 = 16N

2N = 36

N = $18

Cost per member could be $18 in the beginning, and there could be 16 members at the beginning

the cost per member would drop to $16 dollars when there are 18 members

I is possible

Testing II -- 17 members

N = C/17

C = 17N

N - $2 = 17N / 19

19N - 38 = 17N

2N = 38

N = 19

the original cost per member could be $19 per member when there is 17 members

the cost per member will drop to $17 per member when there is 19 members

III. 18

n = C/18

C = 18n

n -$2 = 18n / 20

20n - 40 = 18n

2n = 40

n = 20

would cost $20 per member when there was 18 members

then would cost $18 per member when there was 20 members

I, II, and III all work

-E-

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