Bunuel wrote:
An organization offers only three types of annual memberships. Basic memberships cost $50 each, standard memberships cost $70 each, and premium memberships cost $100 each. Last year, a total of 200 memberships were purchased at an average arithmetic mean) cost of $80 per membership. What was the number of premium memberships purchased last year?
(1) Last year, the number of basic memberships purchased was equal to the number of standard memberships purchased.
(2) Last year, the number of premium memberships purchased was twice the number of basic memberships purchased.
This question can be solved using observation and using logic / reasoning (Though I used a bit of algebra, we can avoid that as well).
Let the number of basic membership, standard membership and premium membership sold last year be x, y and z respectively.
Average = \(\frac{50x + 70y + 100z}{ x + y+ z }\)
\(\frac{50x + 70y + 100z}{ x + y+ z }\) = 80
Upon solving we get -
2z - 3x - y = 0 ------------------------ (1)
Given: x + y + z = 200 --------------- (2)
Statement 1Last year, the number of basic memberships purchased was equal to the number of standard memberships purchased.As the number of basic membership and the number of standard membership purchased was equal, the average selling price of both the membership is the average of $50 and $70, i.e. $60
For the sake of simplicity, let's take the basic membership + standard membership together as one unit ⇒ let's term this combined unit as "non premium" membership in the rest of our analysis.
The questions stem states that combined average of all the three membership was $80. So we can infer that the combined average of premium membership and "non premium" membership was $80.
The cost of "non premium" membership is $60
The cost of "premium" membership given is $100
As $80, is the mid point of $60 and $100, we can infer that the number of "non premium" membership sold = number of "premium" membership sold.
The question stem gives us the information of the total (premium + non premium) membership sold = 200.
Hence the number of premium membership sold is half of 200, i.e. 100.
The statement is sufficient and we can eliminate B, C and E.
Statement 2Last year, the number of premium memberships purchased was twice the number of basic memberships purchased.z = 2x
We can substitute this in equation (1) ⇒ 2z - 3x - y = 0
2(2x) - 3x - y = 0
x - y = 0
x = y
Hence the number of basic memberships purchased was equal to the number of standard memberships purchased.
This is the same information that was presented to us in statement 1, and we have already determined that Statement 1 is sufficient.
Hence statement 2 is also sufficient.
Option D
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