Data Sufficiency

The average of x number of exams is y. When an additional exam of score z is added in, does the average score of the exams increase by 50% ?

A) 3x = y
B) 2z-3y = xy

Let's rephrase the question. Our initial average was y, and our initial number of terms was x, so our initial sum was yx. If we add 'z', the new sum is yx + z, and the new number of terms is x + 1, so the new average would be (xy + z)/(x +1). We want to know if this is 50% more than the initial average of y, or 1.5y. In other words is (xy + z)/(x +1) = 1.5y? We can simplify

xy + z = 1.5y(x+1) --> xy + z = 1.5yx + 1.5y ---> z = .5yx + 1.5y. Rephrased question: Is z = .5yx + 1.5y?

S1: tells us nothing about z. Not sufficient.

S2: 2z - 3y = xy --> 2z = xy + 3y --> z = .5xy + 1.5y. Statement 2 alone is sufficient. Answer is B

aditiniyer wrote:The average of x number of exams is y. When an additional exam of score z is added in, does the average score of the exams increase by 50% ?

A) 3x = y
B) 2z-3y = xy

Statement 1:
No information about z.
INSUFFICIENT.

Statement 2:
Case 1: x=1 and y=1, implying that 1 exam has an average score of 1
Sum of the scores = (number of exams)(average score) = 1*1 = 1.
Plugging x=1 and y=1 into 2z-3y = xy, we get:
2z - 3*1 = 1*1
2z = 4
z = 2.
When the additional exam with a score of 2 is included, the new average for the resulting 2 exams = (old sum + z)/2 = (1 + 2)/2 = 1.5.
Since the average increases from 1 to 1.5, it increases by exactly 50%.

Case 2: x=3 and y=2, implying that 3 exams have an average score of 2
Sum of the scores = (number of exams)(average score) = 3*2 = 6.
Plugging x=3 and y=2 into 2z-3y = xy, we get:
2z - 3*2 = 3*2
2z = 12
z = 6.
When the additional exam with a score of 6 is included, the new average for the resulting 4 exams = (old sum + z)/4 = (6 + 6)/4 = 3.
Since the average increases from 2 to 3, it increases by exactly 50%.

The two cases above illustrate that -- for any positive values x and y -- the average will increase by exactly 50%.
SUFFICIENT.

The correct answer is B.