ardz24 wrote:Martha obtained an average score of y in a total of x mandatory papers. She also obtained a score of z in an additional optional paper. Does Martha's average score on all the x + 1 papers exceed her average score on the x mandatory papers by more than 50%?
(1) 3x = y
(2) 2z - 3y = xy
To simplify the algebra, plug in a value for y and solve for x and z.
Let y=2, implying that 2 is the average score for x papers and that the sum of the scores for x papers = (number of papers)(average score) = (x)(2) = 2x.
For the y=2 average to increase by more than 50%, the new average must be greater than 3.
When an additional paper earns a score of z, the new sum = 2x+z, and the new average for the x+1 papers = (new sum)/(new number of papers) =
(2x+z)/(x+1).
Since the answer to the question stem will be YES only if the expression in blue is greater than 3, we get:
(2x + z)/(x+1) > 3
2x + z > 3x + 3
z > x + 3.
Question stem, rephrased:
If y=2, is z > x+3?
Statement 1:
No information about z.
INSUFFICIENT.
Statement 2:
Substituting y=2 into 2z - 3y = xy, we get:
2z - (3*2) = 2x
2z - 6 = 2x
z - 3 = x
z = x+3.
Thus, the answer to the question stem is NO.
SUFFICIENT.
The correct answer is
B.
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