Q. The average score 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%?
(1) 3x = y (2) 2z - 3y = xy
Please opine on how to resolve the above problem.
Average score
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Sum/count = average
xy / x = y
The question asks: (xy + z)/(x + 1) = 1.5y?
xy + z = 1.5xy + 1.5y
z - 1.5y = 0.5xy?
2z - 3y = xy?
Statement 1) 3x = y
2z - 3(3x) = x(3x)?
x and z are unknown, so this statement is insufficient.
Statement 2) 2z - 3y = xy
This is the rephrased question stem exactly. Sufficient.
B
xy / x = y
The question asks: (xy + z)/(x + 1) = 1.5y?
xy + z = 1.5xy + 1.5y
z - 1.5y = 0.5xy?
2z - 3y = xy?
Statement 1) 3x = y
2z - 3(3x) = x(3x)?
x and z are unknown, so this statement is insufficient.
Statement 2) 2z - 3y = xy
This is the rephrased question stem exactly. Sufficient.
B
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Agree with B.MI3 wrote:Q. The average score 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%?
(1) 3x = y (2) 2z - 3y = xy
Please opine on how to resolve the above problem.
The question in the sense of variables is: Whether z = y + xy/2 ?
or is 2z - 2y = xy ?
Statement 2 straightaway rules out this equality rendering 'No' as an answer to the main question.
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Here's an approach that requires very little algebra.MI3 wrote:Q. The average score 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%?
(1) 3x = y (2) 2z - 3y = xy
Please opine on how to resolve the above problem.
Statement 1: 3x = y.
No information about z.
Insufficient.
Statement 2: 2z - 3y = xy.
2z = 3y + xy.
Let x=2, y=4.
Sum of the 2 scores = number*average = 2*4 = 8.
2z = 3*4 + 2*4 = 20, so z = 10.
With z included, average of the 3 scores = (8+10)/3 = 6.
Percent increase from 4 to 6 = 50%.
Let x=7, y=10.
Sum of the 7 scores = number*average = 7*10 = 70.
2z = 3*10 + 7*10 = 100, so z = 50.
With z included, average of the 8 scores = (70+50)/8 = 15.
Percent increase from 10 to 15 = 50%.
Since in each case the average increases by 50% -- and the 2 cases employ very different numbers -- sufficient.
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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