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In a museum X, a total entrance fee is $x plus $z each perso

by Max@Math Revolution » Wed Feb 10, 2016 9:00 pm
In a museum X, a total entrance fee is $x plus $z each person greater than 20. In a museum Y, a total entrance fee is $y plus $w each person greater than 30. When 40 persons enter in museums X and Y, is a total entrance fee of the museum X smaller than that of the museum Y?

1) x<y
2) z<w


* A solution will be posted in two days.
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Source: — Data Sufficiency |
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by VivianKerr » Thu Feb 11, 2016 4:18 pm
What is the origin of this question? It does not feel very "GMAT-like" to me, because there is ambiguity in the given information.

"$x plus $z each person greater than 20" could be interpreted to mean "20(x + z)" or "x + 20z."

Here's how to set up the problem assuming the latter:

Image

Each statement is clearly insufficient, and it's easy to pick numbers to show that combined we do not have enough information:

If x = 1, y = 2, z = 1, and w = 2, the answer is YES.
If x = 1, y = 2, z = 1, and w = 1.50, the answer is NO.

The answer is [spoiler](E)[/spoiler].
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by Max@Math Revolution » Sat Feb 13, 2016 9:34 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

In a museum X, a total entrance fee is $x plus $z each person greater than 20. In a museum Y, a total entrance fee is $y plus $w each person greater than 30. When 40 persons enter in museums X and Y, is a total entrance fee of the museum X smaller than that of the museum Y?

1) x<y
2) z<w


When you modify the original condition and the question, x+20z<y+10w? becomes a question. There are 4 variables(w,x,y,z), which should match with the number of equations. So you need 4 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer.
When 1) & 2), x=z=1, y=w=2 -> yes and x=z=2, y=w=3 -> no, which is not sufficient.
Therefore, the answer is E.


� For cases where we need 3 more equations, such as original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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