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by mschling52 » Fri Sep 28, 2007 9:34 am
I think this one is C - both combined are sufficient.

We are dealing with 2 unknown variables (number of people (n), and the admission fee (f)). Note the total revenue generated will be T = n*f.

(1) gives us T = nf = (n+100)(f-0.75). Expanding the right side, we get

nf = nf -0.75n + 100f - 75
0.75n = 100f -75

This is an equation with 2 unknowns, which we can't solve for n.

(2) is similar. It gives us T = nf = (n-100)(f+1.50). Expanding gives us

nf = nf +1.50n -100f -150
-1.50n = -100f - 150

Again this is an equation with 2 unknowns.

However, if we combine (1) and (2), we get 2 linear equations for the 2 unknowns, so we should be able to solve for each variable. Adding the equation from (2) to the equation from (1) gives us

(0.75n = 100f - 75)
+(-1.50n = -100f -150)
= -0.75n = -225

This we can solve for n=300. Then, plugging n=300 back into one of the original equations gives us f=3.
Source: — Data Sufficiency |