mikemcgarry wrote:
In a certain apartment building, there are one-bedroom and two-bedroom apartments. The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments. Let R be the average rental price for all apartments in the building. If R is $5,600 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $10,400 higher that R, then what percentage of apartments in the building are two-bedroom apartments?
(A) 26%
(B) 35%
(C) 39%
(D) 42%
(E) 52%
Use ALLIGATION -- a great way to handle weighted average problems.
Let S = single-room apartments, T = two-room apartments.
Step 1: Draw a number line, with the two apartment types (S and T) on the ends and the mixture of apartments (R) in the middle:
S-----------------R-----------------T
Step 2: Calculate the distances between the averages.
Since the average for S is 5600 less than the average for R, and the average for T is 10,400 more than the average for R, we get the following distances between the averages:
S-----
5600------R----
10400------T
Step 3: Determine the ratio in the mixture.
The ratio of S to T is equal to the RECIPROCAL of the distances in red.
S:T = 10400:5600 = 13:7.
Since S:T = 13:7, there are 13 single-room apartments for every 7 two-room apartments.
Thus, of every 20 apartments, 7 are two-room:
\(\frac{7}{20} = \frac{35}{100}= 35\)%
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