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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