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In a certain apartment building, there are one-bedroom and t

by Mike@Magoosh » Mon Feb 23, 2015 3:50 pm

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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%

One tricky category of problems on the GMAT Quant section are Weighted Average problems. For a full discussion of very strategies to use on these problems, as well as the OA & OE for this problem, see
https://magoosh.com/gmat/2015/gmat-math ... -averages/

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by ravihanda » Mon Feb 23, 2015 6:54 pm
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%
Let us assume that the percentage of two-bedroom apartments is 'x'.
For calculation purpose, let us assume there are 100 apartments
Rent received from 1 bed-room apartments = (100 - x)(R - 5600) = 100R - Rx - 560000 + 5600x
Rent received from 2 bed-room apartments = x(R + 10400) = Rx + 10400x
Sum of the above two values = Total Rent Received = 100R - 560000 + 16000x
Total Rent = Average Rent*Number of Apartments = 100R
=> 100R = 100R - 560000 + 16000x
=> 16000x = 560000
=> [spoiler]x = 35. Option B[/spoiler]
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by Brent@GMATPrepNow » Tue Feb 24, 2015 10:42 am
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%
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

R = Weighted average of BOTH types
R - 5600 = average rent for 1-bedroom apartments
R + 10400 = average rent for 2-bedroom apartments

Let P = percentage of apartments that are two-bedroom apartments. This means P/100 represents to PROPORTION of apartments that are 2-bedroom.
So, 100 - P = percentage of apartments that are one-bedroom apartments. This means (100-P)/100 represents to PROPORTION of apartments that are 1-bedroom.

We can now plug all of this information into the formula to get:

R = (P/100)(R + 10400) + [(100-P)/100](R - 5600)
Multiply both sides by 100 to get: 100R = (P)(R + 10400) + (100-P)(R - 5600)
Expand to get: 100R = PR + 10400P + 100R - 560000 - PR + 5600P
Simplify to get: 0 = 16000P - 560000
Rearrange to get: 16000P = 560000
Divide both sides by 1000 to get: 16P = 560
Divide both sides by 16 to get: P = 35

Answer: B

-----------------------------------------

For more information on weighted averages, you can watch this free GMAT Prep Now video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Here are some additional practice questions related to weighted averages:
- https://www.beatthegmat.com/weighted-ave ... 17237.html
- https://www.beatthegmat.com/weighted-ave ... 14506.html
- https://www.beatthegmat.com/average-weig ... 57853.html
- https://www.beatthegmat.com/averages-que ... 87118.html

Cheers,
Brent
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by GMATGuruNY » Tue Feb 24, 2015 11:18 am
Mike@Magoosh 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%
Let R = 10,000.
Let O = the number of one-bedroom apartments and T = the number of two-bedroom apartments.

R is $5,600 higher than the average rental price for all one-bedroom apartments.
Average price for O = R - 5600 = 10,000 - 5600 = 4400.
The average rental price for all two-bedroom apartments is $10,400 higher that R.
Average price for T = R + 10,400 = 10,000 + 10,400 = 20,400.

This is a MIXTURE problem:
Average price for O = 4400.
Average price for T = 20,400.
Average price for R = average price for the MIXTURE of O and T = 10,000.

To determine the ratio of O to T in the mixture, we can use ALLIGATION.

Step 1: Plot the 3 averages on a number line, with the averages for O and T on the ends and the average for R in the middle.
O 4400-------------------10000------------------20,400 T

Step 2: Calculate the distances between the averages.
O 4400-------5600-------10000------10,400-------20,400 T

Step 3: Determine the ratio in the mixture.
The required ratio of O to T is equal to the RECIPROCAL of the distances in red.
O : T = 10400 : 5600 = 13:7.

Implication:
Of every 20 apartments, 13 are O and 7 are T.
Thus, T is equal to 7/20 of the total number of apartments:
7/20 = 35/100 = 35%.

The correct answer is B.
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by GMATGuruNY » Tue Feb 24, 2015 2:58 pm
A follow-up to my post above:

The reason we can plug in a random value for R is that the value of R is irrelevant.
All that matters here is the DISTANCES between the averages.
An even faster way to determine the ratio of O to T:

Step 1: Plot the 3 averages on a number line, with O and T on the ends and R in the middle.
O-------------------R------------------T

Step 2: Enter the distances between the averages.
Since O is 5600 less than R, and T is 10,400 greater than R, we get:
O-------5600-------R------10,400-------T

Step 3: Determine the ratio in the mixture.
The required ratio of O to T is equal to the RECIPROCAL of the distances in red.
O : T = 10400 : 5600 = 13:7.

Implication:
Of every 20 apartments, 13 are O and 7 are T.
Thus, T is equal to 7/20 of the total number of apartments:
7/20 = 35/100 = 35%.

The correct answer is B.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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by Mike@Magoosh » Tue Feb 24, 2015 3:37 pm
With all due respect to my esteemed colleague Brent at GMAT Prep Now, I would say that the solution of Mitch, the GMAT Guru of NY, is the most elegant, the solution I recommended in the OE. I would urge ravihanda to study this solution involving proportional placement of the weighted average. Mitch demonstrates an application here, and I give a detailed explanation of it in the blog link in the top post of this thread.
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by Brent@GMATPrepNow » Tue Feb 24, 2015 4:06 pm
Mike, surely you're not discouraging multiple approaches.
The great thing about GMAT math questions is that they typically can be solved using a variety of approaches.
I'd say that any approach that enables one to solve a question in a timely manner is a valid approach.
I don't doubt that many students prefer using alligation, but I'd be more inclined to let students determine which approach works best for them, rather than impose any personal ranking on their behalf.

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