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In a certain apartment building, there are one-bedroom and two-bedroom
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23 Feb 2015, 15:52

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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 OE for this problem, see https://magoosh.com/gmat/2015/gmat-math ... -averages/

In a certain apartment building, there are one-bedroom and two-bedroom
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29 Jun 2017, 03:50

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Hi Amber Bajaj,

Initially it looked strange for me as well to see 5600/10400, but while working out from scratch you will get it. Average Rent = R Rent of 1BHK = R1 Rent of 2BHK = R2 Avg. Rent R = [X (R1) + Y (R2)]/X+Y From the question stem we know, Avg. Rent R = 5600 + R1 Avg. Rent R + 10400 = R2 Avg. Rent R will look like R = [X (R-5600) + Y (R+10400)]/ X+Y RX + RY = RX - 5600X + RY + 10400Y 5600X = 10400Y Since we need ratio of 2BHK Y/X = 5600/10400 = 7/13 Hence Percentage of 2BHK becomes (7/20)*100 = 35% - where 20 is total parts Hope it answers your doubt. _________________

In a certain apartment building, there are one-bedroom and two-bedroom
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23 Feb 2015, 23:05

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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%

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 OE for this problem, see https://magoosh.com/gmat/2015/gmat-math ... -averages/

Mike

hi, straight forward question of weighted mixture... here we have the price of two BR:one BR away from average by 10400:5600.... therefore the number of two is 5600/16000 *100= 35% ans B
_________________

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28 Sep 2016, 14:31

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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%

Another approach:

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

Re: In a certain apartment building, there are one-bedroom and two-bedroom
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21 Jun 2017, 04:51

chetan2u wrote:

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%

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 OE for this problem, see https://magoosh.com/gmat/2015/gmat-math ... -averages/

Mike

hi, straight forward question of weighted mixture... here we have the price of two BR:one BR away from average by 10400:5600.... therefore the number of two is 5600/16000 *100= 35% ans B

Hi, Here divident is 5600. Please help me to understand why we did not take 10400 as divident.

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01 Oct 2018, 03:13

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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Re: In a certain apartment building, there are one-bedroom and two-bedroom &nbs
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