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

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In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 23 Feb 2015, 16:52
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A
B
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D
E

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  65% (hard)

Question Stats:

67% (02:47) correct 33% (02:59) wrong based on 265 sessions

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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/

Mike :-)

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In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 29 Jun 2017, 04:50
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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.

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In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 24 Feb 2015, 00:05
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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Re: In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 24 Feb 2015, 01:43
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Correct answer is B.

Ratio of 2 Bedroom Apartment: 1 Bedroom Apartment = 5400 : 104000 -----> 7 : 13

Let total number of Apartments be X

No. of 2 Bedroom Apartment = (7 / 20) * X

percentage of apartments in the building are two-bedroom apartments ---->

(7/20) * 100 ---> 35%

option B is correct
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Re: In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 25 Feb 2015, 04:50
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Differential approach

1room<-------R--------------->2rooms

5600x=10400y

x/y=10400/5600=1.8/1

(y/x+y)*100=1/2.8=0.35*100=35%

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Re: In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 28 Sep 2016, 15:31
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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

Answer:

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Re: In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 21 Jun 2017, 05:51
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.

5600/16000 *100= 35%

Regards
AMber Bajaj
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Re: In a certain apartment building, there are one-bedroom and two-bedroom  [#permalink]

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New post 01 Oct 2018, 04:13
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 [#permalink] 01 Oct 2018, 04:13
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