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

Question

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%

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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/

sarugiri · Accepted Answer

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+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+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.

Temurkhon · Answer

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%

B

BrentGMATPrepNow · Answer

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)  +   (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

RELATED VIDEO
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chetan2u · Answer

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

DesiGmat · Answer

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

Amby02 · Answer

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

GMATGuruNY · Answer

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 %
 
 B

yazlal25 · Answer

We can apply the weighted average formula here which is as follows

W1 / W2 = (Avg 2 - Avg) / (Avg - Avg 1)

Let W1 = no of 2 BH apartments and W2 = no of 1 BH apartments

Therefore, W1/W2 = (R+10400 - R) / (R - (R - 5600))

W1/W2 = 10400 / 5600 = 13 / 7

% of 2 BH apts = W1 / (W1 + W2) * 100 = 13 / (13+7) * 100 = 35%.........Option B.

luisdicampo · Answer

Deconstructing the Question

Let  R  be the overall average rent. 
One-bedroom average is  R - 5600 . 
Two-bedroom average is  R + 10400 . 
Let  p  be the fraction of apartments that are two-bedroom. 
We need  p  as a percent.

Step-by-step

Weighted average:

R = (1-p)(R-5600) + p(R+10400)

Expand:

R = (1-p)R - 5600(1-p) + pR + 10400p

Combine the R-terms:

(1-p)R + pR = R

So:

R = R - 5600(1-p) + 10400p

Subtract R:

0 = -5600(1-p) + 10400p

Expand:

0 = -5600 + 5600p + 10400p

0 = -5600 + 16000p

Solve:

16000p = 5600

p = \frac{5600}{16000} = 0.35

Convert to percent:

35\%

Answer: 35%

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

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GMAT Problem Solving (PS) Questions

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

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