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Monthly rent for units in a certain apartment building is de

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GRE Prep Club Legend
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Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Monthly rent for units in a certain apartment building is de [#permalink] New post 30 Jul 2018, 09:38
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Question Stats:

81% (01:37) correct 18% (01:45) wrong based on 11 sessions
Monthly rent for units in a certain apartment building is determined by the formula \(k\frac{5r^2+10t}{f+5}\) where k is a constant, r and t are the number of bedrooms and bathrooms in the unit, respectively, and f is the floor number of the unit. A 2-bedroom, 2-bathroom unit on the first floor is going for $800/month. How much is the monthly rent on a 3-bedroom unit with 1 bathroom on the 3rd floor?

(A) $825
(B) $875
(C) $900
(D) $925
(E) $1,000
[Reveal] Spoiler: OA

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Sandy
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GRE Prep Club Legend
GRE Prep Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 118

Kudos [?]: 1897 [0], given: 397

Re: Monthly rent for units in a certain apartment building is de [#permalink] New post 12 Aug 2018, 05:58
Expert's post
Explanation

First, solve for the constant k using the price information of the 2-bedroom, 2-bath unit (m = 800, r = t = 2, and f = 1):

\(800 = k(\frac{5^2+10\times 2}{1+5})\)

\(800 = k(\frac{20 + 20}{6})\)

\(800 = k(\frac{40}{6})\)

\(120 = k\)

Next, solve for the rent on the 3-bedroom, 1-bath unit on the 3rd floor (r = 3, t = 1, and f = 3):

\(m = 120(\frac{5(3)^2+10\times 1}{3+5})\)

\(m = 120(\frac{45+10}{3+5})\)

m = 825
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Re: Monthly rent for units in a certain apartment building is de [#permalink] New post 01 May 2019, 16:19
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sandy wrote:
Monthly rent for units in a certain apartment building is determined by the formula \(k\frac{5r^2+10t}{f+5}\) where k is a constant, r and t are the number of bedrooms and bathrooms in the unit, respectively, and f is the floor number of the unit. A 2-bedroom, 2-bathroom unit on the first floor is going for $800/month. How much is the monthly rent on a 3-bedroom unit with 1 bathroom on the 3rd floor?

(A) $825
(B) $875
(C) $900
(D) $925
(E) $1,000


First we need to determine the value of k, using the following equation based on r = 2, t = 2 and f = 1:

k * [5(2)^2 + 10(2)]/(1 + 5) = 800

k * 40/6 = 800

k = 800 * 6/40

k = 120

Now, using k = 120, r = 3, t = 1 and f = 3, the monthly rent for a 3-bedroom unit with 1 bathroom on the 3rd floor is:

120 * [5(3)^2 + 10(1)]/(3 + 5)

120 * 55/8

15 * 55

825

Answer: A
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Re: Monthly rent for units in a certain apartment building is de   [#permalink] 01 May 2019, 16:19
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