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Retired Moderator Joined: 07 Jun 2014
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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]
Expert's post 00:00

Question Stats: 89% (01:30) correct 10% (01:45) wrong based on 19 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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Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 171

Kudos [?]: 2911 , given: 394

Re: Monthly rent for units in a certain apartment building is de [#permalink]
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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Sandy
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Re: Monthly rent for units in a certain apartment building is de [#permalink]
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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

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# Scott Woodbury-Stewart

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Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GRE quant course on GRE Prep Club. Read Our Reviews Re: Monthly rent for units in a certain apartment building is de   [#permalink] 01 May 2019, 16:19
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