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Monthly rent for units in a certain apartment building is de
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30 Jul 2018, 09:38
30 Jul 2018, 09:38
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Question Stats:
90% (02:07) correct
9% (02:43) wrong based on 32 sessions
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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
(A) $825
(B) $875
(C) $900
(D) $925
(E) $1,000
Re: Monthly rent for units in a certain apartment building is de
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12 Aug 2018, 05:58
12 Aug 2018, 05:58
Expert Reply
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
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
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01 May 2019, 16:19
01 May 2019, 16:19
1
Expert Reply
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
(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
