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Bunuel
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


\(k*\frac{5r^2+10t}{f+5}\) replaceing the values for r=2, t=2 and f=1 and equating to 800 gives k =120

k[(5*4+10*2)/(1+5)] = 800
k = 120

Now
for r = 3, t= 1 and f=3
we get
120 [(5*9 + 10*10/(3+5)] = 825
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Bunuel
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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