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Monthly rent for units in a certain apartment building is determined

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Monthly rent for units in a certain apartment building is determined  [#permalink]

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New post 31 Jul 2018, 00:47
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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

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

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New post 31 Jul 2018, 01:24
Bunuel 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


As we're given all the data we need and just need to plug things in, we'll jump straight into the calculation.
This is a Precise approach.

Since r = 2, t = 2, f = 1 give a rent of 800, then plugging in the numbers gives
800 = k * (5*4 + 20)/(1+5) --> 800 = 40k/6 --> 20 = k/6 --> k = 120

So, if r = 3, t = 1, and f = 3 then the rent is
120(5*9+10)/(3+5) = 120*55/8 = 15*55 = 550 + 275 = 825

(A) is our answer.
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Re: Monthly rent for units in a certain apartment building is determined  [#permalink]

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New post 31 Jul 2018, 03:53
Bunuel 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?


First case :
r = 2, t = 2, f = 1, Rent = 800
On solving we get:
800 = (k * (5 * 4 + 20 )) / ( 1 + 5 )
800 = 40k / 6
40k = 800 * 6
k = 4800 / 40
k = 120


Second case :
r = 3, t = 1, f = 3, k = 120
On solving we get:
Rent = (120 * ( 5 * 9 + 10 )) / ( 3 + 5 )
Rent = (120 * 55) / 8
Rent = 6600 / 8 = 825

Hence, A.
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Re: Monthly rent for units in a certain apartment building is determined  [#permalink]

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New post 31 Jul 2018, 03:57
1
Bunuel 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



\(k*\frac{5r^2+10t}{f+5}\) 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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Re: Monthly rent for units in a certain apartment building is determined &nbs [#permalink] 31 Jul 2018, 03:57
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