Of 120 hotel rooms rented one night, some were suits rented for $115 each and the rest were double rooms rented for $85 each. If the total revenue from the room rentals ffor this night was $10890, how many suits were rented?
A) 23
B) 35
C) 54
D) 94
E) 99
OA: A
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We can solve this equation using 1 variable or 2 variables.Mo2men wrote:Of 120 hotel rooms rented one night, some were suites rented for $115 each and the rest were double rooms rented for $85 each. If the total revenue from the room rentals for this night was $10,890, how many suites were rented?
A) 23
B) 35
C) 54
D) 94
E) 99
Let's use 1 variable.
Let x = # of suites rented
Since 120 rooms were rented in total, 120-x = # of double rooms rented
Suites rented for $115 each
So, revenue from the x suites = 115x
Double rooms rented for $85 each
So, revenue from the 120-x double rooms = 85(120-x)
TOTAL revenue was $10,890
So, 115x + 85(120-x) = 10,890
Expand to get: 115x + 10,200 - 85x = 10,890
Simplify left side: 30x + 10,200 = 10,890
Subtract 10,200 from both sides to get: 30x = 690
Divide both sides by 30 to get: x = 23
Answer: A
Cheers,
Brent
The rooms of the hotel are suites or double rooms. The total of rooms is 120.
So we can set:
S+D=120. (equation 1)
where
S - suites
D - double rooms.
The revenue for renting the suites is S*115$ and the revenue for renting the double rooms is D*85$.
The revenue for this night is 10890$. So
S*115$+D*85$=10890$. (equation 2)
From (equation 1) we can get: D=120$-S and introducing D in (equation 2) we will get
S*115$+(120$-S)*85$ = 10890$
S*115$+10200$-S*85$ = 10890$
S*(115$-85$) = 10890$-10200$
S*30$ = 690$
S= (690$/30$)
S = 23
So, this night were rented 23 suites.
PD: we can solve (equation 1) for S, but at the end we will get the number of Double Rooms rented, then we should introduce the value of D in (equation 1) to get the number of suites rented.
So we can set:
S+D=120. (equation 1)
where
S - suites
D - double rooms.
The revenue for renting the suites is S*115$ and the revenue for renting the double rooms is D*85$.
The revenue for this night is 10890$. So
S*115$+D*85$=10890$. (equation 2)
From (equation 1) we can get: D=120$-S and introducing D in (equation 2) we will get
S*115$+(120$-S)*85$ = 10890$
S*115$+10200$-S*85$ = 10890$
S*(115$-85$) = 10890$-10200$
S*30$ = 690$
S= (690$/30$)
S = 23
So, this night were rented 23 suites.
PD: we can solve (equation 1) for S, but at the end we will get the number of Double Rooms rented, then we should introduce the value of D in (equation 1) to get the number of suites rented.
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We let s = the number of suites rented and d = the number of double rooms rented. We can create the equations:Mo2men wrote: ↑Fri Sep 01, 2017 11:36 amOf 120 hotel rooms rented one night, some were suits rented for $115 each and the rest were double rooms rented for $85 each. If the total revenue from the room rentals ffor this night was $10890, how many suits were rented?
A) 23
B) 35
C) 54
D) 94
E) 99
OA: A
s + d = 120
and
115s + 85d = 10,890
Multiplying the first equation by 115, we have:
115s + 115d = 13,800
Subtracting the second equation from the new equation, we have:
30d = 2910
d = 97
Since s + d = 120 and we know that d = 97, then we have:
s + 97 = 120
s = 23
Answer: A
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