Respuesta :
Number of single rooms rented is 9 and number of double rooms rented is 16
Solution:
Let "d" be the number of double rooms rented
Let "s" be the number of single rooms rented
Given that,
cost of double room per day = $ 33
cost of single room per day = $ 27
25 rooms were rented one day for a total of $771
We can frame a equation as:
number of single rooms rented + number of double rooms rented = 25
s + d = 25 --------- eqn 1
number of single rooms rented x cost of single room per day + number of double rooms rented x cost of double room per day = 771
[tex]s \times 27 + d \times 33 = 771[/tex]
27s + 33d = 771 --------- eqn 2
Let us solve eqn 1 and eqn 2 to find values of "s" and "d"
From eqn 1,
s = 25 - d ------ eqn 3
Substitute eqn 3 in eqn 2
27(25 - d) + 33d = 771
675 - 27d + 33d = 771
675 + 6d = 771
6d = 771 - 675
6d = 96
d = 16
Substitute d = 16 in eqn 3
s = 25 - 16 = 9
s = 9
Thus number of single rooms rented is 9 and number of double rooms rented is 16
Answer:
Number of Single room rented = 9
Number of Double room rented = 16
Step-by-step explanation:
Given:
Cost of motel rents for double rooms = $33 per day
Cost of motel rents for single rooms = $27 per day
Total number of rooms rented = 25
Cost of the 25 rooms = 771
To Find:
Number of double rooms and number of single rooms that were rented = ?
Solution:
Let
The number of Single rooms be x
The number of double rooms be y
total rooms rented = 25
x+ y = 25
x = 25 –y-------------------------------------------------------------(1)
Now the total cost
(Number of single rooms X Rent per single room) + (Number of double rooms X Rent per double room) = 771
[tex](x \times 27) +( y \times 33) = 771[/tex]-------------------------------------(2)
Substituting the value of (1) in (2), we get
[tex]((25-y) \times 27) +( y \times 33) = 771[/tex]
675-27y+ 33y = 771
33y -27 y = 771-675
6y = 96
y = [tex]\frac{96}{6}[/tex]
y= 16-------------------------------------------------------------------------------(3)
Substituting (3) in (1)
x = 25 –16
x= 9
