Question
A car rental agency rents compact, midsize, and luxury cars. Its goal is to purchase 60 cars with a total of $1 comma 400 comma 000 and to earn a daily rental of $2000 from all the cars. The compact cars cost $15 comma 000 and earn $20 per day in rental, the midsize cars cost $25 comma 000 and earn $40 per day, and the luxury cars cost $45 comma 000 and earn $60 per day. To find the number of each type of car the agency should purchase, solve the system of equations below, where x, y, and z are the numbers of compact cars, midsize cars, and luxury cars, respectively.
StartSet Start 3 By 9 Matrix 1st Row 1st Column x 2nd Column plus 3rd Column y 4st Column plus 5st Column z 6st Column equals 7st Column 60 8st Column 9st Column left parenthesis 1 right parenthesis 2nd Row 1st Column 15 comma 000 x 2nd Column plus 3rd Column 25 comma 000 y 4st Column plus 5st Column 45 comma 000 z 6st Column equals 7st Column 1 comma 400 comma 000 8st Column 9st Column left parenthesis 2 right parenthesis 3rd Row 1st Column 20 x 2nd Column plus 3rd Column 40 y 4st Column plus 5st Column 60 z 6st Column equals 7st Column 2000 8st Column 9st Column left parenthesis 3 right parenthesis EndMatrix
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes within your choice.
A.
There is one solution. The agency should purchase
enter your response here compact cars,
enter your response here midsize cars, and
enter your response here luxury cars.
(Simplify your answers.)
B.
There are infinitely many solutions. If the agency purchases z luxury cars, then the agency should purchase
enter your response here compact cars and
enter your response here midsize cars.
(Type expressions using z as the variable.)
C.
There is no solution.
StartSet Start 3 By 9 Matrix 1st Row 1st Column x 2nd Column plus 3rd Column y 4st Column plus 5st Column z 6st Column equals 7st Column 60 8st Column 9st Column left parenthesis 1 right parenthesis 2nd Row 1st Column 15 comma 000 x 2nd Column plus 3rd Column 25 comma 000 y 4st Column plus 5st Column 45 comma 000 z 6st Column equals 7st Column 1 comma 400 comma 000 8st Column 9st Column left parenthesis 2 right parenthesis 3rd Row 1st Column 20 x 2nd Column plus 3rd Column 40 y 4st Column plus 5st Column 60 z 6st Column equals 7st Column 2000 8st Column 9st Column left parenthesis 3 right parenthesis EndMatrix
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes within your choice.
A.
There is one solution. The agency should purchase
enter your response here compact cars,
enter your response here midsize cars, and
enter your response here luxury cars.
(Simplify your answers.)
B.
There are infinitely many solutions. If the agency purchases z luxury cars, then the agency should purchase
enter your response here compact cars and
enter your response here midsize cars.
(Type expressions using z as the variable.)
C.
There is no solution.
Answers
GPT-4o mini
To find the number of each type of car the agency should purchase, we will solve the system of equations represented by the given conditions.
The equations based on the car rental agency's conditions are:
1. \( x + y + z = 60 \) (total number of cars)
2. \( 15000x + 25000y + 45000z = 1400000 \) (total cost of cars)
3. \( 20x + 40y + 60z = 2000 \) (total daily rental income)
Here, \( x \) is the number of compact cars, \( y \) is the number of midsize cars, and \( z \) is the number of luxury cars.
Let's simplify the equations step by step:
- **Equation 3** can be simplified by dividing everything by 20:
\[
x + 2y + 3z = 100 \quad (4)
\]
Now we have the following simplified system:
1. \( x + y + z = 60 \) (equation 1)
2. \( 15000x + 25000y + 45000z = 1400000 \) (equation 2)
3. \( x + 2y + 3z = 100 \) (equation 4)
Next, we can solve this system of equations. We will first isolate one variable in terms of the others using equation 1.
From equation 1:
\[
x = 60 - y - z \quad (5)
\]
Now substitute equation (5) into equations (2) and (4).
**Substituting into equation 2**:
\[
15000(60 - y - z) + 25000y + 45000z = 1400000
\]
Distributing the \( 15000 \):
\[
900000 - 15000y - 15000z + 25000y + 45000z = 1400000
\]
Combine like terms:
\[
900000 + 10000y + 30000z = 1400000
\]
Subtract \( 900000 \) from both sides:
\[
10000y + 30000z = 500000
\]
Divide everything by \( 10000 \):
\[
y + 3z = 50 \quad (6)
\]
**Substituting into equation 4**:
Using equation (5) with equation (4):
\[
(60 - y - z) + 2y + 3z = 100
\]
Combine like terms:
\[
60 + y + 2z = 100
\]
Subtract 60 from both sides:
\[
y + 2z = 40 \quad (7)
\]
Now we have two new equations from equations (6) and (7):
1. \( y + 3z = 50 \) (equation 6)
2. \( y + 2z = 40 \) (equation 7)
Now subtract equation (7) from equation (6):
\[
(y + 3z) - (y + 2z) = 50 - 40
\]
This simplifies to:
\[
z = 10
\]
Now, substitute \( z = 10 \) back into equation (7):
\[
y + 2(10) = 40
\]
\[
y + 20 = 40
\]
So,
\[
y = 20
\]
Finally, substitute \( y = 20 \) and \( z = 10 \) into equation (5):
\[
x = 60 - 20 - 10
\]
\[
x = 30
\]
So the solution is:
- \( x = 30 \) (compact cars)
- \( y = 20 \) (midsize cars)
- \( z = 10 \) (luxury cars)
### Final Answer:
A. There is one solution. The agency should purchase:
- 30 compact cars,
- 20 midsize cars,
- and 10 luxury cars.
The equations based on the car rental agency's conditions are:
1. \( x + y + z = 60 \) (total number of cars)
2. \( 15000x + 25000y + 45000z = 1400000 \) (total cost of cars)
3. \( 20x + 40y + 60z = 2000 \) (total daily rental income)
Here, \( x \) is the number of compact cars, \( y \) is the number of midsize cars, and \( z \) is the number of luxury cars.
Let's simplify the equations step by step:
- **Equation 3** can be simplified by dividing everything by 20:
\[
x + 2y + 3z = 100 \quad (4)
\]
Now we have the following simplified system:
1. \( x + y + z = 60 \) (equation 1)
2. \( 15000x + 25000y + 45000z = 1400000 \) (equation 2)
3. \( x + 2y + 3z = 100 \) (equation 4)
Next, we can solve this system of equations. We will first isolate one variable in terms of the others using equation 1.
From equation 1:
\[
x = 60 - y - z \quad (5)
\]
Now substitute equation (5) into equations (2) and (4).
**Substituting into equation 2**:
\[
15000(60 - y - z) + 25000y + 45000z = 1400000
\]
Distributing the \( 15000 \):
\[
900000 - 15000y - 15000z + 25000y + 45000z = 1400000
\]
Combine like terms:
\[
900000 + 10000y + 30000z = 1400000
\]
Subtract \( 900000 \) from both sides:
\[
10000y + 30000z = 500000
\]
Divide everything by \( 10000 \):
\[
y + 3z = 50 \quad (6)
\]
**Substituting into equation 4**:
Using equation (5) with equation (4):
\[
(60 - y - z) + 2y + 3z = 100
\]
Combine like terms:
\[
60 + y + 2z = 100
\]
Subtract 60 from both sides:
\[
y + 2z = 40 \quad (7)
\]
Now we have two new equations from equations (6) and (7):
1. \( y + 3z = 50 \) (equation 6)
2. \( y + 2z = 40 \) (equation 7)
Now subtract equation (7) from equation (6):
\[
(y + 3z) - (y + 2z) = 50 - 40
\]
This simplifies to:
\[
z = 10
\]
Now, substitute \( z = 10 \) back into equation (7):
\[
y + 2(10) = 40
\]
\[
y + 20 = 40
\]
So,
\[
y = 20
\]
Finally, substitute \( y = 20 \) and \( z = 10 \) into equation (5):
\[
x = 60 - 20 - 10
\]
\[
x = 30
\]
So the solution is:
- \( x = 30 \) (compact cars)
- \( y = 20 \) (midsize cars)
- \( z = 10 \) (luxury cars)
### Final Answer:
A. There is one solution. The agency should purchase:
- 30 compact cars,
- 20 midsize cars,
- and 10 luxury cars.
Related Questions
Assume that you are to pick 3 cars from the motor pool, which contains 6 subcompact cars, 6 compact...
Ricardo's car rental offers five different types of vehicles, compact cars (CC), midsize cars (MC),...
At a certain car rental agency a compact car rents for $40 a day and 20¢ a mile.
(a) How much does...