Question 1196509: I have been struggling with a word problem. The problem is "Best rentals charges a daily fee plus a milage fee for renting its cars. Mateo was charged $111.00 for 3 days and 300 miles, while Dara was charged $207.00 for 5 days and 600 miles. What does Best Rentals charge per day and per mile?"
I have tried to come up with an equation for this, but I've only managed to confuse myself. I have tried putting the time in as "x" and miles in as "y" and putting them equal to the total rental charge, then solving for "x" and "y". Obviously I'm not solving anything with that arrangement. I have tried making a "tic tac toe" table with the given data, should I be adding all my data up and using that to make a bigger equation. I am just confused with how to set this up. Any help is greatly appreciated, thank you.
Found 4 solutions by math_tutor2020, ikleyn, josgarithmetic, MathTherapy:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I appreciate you showing your work and thought process, rather than just copy/pasting the problem only.
x = cost per day
y = cost per mile
Both costs are in dollars
Let's list the given facts
[1] Mateo was charged $111.00 for 3 days and 300 miles
[2] Dara was charged $207.00 for 5 days and 600 miles
Fact [1] gives us the equation
3x+300y = 111
since
3x = 3 times x = cost for the 3 days
300y = cost for the 300 miles
3x+300y = total cost
Fact [2] yields the equation 5x+600y = 207 through similar reasoning.
Here's our system of equations
3x+300y = 111
5x+600y = 207
Let's double the first equation to go from 3x+300y=111 to 6x+600y = 222
We arrive at this equivalent system of equations
6x+600y = 222
5x+600y = 207
Notice the 600y terms matching up. This allows us to cancel out the y terms if we subtract straight down
The x terms combine to 6x-5x = 1x = x
The right hand sides combine to 222-207 = 15
Therefore, x = 15
Then we can determine y
3x+300y = 111
3*15+300y = 111
45+300y = 111
300y = 111-45
300y = 66
y = 66/300
y = 0.22
You could use other equations previously mentioned involving x and y.
We found that: x = 15 and y = 0.22
Therefore, it costs $15 per day and $0.22 per mile (aka 22 cents per mile).
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Check:
Mateo:
1 day = 15 dollars
3 days = 15*3 = 45 dollars so far
1 mile = 0.22 dollars
300 miles = 300*0.22 = 66 dollars added on
45+66 = 111 dollars total for Mateo
Or we could say
3x+300y = 111
3*15+300*0.22 = 111
45+66 = 111
111 = 111
which confirms Mateo's equation
Let's follow a similar idea for Dara
5x+600y = 207
5*15+600*0.22 = 207
75+132 = 207
207 = 207
Her equation is confirmed as well.
Both equations are true for (x,y) = (15, 0.22)
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Answers:
$15 per day
$0.22 per mile (aka 22 cents per mile)
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
Best rentals charges a daily fee plus a milage fee for renting its cars.
Mateo was charged $111.00 for 3 days and 300 miles, while Dara was charged $207.00 for 5 days and 600 miles.
What does Best Rentals charge per day and per mile?
~~~~~~~~~~~~~~~~~~~~
x = charge per day
y = charge per mile.
The problem is described by the system of 2 equations in 2 unknowns
3x + 300y = 111 (1) (Mateo case)
5x + 600y = 207 (2) (Dara case)
To solve, multiply equation (1) by 2; keep equation (2) as is. You will get
6x + 600y = 222 (3)
5x + 600y = 207 (4)
Now subtract equation (4) from equation (3). You will get
x = 222 - 207 = 15 (the terms with y will cancel each other).
Thus charge per day is 15 dollars. <<<---=== ANSWER
Now from equation (1)
3*15 + 300*y = 111,
which gives
300y = 111 - 45 = 66, y = 66/300 = 22/100 = 0.22.
So, charge per 1 mile is $0.22, or 22 cents. <<<---=== ANSWER
Solved.
Answer by josgarithmetic(39630) (Show Source):
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
I have been struggling with a word problem. The problem is "Best rentals charges a daily fee plus a milage fee for renting its cars. Mateo was charged $111.00 for 3 days and 300 miles, while Dara was charged $207.00 for 5 days and 600 miles. What does Best Rentals charge per day and per mile?"
I have tried to come up with an equation for this, but I've only managed to confuse myself. I have tried putting the time in as "x" and miles in as "y" and putting them equal to the total rental charge, then solving for "x" and "y". Obviously I'm not solving anything with that arrangement. I have tried making a "tic tac toe" table with the given data, should I be adding all my data up and using that to make a bigger equation. I am just confused with how to set this up. Any help is greatly appreciated, thank you.
Let daily and mileage fees be d, and m, respectively
We then get: 3d + 300m = 111 ---- eq (i)
Also, 5d + 600m = 207 ---- eq (ii)
6d + 600m = 222 ---- Multiplying eq (i) by 2 ------ eq (iii)
Subtracting eq (ii) from eq (iii), we get the daily fee, or 
3(15) + 300m = 111 -----Substituting 15 for d in eq (i)
15 + 100m = 37 ----- Dividing each expression by GCF, 3
100m = 22
Mileage fee, or 
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