Andrea's Sequoia gets 12 miles per gallon while driving in the city, and 20 miles per gallon while on the highway. The last time she filled up the car it took 16 gallons of gas, and she had driven 280 miles since the previous fill-up. How many miles of that 280 were city driving?
[Math] System of equations word problem
algebra-precalculus
Related Solutions
I would recommend learning to set up problems in a clear and concise way.
For example:
Distance = D = 38.91 Miles
Gallons per Mile = G = 22 Miles / Gallon
Cost Gas = C = 3.699 Dollars / Gallon
Now, how would you describe in words how to calculate how many Gallons you used?
Knowing that, how would you calculate your total cost?
I know this looks trivial, but trust me, it will be a huge help in the future with easy and much harder problems!
HTH ~A
The easiest way, if you wish to distribute equitably, is to compute the number of person-days in the month.
If you, your wife, and Bob all live there 100% of the time, for 45 days in a billing period, that's 135 person-days. If his three kids are there for 14 days each in the billing period, that's another 42 person-days.
Thus, the bill is going to be $\frac{\$161.21}{135+42} = \frac{\$161.21}{177} = \$0.911$ per person-day.
Then, Bob can be attributed 45+42 = 87 person-days.
So his share of the bill is $87 \cdot \$0.911 = \$79.24$.
Edit: updated to reflect 45 day billing period
This process is a standard management/accounting practice. If you've ever done project management, you've almost certainly had to estimate "man-months" or "man-hours". This is the same idea.
When you're computing your bill, if you're a bachelor, you might only be concerned with "dollars per day." Once you start living with more people, then it might be sensible to start looking at "dollars per person per day" or "dollars per day per person". Are these two quantities different? How can we find out?
It turns out that units can be manipulated mathematically through multiplication and division: if you walk 5 meters per 10 seconds, you walk $\frac{5 \textrm{meters}}{10 \textrm{seconds}} = \frac12 \textrm{m/s}$. If you do that for 100 seconds, you multiply:
$$\require{cancel} \frac12 \frac{\textrm{meters}}{\cancel{\textrm{seconds}}} \cdot 100 \cancel{\textrm{seconds}} = 50 \textrm{meters}.$$
In your example, if you compute "dollars per person per day", you have
$$\frac{\left(\frac{161.21 \textrm{ dollars}}{N \textrm{ people}}\right)}{45 \textrm{ days}} = \frac{161.21 \textrm{ dollars}}{45N \textrm{ people}\cdot \textrm{days}}.$$
Otherwise, if you compute "dollars per day per person", you have
$$\frac{\left(\frac{161.21 \textrm{ dollars}}{45 \textrm{ days}}\right)}{45 \textrm{ people}} = \frac{161.21 \textrm{ dollars}}{45N \textrm{ people}\cdot \textrm{days}}.$$
These quantities are the same.
Your only challenge is to compute $N$ -- the number of people. Since each kid is only there 30% of the time, they only effectively count as 30% of a person. So your $N$ can be computed as
$$N = 1_{\textrm{you}} + 1_{\textrm{wife}} + 1_{\textrm{Bob}} + 3\cdot 0.3_{\textrm{kid}} = 3.9$$
Then, now that you can compute "dollars per person-day", you multiply it by each party's attributable person-days: 87 for Bob, 90 for you and your wife. And bam, you get your bill.
(One side note: in my earlier example, I rounded the number of days of kids up to an integer. In this example, 45*0.3 is not an integer number of days, so the result will be different by a handful of cents).
Best Answer
This is a standard type of problem where you're given enough information to form two linear equations, with two unknowns. And the process outlined below is one that will generalize to many such situations. So this is a good time to learn and understand how you can approach such a problem:
Let $\,c =$ (the number of miles that Andrea drove in the city since her last fill-up).
Let $\,h = $ (the number of miles she drove on the highway since her last fill-up.)
So one equation relating $c$ and $h$ is
From equation $(1)$, we can express miles of city driving, $c$ as follows: $$c + h = 280 \implies c = 280 - h.\tag{c}$$
Now simply substitute $c = 280 - h$ into $c$ of equation $(2)$, and solve for $h$. Then you can easily determine what $c$ must be by using the equation for $(c)$.