Working out "break even" point Please bear with me, my brain is hopeless at math.
My colleague has a Jeep Grand Cherokee. He had a chip installed which cost him $\textrm{ZAR}3500$. He wants to know when his purchase of the chip will break even whether it's distance traveled or money spent on diesel.
His Jeep was consuming $11.8$ litres per $100\,\textrm{km}$ and now consumes $10.6$ litres per $100\textrm{km}$. The size of the tank is $55$ litres.
EDIT: Forgot to add the cost of diesel $\textrm{ZAR}11.18$ per litre.
What we've worked out so far is the following
Distance he can travel on a full tank of diesel:
Before chip: $466.10\textrm{km}$
After chip: $518.86\textrm{km}$.
Distance per litre:
Before chip: $8.47\textrm{km/L}$
After chip: $9.4\textrm{km/L}$
That's as far as we've gotten. We don't know how we would work out his "break even" point. Any help would be appreciated.
 A: This is just a problem of managing units. You have the following relations:


*

*Cost of fuel: $\frac{\textrm{ZAR}}{\textrm{liter}};$

*Fuel efficiency: $\frac{\textrm{km}}{\textrm{liter}}.$


You need to figure out how the increase in fuel efficiency earns you back money in terms of savings, by estimating the cost of fuel for that efficiency gained.
We'll denote your gain in fuel efficiency by the following symbol: $\Delta E$.
Now, $\Delta E = E_{\textrm{post-chip}} - E_{\textrm{pre-chip}} = \frac{100\textrm{ km}}{10.6\textrm{ L}} - \frac{100\textrm{ km}}{11.8\textrm{ L}} \approx 0.959 \frac{\textrm{km}}{\textrm{L}}.$
So by purchasing and installing the chip, you can drive about 959 meters farther per liter of fuel consumed. Not bad!
Next, we must translate this into savings. Since those 959 extra meters are essentially "free" (excluding the cost of the chip), we will compute the cost of the fuel that we would have burned to go that far were it not for the chip.
This can be computed as
$$\textrm{cost savings} = 959 \textrm{ m} \times \textrm{old fuel efficiency} \times \textrm{cost of fuel per liter}$$
You didn't give us a fuel cost, so I looked it up here: http://www.aa.co.za/on-the-road/calculator-tools/fuel-pricing.html. We'll use $11.18 \frac{\textrm{ZAR}}{\textrm{L}}$ as a baseline.
This gives us
$$\textrm{cost savings} = 0.959\textrm{ km} \times \frac{10.6\textrm{ L}}{100\textrm{ km}} \times \frac{11.18\textrm{ ZAR}}{\textrm{L}} \approx 1.14 \textrm{ ZAR}.$$
Therefore, we get a cost savings of about $1.14$ ZAR per $0.959$ km driven, or about $1.19$ ZAR per km.
Now we must ask ourselves, "how far must our friend drive before he makes back his investment?" To compute that, we simply solve
$$\textrm{cost savings per kilometer} \times \textrm{kilometers driven} = \textrm{cost of investment}$$
Let $x$ denote kilometers driven, and let $3500$ be the cost of investment; the resulting equation is
$$3500 = 1.19 x.$$
This is easily solved as
$$x = \frac{3500}{1.19} \approx 2941 \textrm{ km}.$$
Given that he now drives $518.86$ km per fill-up, he will require about $5.67$ fill-ups to make back his investment.
We can check our work:
$$5.67 \textrm{ full tanks} \times \frac{55\textrm{ liters}}{\textrm{full tank}} \times \frac{11.18\textrm{ ZAR}}{\textrm{liter}} = 3486 \textrm{ ZAR}.$$
Note -- the solution doesn't come to 3500 exactly because I rounded off several times in each of my steps.
Noting this, we can now generalize our computation:
$$\textrm{number of fillups until break-even} = \frac{\textrm{cost of initial investment}}{\textrm{capacity of tank } \times \textrm{ cost of fuel per liter}}$$
