Question

Nasser's 2015 car is quite functional, but it only averages 20 kilometers per Liter (km/). He has found a somewhat newer vehicle (roughly the same functionality) that averages 26 km/l. He can sell his current car for KD 2800 and purchase the newer vehicle for K 4,000. Assu Jim drive if he wants to recover his investment in three years? Assume an interest rate of 6%, zero salvage value for either vehicle after three years, and identical maintenance cost. me a cost of gasoline KD 4.00 per liter How many Kilometers per year must

Answer 1

Let x be the number of Kilometre Nasser will drive every year.

The cost of running old car can be written as follows

PV (old car) = -2800 -

$\large PV_{Old Car} = -2800 - \frac{0.2x}{1+i} - \frac{0.2x}{(1+i)^2} -\frac{0.2x}{(1+i)^3}$

$\large PV_{Old Car} = -2800 - \frac{0.2x}{i}[1-\frac{1}{(1+i)^3}]$

Here price of fuel is 4 and mileage is 20 km/l. Thus total km covered divided by mileage we will get total fuel consumption and by multiplying it with fuel price we can determine the total running cost.

Cost of fuel per year = 4*(x/20) = 0.2x

Similarly, for new car it can be written as

Here fuel cost = 4*(x/26) = 0.153846x

$\large PV_{New Car} = -4000 - \frac{0.153846x}{1+i} - \frac{0.153846x}{(1+i)^2} -\frac{0.15386x}{(1+i)^3}$

$\large PV_{New Car} = -4000 - \frac{0.153846x}{i}[1-\frac{1}{(1+i)^3}]$

Now equate these to equations we get

$\large -2800 - \frac{0.2x}{i}[1-\frac{1}{(1+i)^3}] = - 4000 - \frac{0.153846x}{i}[1-\frac{1}{(1+i)^3}]$

$\large 4000 -2800 = \frac{0.2x}{i}[1-\frac{1}{(1+i)^3}] - \frac{0.153846x}{i}[1-\frac{1}{(1+i)^3}]$

$\large 1200 = \frac{x}{i}[1-\frac{1}{(1+i)^3}] (0.2 - 0.153846)$

Here i = 6%

$\large 1200 = \frac{x}{0.06}[1-\frac{1}{(1+0.06)^3}] *0.046153$

$\large \frac{1200}{0.046153} = 2.673011x$

$\large 2.673011x = 25,999.91$

$\large x = \frac{25,999.91}{2.673011} = 9726.822$

In order to have equal value he must drive 9726.82 km every year.

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