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I am struggling on a financial mathematics question in a practice paper. The question is: Mr Bakele owns a courier company. His first truck costs R550 000 and it depreciates 35% per annum on a reducing balance, In 4 years the truck must be replaced. New truck prices appreciate at 15% per year. What will he pay in after four years if the current one is a trade in?

What I have assumed so far is that the truck costs R550 000 now. Every year it will go up 15%, meaning that when he is going to buyhis new truck in four years it will cost R880 000?

Update: I have the answer The new truck will cost R880 000. He will be able to trade in the old truck for R98178,44 Therefore he will pay R781821.56 for the new truck.

Nicole Carr
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  • If it depreciates 35% in four years he will lose 220 000? – Nicole Carr Nov 09 '16 at 07:08
  • I don't think so. Check my answer and try to understand the formulas. Then use them to calculate the truck's new price and how much the truck you own will be worth. – RGS Nov 09 '16 at 07:19
  • I don´t fully understand the exercise. But if it depreciates 35% per year then the value of the truck after four years is $550,000\cdot (1-0.35)^4=500,000\cdot (0.65)^4=98,178.44 $ And the new truck price after four years is $550,000\cdot 1.15^4$ – callculus42 Nov 09 '16 at 07:21
  • @NicoleCarr The value for the old truck is right (see my comment). The price for the new truck is not right. What have you calculated ? – callculus42 Nov 09 '16 at 07:36

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When an item's price appreciates at a rate of $x $% / year, it means that every year its price is multiplied by $r = \frac{100+x}{100}$. Let $p_i $ be the price of the item after $i $ years. Then set $p_0 = p $. Now after 1 year the item appreciates so we get $p_1 = rp_0$. Then its new price is $rp_0$. After another year, it appreciates again, giving $p_2 = rp_1 = rrp_0 = r^2p_0$ as its new price. Repeating this process we get $p_i = r^ip_0$.

Thus after 4 years, and at a rate of $15$% / year, its price will be $p_4 = r^4p_0 = (\frac{100+15}{100})^4\times 550000 = 1.15^4\times 550000 \not= 880000$

The same thing goes for ammortization, with $r = \frac{100-x}{100}$.

RGS
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