# A certain car model costs 21,000 euro now, and it devaluates 5% every year. How much will it cost in 6 years?

A certain car brand estimates that a certain model, that now costs 21000 euro, will devaluate 5% per year.

How much will this car cost in 6 years?

I did:

$$21000 \cdot (1+.05)^6 \approx 28142$$

Now I remove the initial value from this one to get the compound devaluation

$$28142-21000 = 7142$$

Then I subtract this value from the initial value

$$21000-7142 = 13858$$

But my book says the solution is 15,437 euro.

What did I do wrong?

• Sounds related to Christina's car question math.stackexchange.com/q/2029665/35369 :-) Nov 25, 2016 at 4:20
• "Devaluate" means that the car loses value! Nov 25, 2016 at 9:07
• @NajibIdrissi It does indeed. And 13858 is less than 21000. Nov 25, 2016 at 9:08
• @Taemyr And how was this number computed? OP first assumed that the car gained 5% per year, obtained a new (bigger) price, subtracted the old price from the new price to obtain the net increase... And then subtracted the increase from the old price to get a number that means nothing! Try to plug in more than 6 years in this procedure and you'll eventually get a negative price (after 15 years). How is this supposed to mean something? Nov 25, 2016 at 10:40
• I fail to understand why this problem is shown as "Hot Network Questions", if not getting downvoted. Nov 26, 2016 at 12:25

You're adding $5\%$ to the value each year when you multiply by $(1+.05)^6$. To subtract $5\%$ each year, use $(1-.05)^6$. It's not the same, because after the first year, you're taking $5\%$ from a smaller value each year. If it grows, you're adding $5\%$ to a larger value each year.

Don't rote the formula. It is very easy to derive it.

The sentence that "a certain model, that [in year t] costs [f(t)] euro, will devaluate 5% per year" can be formalized mathematically as:

$$f(t+1) = f(t) \cdot (100\% - 5\%)$$

Substitute $f(t+1)$ into $f(t)$, and we get:

\begin{align*} f(t+2) &= f(t+1) \cdot (100\% - 5\%)\\ &= f(t) \cdot (100\% - 5\%) \cdot (100\% - 5\%)\\ &= f(t) \cdot (100\% - 5\%)^2 \end{align*}

And so on and on, for 5 times, and voila, here's the formula for "[h]ow much will this car cost in 6 years". \begin{align*} &...\\ f(t+6) &= f(t) \cdot (100\% - 5\%)^6 \end{align*} If you forget the compound interest formula again, you can derive it back quickly, or use this method to check if your formula is correct.

Devaluate means value will decrease. So price will be $(1-0.05)$ times previous year's price.

So, new value = $$21000 \cdot (1-0.05)^6 \approx 15436.93$$

Understood?

If the care loses $5\%$ value each year, then its value is $(100\%-5\%)=95\%$ of its value the previous year.

You're supposed to use $0.95^5\approx 0.7351$ which gives the correct number. You DECREASE by 5 percent so $100\%-5\%=0.95$

A different way of organizing what was already said:

$\$ 7142$would be the total increment in the value if it was increasing year after year. But with the value increasing, the increment is each year greater. So this results greater than the correct total decrement, which is $$21000- 15437 = 5563$$ The number$15437$is calculated directly, considering that each year the devaluation is represented by multiplication by$1-0.05$(as this is less than$1\$ the number effectively decreases, as we want).

Therefore after six years, $$21000 \cdot (1-.05)^6 \approx 15437$$