A car is worth $24000 and loses 12.5% of its value each year. Write… Step by Step

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The question

Question: A car is worth $24000 and loses 12.5% of its value each year.

Write the model.

1 Find the start

t is the time in years.

We call the value V.

V = value, t = years

At the start no time has passed: t = 0.

Then the value is $24000.

That is the start.

start, t = 0: V = $24000

2 Find the factor

Each year, 12.5% of the value is lost.

What is left is 100% take away 12.5%: 87.5%.

100% − 12.5% = 87.5%

Percent means out of 100.

So 87.5% is 87.5/100: 0.875.

87.5% = 87.5100 = 0.875

So the factor is 0.875: each year, we multiply by 0.875.

factor: × 0.875 each year

3 Make a small table

A small table: t = 0, 1, 2.

Each step, we multiply by 0.875: 24000 times 0.875 is 21000.

Times 0.875 again is 18375.

After 2 years, we multiplied by 0.875 two times.

Two times is the power 2: 0.875 squared.

t = 2: 24000 × 0.875 × 0.875 = 24000 × 0.875²

After t years, we multiply t times.

So the power is t.

after t years: × 0.875, t times = × 0.875^t

4 Write the model

Start times the factor, to the power t.

So the model is V = 24000 times 0.875 to the power t.

V = 24000 × 0.875^t

Careful!

Not 0.125.

That is the part that is lost.

We multiply by the part that is left: 0.875.

Careful: not V = 24000 × 0.125^t ✗

5 Check

Let's check the first year the story way.

12.5% of 24000 is 0.125 times 24000: 3000.

12.5% of 24000 = 0.125 × 24000 = 3000

It loses 3000: 24000 minus 3000 is 21000.

24000 − 3000 = 21000

The model at t = 1 gives 24000 times 0.875: 21000 too.

It works!

model, t = 1: 24000 × 0.875¹ = 21000

Let's look back.

First: find the start.

Then: find the factor.

Then: make a small table.

Then: write the model.

Last: check.

Answer