The value, V, of a $25000 vehicle after y years is given by V = 25000 (0.85)y
What is the rate of depreciation? (1 mark)
What will the car be worth after 5 years? (1 mark)
To the nearest month, how long would it take to reduce the vehicle’s value to 10% of its original amount? (3 marks)
you probably meant
V = 25000 (.85)^y
a) replace y with 5
V = 25000(.85)^5
= ....
b) when is it worth 2500 ?
(.85)^y = .10
log both sides
log (.85^t) = log .10
t log .85 = log .10
t = log .10/log .85 = 14.168 years
or 14 years and 2 months
To find the rate of depreciation, we will use the formula V = P(1-d)^t, where V is the value of the vehicle after t years, P is the initial value of the vehicle, and d is the rate of depreciation.
We can compare this formula to the given equation V = 25000 (0.85)^y to determine that the rate of depreciation is 1 - 0.85 = 0.15 or 15%.
To find the value of the vehicle after 5 years, we can evaluate the given equation for y = 5:
V = 25000 (0.85)^5
V ≈ 25000 (0.443705)
V ≈ 11,092.63
So, the car will be worth approximately $11,092.63 after 5 years.
To find how long it would take for the vehicle's value to reduce to 10% of its original amount, we need to solve the equation:
0.10P = 25000 (0.85)^y
Dividing both sides by 25000, we get:
0.10 = 0.85^y
To solve for y, we need to take the logarithm of both sides:
log(0.10) = y log(0.85).
Using a logarithm calculator, we find:
y ≈ log(0.10) / log(0.85)
y ≈ 15.363.
To the nearest month, it would take approximately 15 months for the vehicle's value to reduce to 10% of its original amount.