Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%. The second car depreciates at an annual rate of 15%. What is the approximate difference in the ages of the two cars?
1st car ... .60 = (1 - .10)^(t1)
2nd car ... .60 = (1 - .15)^(t2)
solve for t1 and t2
Let 'a' be the original value of the first car,
Let 'b' be the original value of the second car
Car A:
Final = Original(Rate)^t1
0.6a = a(0.9)^t1
0.6 = 0.9^t1
t1 = (Log0.6)/(Log0.9)
t1 = 4.85
Car B:
0.6b = b(0.85)^t2
0.6 = 0.85^t2
t2 = (Log0.6)/(Log0.85)
t2 = 3.14
t1 - t2 = 4.85 - 3.14 = 1.71
Therefore the difference in the ages of the two cars is approximately 1.7 years
To find the approximate difference in the ages of the two cars, we need to determine how many years it takes for each car to depreciate to 60% of its original value.
Let's assume the original value of both cars is 100 units.
For the first car, which depreciates at an annual rate of 10%, we can set up the equation:
100 * (1 - 0.10)^n = 60
Where 'n' represents the number of years it takes for the car to depreciate to 60% of its original value.
Simplifying the equation, we have:
(0.9)^n = 0.60
To solve for 'n', we can take the logarithm of both sides of the equation:
log(0.9)^n = log(0.60)
n * log(0.9) = log(0.60)
n = log(0.60) / log(0.9)
n ≈ 2.078
Therefore, it takes approximately 2.078 years for the first car to depreciate to 60% of its original value.
Now, let's do the same calculation for the second car, which depreciates at an annual rate of 15%:
100 * (1 - 0.15)^n = 60
Simplifying this equation, we have:
(0.85)^n = 0.60
Applying similar steps as before, we find that it takes approximately 2.761 years for the second car to depreciate to 60% of its original value.
The approximate difference in the ages of the two cars is the difference between these two time periods:
2.761 - 2.078 ≈ 0.683 years or about 8 months.
Therefore, the approximate difference in the ages of the two cars is approximately 8 months.