two different cars each depreciate to 60% of their respective original values. the first car depreciates at…

Answer:

A. 1.7 years

Explanation:

Step1: Use depreciation formula

The depreciation formula is $A = P(1 - r)^t$, where $A$ is the final - value, $P$ is the initial value, $r$ is the rate of depreciation, and $t$ is the time in years. Since $A = 0.6P$ for both cars. For the first car with $r_1=0.1$: $0.6P = P(1 - 0.1)^t_1$ $0.6=(0.9)^{t_1}$ Take the natural - logarithm of both sides: $\ln(0.6)=t_1\ln(0.9)$ $t_1=\frac{\ln(0.6)}{\ln(0.9)}\approx\frac{- 0.5108}{-0.1054}\approx4.85$ years.

Step2: For the second car with $r_2 = 0.15$

$0.6P = P(1 - 0.15)^t_2$ $0.6=(0.85)^{t_2}$ Take the natural - logarithm of both sides: $\ln(0.6)=t_2\ln(0.85)$ $t_2=\frac{\ln(0.6)}{\ln(0.85)}\approx\frac{-0.5108}{-0.1625}\approx3.14$ years.

Step3: Calculate the difference in ages

$\Delta t=t_1 - t_2$ $\Delta t\approx4.85−3.14 = 1.71\approx1.7$ years.