Respuesta :
Answer:
A. 1.7 years
Step-by-step explanation:
Let [tex]P_1[/tex] be the original value of first car,
Since, the car depreciates at an annual rate of 10%,
Let after [tex]t_1[/tex] years the value of car is depreciated to 60%,
That is,
[tex]P_1(1-\frac{10}{100})^{t_1}=60\%\text{ of }P_1[/tex]
[tex]P_1(1-0.1)^{t_1}=0.6P_1[/tex]
[tex]0.9^{t_1}=0.6[/tex]
Taking ln on both sides,
[tex]t_1ln(0.9) = ln(0.6)[/tex]
[tex]\implies t_1=\frac{ln(0.6)}{ln(0.9)}[/tex]
Now, let [tex]P_2[/tex] is the original value of second car,
Since, the car depreciates at an annual rate of 15%
Suppose after [tex]t_2[/tex] years it is depreciated to 60%,
[tex]P_2(1-\frac{15}{100})^{t_2}=60\%\text{ of }P_2[/tex]
[tex]P_2(1-0.15)^{t_2}=0.6P_2[/tex]
[tex]0.85^{t_2}=0.6[/tex]
Taking ln on both sides,
[tex]t_2ln(0.85) = ln(0.6)[/tex]
[tex]\implies t_2=\frac{ln(0.6)}{ln(0.85)}[/tex]
[tex]\because t_1-t_2=\frac{ln(0.6)}{ln(0.90)}-\frac{ln(0.6)}{ln(0.85)}[/tex]
[tex]=1.70518303046[/tex]
[tex]\approx 1.7[/tex]
Hence, the approximate difference in the ages of the two cars is 1.7 years,
Option 'A' is correct.
