Experience the power of community-driven knowledge on IDNLearn.com. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
What is depreciation?
Depreciation is to decrease in the value of a product in a period of time. This can be given as,
[tex]FV=P\left(1-\dfrac{r}{100}\right)^n[/tex]
Here, (P) is the price of the product, (r) is the rate of annual depreciation and (n) is the number of years.
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.
Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is [tex]n_1[/tex]. Thus, by the above formula for the first car,
[tex]0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}[/tex]
Take log both the sides as,
[tex]\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85[/tex]
Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.
Thus, the depreciation price of the car is 0.6y. Let the number of year is [tex]n_2[/tex]. Thus, by the above formula for the second car,
[tex]0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}[/tex]
Take log both the sides as,
[tex]\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14[/tex]
The difference in the ages of the two cars is,
[tex]d=4.85-3.14\\d=1.71\rm years[/tex]
Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
Learn more about the depreciation here;
https://brainly.com/question/25297296
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
