Answer: 1
Step-by-step explanation:
Let the original price of first car = x dollars,
If after t years the price of car by decreasing 10% per year = 60 % of the original price = 60 % of x = 0.6 x
Hence, For first car,
\(0.6x = x (1-\frac{10}{100})^{t_1}\)
\(\implies 0.6 = (1-0.1)^{t_1}\)
\(\implies 0.6 = 0.9^{t_1}\)
By taking log on both sides,
\(\implies log(0.6) = t_1 log(0.9)\)
\(t_1=\frac{log(0.6)}{log(0.9)}=4.84835918443\approx 4.85\)
Now,
Let the original price of second car = y dollars,
If after t years the price of car by decreasing 15% per year = 60 % of the original price = 60 % of y = 0.6 y
Hence, For second car,
\(0.6x = x (1-\frac{15}{100})^{t_2}\)
\(\implies 0.6 = (1-0.15)^{t_2}\)
\(\implies 0.6 = 0.85^{t_2}\)
By taking log on both sides,
\(\implies log(0.6) = t_2 log(0.85)\)
\(t_2=\frac{log(0.6)}{log(0.85)}=3.14317615397\approx 3.14\)
Hence, the difference in the ages of the two cars = 4.85 - 3.14=1.71 years.
Thus, the difference in the ages of the two cars is approximately 1 years.
