Help with difficult algebra word problem!? A new car begins to decrease in value from the moment it's purchased. Suppose a new car purchased for $25,000 in July of 2013 depreciates by 17% each...
Help with difficult algebra word problem!?
A new car begins to decrease in value from the moment it's purchased. Suppose a new car purchased for $25,000 in July of 2013 depreciates by 17% each year. When the value reaches $500, depreciation stops and the car is worth $500 from then on.
- In what year will the value of the car reach $500? I believe the answer to this is the year 2034, is this correct?
- Find the rate of depreciation for a car that takes 7 years to lose two-thirds of its value.
- In 2015, Sally's truck, which she bought used in 2009 for $13,995.36, is worth $4477.31. Find the price paid by the original owner in 2004.
We are given a depreciation rate of 17% annually, until the value reaches $500.
(a) The car is worth 25,000 originally. The formula for compound depreciation is A=P(1-r)^t where A is the amount the car is worth at time t, P is the original price, r is the annual depreciation rate, and t is the time in years.
Here A=500,P=25000,r=.17 and we are solving for t:
.02=.83^t Take a logarithm (I will use the natural logarithm but it does not matter) of each side.
ln(.02)=ln(.83^t) Use a property of logs:
t=[ln(.02)]/[ln(.83)] or t is approximately 20.995 or 21 years.
The car will be worth $500 in 2034.
(b)Find the depreciation rate so that a car loses 2/3 of its value in 7 years.
Use the same formula with A=2/3, P=1, r unknown, and t=7:
2/3=(1-r)^7 Take the 7th root of both sides (or raise both sides to the 1/7 power) to get:
(2/3)^(1/7)=1-r so r=1-(2/3)^(1/7). Thus r is approximately .0563.
r is about 5.63%
(c) P=13,995.36, A=4477.31, r is unknown and t=6 (2015-2009)
.3199=(1-r)^6 Take the 6th root of both sides:
.827=1-r ==> r is about .173 so r is 17.3%.
Now we have A=13995.36, P is unknown, the depreciation rate r is .173 and t is 5 (2009-2004)
P is about 36179 .
The original owner paid $36,179