You are thinking about buying a beautiful new hybrid sedan. The car is on sale and costs $20,000. You know you are only keeping the car for a couple of years, probably just enough to get yourself through college but it could be longer if you don't get your dream job right out of college.
You are fortunate enough to be able to pay cash for the car, so there are no pesky payments or interest to worry about when buying your car. The salesman tells you the car will decreases in value at the rate of 20 percent each year, based on the value at the beginning of the year. At the end of how many years will the value of the car first be less than half the price you are paying for it? You will need to include your process for determining when it will reach half the value of the purchase price as well as your solution. There are a couple of ways to detemine the solution.
Further, how long will you keep your car before selling it? Make sure to include a rational for your decision. (Keep in mind by the time you buy your next car it will definetly be more expensive than the price you are paying today.)
CCG
ReplyDelete$20,000
20000(1-.20)t
20000(.80)t
1. $16000
2. $12800
3. $10240
4. $8192
5. $6553.60
6. $5242.88
7. $4194.30
8. $3355.44
1. The car will be less than half the price at the fourth year of owning the car.
2. I probably would keep the car for four years and sell it on the fifth because by then the prices would be low and about 1/3 of the original and after that the prices are very low ant the car would not be worth being sold.
AC
ReplyDeleteY=20000(1-.20)^t
This is the equation that someone would use to find the cars price after a set amount of years. After finding the equation it’s just a guess and check to see what year it will be less then half its original price.
Years Price
0 $20000
1 $16000
2 $12800
3 $10240
4 $8192
5 $6553.60
Based on my data the car will be less than half its original price after 4 years. The quick decrease in the car value is bad and if I want any down payment for a new car I would want to sell it somewhat quickly. I would wait 5 years, till I was out of college and on my feet. I would at lest have $6500 dollars to buy a new, better car. That $6500 could help me reduce the price on a better car (down- payment) also decreasing my monthly payment.
AC
ReplyDelete$20,000
20,000(1-.20)
Exponents
1. $16000
2. $12800
3. $10240
4. $8192
5. $6553.60
6. $5242.88
7. $4194.30
8. $3355.44
•It will become half price after the fourth year of owning the car.
•I would keep that car until the fifth year because the other prices below it become smaller and smaller, and during the fifth year it is 1/3 the price, while the others are 1/4, 1/5, 1/6, e.t.c.
DC
ReplyDeleteThe equation for the decrease of the car would be “20,000(0.8)t. If he is only keeping for a few years than I can assume that it means 4. So I can make a table up to four years of owning the car.
Year Value of Car
0 $20,000
1 $16,000
2 $12,800
3 $10,240
4 $8,192
That means the car will be less than half of its original price in the fourth year of ownership. So if I was going to sell the car I would sell it in my 4th year of ownership because I would be out of college and I would have $8,000 to spend on a new car. Since I was in college I would have a job and I would have at least that amount of money to put down on the car. With my new job I would be eventually able to pay off the new car and I wouldn’t have to pay as much because I put the money from my old car into the new one.
y = c(1-r)t
ReplyDeletey = 20,000(1-.20)3
y = 20,000(0.80)3
y = 20,000(0.512)
y = $10,240
y = 20,000(1-.20)4
y = 20,000(0.80)4
y = 20,000(0.410)
y = $8192
y = 20,000(1-.20)5
y = 20,000(0.80)5
y = 20,000(0.327)
y = $6553.60
I would keep the car until 3 years after I bought it because I could sell it for more then half of it’s original value.
F.P.
Y= c(1-r)
ReplyDeleteY=20000(1-.20)4
(.80)
20000 (.409)
Y=8180$
I would do keep the car for 3 years because, it’ll give me a little over halve of what I payed for and that will help me buy a new car.
AM
P.H
ReplyDeleteIn order for the car to be half the price of when it was bought I used the equation 20,000(1- 0.2) t I made a table up to then years to see by how much the amount of the car went down every time. Once I had gotten to ten I looked at the table and tried to find the closest amount to 10,000 (half of the starting amount). The closets number to 10,000 is 10,240 which occurred after 3 years.
Years Price
1 16,000
2 12,800
3 10,240
4 8,192
5 6,553.60
6 5,242.88
7 4,194.30
8 3,355.44
9 2,684.35
10 2,147.48
G.F.
ReplyDeleteThe first thing I did was make an exponential decay equation: $20,000(1-0.2)^t. Next, I found out what half of 20,000 is: 10,000. On a seperate piece of paper, I made a table following my equation. I replaced "t" with 1 and got 16,000, 2 and got 12,800, 3 and got 10,240, 4 and got 8,192. My final answer is four years because the value of the car is $8,192.00 and it is less than half the price of the car when I first bought it.
A.E.,
ReplyDeleteI started with 20,000 and half of that is 10,000. After, I formulated an equation for the problem which was 20,000(1-.20)^t. Using this euation, my solution was it would take about 3.5 to 4 years for the car to be worth half of it's initial price.I mainly used the trial and error method:
1:16,000
2:12,800
3:10,240
4:8,192
I would sell sell my car after about a year of two because the price is dropping quickly and if I wanted to get a better car,I would need all the money I could get.
I used the equation C(1-r)(to the power of T)
ReplyDeleteI plugged in differint numbers for T until i found the year that the car is less then half of the original price.
the answer is four.
20000(.8)4
20000 (.41)
8200
four years is the time period it takes to go to a traditional college. it would drop half of its value by the time the car would be sold.
three years would not be the time period because the result would not be less then half of the original price.
CL.
E.G.
ReplyDeletemy answer was 4 years and i solved it with this equation c(1-r)to power of t and i solved it like this:
time period(t)=4
20,000(0.8)to the power of 4
20,000(0.41)
8,200
The equation i used to solve this problem was
ReplyDeleteC(1-r)to the power of t.
C= 20,000
r= 0.2
t= the number of years( i used the number of years he was in college which is 4.)
So the equation was:
20,000(1-0.2) to the power of 4.
20,000(0.8) to the power of 4
20,000(.41)
8,200
so after 4 years the car would be under 10,000 dollars.
AA