Answers:
- x = 12
- It will be worth 22490 pounds after four years
- 8 years will be when it has fallen below 14520 pounds
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Step-by-step explanation:
The car is initially worth 44000 pounds.
It loses 25% of its value after the first year, but keeps the remaining 75% of it.
75% of 44000 = 0.75*44000 = 33000
The car is worth 33000 pounds at the end of year 1. This is effectively the "starting" value when we change the rate of depreciation.
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The new percentage of the depreciation rate is x%
The car loses x% of its value each year
But it keeps the remaining (100-x)% of the value from year to year. Refer to the previous section.
x% = x/100
(100-x)% = (100-x)/100 = 100/100 - x/100 = 1 - 0.01x
The 1 - 0.01x is the value of b in the exponential equation y = a*b^t
The value of 'a' is the "starting" value of 33000 pounds.
The t represents the number of years after that first year (so that means we have t+1 years total).
The equation we have is
y = a*b^t
y = 33000(1 - 0.01x)^t
The teacher mentions "after 3 years" which means we'll have t+1 = 3 solve to t = 2 which is what we'll plug in along with y = 25555
Let's solve for x.
y = 33000(1 - 0.01x)^t
25555 = 33000(1 - 0.01x)^2
25555/33000 = (1 - 0.01x)^2
0.77439393939393 = (1 - 0.01x)^2
(1 - 0.01x)^2 = 0.77439393939393
1 - 0.01x = sqrt(0.77439393939393)
1 - 0.01x = 0.87999655646709
-0.01x = 0.87999655646709-1
-0.01x = -0.1200034435329
x = -0.1200034435329/(-0.01)
x = 12.00034435329
Due to rounding error, it appears that it should be x = 12 since your teacher mentions x is an integer.
Also, 12.00034435329 is pretty close to 12.
If it decays at 12% per year, then the car keeps the remaining 88% since 100-12 = 88
Therefore 0.88 is the base of the exponential, it's the value of b.
Computing 33000(0.88)^2 gets us 25,555.2 which isn't too far from the goal of 25,555 pounds.
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We found the exponential decay equation to be
y = 33000(0.88)^t
where t is the number of years after that first year. We have t+1 years total.
Your teacher asks how much the car will be worth after 4 years, so we'll plug in t = 3 because it's the solution to t+1 = 4.
In other words, we're working 3 years after that first year is up, so 3+1 = 4 total years.
Plug in t = 3 to find the following
y = 33000(0.88)^t
y = 33000(0.88)^3
y = 22,488.576
Rounding to the nearest pence gets us 22,489.58 which is the value of the car after 4 years. You can verify this with a spreadsheet table as I have shown in the diagram below. The highlighted rows mention values on your screenshot.
I'll round to the nearest pound since the other car values mentioned were whole numbers rather than decimal ones.
The 22489.58 rounds to 22490
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For the final part, we'll need logarithms to solve the equation. Logarithms are useful to isolate the exponent.
We'll plug in y = 14520 and solve for t
y = 33000(0.88)^t
14520 = 33000(0.88)^t
14520/33000 = (0.88)^t
0.44 = (0.88)^t
Log[ 0.88^t ] = Log[ 0.44 ]
t*Log[ 0.88 ] = Log[ 0.44 ]
t = Log[ 0.44 ]/Log[ 0.88 ]
t = 6.42227097958021
This rounds up to t = 7
Round up instead of down because we want to clear the hurdle of passing the 14520 marker.
Recall once again that t is the number of years after the first year.
So t = 7 represents t+1 = 7+1 = 8 years in total that have passed by. Therefore, after 8 years is when the car's value is below 14520 pounds. Refer to the table diagram below.