SOLUTION: The buildings in a factory depreciate at the rate of 2.5% per year. Hence the value A of the factory buildings after t years and a depreciation rate of r% can be modelled by: A = P

Question 1048642: The buildings in a factory depreciate at the rate of 2.5% per year. Hence the value A of the factory buildings after t years and a depreciation rate of r% can be modelled by: A = P * R^t where R = 1 - (r/100)
The factory buildings were initially valued at $450000
i) After how many years will the factory buildings be valued at half of their original value?
ii) If the rate of depreciation is changed to half of the original rate after eleven years, then how many more years will it take for the value of the factory buildings to have reduced to half their original value?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
formula for future value of a present amount using compound interest is:

f = p * (1+r)^n

f is the future value, consistent with A in the formula you were provided.
p is the present value, consistent with P in the formula you were provided.
r is the interest rate per time period, consistent with R in the formula you were provided.
n is the number of time periods, consistent with t in the formula you were provided.

bottom line is:

your formula of A = P * (1+R)^t becomes my formula of f = p * (1+r)^n

assuming a rate of -.025, then 1+r becomes 1+(-.025) which becomes 1-.025.

the factor buildings were initially valued at 450,000.
that would be the value of p.

the interest rate per year is -2.5%.
that's in percent form.
in decimal form, the interest rate per year is -.025.
that would be the value of r.

you want to know in how many years, the value of the factory building will be half of what it is now.
that would be 450,000 / 2 = 225,000.
that would be the value of f.

your formula becomes:

225,000 = 450,000 * (1 - .025)^n

225,000 is equal to f.
450,000 is equal to 0
-.025 is equal to r
n is what you want to find
the time period is in years.

start with:
225,000 = 450,000 * (1 - .025)^n
divide both sides of the equation by 450,000 to get:
225,000 / 450,000 = (1 - .025)^n
take the log of both sides of the equation to get:
log(1/2) = log((1-.025)^n)
since log(a^b) = b * log(a), the equation becomes:
log(1/2) = n * log(1-.025)
divide both sides of the equation by log(1-.025) to get:
log(1/2) / log(1-.025) = n
solve for n to get:
n = 27.37785123 years

it would take 27.37785123 years for the original value to become half at a depreciation rate of .025 per year.

replace n with 27.37785123 in the original equation to get:
225,000 = 450,000 * (1 - .025)^27.37785123
simplify to get:
225,000 = 225,000

this confirms the solution is correct.

assuming .025 depreciation rate for the first 11 years and then half of that for the remaining years, the following is what will occur.

same formula of f = p * (1+r)^n
p = 450,000
r = -.025
n = 11

formula becomes:
f = 450,000 * (1-.025)^11
solve for f to get:
f = 340614.6212

the value of the property is 340,614.6212 after 11 years.

formula to get to half of the original value now becomes:

225,000 = 340,614.6212 * (1 - .0125)^n
divide both sides of this equation by 340,614.6212 to get:
225,000 / 340614.6212 = (1 - .0125)^n
take the log of both sides of this equation to get:
log(225,000 / 340,614.6212) = log((1 - .0125)^n)
since log(a^b) = b*log(a), this equation becomes:
log(225,000 / 340,614.6212) = n * log(1 - .0125)
divide both sides of this equation by log(1 - .0125) to get:
log(225,000 / 340,614.6212) / log(1 - .0125) = n
solve for n to get:
n = 32.96434313.

replace n with 32.96434313 in the original equation to get:
225,000 = 340614.6212 * (1 - .0125)^32.96434312
simplify to get:
225,000 = 225,000

this confirms the solution is correct.

it would take 11 years for the value to become 340,614.6212.
it would take an additional 32.96434312 years for the value to become 225,000.

the total time for the original value to become half would be 32 + 11 = 43.96434312 years, assuming a depreciation rate of -.025 for the first 11 year and then a depreciation rate of -.0125 for the next 32.96... years.

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