Don has been tracking the population of a town. The population was 7,220 last year and has been increasing by 1.45 times each ye
2 answers:
Answer: He will use exponential growth function to model the situation.
Step-to step explanation:-
Given: The population in last year = 7,220
The growth factor of population = 1.45 each year
To calculate the population in coming years Don must use exponential growth function.
The exponential growth equation is given by :-
, where A is the initial population, b is the growth factor and x is the time period.
Then, the function model the given situation will be
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9514 1404 393
Answer:
a) RM2256.09 . . . principal paid by 8th payment
b) RM1791.10 . . . . interest paid by 12th payment
Step-by-step explanation:
First of all, we need to find the payments.
The payment amount is the amount that makes the future value of the series of payments equal to the future value of the loan at the given interest rate.
The future value of a single amount is ...
FV = P(1 +r)^n . . . . . where r is the annual rate, and n is the number of years in the future
The future value of a series of payments is ...
FV = P((1 +r)^n -1)/r . . . . . where n is the number of payments of P earning annual rate r
For payments in a series that does not end at the end of the loan, the future value is the product of that of the series and the effect of the accumulation of interest for the remaining time.
__
The first 4 payments will have a future value at the end of the loan period of ...
s1 = X((1 +0.05)^4 -1)/0.05×(1 +0.05)^11 = X(1.05^15 -1.05^11)/0.05
s1 = 7.3717764259X
The next 5 payments will have a future value at the end of the loan period of ...
s2 = 2X((1 +0.05)^5 -1)/0.05×(1 +0.05)^6 = 2X(1.05^11 -1.05^6)/0.05
s2 = 14.8097486997X
The last 6 payments will have a future value at the end of the loan period of ...
s3 = 4X((1 +0.05)^6 -1)/0.05 = 27.20765125X
So, the total future value of the series of payments is ...
payment value = 7.3717764259X +14.8097486997X +27.20765125X
= 49.3891763756X
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The future value of the loan amount after 15 years is ...
loan value = 60,000(1 +0.05)^15 = 124,735.69
In order for these amounts to be the same, we must have ...
49.3891763756X = 124,735.69
X = 124,735.69/49.3891763756 = 2,525.57
__
At this point, it is convenient to use a spreadsheet to find the interest and principal portions of each of the loan payments. (We find the interest charge to be greater than the payment amount for the first 4 payments. So, the loan balance is increasing during those years.)
In the attached, we have shown the interest on the beginning balance, and the principal that changes the beginning balance to the ending balance after each payment. (That is, the interest portion of the payment is on the row above the payment number.)
The spreadsheet tells us ...
A) the principal repaid in the 8th payment is RM2,256.09
B) the interest paid in the 12th payment is RM1,791.10
_____
<em>Additional comment</em>
The spreadsheet "goal seek" function could be used to find the payment amount that makes the loan balance zero at the end of the term.
We have used rounding to sen (RM0.01) in the calculation of interest payments. The effect of that is that the "goal seek" solution is a payment value of 2525.56707 instead of the 2525.56734 that we calculated above. The value rounded to RM0.01 is the same in each case: 2525.57.
