Respuesta :
Answer:
The amount at the end of 30 years is $174,952.
Explanation:
In this problem we first need to determine the future value after making A = $100 investment for n = 10 years at r = 11% per year compounded monthly.
Then we need to compute the compound interest on this future value for 20 years at 11% interest compounded annually.
The future value formula is:
[tex]FV=A\times [\frac{(1+r)^{n}-1}{r}][/tex]
The amount is compounded monthly.
The rate of interest per month is:
[tex]r=\frac{11}{12}\%= 0.9167\%[/tex]
The number of periods is: n = 10 × 12 = 120 months.
Determine the future value as follows:
[tex]FV=100\times [\frac{(1+0.009167)^{120}-1}{0.009167}]=21700[/tex]
Thus, the amount at the end of 10 years is $21700.
Now this amount is kept the account for t = 20 years and earns an interest at the rate of 11% compounded annually.
Amount at the end of 30 years = [tex]FV(1+r)^{t}[/tex]
[tex]=21700\times(1+0.11)^{20}\\=174952[/tex]
Thus, the amount at the end of 30 years is $174,952.
Answer:
$174,950.7
Explanation:
Given:
Monthly payment(C) = $100
Rate (I) = 11% yearly = 11/12 = 0.916667% monthly = 0.00916667
number of annuity = 10 x 12 = 120
Calculation:
[tex]Future\ investment = C[\frac{(1+I)^N-1}{I} ]\\=100[\frac{\ (1+0.00916667)^{120}-1}{0.00916667} ]\\=100[\frac{(1.00916667)^{120}-1}{0.00916667} ]\\=100[\frac{2.98915079-1}{0.00916667} ]\\=100[\frac{1.98915079\\}{0.00916667} ]\\=100[216.998189]\\=21699.8189[/tex]
Reinvest $21699.8189, when n=20 r= 11%
[tex]A = P(1+r)^n\\=21699.8189(1+.11)^{20\\}\\=21699.8189(1.11)^{20}\\=21699.8189(8.06231154)\\=174,950.7 \\[/tex]
Total Amount he get is $174,950.7
