Respuesta :
Answer:
[tex]n = 1.42[/tex]
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]I = P*r*t[/tex]
In which I is the amount earned with interest, P is the money deposited(the principal), t is the time in years and r is the annual interest rate, as a decimal.
At the end of the period, the total amount is:
[tex]A = P + I[/tex]
Joe deposits 10 today and another 30 in five years into a fund paying simple interest of 11% per year.
So, he is going to have two interests.
For the first one: He deposits 10, so [tex]P = 10[/tex]. The period is 10 years, so [tex]t = 10[/tex]. The annual rate is 11%, so [tex]r = 0.11[/tex]
His first interest is:
[tex]I_{1} = P*r*t = 10*0.11*10 = 11[/tex]
From the first interest, Joe's amount is
[tex]A_{1} = P + I_{1} = 10 + 11 = 21[/tex]
The second one: He deposits 30, so [tex]P = 30[/tex]. The period goes from the 5th to the 10th years, so [tex]t = 10 - 5 = 5[/tex]. The annual rate is 11%, so [tex]r = 0.11[/tex].
His second interest is:
[tex]I_{2} = P*r*t = 30*0.11*5 = 16.5[/tex]
From the second interest, his amount is:
[tex]A_{2} = P + I_{2} = 30 + 16.5 = 46.5[/tex]
Joe's total amount is:
[tex]A = A_{1} + A_{2} = 21 + 46.5 = 67.5[/tex]
Now for Tina
Tina will make the same two deposits, but the 10 will be deposited n years from today and the 30 will be deposited 2n years from today. Tina’s deposits earn an annual effective rate of 9.15%. At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits.
Tina has two deposits, each one with an amount, [tex]A_{1}[/tex] for the first deposit and [tex]A_{2}[/tex] for the second deposit. Her accumulated amount is equal to Joe's. Joe is 67.5. So:
[tex]A_{1} + A_{2} = 67.5[/tex]
[tex]A_{1}[/tex] is the principal of 10 plus the interest. So [tex]A_{1} = 10 + I_{1}[/tex]
[tex]A_{2}[/tex] is the principal of 30 plus the interest. So [tex]A_{2} = 30 + I_{2}[/tex]
[tex]A_{1} + A_{2} = 67.5[/tex]
[tex]10 + I_{1} + 30 + I_{2} = 67.5[/tex]
[tex]I_{1} + I_{2} = 27.5[/tex]
[tex]I_{1}[/tex] is the interest earned from her deposit of 10. This means that [tex]P = 10[/tex]. The interest rate is 9.15%. So [tex]r = 0.0915[/tex]. The deposit will happen n years from today and will end in 10 years. So [tex]t = 10 - n[/tex]. So
[tex]I_{1} = P*r*t = 10*0.0915(10 - n) = 0.915(10 - n) = 9.15 - 0.915n[/tex]
[tex]I_{1}[/tex] is the interest earned from her deposit of 30. This means that [tex]P = 30[/tex]. The interest rate is 9.15%. So [tex]r = 0.0915[/tex]. The deposit will happen 2n years from today and will end in 10 years. So [tex]t = 10 - 2n[/tex]. So
[tex]I_{2} = P*r*t = 30*0.0915(10 - 2n) = 2.745(10 - 2n) = 27.45 - 5.49n[/tex]
We have that:
[tex]I_{1} + I_{2} = 27.5[/tex]
[tex]9.15 - 0.915n + 27.45 - 5.49n = 27.5[/tex]
[tex]-6.405n = -9.10[/tex] *(-1)
[tex]6.405n = 9.10[/tex]
[tex]n = \frac{9.10}{6.405}[/tex]
[tex]n = 1.42[/tex]
