Mathematics, 23.09.2019 20:10 igivehead666

Joe deposits 10 today and another 30 in five years into a fund paying simple interest of 11% per year. 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. calculate n

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Answer from: olamagdziak2007

n = 1.42

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

I = P*r*t

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:

A = P + I

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 P = 10 . The period is 10 years, so t = 10 . The annual rate is 11%, so r = 0.11

His first interest is:

$I_{1} = P*r*t = 10*0.11*10 = 11$

From the first interest, Joe's amount is

$A_{1} = P + I_{1} = 10 + 11 = 21$

The second one: He deposits 30, so P = 30 . The period goes from the 5th to the 10th years, so t = 10 - 5 = 5 . The annual rate is 11%, so r = 0.11 .

His second interest is:

$I_{2} = P*r*t = 30*0.11*5 = 16.5$

From the second interest, his amount is:

$A_{2} = P + I_{2} = 30 + 16.5 = 46.5$

Joe's total amount is:

$A = A_{1} + A_{2} = 21 + 46.5 = 67.5$

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, $A_{1}$ for the first deposit and $A_{2}$ for the second deposit. Her accumulated amount is equal to Joe's. Joe is 67.5. So:

$A_{1} + A_{2} = 67.5$

$A_{1}$ is the principal of 10 plus the interest. So $A_{1} = 10 + I_{1}$

$A_{2}$ is the principal of 30 plus the interest. So $A_{2} = 30 + I_{2}$

$A_{1} + A_{2} = 67.5$

$10 + I_{1} + 30 + I_{2} = 67.5$

$I_{1} + I_{2} = 27.5$

$I_{1}$ is the interest earned from her deposit of 10. This means that P = 10 . The interest rate is 9.15%. So r = 0.0915 . The deposit will happen n years from today and will end in 10 years. So t = 10 - n . So

$I_{1} = P*r*t = 10*0.0915(10 - n) = 0.915(10 - n) = 9.15 - 0.915n$

$I_{1}$ is the interest earned from her deposit of 30. This means that P = 30 . The interest rate is 9.15%. So r = 0.0915 . The deposit will happen 2n years from today and will end in 10 years. So t = 10 - 2n . So

$I_{2} = P*r*t = 30*0.0915(10 - 2n) = 2.745(10 - 2n) = 27.45 - 5.49n$

We have that:

$I_{1} + I_{2} = 27.5$

9.15 - 0.915n + 27.45 - 5.49n = 27.5

-6.405n = -9.10 *(-1)

6.405n = 9.10

$n = \frac{9.10}{6.405}$

n = 1.42

Answer from: Quest

step-by-step explanation:

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Answer from: Quest

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