Noha is investing $2517 in an account compounded monthly. She wants to have $3000 in 3 years for a trip. What interest rate, compounded monthly,does she need?To the nearest hundredth of a precent

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Noha is investing $2517 in an account. Let the rate at which the amount is compounded monthly so that it is equal to $3000 after 3 years be r.

If the initial amount is P and the rate of interest is r, the amount after t billing cycles is equal to `P*(1 + r)^t` . Substituting the values given in the question:

`3000 = 2517*(1 + r)^36`

=> `(1 +r)^36 = 3000/2517`

Take the log of both the sides

=> `log(1 + r) = (log(3000/2517))/36`

=> `log(1 +r) ~~ 2.1177*10^-3`

Converting the logarithm to exponential function

`1 + r ~~ 1.004888`

=> `r ~~ 0.00488`

Converting this to an annualized rate of interest gives `0.00488*12*100 ~~ 5.87%`

The annual rate of interest required is 5.87%

mathlete13's profile pic

mathlete13 | High School Teacher | (Level 2) Adjunct Educator

Posted on

Begin with the compound interest formula, A = P(1 + r/n)^nt, where A is the amount in the account after some time period of investment, P is the amount invested, r is the interest rate, n is the number of times per year interest is earned, and t is the length of time of the investment. In this problem, A = $3000, P = $2517, n = 12, and t = 3, which leaves r for us to find. Substitute in the known values and solve for r:

3000 = 2517(1 + r/12)^(12*3)

3000/2517 = (1 + r/12)^36

(3000/2517)^(1/36) = 1 + r/12

(3000/2517)^(1/36) - 1 = r/12

[(3000/2517)^(1/36) - 1]*12 = r

To the nearest hundredth of a percent, we find r to be approximately 5.87%.

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