1- Find the principal needed to get $1200 in 6 years at 12% compounded monthly.

Question

1- Find the principal needed to get $1200 in 6 years at 12% compounded monthly.

Principal = $

2- In 3 years Harry and Sally would like to have $20000 for a down payment on a house. How much should they deposit each month into an account paying 5% compounded monthly?

Answer = $

3- Find the amount after 7 years if $500 is invested at 10% compounded:

a) Annually
Amount = $
b) Semiannually
Amount= $
c) Quarterly
Amount= $
d) Monthly
Amount= $

a) The cost of the stereo.
Answer = $
Hint: Make sure you are using the MONTHLY interest rate, meaning 2.75 % is APR/12.

b) The total amount of interest paid.
Answer = $

Rate = %

Amount = $

Answer = $

yearly payment = $

At the end of 10 years he ceased the IRA payments, but continued to invest his accumulated amount at 12% compounded monthly for the next 3 years.
a) What was the value of his IRA at the end of 10 years?
Answer = $

Explanation / Answer

1.

1200 = P(1+0.01)^72

P = 586 $

2.

20000 = P(1+0.05/12)^36+PP(1+0.05/12)^35+P(1+0.05/12)^34.....P(1+0.05/12)^1

20000 = P(1+0.05/12)((1+0.05/12)^36 - 1)/(1+0.05/12 - 1)

P=513.94 $

3.

a.) A = 500(1+0.1)^7 = 974.4 $

b.) A = 500(1+0.05)^14 = 990 $

c.) A = 500(1+0.025)^28 = 998$

d.) A = 500(1+0.1/12)^84 = 1004 $

4.

cost of stereo + 500+30(1.0275)^36+30(1.0275)^35+30(1.0275)^34.......30(1.0275)=500+30(1.0275)(1.0275^36 - 1)/(1.0275-1)= 500+1937.5 = 2437.5 $

interest pain =2437.5 - 500-(30*36) = 857.5 $

5. 2P = P(1+r)^9

=> 9log(1+r) = log2

=> log(1+r) = log2/9

=> 1+r = 1.08

=> r = 8%

7. A = 1100(1+0.03/12)^120+1100(1+0.03/12)^119+1100(1+0.03/12)^118.....1100(1+0.03/12)^1=1100(1+0.03/12)((1+0.03/12)^120-1)/(1+0.03/12-1) = 154099.85 $

8. 29000 = P(1+0.09)^16+P(1+0.09)^15+P(1+0.09)^14+....P(1+0.09)^1=P(1+0.09)((1+0.09)^16-1)/(1+0.09-1)

P = 806 $

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