Question

1) How much will you have accumulated over a period of 30 years if, in an IRA which has a 10% interest rate compounded quarterly, you annually invest:

a. $1 b. $4000 c. $10,000 d. Part (a) is called the effective yield of an account. How could Part (a) be used to determine Parts (b) and (c)? (Your answer should be in complete sentences free of grammar, spelling, and punctuation mistakes.)

2) How much will you have accumulated, if you annually invest $1,500 into an IRA at 8% interest compounded monthly for: a. 5 year b. 20 years c. 40 years d. How long will it take to earn your first million dollars? Your answer should be exact rounded within 2 decimal places. Please use logarithms to solve.

Answer 1

1). The formula for compounding of interest is:
A = P (1 + r/n)^nt , where A is the future value of the investment, including interest, P is the principal/ initial amount invested, r is the annual interest rate in decimals, n is the number of times that interest is compounded in a year and t is the number of years the money is invested. Here, r = 0.10, and n = 4.

a. Here, P = $ 1 so that A = 1(1+0.10/4)^4*30 = (1.0025)¹²⁰ = $ 1.349353547 = $ 1.35 ( on rounding off to the nearest cent).

b. If P = $ 4000, then A = 4000*(1+0.10/4)^4*30 = 4000(1.0025)¹²⁰ = $ 5397.41( on rounding off to the nearest cent).

c. If P = $ 10000, then A = 10000*(1+0.10/4)^4*30 = 10000(1.0025)¹²⁰ = $ 13493.54( on rounding off to the nearest cent).

d. If we do not round off the effective yield in part(a), and instead, round off after the final calculation, then the result in part a has to be simply multiplied by the principal to get the maturity value of the investment.

2).The formula for compounding of interest is:
A = P (1 + r/n)^nt, where A is the future value of the investment, including interest, P is the principal/ initial amount invested, r is the annual interest rate in decimals, n is the number of times that interest is compounded in a year and t is the number of years the money is invested. Here, P = $ 1500, r = 0.08,t = 30 and n = 12.

a. Here, t = 5 so that A = 1500(1+0.08/12)⁶⁰ = 1500*1.489845709 = 2234.77 ( on rounding off to the nearest cent).

b. Here, t = 20 so that A = 1500(1+0.08/12)²⁴⁰ = 1500*4.926802775 = $ 7390.20( on rounding off to the nearest cent).

c. Here, t = 40 so that A = 1500(1+0.08/12)⁴⁸⁰ = 1500*24.27338558 = $ 36410.08( on rounding off to the nearest cent).

d. Let the amount increase to $ 1000000 in t years. Then 1000000 = 1500(1+0.08/12)^12t or, (12.08/12)^12t = 1000000/1500 = 2000/3. Now, on taking log of both the sides, we get log (2000/3) = 12t(log 12.08/12) or, t=(log2000-log3)/12(log12.08-log12)=(3.301029996-0.477121254)/12*(1.082066934-1.079181246)= 2.823908742/12*0.00288568829 =2.823908742/0.034628259 = 81.54925554 = 81.55 years ( on rounding off to 2 decimal places).