Expert: Sherman D. - 2/27/2005

Question
complete the table for each question to determine the balance for A and P dollars investd at a rate r for t years and compund n times per year.   
_______________________________________
n | 1 | 2 | 4 | 12 | 365 | continuous |      
---------------------------------------
A |   |   |   |    |     |            |
_______________________________________

44. P = $1000, r = 6%, t= 10 years

46. p = %1000, r = 6%, t = 40 years

Complete the table to determine the balance A for $12,000 invested at rate r for t years, compound continuosly.

______________________________
| t | 10 | 20 | 30 | 40 | 50 |
------------------------------
| A |    |    |    |    |    |
______________________________

48. r= 6%
50. r= 7.5%

52. A deposit of $5000 is made in a trust fund that pays 7.5% interest, compounded continuously. It is speciied that the balance will be givento the college from which the donor graduated after the money has earned interest for 50 years. How much will the college recieve?

Answer
For n = 1,2,4,12,365 use I = P(1 + (r/n))^nt
For n = Continous, use I = Pe^rt

44.) P = $1000, r = 6%, t= 10 years
A = P(1 + (r/n))^nt

n = 1
A = 1000(1 + (.06/1))^(1 * 10)
A = 1000(1 + .06)^10
A = 1000(1.06)^10
A = $1,790.85

n = 2
A = 1000(1 + (.06/2))^(2 * 10)
A = 1000(1 + .03)^20
A = 1000(1.03)^20
A = $1,806.11

n = 4
A = 1000(1 + (.06/4))^(4 * 10)
A = 1000(1 + .015)^40
A = 1000(1.015)^40
A = $1,814.02

n = 12
A = 1000(1 + (.06/12))^(12 * 10)
A = 1000(1 + .005)^120
A = 1000(1.005)^120
A = $1,819.40

n = 365
A = 1000(1 + (.06/365))^(365 * 10)
A = 1000(1 + .000164383)^3650
A = 1000(1.000164383)^3650
A = $1,822.03

n = Continous
A = 1000e^(.06 * 10)
A = 1000e^(.6)
A = $1,822.12

$1,790.85, $1,806.11, $1,814.02, $1,819.40, $1,822.03, $1,822.12

-----------------------------------------------------------

46.) p = %1000, r = 6%, t = 40 years

n = 1
A = 1000(1 + (.06/1))^(1 * 40)
A = 1000(1 + .06)^40
A = 1000(1.06)^40
A = $10,285.72

n = 2
A = 1000(1 + (.06/2))^(2 * 40)
A = 1000(1 + .03)^80
A = 1000(1.03)^80
A = $10,640.89

n = 4
A = 1000(1 + (.06/4))^(4 * 40)
A = 1000(1 + .015)^160
A = 1000(1.015)^160
A = $10,828.46

n = 12
A = 1000(1 + (.06/12))^(12 * 40)
A = 1000(1 + .005)^480
A = 1000(1.005)^480
A = $10,957.45

n = 365
A = 1000(1 + (.06/365))^(40 * 365)
A = 1000(1 + .000164383)^14600
A = 1000(1.000164383)^14600
A = $11,021.00

n = Continous
A = 1000e^(.06 * 40)
A = 1000e^(2.4)
A = $11,023.18

$10,285.72, $10,640.89, $10,828.46, $10,957.45, $11,021.00, $11,023.18

-----------------------------------------------------------

P = 12000
t = 10, 20, 30, 40, 50

48.) r = 6%
A = Pe^rt

t = 10
A = 12000e^(.06 * 10)
A = 12000e^(.6)
A = $21,865.43

t = 20
A = 12000e^(.06 * 20)
A = 12000e^(1.2)
A = $39,841.40

t = 30
A = 12000e^(.06 * 30)
A = 12000e^(1.8)
A = $70,534.44

t = 40
A = 12000e^(.06 * 40)
A = 12000e^(2.4)
A = $132,278.12

t = 50
A = 12000e^(.06 * 50)
A = 12000e^(3.0)
A = $241,026.44,

$21,865.43, $39,841.40, $70,534.44, $132,278.12, $241,026.44

-----------------------------------------------------------

50.) r = 7.5%

t = 10
A = 12000e^(.075 * 10)
A = 12000e^(.75)
A = $25,404.00

t = 20
A = 12000e^(.075 * 20)
A = 12000e^(1.5)
A = $53,780.27

t = 30
A = 12000e^(.075 * 30)
A = 12000e^(2.25)
A = $113,852.83

t = 40
A = 12000e^(.075 * 40)
A = 12000e^(3)
A = $241,026.44

t = 50
A = 12000e^(.075 * 50)
A = 12000e^(3.75)
A = $510,252.98

$25,404.00, $53,780.27, $113,852.83, $241,026.44, $510,252.98

-----------------------------------------------------------

52.)
A = Pe^rt
A = 5000e^(.075 * 50)
A = 5000e^(3.75)
A = $212,605.41

Let me know if that isn't what they got. Also you can find a calcultor for Compound Interest and Continous Compound Interest at www.moneychimp.com/articles/finworks/fmfutval.htm