How much money you must invest in order to withdraw $2000 annually for 10 years if the interest rate is 9%? can you please show me the formula and solution to this problem..thanks...
2007-07-24 03:48:47 · 6 answers · asked by Dariel J 2 in Science & Mathematics ➔ Mathematics
The answer is $12,835.31.
Proof
Year---9% Int----W/draw-----Balance
0-----------------------------------12,835.31
1------1,155.18--2,000.00--11,990.49
2------1,079.14--2,000.00--11,069.63
3------996.27-----2,000.00--10,065.90
4------905.93-----2,000.00----8,971.83
5------807.46-----2,000.00----7,779.29
6------700.14-----2,000.00----6,479.43
7------583.15-----2,000.00----5,062.58
8------455.63-----2,000.00----3,518.21
9------316.64-----2,000.00----1,834.85
10----165.14-----2,000.00-----(0.01)
2007-07-27 22:58:37 · answer #1 · answered by Jun Agruda 7 · 2⤊ 0⤋
Basically - you want to draw out $2000 dollars per year and but still have enough left in to make sure you can make enough interest to get another $2000 out within one year.
You need to earn $2000 at an interest rate of 9%
So your initial investment must be atleast $22,222.
This is because 9% of your investment = $2000
divide $2000 by 9 to find out what 1% is, then multiply it by 100 to find out your initial investment.
P.s. if you find a high street bank offering 9% tell me :-)
2007-07-24 03:58:16 · answer #2 · answered by Chris A 2 · 0⤊ 0⤋
interest= investment* annual rate * years
2000=x*9/100*10
20000/9
2007-07-24 03:59:07 · answer #3 · answered by Anonymous · 0⤊ 0⤋
the Equation for interest compounded continualy (the way most banks compound interest) is P=P(not)*e^(rt) which means that
2000t=(P(not)*e^(.09)(t)
20000=(P(not))*e^.9
20000/e^.9=P(not)
P(not)=8131.393194811982
2007-07-24 04:06:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
20000 = P(1+.09)^10
P = 20,000/(1.09)^10 = $8,448.22
2007-07-24 03:56:53 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋
a little over 20,000
2007-07-24 03:56:27 · answer #6 · answered by Meh 3 · 0⤊ 0⤋