Suppose $9,500 is invested on the 1st January of a certain year at 12% compounded annually and $800 is withdrawn at the end of each year. How much would remain after 12 years? Hint: find the total future values of the amount withdrawn at the end of each year.
2006-11-23 19:08:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
A withdrawal of $w equivalent to $1.12w next year.
This $w is equivalent to w.(1.12)^y after y years.
w remains withdrawn for 11 years,
another w amount remains withdrawn for 10 years,
and so on...
Total effective withdrawn amount at the end of 12 years,
w (1.12^11+ 1.12^10+ 1.12^9 + ..... + 1)
= w (1.12^12 -1)/(1.12 -1)
= 24.133 w
Total value of $9500 in 12 years
= P(1+r/100)^N
= 9500(1.12)^12
=9500*3.896
Net future value after 12 years
= 9500(1.12)^12 - 800*(1.12^12)/0.12
= 9500*3.896 - 800*24.133
= $17,705.4
2006-11-23 20:24:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
So, every year we have to multiply by 1.12 (100% + 12%), and then subtract 800.
Do that 12 times..
9500 - 9840 - 10220.8 - ... - 16522.56 - 17705.27.
Just type in 9500 in a calculator, then type "Ans * 1.12 - 800" and push equals 12 times (or whatever the same thing is for your calculator.
2006-11-24 03:13:12 · answer #2 · answered by stephen m 4 · 0⤊ 1⤋
enough to pay back your student loan!
2006-11-24 03:10:02 · answer #3 · answered by ...n... 3 · 0⤊ 0⤋
$14,480
2006-11-24 03:46:12 · answer #4 · answered by billy joe j 1 · 0⤊ 1⤋