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Jason is a new college student who has saved $10000 when he was a teenager and is invested at 12 percent in a financial assent that pays interest monthly. He expects to graduate 4 years from today.
a. How much can Jason withdraw every month while he is in college if the first withdrawal occurs today ? (Answer 260.73)
b. How much can Jason withdraw every month while he is in college if he waits until the end of this month to make the first withdrawal ? ( Answer 263.34)
 2 years ago
 2 years ago
Jason is a new college student who has saved $10000 when he was a teenager and is invested at 12 percent in a financial assent that pays interest monthly. He expects to graduate 4 years from today. a. How much can Jason withdraw every month while he is in college if the first withdrawal occurs today ? (Answer 260.73) b. How much can Jason withdraw every month while he is in college if he waits until the end of this month to make the first withdrawal ? ( Answer 263.34)
 2 years ago
 2 years ago

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amistre64Best ResponseYou've already chosen the best response.3
I cant tell if the information is saying: 12% annually is divided up into monthly installments (1+(.12/12)); or if it means that the account pays out 12% each month. (1+ .12). But if I see it right the equations go along the lines of: a. B{n} = [B{n1}  withdrawal] * (1 + interest) and b. B{n} = [B{n1} * (1 + interest)]  withdrawal
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
i got (b) worked out :) \begin{align} B_n&=B_{n1}(1.01)w\\ &=B_{n2}(1.01)^2w(1.01)w\\ &=B_{n3}(1.01)^3w(1.01)^2w(1.01)w\\ &\ or \ simply\\ B_{n}&=B_{n3}(1.01)^3w(1.01^2+1.01+1)\\ &=B_{n4}(1.01)^4w(1.01^3+1.01^2+1.01+1)\\ &=B_{nr}(1.01)^rw(1.01^{r1}+1.01^{r2}+...+1.01^2+1.01+1)\\ \end{align} \(B_{nr}=B_{0}\); when r=n \begin{align} B_{n} = B_{0}(1.01)^n  w(1.01^{n1}+1.01^{n2}+...+1.01^2+1.01+1)\\ \end{align} We can clean this up by suming up those \((1.01^{n1}...+1)\) parts like this. \begin{align} A&=\{\hspace{3.5em}1.01^{n1}+1.01^{n2}+...+1.01^2+1.01+1\}\\ (1.01)A&=\{1.01^{n}+1.01^{n1}+1.01^{n2+}...+1.01^2+1.01\hspace{2em}\}\\ \end{align} \begin{align} A&=\{\hspace{3.5em}1.01^{n1}+1.01^{n2}+...+1.01^2+1.01+1\}\\ (1.01)A&=\{1.01^{n}+1.01^{n1}+1.01^{n2}+...+1.01^2+1.01\hspace{2em}\}\\ &=\\ .01A&=1  1.01^n\\ &\\ A &= \frac{11.01^n}{.01} \end{align} \begin{align} B_{n} &= B_{0}(1.01)^n w(A)\\ B_nB_0(1.01)^n&=w(A)\\ \end{align} \[w=\frac{B_nB_0(1.01)^n}{A}\] \[w=\frac{B_{48}10000(1.01)^{48}}{A}\] Thats one part of it :)
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
We want B{48} to equal 0 which makes this equation simply, and I use that term loosely :) \[w = \frac{10000(1.01)^{48}(.01)}{11.01^{48}}=263.338...\]
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
And the other one I finally worked out to be: \[w = \frac{10000(1.01)^{48}(.01)}{1.01^{48}(.01)+1.01^{48}1.01}=260.73104....\] ended up with a negative in there somehow, butI think its pretty close :)
 2 years ago

aama100Best ResponseYou've already chosen the best response.0
Marvelous work! Thank you!
 2 years ago

aama100Best ResponseYou've already chosen the best response.0
so from where did you get 1.01 and .01 ?
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
the 1.01 is just the interest rate of 12% compounded monthly; 1 + .12/12 = 1.01 the ".01" comes in as a result of summing up the interest over any given amount of time. The equations start out as recursions and can be manipulated into functions with a great deal of effort :)
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
the equations are by hand so they dont follow any sort of given formula as is, but they should be close since I pretty much had to reinvent the wheel to solve them :)
 2 years ago
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