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Masteriza [31]
3 years ago
6

Suppose you accumulated $500,000, perhaps from many years of saving. You put the money in a savings plan earning 6% compounded m

onthly. If you want to withdraw $4,000 at the beginning of each month, how long before the savings plan is exhausted?
Mathematics
1 answer:
Brut [27]3 years ago
4 0
The amortization formua I'm familiar with assumes payments are made at the end of the period, so we'll use it for the part after the first payment has already been made.
.. A = 4,000
.. P = 500,000 -4000 = 496,000
.. i = 0.06
.. n = 12
.. t = to be determined
And the formula is
.. A = Pi/(n(1 -(1 +i/n)^(-nt))) . . . . . amortization formula with payments at the end of the period
.. 1 -(1 +i/n)^(-nt) = Pi/(An) . . . . . . rearrange to get "t" factor in numerator
.. 1 -Pi/(An) = (1 +i/n)^(-nt) . . . . . . get "t" factor by itself
.. log(1 -Pi/(An)) = -nt*log(1 +i/n) . . . . use logarithms to make the exponential equation into a linear equation
.. log(1 -Pi/(An))/(-n*log(1 +i/n)) = t . . . . divide by the coefficient of t
.. t = 16.1667 . . . . . years (after the first monthly withdrawal)

The plan will support withdrawals for 16 years and 3 months (195 payments).
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The system of equations have no solution.

Explanation:

The equations are 4r+\frac{1}{2}s=12 and 2r+\frac{1}{4} s=8

To find the two numbers, let us solve the equations using substitution method.

From the equation 2r+\frac{1}{4} s=8, let us determine the value of r,

$\begin{aligned} 2 r+\frac{1}{4} s &=8 \\ 2 r &=8-\frac{1}{4} s \\ r &=\frac{1}{2}\left(8-\frac{1}{4} s\right) \\ r &=4-\frac{1}{8} s \end{aligned}$

Let us substitute the value of r in 4r+\frac{1}{2}s=12, we get,

$\begin{aligned} 4\left(4-\frac{1}{8} s\right)+\frac{1}{2} s &=12 \\ 16-\frac{1}{2} s+\frac{1}{2} s &=12 \\ 16 &=12 \end{aligned}$

which is not possible.

This means that the system has no solution.

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