How much money will I need to have at retirement so I can withdraw $60,000 year for 20 years from an account earning 8% compounded annually? a) How much do you need in your account at the beginning b)How much total money will you pull out of the account? c) How much of that money is interest?

a) $ 589,088.84

b) $ 1,200,000

c) $ 610,911.16

Step-by-step explanation:

We have that the time is 20 years and a rate of 8%, that is to say 0.08

We solve each point:

a) How much do you need in your account at the beginning, we can calculate it using the present value:

PV = 60000 * (1 - (1 + 0.08) ^ - 20) /0.08

PV = 589088.84

In other words, you need in your account at the beginning $ 589,088.84

b) How much total money will you pull out of the account?

Total money would be:

20 * 60000 = 1200000

In other words, the total money will be $ 1,200,000

c) How much of that money is interest?

It is the subtraction of the total money and the one that you need in your account at the beginning, that is:

1200000 - 589088.84 = 610911.16

In other words, $ 610,911.16 would be in interest

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.