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Can someone help me with this problem?
"at the end of each yr for the next 10 yrs you will receive $50. Initial investment is $320. what is the rate of return expected?"
I started by by looking at it this way
320 = 50 * (1-(1/(1+r)^10))/r
but impossible to solve...
p.s. i'm not allowed to use a financial calculator
do i just estimate it ..that will take forever and i'm not sure if there will be tables on the exam
Any suggestions?
Thanks.
If at the end of each year you will receive only $50, then the investment is in simple interest only. It is not compounding.Quote:
Originally Posted by reneebu
A = P + r*P*t
A = P +(r*P)t
A = P(1 +rt)
where
A = final amount after t years.
P = principal or initial amount.
r = interest rate per annum, or yearly rate of return.
t = time in years
The "r*P" in your question is $50.
So,
r*P = 50
r(320) = 50
r = 50/320 = 0.15625
r = 15.625 percent per annum ---------------answer.
What you have looks OK to me.Quote:
Originally Posted by reneebu
The trial and error should find the answer (approximatly fairly quickly)
We know the interest rate is less than 50/320 ~= 0.156, so for a first
guess take r=0.1, and plug it into:
A=50 * (1-(1/(1+r)^10))/r
to get A=307.2, so we are on the right track.
We also try r=0.11, and get A=294.5, so we are going the wrong way.
So now try r=0.09, and we get A=320.88, which is close.
(as an alternative to what is described below, at this point you could draw
a graph of A against r using these three points and estimate the r that
will give A=320 from that)
At this point we can accept r=0.09, or do a little linear interpolation
(or guess the next correction). Changing r from 0.1 to 0.09 chenges A
from ~307 to ~320, or a change or ~13, we want less change (about 1/15th
of this change) so we try 0.09 + 0.01/15=0.09067, which gives A=319.9.
So we find the intrest rate ~0.0907, or 9.1% after rounding to the nearest
tenth of a percent.
RonL