Re: Spring 1998 exam Question 5

stephen.couzens@milliman.com
Thu, 24 Sep 1998 13:20:58 -0800

The value of the series of deposits at time t=3D6 is $2,731=2E39=2E

The value of the three $1,500 payments at time t=3D6 is $3,936=2E47=2E

Therefore the value of the -Q and -2Q payments at time t=3D6 is=20=
$2,731=2E39-$3,936=2E47=3D-$1,205=2E08=2E
Or, $1,205=2E08=3DQ(v^2)+2Q(v^3)=2E With a little algebra,=20=
Q=3D$1,205=2E08*(1=2E07)^2/(1+2/1=2E07)=3D$480=2E87=2E

_______________________
The question reads:

Leo will make an initial deposit into an account of $1000 at time t=3D0,=20=
followed by five
annual deposits of $200 at times t =3D 1, 2, 3, 4, 5=2E Leo will receive paymen=
ts=20=
from the
account at times t =3D 7, 8, 9, starting at $1500 and decreasing by $Q per year=2E=
=20=
The
balance in the account after the last payment is $0=2E

You may assume this account earns an effective annual interest rate of 7%=2E

Determine Q=2E

A=2E Less than $450
B=2E At least $450, but less than $490
C=2E At least $490, but less than $530
D=2E At least $530, but less than $570
E=2E $570 or more

I keep getting an answer of Q =3D $621=2E82 but the answer is B=2E

Can someone help me please?!?!?

Thanks,
Erica Szeto