Actuarial Outpost Some Sample Problems for y'all
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 Financial Mathematics Old FM Forum

#1
10-25-2006, 10:29 AM
 MyKenk Note Contributor CAS AAA Join Date: Nov 2005 Posts: 8,596
Some Sample Problems for y'all

In case you are out of problems ():

1-> Deposits are made into a fund at the end of each year starting one year from today. The first payment is \$10,000 and payments increase by \$500 per year in the second through the tenth year. All subsequent payments increase by 3.5% over the prior payment. If i = .07, find the value of the fund at the end of 20 years

a. 563,960
b. 564,010
c. 564,060
d. 564,110
e. 564,160

2-> Let A equal the accumulated value on December 31, 2006 of \$500 invested at the end of each month during 2006 @ i(4) = .08

Let B equal the present value on January 1, 2006 of A at d(2) = .06

Let C equal the accumulated value on December 31, 2006 of \$1,500 invested at the end of each quarter during 2006 at d(12) = .06

Let D equal the present value on January 1, 2006 of C at a nominal annual interest rate of j, convertible once every two years.

If B=D, find j.

a. .0481
b. .0484
c. .0487
d. .0490
e. .0493
Spoiler:

i(4) = .08 => i(12)/12 = .00662271

A=500*a-angle-12@.00662271 = 6223.4467
B=6223.4467(1-.06/2)^2 = 5855.41

d(12) = .06 => i(4)/4 = .015151259

C = 1500*a-angle-4@.015151259 = 6137.74
D = 5855.41 = 6137.74 *(1+2j)^-.5 => j = .0493794 => Answer E.

3-> A Loan is to be repaid with level annual payments at the end of each year for 20 years, with the first payment one year from today.

The principal portion of the 13th payment is \$54.40. The principal portion of the 18th payment is \$70.09

Find the total amount of interest paid over the term of the loan

a. 632
b. 634
c. 636
d. 638
e. 640
Spoiler:
We are given P(13) and P(18).
= 29.6115
From these, we can directly find i.

P(13)*(1+i)^5 = P(18) => i =5.19865

P(1) =54.4*v^12 = 29.6115

The sum of all the principal payments will give us the amount of the loan, denoted by L:

L = 29.6115[1+1.0519865+1.0519865^2+...+1.0519865^19] = 1000

Then, using the BA2+

N=20, I/Y = 5.19865, PV = 1000

<CPT> <PMT> = 81.60

<2nd><AMORT>

P1 = 1 <ENTER>
P2 = 20<ENTER>

Scroll down to INT = 631.993 => Ans A

4-> An annuity with 2n annual payments of 1 has it's first payment n+1 years from today. Its present value is 8.00407.

An annuity with 2n+1 payments of 1 has it's first payment n years from today. Its present value is 8.63279.

Find the annual effective rate of interest

a. .0475
b. .0480
c. .0485
d. .0490
e. .0495
Spoiler:

Annuity 1 has a PV of 8.00407
Annuity 2 has a PV of 8.64279

If you plot these on a timeline, you will see that the only difference between these two annuities is a payment of 1 at time n. Therefore:

8.64279 - 8.00407 = v^n = .62872

Then, since Annuity one gives you 8.00407 = a-angle-2n*v^n,

8.00407 = .62872*(1-.62872^2)/i

Solving one equation, one unknown for i gives you .047500117 => Answer A.

5-> Over a 3 year period, a series of deposits are made to a savings account. All deposits within a given year are equal in size and are made at the beginning of each relevant period. Deposits for each year total \$1,200. The following chart shows the frequency of deposits and the interest rate credited for each year:

Year..........Frequency of Deposits........Interest Rate Credited During Year
..1.............Semi-Annually...................d(12) = 6%
..2.............Quarterly.........................i(3)=8%
..3.............Every 2 months.................delta=7%

Find the balance of the account at the end of the 3rd year

a. 4050
b. 4055
c. 4060
d. 4065
e. 4070

6-> John is to receive an annuity payable at the end of each quarter for 10 years. The first payment, to be made on March 31st, 2006, is \$1,000. Each subsequent payment is 1% larger than the previous payment.

Jane is to receive an annuity payable at the end of each year for 15 years. The first payment, to be made on December 31st, 2006, is R. Each subsequent payment is \$25 less than the previous payment.

On January 1, 2008, the present values of the remaining payments to John and Jane are equal.

If i=.07, find R.

a. 3764
b. 3784
c. 3804
d. 3824
e. 3844

7-> A \$75,000 loan to be repaid over a 5 year period, may be repaid by Method A or Method B.

In Method A, level annual payments are made at the beginning of each year.

In method B, level semi-annual payments are made at the end of each 6-month period.

If d(4)=.076225, find the absolute difference in the payments made each year under the two methods.

a. 1024
b. 1026
c. 1028
d. 1030
e. 1032
Spoiler:
Clearly, in order to use the calculator for this problem, we'll need i, and the equivalent i(2)/2. we are given d(4), from that, we find that i=8%, and i(2)/2 = 3.923%.

Using the BA2+

Method A:
Set to BGN Mode
PV = 75000
N = 5
I/Y = 8
FV = 0
CPT PMT = 17392.81

Method B:
Set to END Mode
PV = 75000
N = 10
I/Y = 3.923
FV = 0
CPT PMT = 9211.41

|17392.81-9211.41*2|=1030.01 => Ans. D

8-> A \$1000, 8% 15-year bond is purchase to yield i=.09. Coupons are paid semi-annually.

With 10 years remaining to maturity, the bond is sold at a price which yields i=.0825 to the buyer.

Find the effective annual rate of return on the investment of the original bondholder.

a. .0961
b. .0967
c. .0972
d. .0978
e. .0984
Spoiler:
Given an annual effective rate of i=.09 => i(2)/2 = .0440306.

We have 1000*.08/2 = 40 dollar coupons, for 30 years. From this, we can find the price paid by the original bond holder.

N = 30
PMT = -40
FV = -1000
I = 4.40306
CPT PMT = 933.5906

Now, the bond is purchased 5 years (or 10 coupon payment periods) later. It will yield the new bondholder 8.25%. i=.0825 => i(2)/2 = 4.04326%

so, the price paid by the new buyer can be found:

N=20
PMT = -40
FV = -1000
I = 4.04326
CPT PV = 994.143

Now, we can calculate the actual yield for the original bond holder, taking the present value of what he recieved, and solving for the yield rate:

N=10
PMT = -40
FV = -994.143
PV = 933.5906
CPT I/Y = 4.8049

Since this is a semi-annual effective rate, add 1 and square it to get the 1 year effective rate of i=.0984 => answer E.

9-> A 20 year loan is to be repaid with monthly payments beginning one month from today. Just after the 43rd payment has been made, the present value of the remaining payments is B.

N is the number of the payment after which the present value of the remaining payments is less than B/2 for the first time.

If d(4) = .08, find N.

a. 172
b. 173
c. 174
d. 175
e. 176
Spoiler:
d(4)=.08 => i(12)/12 = .006756962

After the 43rd payment, there are 197 payments left.

So, B = a-angle-197 = 108.619 => B/2 = 54.5096

Using the BA2+ to solve 54.5096 = a-angle-n gives N = 67.93.

because at time 67.93, the present value of future payments is exactly B/2, and we need the answer on a payment date, we know that the answer is either 240 - 67 =173, or 240-68 = 172. So, we are down to A and B.

To ensure you get it right, use the TVM registers with N = 68 and N=67. With N=68, PV = 54.35339, with N=67, PV = 53.72. So, the first time that the PV of future payments drops below B/2 is with 67 payments remaining, or with payment number 240-67 = 173 = > Answer B

10-> A \$10,000 loan is to be repaid with 40 annual payments at i=.07. The first payment is to be made one year from today.

Let A equal the sum of the interest paid in the even numbered payments.

Let B equal the sum of the principal paid in the odd numbered payments.

Find A+B.

a. 14660
b. 14680
c. 14700
d. 14720
e. 14740
Spoiler:

We know that the payments are 750.09, from the BA2+ TVM registers.

From the amortization worksheet, or the P(t)=Rv^(n-t+1) formula, we know that the principal repaid in the first payment is 50.09.

Because the amount of prinicipal repaid increases geometrically with the interest rate, we know all the amounts of principal repaid in every period.

B is the sum of the principal repaid in the odd numbered payments, P1, P3, ..., P39

Using the fact that these amounts are in geometric progression,

B=50.09 + 50.09(1.07)^2 + ... + 50.09(1.07)^38 = 50.09(1+1.07^2+1.07^4+....+1.07^38) = 50.09*(1-1.07^40)/(1-1.07^2) = 4830.784

Now, the amounts of interest paid in each period are NOT in any sort of common progression, so we'll back into A.

First, find the sum of the principal repaid in the even periods, which are, like the odd payments, in geometric progression:

P2 = P1(1.07) = 53.5963

SUM = 53.5963 + 53.5963(1.07)^2 +....+53.5963(1.07)^38
=53.5963*(1-1.07^40)/(1-1.07^2) = 5168.9388

realize that the amount of interest paid in the even number periods is equal to the total amount paid in the even periods minus the amount of principal repaid in these periods.

A = 750.09*20 - 5168.9388 = 9832.8612

A+B = 14,663 => Answer A.

-----------------------------------
Spoiler:
C E A A C E D E B A

Good Luck!

Last edited by MyKenk; 10-29-2006 at 03:22 PM.. Reason: Added Solutions to some of the problems
#2
10-28-2006, 11:40 AM
 stbaugh13 Member SOA Join Date: Jul 2006 Location: Melbourne, Australia Posts: 136

How do you do #4 and #10. Also for #8 are the interest rates quoted nominal or effective. Thanks much.
#3
10-28-2006, 11:46 AM
 MyKenk Note Contributor CAS AAA Join Date: Nov 2005 Posts: 8,596

Honestly, i haven't gone through them yet, but I will by the end of the weekend. I'm guessing for number 8 that the rates are nominal, as that is generally what we see with bonds. (9% bond with semi-annual coupons has 2 coupons per year that are 4.5% of the Face Value each).

I'll let you know once I get through them if I figure out 4 & 10, I don't have solutions to these problems, just the answer key. They are old EA-1 questions, so if you're really desperate, you could look through the old EA-1 exams and try to find the questions, but I'm not even sure they release those.

#4
10-28-2006, 02:08 PM
 MyKenk Note Contributor CAS AAA Join Date: Nov 2005 Posts: 8,596

As I work these out, I am adding my solutions to the original post, in SPOILER tags for those who want to do it on their own.

Last edited by MyKenk; 10-28-2006 at 03:13 PM..
#5
10-29-2006, 07:51 AM
 stbaugh13 Member SOA Join Date: Jul 2006 Location: Melbourne, Australia Posts: 136

Hey Mykenk don't worry about #10 I worked that one out.
#6
10-29-2006, 08:43 AM
 MyKenk Note Contributor CAS AAA Join Date: Nov 2005 Posts: 8,596

already added number 10, glad you got it on your own. If you have a better solution, let me know, and I will post the most efficient solutions in the original post for future users
#7
10-29-2006, 12:54 PM
 YoungestEverFSA Member Join Date: Oct 2006 Posts: 178

Thanks for the questions. Should 6d be 3824?
#8
10-29-2006, 01:03 PM
 MyKenk Note Contributor CAS AAA Join Date: Nov 2005 Posts: 8,596

yes, it should, but it's wrong anyways I'll fix it
#9
05-21-2007, 12:51 AM
 GuineaPig Member Join Date: Sep 2006 Studying for ever Posts: 20,043

I thought we could bring back some practice questions for all of us spring takers.
__________________

Quote:
 Originally Posted by Guerilla poster Rickson is right
#10
05-21-2007, 01:51 PM
 ashleyw Member CAS Join Date: May 2007 Posts: 212

will we not be able to see the solution to #6?