ANSWERS TO EXAM SAMPLE QUESTIONS

1) Solution: D

Let N(C) denote the number of policyholders in classification C . Then N(Young ) Female ) Single) = N(Young ) Female) -- N(Young ) Female ) Married) = N(Young) -- N(Young ) Male) -- [N(Young ) Married) -- N(Young ) Married ) Male)] = 3000 -- 1320 -- (1400 -- 600) = 880.

4) Solution: A
Let X and Y be the monthly profits of Company I and Company II, respectively. We are given that the pdf of X is f . Let us also take g to be the pdf of Y and take F and G to be the distribution functions corresponding to f and g. Then G(y) = Pr[Y d y] = P[2X d y] = P[X d y/2] = F(y/2) and g(y) = G2(y) = d/dy F(y/2) = = F2(y/2) = = f(y/2) .

5) Solution: E
Observe that the bus driver collects 21x$50 = $1,050 for the 21 tickets he sells. However, he may be required to refund $100 to one passenger if all 21 ticket holders show up. Since passengers show up or do not show up independently of one another, the probability that all 21 passengers will show up is (10.02)21 = (0.98)21 = 0.65. Therefore, the tour operator's expected revenue is $1,050 -- ($100)(0.65) = $985.

Answers to sample questions for Exam 2/FM

1) Solution: A
The payment using the amortization method is $1,627.45. The periodic interest is 0.10($10,000) = $1,000. Thus, deposits into the sinking fund are $1,627.45-$1,000 = $627.45 .
Then, the amount in sinking fund at end of 10 years is $627.45 10|.14 s

Using BA II Plus calculator keystrokes: 2nd FV (to clear registers) 10 N, 14 I/Y, $627.45 PMT, CPT FV +/- $10,000 = yields $2,133.18 (Using BA 35 Solar keystrokes are AC/ON (to clear registers) 10 N 14 %i 627.45 PMT CPT FV +/-$10,000 =)

3) Solution: A
Value of initial perpetuity immediately after the fifth payment (or any other time) = 100 (1/i) = $100/.08 = $1,250. Exchange for 25-year annuity-immediate paying X at the end of the first year, with each subsequent payment increasing by 8 percent, implies $1,250 (value of the perpetuity) must = X (v + 1.08 v2 + 1.082 v3 + &..1.0824 v25) (value of 25-year annuity-immediate) = X (1.08-1 + 1.08 (1.08)-2 + 1.082 (1.08)-3 + 1.0824 (1.08)-25) (because the annual effective rate of interest is 8 percent) = X (1.08-1 +1.08-1 + & 1.08-1) = X [25(1.08-1)]. So, $1,250 (1.08) = 25 X or X = 54

4) Solution: A
Equate present values: 100 + 200 vn + 300 v2n = 600 v10vn=.76
100 + 152 + 173.28 = 425.28. Thus, v10 = 425.28/600 = 0.7088 => i = 3.5 percent

5) Solution: B
Chris' equation of value at end of year:(Value of contributions) .5 P(1+i) + 760 = P* + .5 P (1.06) (Value of returns), where P = price stock sold for. (*Proceeds received for stock sale at beginning of year are in non-interest bearing account per government regulations.) Thus, Chris' yield i = ((1.03) P - 760)/(.5 P).

Jose's equation of value at end of year:
(Value of contributions) .5 P(1+j) + 760 + 32 = P* + .5 P (1.06) (Value of returns), Thus, Jose's yield j = ((1.03) P - $792)/(.5 P).

It is given that Chris' yield i = twice Jose's yield j. So ((1.03) P - $760)/(.5 P) = 2 ((1.03) P - $792)/(.5 P) or P = $824/(1.03) = $800. Thus, Chris' yield i = ((1.03) 800 - 760)/400 = 64/400 = .16 or 16 percent.

Answers to sample questions from Exam M

1) Solution: A
B = {c (400-x) x <>
0 x > 400

100 = E (B) = c.400 = cE (x 400)
= c.400 - c.300 (1-300/300+400)
= c (400-300.4/7)
C = 100/228.6 = 0.44

2) Solution: C.
http://www.soa.org/ccm/cms-service/stream/asset/?asset_id=8576040 (page 24, question 27)

3) Solution: E
http://www.soa.org/ccm/cms-service/stream/asset/?asset_id=8576040 (page 48, question 61)

4) Solution:
http://www.soa.org/ccm/cms-service/stream/asset/?asset_id=8576040 (page 55, question 76)

5) Solution: B
http://www.soa.org/ccm/cms-service/stream/asset/?asset_id=8576040 (page 73, question 108)

Answers to Sample Questions from Exam 4/M

Answers may be found at http://www.soa.org/ccm/cms-service/stream/asset/?asset_id=10904097

1)Solution: B (page 2, question 3)
3)Solution: E (page 10, question 30)
4)Solution: A (page 21, question 61)
5)Solution: E (page 27, question 77)