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1997 Exam, Question 3

 

Key Concept:             The reduction in the total interest paid is equal to the difference between the payments that would have been made after 1/1/97 (had the loan not been refinanced) and the payments actually made.  Note that in either case the principal repaid must be the same, so the difference must be interest.

 

            Initial payment = 100,000/ = 877.57

            Outstanding Balance on 1/1/97 = 877.57 = 96,574

            Revised payment (effective 1/1/97) = 96,574/ = 979.52

            Total payments after 1/1/97 (original) = (877.57)(300 payments) = 263,271

            Total payments after 1/1/97 (revised) = (979.52)(180 payments) = 176,314

            Reduction in total interest = 263,271 - 176,314 = 86,957

 

Answer is C.

 

 

1997 Exam, Question 11

 

Key Concept:             The increases in each deposit occur annually, but the payments occur monthly.  Determine an annual deposit equivalent to the monthly deposits so that the deposits and the increases occur over the same time period.

 

Step I:   Determine the annual payment made on 1/1/81 that would be equivalent to the 1981 monthly deposits.

                   

            1/1/81 Payment = 25 = 281.38

 

Note that the monthly effective rate of interest is 1% since the annual rate of interest compounded monthly is 12%.

 

Step II:  Calculate the value of the savings account as of 1/1/99.

 

The annual effective rate of interest is:

 

            i = 1.0112 - 1 = .126825

 

            Accumulated value = (281.38)(1.126825)18 + [(281.38)(1.12)](1.126825)17

                                    +  [(281.38)(1.12)2](1.126825)16 + ... +  [(281.38)(1.12)17](1.126825)

                                    = (281.38)(1.126825)18[1 + (1.12/1.126825) + ... + (1.12/1.126825)17]

                                    = (281.38)(1.126825)18  (where j = 1.126825/1.12 - 1 = .0060938)

                                    = 41,283

 

Answer is D.

 

1998 Exam, Question 13

 

Step I: Determine the probability of death and withdrawal.

 

Using standard approximations:

 

            q'(d) = q(d)/[1 - ½q(w)]

 

            .035 = q(d)/[1 - ½(5q(d))]

 

            q(d) = .032184

 

            q(w) = 5q(d) = .160920

 

Step II:  Determine the absolute rate of withdrawal.

 

            q'(w) = q(w)/[1 - ½q(d)] = .160920/[1 - ½(.032184)] = .163549

 

Answer is B.

 

 

Problem 24

 

Using first principles, the equation of value (where P is the annual payment) is:

 

            100,000 = P × [v(l[65]+1/l[65]) + v2(l[65]+2/l[65]) + v3(l65+3/l[65]) + v4(l66+3/l[65])]

                         = P × 3.197833

 

            P = 31,271

 

Answer is B.

 

 

2000 Exam, Question 7

 

The outstanding balance is equal to the difference between the accumulated loan amount and the accumulated payments.

 

2% of Smith's monthly 2000 salary is $60 ($3,000 × .02).  2% of the increase in Smith's monthly salary in 2001 is $3 ($3,000 × .05 × .02).

 

The outstanding balance as of 1/1/2002 is:

 

[5,000 × (1.01)24] - 60 - 3 = 4,692

                  

Answer is C.