MATH 373 Test 2 Fall 2017

October 26, 2020 Copyright Jeffrey Beckley 2017, 2020

MATH 373

Test 2

Fall 2017 October 19, 2017

1. Kylie invests P at the beginning of each year into an account which earns an annual effective

interest rate of 7.2%.

Kylie’s balance at the end of 13 years is 100,000.

Determine P .

Solution:

13

13

13

100,000

100,000 100,0004571.91

(1.072) 1(1.072)

0.072

Ps

Ps


2. An annuity makes payments at the end of each year. The first payment is 10,000. The second

payment is 9980. The third payment is 9960. The payments continue until the last payment is

8000.

Using an annual effective interest rate of 8%, calculate the present value of this annuity.

Solution:

The hard part here is determining the number of payments. There are 101 payments. One way

to see this that there are there are 100 decreases (10000-8000)/20=100. But this does not include

the first

101

101 101

101

101

payment so there are 101 payments.

2010,000 101(1.08)

0.08

1 (1.08)12.49473849

0.08

121,834.33

PV a a

a

PV


3. You are given that 1000 12,676.16n

a and 1000 45,068.90n

s .

Both annuity values are calculated using the same interest rate.

Determine n .

Solution:

1 1

1000 12,676.16 12.67616 and 1000 45,068.90 45.06890

1 10.0567

12.67616 45.06890

If you do not remember the formula above, you still find using algebra.

a (1 ) 12.67

n n

n n n n

n

n n

ia s

a a s s

i i

i

i s

45.06890

616(1 ) 45.06890 (1 ) 3.55540636912.67616

(1 ) 1 3.55506369 1 2.5550636945.06890 0.0567

45.06890

(1.0567) 145.0689 (1.0567) (45.0689)(0.0567) 1 3.55540663

0.0567

ln(3.

n n

n

n

nn

i i

is i

i i

Then

n

55540663)

23ln(1.0567)


4. Giacomo Inc. borrows 60,000 which will be repaid with monthly payments of 500 plus a balloon

payment of B . The interest rate on the loan is 5.8% compounded monthly.

Freija has an identical loan except her loan calls for the final payment to be a drop payment

of D .

Determine .B D

Solution:

Use your calculator to find n.

60,000; / 5.8 /12 0.48333333; 500

179.916 round down to 179

2 1 1; 2 179; 456.0271744

456.03 500 956.03

( )(1.00483333333) 458.2

PV I Y PMT

CPT N

nd Amort P P Bal

Balloon Bal Q

Drop Bal

3

956.03 458.23 497.80B D


5. Megan invests 5000 in the Stanley Fund at the end of each year for 20 years. The Stanley Fund

pays an annual effective interest rate of 6%.

At the end of each year, any interest in the Stanley Fund is withdrawn and invested into the

Hahn Fund. The Hahn Fund pays an annual effective interest rate of 5%.

Determine the amount that Megan has at the end of 20 years in both funds combined.

Solution:

Megan will deposit 5000 at the end of each year for 20 years. Since the interest will be removed

at the end of each year, all that will be left in the Stanley Fund at the end of 20 years will be

Megan's deposits (20)(5000) 100,000.

The interest earned in the first year is 0 since the first contribution does not occur until the end of

the first year. The interest earned in the second year is (5000

)(0.06) 300. This is moved to the

Hahn Fund at the end of the second year. The interest earned in the third year is (2)(5000)(0.06)

600. This is moved to the Hahn Fund at the end of the third year.

19 0.05 19 0.05

This pattern continues until

(19)(5000)(0.06) 19(300) is moved to the Hahn Fund at the end of the 20th year.

The amount in the Hahn Fund at the end of 20 years is:

300300 19(1.0

0.05a a

19 19

19 1919 19

5) (1.05)

1 (1.05) 300 1 (1.05)300 19(1.05) (1.05) 78,395.72

0.05 0.05 0.05

100,000 78,395.72 178,395.72Total


6. Qian has a loan of 250,000 which has an annual effective interest rate of 5%. The loan will be

repaid with annual payments for 15 years.

The first payment is 10,000. The second payment is 20,000. The third payment is 30,000. The

fourth payment is 40,000. The fifth payment is 50,000. The payments at the end of the sixth

through the 15th year are a level amount of Q .

Determine the outstanding loan balance right after the 4th payment.

Solution:

4 3 2

4

The easy way to do this is the retrospective approach which is the accumulated value of

past payments.

(250,000)(1.05) (10,000)(1.05) (20,000)(1.05) (30,000)(1.05) 40,000

198,750.31

OLB


7. Qian has a loan of 250,000 which has an annual effective interest rate of 5%. The loan will be

repaid with annual payments for 15 years.

The first payment is 10,000. The second payment is 20,000. The third payment is 30,000. The

fourth payment is 40,000. The fifth payment is 50,000. The payments at the end of the sixth

through the 15th year are a level amount of Q .

Determine Q .

Solution:

There are numerous methods to do this. I will do it multiple ways since this was the hardest

question on the test.

One method is to use our point in time as time 0. Then the loan must equal the prese

1 2 3 4

nt value

of the payments.

250,000 (10,000)(1.05) (20,000)(1.05) (30,000)(1.05) (40,000)(1.05)

(

5 5

10

105

105

5

50,000)(1.05) (1.05)

1 (1.05) 250,000 125,663.93125,663.93 (1.05) 20,550.80

0.05 1 (1.05)(1.05)

0.05

Another method is to set the OLB based on retrospective calculation

Q a

Q Q

5

5 4 3

5

2

must be equal to

OLB based on prospective calculation.

Retrospectively

(250,000)(1.05) (10,000)(1.05) (20,000)(1.05)

(30,000)(1.05) 40,000(1.05) 50,000

OLB

5 4

5 10 10

158,687.83

(1.05) 50,000 158,687.83

158,687.83Prospectively 20,550.80

1 (1.05)

0.05

or OLB OLB

OLB Qa Q


8. Ben is the beneficiary of a trust fund. He (or his heirs) will receive a monthly perpetuity from

the trust fund.

The monthly payments are 1000 at the end of each month for the first 10 years. Thereafter the

payments are each 10 higher than the previous payment. In other words, Ben (or his heirs) will

receive 120 payments of 1000 followed by payments of 1010, 1020, 1030, etc.

Calculate the present value of the payments from the trust fund using an interest rate of 12%

compounded monthly.

Solution:

You have to split these up into two parts. There are numerous ways to do this. This solution will

be 120 payments of 1000. Then beginning with the 121 payment of 1010 it is an arithmetically

increasi

120

2120

(12)

120120

2

ng perpetuity.

1010 101000

0.12where the interest rate use is the monthly effective interest rate of 0.01

12 12

1 (1.01) 1010 101000 (1.01)

0.01 0.01 (0.01)

PV a vi i

i

PV

130,602.47


9. Evan was awarded a scholarship upon his entrance to Purdue. The scholarship makes quarterly

payments for four years at the beginning of each quarter. The first payment was 1200 . The

second payment was 1200(1.1) . The third payment was 21200(1.1) . The payments continue

to increase with each payment being 110% of the prior payment.

Evan takes each payment and invests the payment into an account that earns an annual

effective interest rate of 8%.

Determine the amount that Evan will have at the end of four years.

Solution:

16

This is an annuity with payments in a geometric sequence. The first payment is 1200 and we

want the value at the end of 16 periods so the value is 1200(1 ) . The value at the end of the

second paym

i

15

15 1

16 15

ent is 1200(1.1)(1 ) . The payments continue until the 16th payment is made at

the end of the 15 quarter and the value one quarter later is 1200(1.1) (1 ) .

1200(1 ) 1200(1.1)(1 ) ... 1200

i

i

AV i i

15 1

16 16 0

1

(1.1) (1 )

1200(1 ) 1200(1.1) (1 )

1 1 (1.1)(1 )

Since payments are quarterly, we need the quarterly effective interest rate. We are given the

annual effect

i

FirstTerm NextTermAfterLast i i

ratio i

(4)0.25

16 16 0

1

ive interest rate.

(1.08) 1 0.0194265474

1200(1.019426547) 1200(1.1) (1.019426547)49,107.77

1 (1.1)(1.019426547)

i

PV


10. Adam buys a new car for 30,000. Abbott Bank offers Adam the choice of two loans. Both loans

have an interest rate of 12% compounded monthly.

Loan #1 is a five year loan with 60 monthly payments. This is a standard loan with the first

payment at the end of the first month. The payment on this loan is Q .

Loan #2 is a five and one half year loan with 60 monthly payments. Under this loan, the

payments are deferred for six months which means that the first payment will be made at the

end of the 7th month. The payment on this loan is X .

Determine X Q .

Solution:(12)

(12)(12)

60 60

Payments are made monthly so we need the monthly effective interest rate of . We are12

given that 0.12 0.0112

Loan #1

30,00030,000 667.33

1 (1.01)

0.01

Or using your c

i

ii

Qa Q

6

60 606

alculator / 1; 30,000; 60; 667.33

Loan #2

30,00030,000 708.39

1 (1.01)(1.01)

0.01

708.39 667.33 41.06

I Y PV N CPT PMT

Xv a X

X Q


11. Tom’s parents are paying him an annuity for the next 10 years. The annuity makes quarterly

payments at the start of each quarter. The first payment is 1000. The second payment is 1500.

The third payment is 2000. Each payment continues to increase in the same pattern with each

payment being 500 greater than the prior payment.

Calculate the present value of Tom’s payments at an interest rate of 10% compounded

quarterly.

Solution:

40

40 40

40 4040

5001000 40(1 ) (1 )

0.025

1 (1.025) 500 1 (1.025)1000 40(1.025) (1.025)

0.025 0.025 0.025

234,944.12

We need the quarterly effective interes

PV a a i i

(4)

t rate since payments are quarterly.

0.100.025

4 4

We multiply by (1+i) at the end because the payments are at the beginning

of the quarter.

i


12. Mindy has won the lottery! She has the option of the following two payouts:

a. A perpetuity due with monthly payments of 50,000.

b. An annuity due for 35 years with annual payments of P .

The present value of the payments under either option calculated using an annual effective

interest rate of i is 10 million which is 10,000,000.

Determine the payment P under Option b.

Solution:

From Option a, we can get the interest rate:

1 50,00010,000,000 (50,000) 1 0.005025126

9,950,000

This is the mothhly effective rate since payments are monthly. For Option b, we need the annua

ii

12

35

35

l

effective rate since the payments are annual. The annual effective interest rate is:

(1.005025126) 1 0.061996372

1 (1.061996372)10,000,000 (1.061996372) 664,74

0.061996372

i

PV Pa P P

6.82

MATH 373 Test 2 Fall 2017

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