ISM Chap 10 2008 Finite - Stanoyevitchstanoyevitch.net/gateways/MAT105/SM_GFM_10e_Ch10.pdf35. 6 1...

10-1

Chapter 10

Exercises 10.1

1. a. .12

.0112

i = =

(12)(2) 24n = =

b. .08

.024

i = =

(4)(5) 20n = =

c. .10

.052

i = =

(2)(20) 40n = =

2. a. .06

.061

i = =

(1)(3) 3n = =

b. .06

.00512

i = =

(12)(4.5) 54n = =

c. .09

.02254

i = =

(4)(4.75) 15n = =

3. a. .06

.061

i = =

(1)(4) 4n = =

P = $500 F = $631.24

b. .06

.00512

i = =

(12)(10) 120n = =

P = $800 F = $1455.52

c. .04

.022

i = =

(2)(9.5) 19n = =

P = $6177.88 F = $9000

4. a. .045

.00086552

i = ≈

(52)(1) 52n = =

P = $7170.12 F = $7500

b. .06

.00512

i = =

(12)(30) 360n = =

P = $3000 F = $18,067.73

c. .08

.024

i = =

(4)(384) 1536n = =

P = $24 F = $389,112,397,500,000

5. 12 2

.061 ($1000) $1127.16

12

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

6. ( )2 5.04

2

1($10,000) $8203.48

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

7. ( )12 25.06

12

1($100,000) $22,396.57

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

8. 365 1

.0731 ($1000) $1075.72

365

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

9. 12 3

.061 ($6000) $7180.08

12F

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

$7180.08 $6000 $1180.08i F P= − = − =

10. 2 7

.041 ($2000) $2638.96

2F

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

$2638.96 $2000 $638.96i F P= − = − =

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

ISM: Finite Math Chapter 10: The Mathematics of Finance

10-2

11. 4 5

5

.031 ($1000) $1161.18

4B

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

4 4

4

.031 ($1000) $1126.99

4B

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

5 4 $1161.18 $1126.99 $34.19i B B= − = − =

12. 12 6

6

.0361 ($1000) $1240.70

12B

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

12 5

5

.0361 ($1000) $1196.89

12B

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

6 5 $1240.70 $1196.89 $43.81i B B= − = − =

13. ( )4 1.04

4

1($406.04) $10,000

1 1P ×

⎡ ⎤⎢ ⎥= =⎢ ⎥+ −⎣ ⎦

14. ( )12 3.06

12

1($196.68) $1000

1 1P ×

⎡ ⎤⎢ ⎥= =⎢ ⎥+ −⎣ ⎦

15. a. ( )12 2.06

12

1($4000) $3548.74

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

b. $4000 $3548.74 $451.26i F P= − = − =

c. Month Interest Balance

0 $3548.74

1 $17.74 (1.005)($3548.74) = $3566.48

2 $17.83 (1.005)2($3548.74) = $3584.31

3 $17.92 (1.005)3($3548.74) = $3602.23

16. a. ( )4 15.08

4

1($10,000) $3047.82

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

b.

$10,000 $3047.82 $6952.18

i F P= −= − =

c. Quarter Interest Balance

0 $3047.82

1 $60.96 (1.02)($3047.82) = $3108.78

2 $62.17 2(1.02) ($3047.82) $3170.95=

3 $63.42 3(1.02) ($3047.82) $3234.37=

17. 4 6.25

.041 ($10,000) $12,824.32

4

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

18. ( )12 4.06

12

1($10,000) $7870.98

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦



10-3

19. ( )4 3.04

4

1($10,000) $8874.49

1×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

20. 1 20

.061 ($100,000) $320,713.55

1

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

21. a. ($1000.00) $5.00,12

r = so r = .06 = 6%

b. 3(1.005) ($1000.00) $1015.08=

$1015.08 – $1010.03 = $5.05

c. 24(1.005) ($1000.00) $1127.16= 24 23[(1.005) – (1.005) ]($1000.00) $5.61=

22. a. ($10,000.00) $200.00,2

r = so r = .04 and

i = .02. 1.02(10,404.00) = $10,612.08 $10,612.08 – $10,404.00 = $208.08

b. 18(1.02) ($10,000.00) $14,282.46= 18 17[(1.02) – (1.02) ]($10,000) $280.05=

23. For P = $1000, 9

.061 ($1000) $1689.48.

1F

⎛ ⎞= + =⎜ ⎟⎝ ⎠

$1700 in 9 years is more profitable.

24. For P = $7000, 4 9

.041 ($7000) $10,015.38

4F

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

$7000 now is more profitable.

25. 52

eff

.6221 1 .8558

52r

⎛ ⎞= + − ≈⎜ ⎟⎝ ⎠

This interest rate is better than 85%.

26. 365

.0731 1 .0757,

365effr⎛ ⎞= + − ≈⎜ ⎟⎝ ⎠

which is a

7.57% annual increase. 8% compounded annually is better.

27. ($10,000) $100,4

r = so r = .04 and i = .01

12

1($10,000) $8874.49

(1.01)=

28. .044

1($2020) $2000

1=

+

29. a. r = .04 6 1

12 2n = =

P = $500 F = $510

b. r = .05 n = 2 P = $500 F = $550

30. a. r = .06 8 2

12 3n = =

P = $1000 A = $1040

b. r = .04 n = 5 P = $3000 A = $3600

31. (1 3 0.05)($1000) $1150F = + ⋅ =

32. (1 1.5 0.06)($2000) $2180F = + ⋅ =

33. 1

($3000) $2500(1 2 0.10)

P⎡ ⎤

= =⎢ ⎥+ ⋅⎣ ⎦

34. 1

($2000) $1562.50(1 4 0 .07)

P⎡ ⎤

= =⎢ ⎥+ ⋅⎣ ⎦

35. 6

1 ($980) $100012

r⎛ ⎞+ =⎜ ⎟⎝ ⎠

r ≈ .0408 = 4.08%

36. (1 0.07)($1000) $1210,n+ ⋅ = n = 3 years

37. (1 0.06)($500) $800,n+ ⋅ = n = 10 years



10-4

38. ( )1 5 ($1000) $1200r+ =

r = 0.04 = 4%

39. (1 0.05) 2 ,n P P+ ⋅ = n = 20 years

40. F = (1 + nr)P; –1 –F

P F Pr

n nP= =

41. F = (1 + nr)P; 1

FP

nr=

+

42. F = (1 + nr)P; –1 –F

P F Pn

r rP= =

43. (a)

44. (a)

45. 4 1

.041 ($100) $104.06

4

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

$4.06.0406 4.06%

$100= =

46. 12 1

.061 ($100) $106.17

12

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

$6.17.0617 6.17%

$100= =

47. 2

eff

.041 1 .0404

2r

⎛ ⎞= + − =⎜ ⎟⎝ ⎠; 4.04%

48. 52

eff

.08451 1 .0881

52r

⎛ ⎞= + − ≈⎜ ⎟⎝ ⎠; 8.81%

49. 12

eff

.0441 1 .0449

12r

⎛ ⎞= + − ≈⎜ ⎟⎝ ⎠; 4.49%

50. 4

eff

.07951 1 .0819

4r

⎛ ⎞= + − ≈⎜ ⎟⎝ ⎠; 8.19%

51. 4

1 1 .04064

r⎛ ⎞+ − =⎜ ⎟⎝ ⎠; r ≈ .04; 4%

52. 12

1 1 .0312

r⎛ ⎞+ − =⎜ ⎟⎝ ⎠; r ≈ .0296; 2.96%

53. Since we start with $1000 and this amount doubles every six years, we have $2000 at the end of six years, $4000 at the end of twelve years and $8000 at the end of eighteen years. So, it will take 18 years for the investment to grow to $8000.

54. Assume an initial investment of $100. Then after

one year the investment is worth 100 + 0.20(100) = 120. After the second year, the investment is worth 120 + 0.30(120) = 156. So overall, there was a 56% increase, so answer b) is correct.

55. Assume an initial investment of $100.00. Then over a 10 year period, the amount of growth

would be ( )10100 1 0.04 148.02+ = about a

48% increase; so answer d) is correct. 56. False

57.

( )

7

7

7

1500(1 ) 2100

1 1.4

(1 ) 1.4

4.9%

r

r

r

r

+ =

+ =

+ =≈

58.

( )

3

3

3

2000(1 ) 2300

1 1.15

(1 ) 1.15

4.8%

r

r

r

r

+ =

+ =

+ =≈

59. Assume $100 invested initially. Then 100 would increase to 102.5 after the first year. The total amount of the investment after three years would then be 100(1.025)(1.03)(1.084) $114.44= . If

the same amount was invested at an interest rate of r % compounded annually, the amount would be the same , so;

( )

3

3

3

100(1 ) 114.44

1 1.1444

(1 ) 1.1444

4.6%

r

r

r

r

+ =

+ =

+ =≈

60. No 61. After 1 year: 1.065($1000) = $1065.00

After 2 years: 2(1.065) ($1000) $1134.23=

After 3 years: 3(1.065) ($1000) $1207.95=

After 8 years: 8(1.065) ($1000) $1655.00=

After 11 years: 11(1.065) ($1000) $1999.15=

$1065.00; $1134.23; $12076.95; 8; 12



10-5

62. After 5 months: 5(1.005) ($10,000) $10,252.51=

After 10 months: 10(1.005) ($10,000) $10,511.40=

After 15 months: 15(1.005) ($10,000) $10,776.83=

After 30 months: 30(1.005) ($10,000) $11,614.00=

After 53 months: 53(1.005) ($10,000) $13,025.71=

$10,252.51; $10,511.40; $10,776.83; 30; 53

63. (1.08) ($100,000)n passes $1,000,000 when

n = 30 years.

64. 4

.061 ($100)

4

n⎛ ⎞+⎜ ⎟⎝ ⎠

passes $200 when

n = 11.64 years. 65. After year 1, option A would have a value of

(1 1(.08))($1000) $1080F = + = and option B

would have a value of 1(1 .06) ($1000) $1060.F = + = After year 2,

option A would have a value of (1 2(.08))($1000) $1160F = + = and option B

would have a value of 2(1 .06) ($1000) $1123.60.F = + = Continue

until year 11 when option A has a value of (1 11(.08))($1000) $1880F = + = and option

B has a value of 11(1 .06) ($1000) $1898.30.F = + =

66. After year 1, option A would have a value of (1 1(.04))($1000) $1040F = + = and option B

would have a value of

1(365).04(1 ) ($1000) $1040.81

365F = + = After

year 2, option A would have a value of (1 2(.04))($1000) $1080F = + = and option B

would have a value of

2(365).04(1 ) ($1000) $1083.28

365F = + =

Continue until year 11 when option A has a value of (1 11(.04))($1000) $1440F = + =

and option B has a value of

11(365).04(1 ) ($1000) $1552.67

365F = + =

Exercises 10.2

1. a. .06

.00512

i = =

(12)(10) 120n = =

R = $50 F = $8193.97

b. .04

.022

i = =

(2)(10) 20n = =

R = $8231.34 F = $200,000

2. a. .06

.00512

i = =

(12)(20) 240n = =

R = $520 P = $72,582

b. .08

.024

i = =

(4)(5) 20n = =

R = $700 P = $11,446

3. ( )4 5.06

4

.064

1 –1($1000) $23,123.67

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

4. ( )

.042

2 7.042

($10,000) $626.021 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

5. ( )

.084

4 7.084

($100,000) $4698.971 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

6. 101 (1.06)

($1000) $7360.09.06

−⎡ ⎤− =⎢ ⎥⎣ ⎦



10-6

7. a. ( )12 4.06

12

.0612

1 –1($500) $27,048.92

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

b. $27,048.92 – 48($500) = $3048.92

c. Month Interest Balance

1 $500.00

2 .005 500 $2.50× = (1.005)(500) + 500 = $1002.50

3 .005 1002.50 $5.01× =

(1.005)(1.002.50) + 500 = $1507.51

8. a. ( )4 5.06

4

.064

1 –1($800) $18,498.93

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

b. $18,498.93 – 20($800) $2498.93=

c. Quarter Interest Balance

1 $800.00

2 .015 800 $12.00× = (1.015)(800) + 800 = $1612.00

3 .015 1612 $24.18× = (1.015)(1612) + 800 = $2436.18

9. ( )

.0612

12 3.0612

($12,000) $305.061 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

Deposited: 36($305.06) = $10,982.16 Interest: $12,000 – $10,982.16 = $1017.84

10. ( )

.084

4 3.084

($5000) $372.801 – 1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

$5000 – 12($372.80) = $526.40

11. 12

.061 ($2000) $2123.36

12⎛ ⎞+ =⎜ ⎟⎝ ⎠

( )12.0612

.0612

1 –1($200) $2467.11

⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

$200 each month is better, by $343.75.

12. Month Balance

1 5000.00

2 ( ).06121 (5000) – 500 4525.00+ =

3 ( ).06121 (4525) – 500 4047.63+ =

4 ( ).06121 (4047.63) – 500 3567.87+ =

13. ( )

.0422 15.04

2

($1,000,000) $24,649.921 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦



10-7

14. ( )

.1222 10.12

2

($50 ) $1.359 1 – 1

million million×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

15. ( ) 12 15.12

12

.1212

1 1($30 ) $2499 million million

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

or approximately $2.5 billion

16. ( )

.0544 10.05

4

($12,000,000) $233,056.971 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

17. After 10 years, the fund will be worth:

( )12 10.1212

.1212

1 –1($100,000) $23,003,868.95

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦ After 10 years, the cost of the equipment will be:

10(1.06) (13,000,000) $23,281,020.06=

Therefore, the annuity will be short by $23,281,020.06 – $23,003,868.95 or $277,151.11.

18. The cost of the warehouse in 5 years will be: 5(1.1) (8,000,000) $12,884,080=

Therefore, the monthly payments to the annuity should be:

( ).1212

12 5.1212

($12,884,080) $157,758.441 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

19. ( )

.0844 15.08

4

($5,000,000) $43,839.831 –1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

20. ( )12 1.06

12

.0612

1 –1($10) $123.36

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

21. Jack withdraws for 12(.75) = 9 months.

( ) 12 0.75.0612

.0612

1 1($100) $877.91

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

22. ( )

.0612

12 1.0612

($45) $3.871 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

23. ( )12 10.06

12

.0612

1 –1($1000) $163,879.35

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

$1000 at the end of each month is better.

24. ( )4 3.08

4

.084

1 – 1($750) $10,059.07

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

$750 a quarter is better.

25. ( )4 94 9 .08

4

.084

1 –1.081 ($1000) ($100) $7239.32

4

×× ⎡ ⎤+⎛ ⎞ ⎢ ⎥+ + =⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦

26. ( )12 412 5 .06

12

.0612

1 – 1.061 ($100) ($10) $675.86

12

×× ⎡ ⎤+⎛ ⎞ ⎢ ⎥+ + =⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦



10-8

27. ( )12 10 12 3.06

12

.0612

1 –1 .06($100) 1 ($1000) $17,584.62

12

× ×⎡ ⎤+ ⎛ ⎞⎢ ⎥ + + =⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

28. ( )12 10 12 3.06

12

.0612

1 –1 .06($100) – 1 ($1000) $15,191.25

12

× ×⎡ ⎤+ ⎛ ⎞⎢ ⎥ + =⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

29. The future value of the annuity will be 10R + $5725.43, so:

( )101 .06 –1

10 $5725.43.06

10 $5725.43 13.180795

$5725.43 3.180795

$1800

R R

R R

R

R

⎡ ⎤+⎢ ⎥+ =⎢ ⎥⎣ ⎦

+ ===

30. The future value of the annuity will be 20R + $403.80, so:

( )4 5.044

.044

1 –120 $403.80

20 $403.80 22.019004

$403.80 2.019004

$200

R R

R R

R

R

×⎡ ⎤+⎢ ⎥+ =⎢ ⎥⎣ ⎦

+ ===

31. (a)

32. (a)

33. .05P = $1200, so P = $24,000.

34. If the initial deposit is P, iP = R, so .R

Pi

=

35. 7(1.06) ($10,000) $15,036.30P = =

4

.06($15,036.30) $4339.35

1 (1.06)R −

⎡ ⎤= =⎢ ⎥−⎣ ⎦

36. Let P be the amount when he is 17. 41 (1.06)

($10,000) $34,651.06.06

P−⎡ ⎤−= =⎢ ⎥

⎣ ⎦

7

$34,651.06$23,044.93

(1.06)=

37. $60,000

$1,000,000.06

RP

i= = =

9

$1,000,000$591,898.46

(1.06)=

38. Over m interest periods, P grows to (1 ) .mi P+

Then

(1 )(1 )

(1 ) –1

nm

n

i iR i P

i

⎡ ⎤+= +⎢ ⎥+⎣ ⎦

(1 ).

(1 ) –1

n m

n

i iP

i

++= ⋅

+

39. Present value: 30

30

1 (1.015) .04 $5000($5000)

.015 2 (1.015)

−⎡ ⎤− ⎛ ⎞ +⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦

= $5600.40

40. Present value: 601 (1.005)

($500) $25,862.78.005

−⎡ ⎤− =⎢ ⎥⎣ ⎦

41. a.( )

.1812

12 5.1812

($50,000)1 1

$1269.67

− ×

⎡ ⎤⎢ ⎥⎢ ⎥− +⎣ ⎦=

b. Treat as two annuities. The first has a present value of

601 (1.01)(.015)($50,000)

.01

$33,716.28.

−⎡ ⎤−⎢ ⎥⎣ ⎦=

The second has a present value of 60

601 (1.01)($1269.67) (1.01)

.01

$31,418.60.

−⎡ ⎤− ÷⎢ ⎥⎣ ⎦=

The total present value = $65,134.88



10-9

42. 601 (1.005)

($1000) $51,725.56.005

−⎡ ⎤− =⎢ ⎥⎣ ⎦

The benefit to the agency is a current-value savings of $60,000 – $51,725.56 = $8274.44, since the money not paid out right away can be invested at 6% compounded monthly.

43. Formula for a decreasing annuity: (1 )new previousB i B R= + −

44. Option 1:

( )2420,000,000 1.06 $80,978,692.83=

Option 2:

(1 ) 1

$60,000,000$2,400,000

25

niF R

i

R

⎡ ⎤+ −= ⋅⎢ ⎥⎣ ⎦

= =

25(1 0.06) 1$2,400,000

0.06

$131,674,828.80

F⎡ ⎤+ −= ⋅⎢ ⎥⎣ ⎦

=

Option 2 is better. 45. Set up a table with

1Y (1.05 ^ X –1) / .05*1000.= After 2, 3, and

4 years, 1Y equals $2050, $3152.50 and

$4310.13. 1Y equals $30,539 after 19 years, and

exceeds $50,000 after 26 years.

46. Set up a table with

1Y (1.02 ^ X –1) / .02*700.= After 1, 2, and

3 quarters, 1Y equals $700, $1414, and

$2142.28. 1Y equals $5204 after 7 quarters, and

exceeds $10,000 after 13 quarters.

47. Use 1Y (1.001^ X –1) / .001*15.= 1Y 503=

after 33 weeks.

48. Use 1Y (1.005 ^ X –1) / .005*100.=

1Y 3228= after 30 months

49. The original $5000 will go through 10 years of interest and 10 years of payments. Therefore, the original money will be worth

10 10$5000(1.06) (0.997) $8689.21.= The

$5000 deposited in the second year will go through 9 years of interest and 9 years of payment, therefore it will be worth

9 9$5000(1.06) (0.997) $8222.03.= Continue

this pattern for the 10 years and adding all the values together will give you a balance of $68,617.21 at the 0.3% payment. Using the same method for the 1.5% payment will yield a balance of $63,882.62 after the 10 years. The difference would then be $4734.59.

Exercises 10.3

1. ( )

.0612

12 5.0612

($10,000) $193.331 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

2. ( )

.1212

12 25.1212

($100,000) $1053.221 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

3. ( ) 2 10.12

2

.122

1 1($1000) $11,469.92

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

4. ( ) 12 3.09

12

.0912

1 1$500 ($100) $3644.68

− ×⎡ ⎤− +⎢ ⎥+ =⎢ ⎥⎣ ⎦

5. a. .12

($58,331) $583.3112

=

b. $600 – $583.31 = $16.69

c. $58,331 – $16.69 = $58,314.31

d. ( ) 12(5).12

12

.1212

1 1($600) $26,973.02

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

e. ( ) 12(4).12

12

.1212

1 1($600) $22,784.38

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

$26,973.02 – $22,784.38 = $4188.65 (with a $.01 discrepancy due to rounding errors)

f. Use result of part (d): .01($26,973.02) = $269.73



10-10

6. a. ( ) 4(7–5).08

4

.084

1 1($1000) $7325.48

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

b. ( ) 4(3).08

4

.084

1 1($1000) $10,575.34

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

4($1000) – ($10,575.34 – $7325.48) = $750.14

c. .08

$1000 – ($21,281.27) $574.374

=

d. (4)(7)($1000) – $21,281.27 = $6718.73

7. a. ( )

.1212

12 3.1212

($8000) $265.711 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. ($265.71)(3)(12) $9565.56=

c. $9565.56 – $8000 = $1565.56

d. ( ) 12(2).12

12

.1212

1 1($265.71) $5644.58

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

e. ( ) 12(1).12

12

.1212

1 1($265.71) $2990.59

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

f. 12($265.71) – ($5644.58 – $2990.59) = $534.53

g.

Payment Amount Interest Applied to Principal Unpaid balance

1 $265.71 $80.00 $185.71 $7814.29

2 265.71 78.14 187.57 7626.72

3 265.71 76.27 189.44 7437.28

4 265.71 74.37 191.34 7245.94

8. a. ( )

.06312

12 25.06312

($370,000 – $70,000) $1988.291 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. $70,000 (12)(25)($1988.29) $666,487+ =

c. $666,487 $370,000 $296,487− =



10-11

d. ( ) 12(25–23).063

12

.06312

1 1($1988.29) $44,724.98

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

e. ( ) 12(25–24).063

12

.06312

1 1($1988.29) $23,064.84

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

f. $44,724.98 – $23,064.84 = $21,660.14 12($1988.29) – $21,660.14 = $2199.34

g.

Payment Amount Interest Applied to principal Unpaid balance

1 1988.29 1575.00 413.29 299,586.71

2 1988.29 1572.83 415.46 299,171.25

3 1988.29 1570.65 417.64 298,753.61

9. .09

1 ($10,000) – $1125 $895012

⎛ ⎞+ =⎜ ⎟⎝ ⎠

10. 1 ($10,000) – $1500 $90004

r⎛ ⎞+ =⎜ ⎟⎝ ⎠

1 1.054

0.054

0.20 20%

r

r

r

+ =

=

= =

11. ( )

( )

2 8.122

2 8.12 .122 2

1 1 1($1000) ($10,000) $14,042.36

1

− ×

×

⎡ ⎤ ⎡ ⎤− +⎢ ⎥ ⎢ ⎥+ =⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦

12. ( )

.1212

12 5.1212

.12($105,504.50) ($10,000) $2446.89

121 1− ×

⎡ ⎤⎢ ⎥ + =⎢ ⎥− +⎣ ⎦

13. ( )

.1212

4.1212

100 256.281 1

−

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

Payment Amount Interest Applied to Principal Unpaid balance

1 $256.28 $10.00 $246.28 $753.72

2 256.28 7.54 248.74 504.98

3 256.28 5.05 251.23 253.74

4 256.28 2.54 253.74 0.00



10-12

14. ( )

.122

2.122

10,000 5454.371 1

R −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

Payment number Amount Interest Applied to Principal Unpaid Balance

1 $5454.37 $600.00 $4854.37 $5145.63

2 $5454.37 $308.74 $5145.63 $0.00

15. ( )

.06312

12 30.06312

($360,000 – $60,000) $1856.921 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

16. ( )

.0612

12 25.0612

($50,000) $322.151 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

per month

( ) 12(25–10).0612

.0612

1 1($322.15) $38,175.91

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

$150,000 – $38,175.91 = $111,824.09

17. ( ) 12 25.09

12

.0912

1 1($1200 – $200) $119,161.62

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

18. Option 1: Pay 500 for first five months. Loan amount is 5000

Monthly payment ( )

.0912

13.0912

($5000) $405.111 1

R −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

Total amount of payments: ($405.11)(13) $500 $5766.43+ =

Option 2: r = 0.06. n = 18 and P = 5500

Monthly payment ( )

.0612

18.0612

($5500) $320.271 1

R −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

Total amount of payments: ($320.27)(18) $5764.86,= so Option 2 costs slightly less.

19. ( ) 12 3.06

12

.0612

1 1($100) $3287.10

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

$6287.10 – ($2000 + $3287.10) = $1000.00 12 3

.061 ($1000) $1196.68

12

×⎛ ⎞+ =⎜ ⎟⎝ ⎠

20. a. ( )

.1212

12 25.1225

($1 million) = $10,532.241 1

− ×

⎡ ⎤⎢ ⎥⎢ ⎥− +⎣ ⎦



10-13

b. ( ) 12(25–5).12

12

.1212

1 1($10,532.24) $956,532.02

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

( ).1212

12 10.1222

($956,532.02) $13,723.461 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

c. ( ) 12(10–5).12

12

.1212

1 1($13,723.46) $616,938.67

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

21. (a)

22. (a)

23. a. 20(1.06) –1

($5000) $183,927.96.06

⎡ ⎤=⎢ ⎥

⎣ ⎦

b. 10

.06($183,927.96) $24,989.92

1 (1.06)−⎡ ⎤

=⎢ ⎥−⎣ ⎦

c. (10–5)1 (1.06)

($24,989.92) $105,266.63.06

−⎡ ⎤− =⎢ ⎥⎣ ⎦

24. In August 2008, their mortgage value would be (0.80)($400,000) = $320,000. The monthly payment would be

( ) 12(30).06712

.06712

1 1320,000

320,000 154.9719

$2064.89

R

R

R

−⎡ ⎤− +⎢ ⎥=⎢ ⎥⎣ ⎦

==

In August 2009, the price of the house will be (0.91)($400,000)=$364,000 and their mortgage value would be (0.80)($364,000)=$291,200. The monthly payment in August 2009 would be

( ) 12(30).07112

.07112

1 1291,200

291,200 148.8024

$1956.96

R

R

R

−⎡ ⎤− +⎢ ⎥=⎢ ⎥⎣ ⎦

==

25. a. $26000

$722.2236

= monthly

payment

b. amount of loan = $26,000 – $1000 = $25,000 Monthly payment

( ).0612

12 3.0612

($25,000) $760.551 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

c. Option a is more favorable.

26. a. $31,000

$645.8348

= monthly

payment

b. amount of loan = $31,000 – $2500 = $28,500 Monthly payment

( ).0612

12 4.0612

($28,500) $669.321 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

c. Option a is more favorable.



10-14

27. Assume you have a $100,000 mortgage, find your payments, and your balance after 15 years. The payment will be

( ).06812

12 30.06812

($100,000) $651.93.1 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦ The balance after 15 years will be

( ) 12 15.06812

.06812

1 1($651.93) $73,441.68.

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦ Therefore, the percent paid is 100,000 73,441.68

0.2656 26.56%.100,000

− = =

28. Assume you have a $100,000 mortgage, find your payments, and your balance after 10 years. The payment will be

( ).06412

12 20.06412

($100,000) $739.70.1 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦ The balance after 10 years will be

( ) 12 10.06412

.06412

1 1($739.70) $65,437.11.

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦ Therefore, the percent paid is 100,000 65,437.11

0.3456 34.56%.100,000

− = =

29. The loan amount (F) is 36 times the payment amount (R) minus the interest. The loan amount is also

( ) 12 3.0612

.0612

1 1

32.8710

F R

F R

− ×⎡ ⎤− +⎢ ⎥=⎢ ⎥⎣ ⎦

=

Therefore, 36 $1085.16 32.8710

3.1290 $1085.16

$346.81

R R

R

R

− ===

30. The loan amount (F) is 20 times the payment amount (R) minus the interest. The loan amount is also

( ) 4 5.084

.084

1 1

16.3514

F R

F R

− ×⎡ ⎤− +⎢ ⎥=⎢ ⎥⎣ ⎦

=

Therefore, 20 $2833.80 16.3514

3.6486 $2833.80

$776.69

R R

R

R

− ===

31. a. 1

1

1

and n n n n

n n

nn

I Q R I iB

iB Q R

R QB

i

−

−

−

+ = =+ =

−=

b. 1

1

1

1

1

1

(1 )

(1 )( )

(1 ) (1 )

(1 )

(1 )

(1 )

n

n

n

n

n

n

R QR Qi R

i ii R Q iR R Q

i R i Q iR R Q

R iR i Q iR R Q

i Q Q

i Q Q

+

+

+

+

+

+

−−+ − =

+ − − = −+ − + − = −+ − + − = −

− + = −+ =

c. The amount of the portion applied to the principle in the next month is equal to the amount of the portion applied to the principle in the previous month multiplied by 1 plus the interest rate.

d. 11 10

11

(1 )

(1 0.01)($100)

(1.01)($100)

$101.00

Q i Q

Q

= += +==

and

12 11

11

(1 )

(1 0.01)($101)

(1.01)($101)

$102.01

Q i Q

Q

= += +==



10-15

32. a. 1

1

1

1

1

1

(1 )

(1 )( )

(1 ) (1 )

(1 )

(1 )

(1 )

n n

n n

n n

n n

n n

n n

Q i Q

R I i R I

R I i R i I

R I R iR i I

I iR i I

I i I iR

+

+

+

+

+

+

= +− = + −− = + − +− = + − +− = − +

= + −

b. The interest portion of the next payment is equal to the interest rate multiplied by the payment subtracted from the interest portion of the current payment multiplied by 1 plus the interest rate.

c. Use

9$400, 0.01, and $100R i I= = =

10 9

10

(1 )

(1 0.01)($100) (0.01)($400)

$101 $4

$97.00

I i I iR

I

= + −= + −= −=

and

11 10

10

(1 )

(1 0.01)($97) (0.01)($400)

$97.97 $4

$93.97

I i I iR

I

= + −= + −= −=

33. After 1 month:

.091 ($2188.91) – $100 $2105.33

12⎛ ⎞+ =⎜ ⎟⎝ ⎠ After 2 months:

.091 ($2105.33) – $100 $2021.12

12⎛ ⎞+ =⎜ ⎟⎝ ⎠ After 3 months:

.091 ($2021.12) – $100 $1936.28

12⎛ ⎞+ =⎜ ⎟⎝ ⎠ The loan will be paid off after 24 months.

34. After 1 year: (1.07)($20,000) – $4195.92 = $17,204.08 After 3 years: Run 1.07 * Ans – 4195.92 twice. $11,011.37 After 5 years: Run 1.07 * Ans – 4195.92 twice more. $3921.39 The loan will be paid off after 6 years.

35. Enter 10000, then run 1.0075 * Ans – 166.68 repeatedly. After 40 iterations, the balance is $5741.79, which means $4258.21 has been paid off. The balance drops below $5000 after 46 months.

36. Enter 4000, then run 1.005 * Ans – 52.57 repeatedly. The balance drops below $3000 after 29 months, below $2000 after 54 months, and below $1000 after 76 months.

37. The balance B must drop to where

.085$250,

12B

⎛ ⎞ <⎜ ⎟⎝ ⎠ or B < $35,294.12.

Let 1 6Y Y (300 – X)*1000= where

6Y ((1 I) ^ X –1) / (I(1 I) ^ X).= + +

Make a table. Then B ≤ $35,294.12 after 260 payments which means that the next payment, after 261 months, is the first one where at least 75% goes toward debt reduction.

38. The balance B must drop to where

.084$300,

12B

⎛ ⎞⋅ <⎜ ⎟⎝ ⎠ or B < $42,857.14.

That happens after 302 months.

39. a. ( )

.02912

12 3.02912

($30,000) $871.111 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. ( )

.07412

12 3.07412

($28,000) $869.691 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

c. Option b is better by $1.42 per month.



10-16

40. a. ( )

.03912

12 3.03912

($21,000) $619.071 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. ( )

.08512

12 3.08512

($19,500) $615.571 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

c. Option b is better by $3.50 per month.

Exercises 10.4

1. deferred

2. free

3. [amount after taxes] (1 .45)(300,000)

$165,000

= −=

4. [amount after taxes] (1 0)(300,000)

$300,000

= −=

5. ( )1 52.06

1

.061

1 1($5000) $1,641,407.11

×⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

6. Example 1 [income tax saved] = $900 [balance after 48 years] = $49,181.62 [amount after taxes] = (1 0.25)($49,181.62) $36,886.22− =

Example 3 [earnings after income tax] = $2100 [balance after 48 years] = $34,427.13 The traditional IRA is more advantageous.

7. If we assume a marginal tax bracket of 20%,

( )1 52.061

.061

1 1(0.8)($5000) $1,313,125.69

×⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

8. Example 1 [income tax saved] = $900 [balance after 48 years] = $49,181.62 [amount after taxes] = (1 0.35)($49,181.62) $31,968.05− =

Example 3 [earnings after income tax] = $2100 [balance after 48 years] = $34,427.13 The Roth IRA is more advantageous



10-17

9. a. For Earl: [earnings after income tax]

[1 tax bracket] [amount]

[.60] [5000]

3000

= − ⋅= ⋅=

( )1 12.061

.061

1 1($3000) $50,609.82

×⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦ This money then earns interest compounded annually for 36 years and grows to

36$50,609.82 (1.06) $50,609.82(8.147252)

$412,330.96

⋅ ==

b. For Larry:

( )1 36.061

.061

1 1($3000) $357,362.60

×⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

c. Earl paid in 12 $3000 $36,000× = while Larry paid in 36 $3000 $108,000.× = Larry paid in more.

d. Earl has $54,968.36 more than Larry.

10. Enid: The balance is given by

( )1 11.061

.061

1 1($5000) $5000 $69,858.21

×⎡ ⎤+ −⎢ ⎥ − =⎢ ⎥⎣ ⎦ Since the contributions are made as early as possible. Lucy:

( )1 10.061

.061

1 1($5000) $65,903.97

×⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

11. (1 ) 4000(1 .10 1)

12 12(1)

$366.67

P rtR

t

+ + ⋅= =

=

12. (1 ) 6000(1 .07 1)

12 12(1)

$535

P rtR

t

+ + ⋅= =

=

13. (1 ) 3000(1 .09 3)

12 12(3)

$105.83

P rtR

t

+ + ⋅= =

=

14. (1 ) 10,000(1 .08 2)

12 12(2)

$483.33

P rtR

t

+ + ⋅= =

=

15. 12 12(171.21)(1) 2000

2000(1)

.0273 or about 2.73%

Rt Pr

Pt

− −= =

≈

16. 12 12(430.33)(1) 5000

5000(1)

.0328 or about 3.28%

Rt Pr

Pt

− −= =

≈

17. 12 12(608.44)(3) 20,000

20,000(3)

.0317 or about 3.17%

Rt Pr

Pt

− −= =

≈

18. 12 12(447.73)(2) 10,000

10,000(2)

.0373 or about 3.73%.

Rt Pr

Pt

− −= =

≈



10-18

19. a. [loan amount] 880

[total repayment]1 1 .06(2)

1000

rt= =

− −=

[total payment] 1000[monthly payment]

12 12(2)

$41.67

t= =

=

b.(1 ) 880(1 .06 2)

12 12(2)

$41.07

P rtR

t

+ + ⋅= =

=

The monthly payment is less.

20. True. The APR takes the cost of the points into account.

21. False. The effective rate differs from the APR only when discount points are involved.

22. True.

23. False. The longer the mortgage will be held, the lower the effective rate.

24. True. Both rates account for up-front fees.

25. False. The up-front fees must change proportionally for there to be no effect on the APR.

26. False. The monthly payment becomes $1609.25 and the balance after 60 months becomes $191,760. The effective mortgage rate becomes 9.25%, an increase of .25%.

27. ( )

.0912

12 25.0912

($250,000) $2097991 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 250,000 – 5000 = 245,000 Using the Excel function 12*RATE(300, –2097.99, 245000, 0) gives .09249 or 9.25%; (d).

28. ( )

.05912

12 30.05912

($250,000) $1482.841 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 250,000 – 2500 = 247,5000 Using the Excel function 12*RATE(360, –1482.84, 247500, 0) gives .05993 or about 6%; (b).

29. ( )

.05512

12 20.05512

($250,000) $1719.721 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 250,000 – 10,000 = 240,000 Using the Excel function 12*RATE(240, –1719.72, 240000, 0) gives .06002 or about 6%; (a).

30. ( )

.0912

12 25.0912

($250,000) $2097.991 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 250,000 – 10,000 = 240,000 Using the Excel function 12*RATE(300, –2097.99, 240000, 0) gives .09507 or about 9.5%; (b).

31. [monthly payment] = $581.03 Using the Excel function 12*RATE(48, –581.03, 99000, –94342.20) gives .05999 or about 6%; (a).



10-19

32. [monthly payment] = $816.08 Using the Excel function 12*RATE(120, –816.08, 97000, –42743.49) gives .06000 or 6%; (b).

33. ( )

.0612

12 30.0612

($100,000) $599.551 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 100,000 – 3000 = 97,000 The mortgage has 360 – 84 = 276 months to go. Therefore,

( ) 276.0612

.0612

1 1($599.55) $89,639.31balance

−⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦

Using the Excel function 12*RATE(84, –599.55, 97000, –89639.39) gives .06560 or about 6.56%; (c).

34. ( )

.0912

12 30.0912

($100,000) $804.621 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

New P = 100,000 – 2000 = 98,000 The mortgage has 360 – 84 = 276 months to go. Therefore,

( ) 276.0912

.0912

1 1($804.62) $93,640.15balance

−⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦

Using the Excel function 12*RATE(84, –804.62, 98000, –93640.46) gives .09401 or about 9.4%; (c).

35. APR: n = 20 · 12 = 240 240

240

.005(1.005)80,000 573.14

1.005 1R = ⋅ =

−

P = 80,000 – .03(80,000) = 77,600 12*RATE(n, R, P, 0) = 12*RATE(240, –573.14, 77,600, 0) = 6.38% effective rate: m = 10 · 12 = 120 R = 573.14, P = 77,600

120

120

1.005 1573.14 51,624.70

.005(1.005)B

−= ⋅ =

12*RATE(m, R, P, B) = 12*RATE(120, –573.14, 77,600, –51,624.70) = 6.47%

36. APR: n = 15 · 12 = 180 180

180

.0075(1.0075)180,000 1825.68

1.0075 1R = ⋅ =

− P = 180,000 – .02(180,000) = 176,400 12*RATE(n, R, P, 0) = 12*RATE(180, –1825.68, 176,400, 0) = 9.35% effective rate: m = 10 · 12 = 120 R = 1825.68, P = 176,400

60

60

1.0075 11825.68 87,949.16

.0075(1.0075)B

−= ⋅ =

12*RATE(m, R, P, B) = 12*RATE(120, –1825.68, 176,400, –87,949.16) = 9.38%



10-20

37. APR: n = 15 · 12 = 180 180

180

.0075(1.0075)120,000 1217.12

1.0075 1R = ⋅ =

− P = 120,000 – .01(120,000) = 118,800

180

180

(1 ) 11217.12 118,800

(1 )

i

i i

+ −⋅ =

+ yields

9.17%. effective rate: m = 5 · 12 = 60 R = 1217.12, P = 118,800

120

120

1.0075 11217.12 96,081.51

.0075(1.0075)B

−= ⋅ =

601217.12 96,081.51 (1217.12 118,000 ) (1 )i i i− = − ⋅ +

yields 9.27%.

38. APR: n = 20 · 12 = 240 240

240

.005(1.005)150,000 1074.65

1.005 1R = ⋅ =

− P = 150,000 – .03(150,000) = 145,500

240

240

(1 ) 11074.65 145,500

(1 )

i

i i

+ −⋅ =

+ yields

6.38%. effective rate: m = 5 · 12 = 60 R = 1074.65, P = 145,500

180

180

1.005 11074.65 127,349.80

.005(1.005)B

−= ⋅ =

601074.65 127,349.80 (1074.65 145,500 ) (1 )i i i− = − ⋅ +

yields 6.76%.

39. The salesman is comparing the future value of the savings account to the sum of the present values of the loan payments at time of payment. A proper comparison would be the future value of the savings account, $1083.14, to the future value of the series of payments (assuming 4% interest)

( )24.0412

.0412

1 1($43.87) $1094.24

⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

40. 1 (1 ) ni i

P Ri R

−⎡ ⎤− + ⎛ ⎞= ⋅⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦

1 (1 )

11

1

11

1

11

1

n

n

n

n

iPi

R

iP

R i

iP

R i

iP

R i

−= − +

⎛ ⎞= − ⎜ ⎟⎝ ⎠+

⎛ ⎞− = − ⎜ ⎟⎝ ⎠+

⎛ ⎞− = ⎜ ⎟⎝ ⎠+

41. a. P = $200,000 and 0.069

0.0057512

i = =

(0.00575)($200,000) $1150Payment iP= = =

b. ( )

.06912

12 10.06912

($200,000) $2311.871 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

42. a. P = $300,000 and 0.072

0.00612

i = =

(0.006)($300,000) $1800Payment iP= = =

b. ( )

.07212

12 10.07212

($300,000) $3514.261 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦



10-21

43. a. For the first 5 years; ( )

.0612

12 25.0612

($250,000) $1610.751 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

b. The balance after 5 years; ( ) 12 20.06

12

.0612

1 1($1610.75) $224,829.73balance

− ×⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦

c. For the sixth year, P = $224,829.73, n = 240, and i = 0.044 + 0.025 = 0.069.

( ).06912

240.06912

($224,829.73) $1729.631 1

Payment −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦


.06312

12 15.06312

($300,000) $2580.451 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦


12

.06312

1 1($2580.45) $229,306.05balance

− ×⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦


( ).0612

120.0612

($229,306.05) $2545.771 1

Payment −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

45. a. The balance after 7 years; ( ) 228.069

12

.06912

1 1($1729.63) $219,418.04balance

−⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦

b. Without the cap, P = $219,418.04, n = 228, and i = 0.077 + 0.025 = 0.102.

( ).10212

228.10212

($219,418.04) $2181.801 1

Payment −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

c. Without the cap, the percentage increase from the sixth year to the seventh year would be 2181.80 1729.63

0.2614 26.14%1729.63

− = =

Since this percentage is greater than the 7 % cap, the monthly payment will be (1.07)($1729.63) $1850.70.=

d. The interest due in the 73rd month would be (0.0085)($219,418.04) $1865.05.=

e. Since the interest owed is more than the payment made, the balance will increase by $1865.05 $1850.70 $14.35.− = Therefore the new balance will be

$219,418.04 $14.35 $219,432.39.+ =



10-22

46. The interest due in the 74th month would be (0.008)($182,674.62) $1461.40.=

Since the interest owed is more than the payment made, the balance will increase by $1461.40 $1440.88 $20.52.− = Therefore the new balance will be $182,674.62 $20.52 $182,695.14.+ =

47.

The APR is about 11.08%.

48.


49.

The APR is about 6.83.%.

50.




10-23

51


52.


53. Mortgage A costs an extra $1750 up front. [difference in monthly payments] = $1181.61 – $1159.84 = $21.77 Using the Excel function NPER(.002, 21.77, –1750) gives 87.72 months.

54. Mortgage B costs an extra $2500 up front. [difference in monthly payments] = $2177.77 – $2123.17 = $54.60 Using the Excel function NPER(.003, 54.60, –2500) gives 49.33 months.



10-24

55. Mortgage A costs an extra $2000 up front. [difference in monthly payments] = $1303.85 – $1283.93 = $19.92 Using the Excel function NPER(.004, 19.92, –2000) gives 128.63 months.

56. Mortgage B costs an extra $4700 up front. [difference in monthly payments] = $1547.71 – $1485.36 = $62.35 Using the Excel function NPER(.00275, 62.35, –4700) gives 84.59 months.

Chapter 10 Supplementary Exercises

1. (d)

2. ( )

.061212 10.06

12

($240,000) $1464.491 – 1

×

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

3. The monthly mortgage payment should not exceed 39,200

(.25) $816.67.12

⎛ ⎞ =⎜ ⎟⎝ ⎠

( ) ( )12 30.09

12

.0912

1 1$816.67 $101,497

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

4. 365

.07350 1 $53.79

365⎛ ⎞+ =⎜ ⎟⎝ ⎠

5. 9% compounded daily yields 365

.091 –1 .0942

365⎛ ⎞+ =⎜ ⎟⎝ ⎠

= 9.42% annually. 10% compounded

annually is better.

6. ( )12 5.06

12

.0612

1 –1($200) $13,954.01

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

7. a. ( )

.1212

12 15.1212

($200,000) $2400.341 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. ( ) 12(15–5).12

12

.1212

1 1($2400.34) $167,304.95

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

8. 120($24,000)(1.005) $43,665.52=

9. ( )12 10.06

12

$50,000$27,481.64

1× =

+

10. ( ) ( )12 2 12 3.06 .06

12 12

$10,000 $5000$13,050.08

1 1× ×+ =

+ +



10-25

11. ( )

.0612

12 4.0612

($12,000 – $3,000) $211.371 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

12. ( )

2 2.042

2 5.042

.041 ($100,000) $12,050.34

21 1

×

− ×

⎡ ⎤ ⎛ ⎞⎢ ⎥ + =⎜ ⎟⎝ ⎠⎢ ⎥− +⎣ ⎦

13. ( )12 15.06

12

$30,000$12,224.47

1× =

+

$105,003.50 – $12,224.47 = $92,779.03

( ).0612

12 15.0612

($92,779.03) $782.921 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

14. ( )12 10.12

12

$100,000$30,299.48

1× =

+

( ).1212

12 10.1212

($509,289.22 – $30,299.48) $6872.111 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

15. ( )12 30.06

12

.0612

1 –1($100) $100,451.50

×⎡ ⎤+⎢ ⎥ =⎢ ⎥⎣ ⎦

16. ( ) 12 10.12

12

.1212

1 1($2000) $139,401.04

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

17. Investment A: 10(1 .06) –1

1000.06

⎡ ⎤+⎢ ⎥⎣ ⎦

= $13,180.79

Investment B: 55000(1 .06) 5000 $11,691.13+ + =

Thus Investment A is the better investment.

18. Present value of annuity is ( ) 12 5.09

12

.0912

1 1($5) $240.87

− ×⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

The present value of $1000 is ( )12 5.09

12

$1000$638.70

1× =

+

Yes, it is a bargain, since the present value is 240.87 + 638.70 = $879.57.

19. 2

.101 –1 .1025 10.25%

2⎛ ⎞+ = =⎜ ⎟⎝ ⎠



10-26

20. 12

.181 –1 .1956 19.56%

12⎛ ⎞+ = =⎜ ⎟⎝ ⎠

21. ( )4 154 15 .08

4

.084

1 –1.081 ($10,000) ($1000)

4

×× ⎡ ⎤+⎛ ⎞ ⎢ ⎥+ +⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦ = $146,861.85

22. ( )

.0612

36.0612

($10,000) $304.221 1

R −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

23. 120

120(1.01) – 1($200)(1.01) $151,843.34

.01

⎡ ⎤=⎢ ⎥

⎣ ⎦

24. ( )

.0612

12 5.0612

($300,000) $5799.841 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

25. ( )

.0912

12 30.0912

($150,000) $1206.931 1

− ×

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

26. a. [amount after taxes] (1 .30)(30,000)

$21,000

= −=

b. 530,000 (1.06) 40,146.77

[amount after taxes] (1 .35)(40,146.77) $26,095.40

⋅ == − =

27. a. [amount after taxes] (1 0)(30,000) $30,000= − =

b. 530,000 (1.06) $40,146.77⋅ =

28. 12 12(228.42)(2) 5000

.048208 4.82%5000(2)

Rt Pr

Pt

− −= = = =

29. Loan A is better, because the monthly payments for Loan B will be 3000(1.06)

$265.12

=

Paym Amount Interest Applied to Principal Unpaid balance

1 $304.22 $50.00 $254.22 $9745.78

2 304.22 48.73 255.49 9490.29

3 304.22 47.45 256.77 9233.52

4 304.22 46.17 258.05 8975.47

5 304.22 44.88 259.34 8716.13

6 304.22 43.58 260.64 8455.49



10-27

30. P = $90,000 – .02($90,000) = $88,200 15 12 180n = ⋅ =

80

.005$90,000 $759.47

1 (1.005)R −= ⋅ =

−

31. 6 12 72m = ⋅ = R = $716.43 P = $100,000 – 0.3($100,000) = $97,000 B = $81,298.32

32. Mortgage A costs an extra $2000 up front. [difference in monthly payments] = $1413.56 – $1350.41 = $63.15 i = .00291667, R = $63.15, P = –$2000

Using the Excel function NPER(.00291667, 63.15, –2000) gives 33.3 months.

33. Through technology, you can find x = .06 on a graphing calculator by finding the intersection of

( )^ ^1 2Y 245000 and Y (1 X /12 240 1) / (X(1 X) 240) 1755.21.= = + − + ∗

34. Solve [ ] 96567.79 (86,837.98) 567.79 (97,000) (1 ) ,i i i− = − + using technology, to find i = .005;

the interest rate is 12i = .06. On a graphing calculator, find the intersection of

1Y 567.79 X(86837.98)= − and ^2Y (567.79 X 97000)(1 X) 96.= − ∗ +

35. a. P = $380,000 and 0.069

0.0057512

i = =

(0.00575)($380,000) $2185Payment iP= = =

b. ( )

.06912

12 15.06912

($380,000) $3394.341 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦


.06312

12 25.06312

($220,000) $1458.081 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦


12

.06312

1 1($1458.08) $198,690.34balance

− ×⎡ ⎤− +⎢ ⎥= =⎢ ⎥⎣ ⎦


( ).0735

12240.0735

12

($198,690.34) $1582.461 1

Payment −

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

Conceptual Exercises

37. The effective rate will be slightly higher than the nominal rate.

38. No, not necessarily. It depends upon the number of compounding periods and the time.

39. No. Much more. For example, a house payment is a decreasing annuity. If you pay an additional 5% on the loan each month, the duration of the loan will decrease significantly more than just 5%.



10-28

40. The payment will decrease because the amount being applied to the principle will decrease. The total amount of interest paid will increase due to the length of time being added to the loan.

41. When you have successive payments, the interest on the loan is re-calculated more frequently. The interest on the loan is always the interest on the unpaid balance. If more frequent payments are made, the interest will decrease faster.

Chapter 10 Chapter Test

1. 9

1 .065 ($500) $524.3812

⎛ ⎞+ × =⎜ ⎟⎝ ⎠

2. ( )1

($5000) $4464.291 3 .04

⎡ ⎤=⎢ ⎥+ ×⎣ ⎦

3.

12.05

1 ($1025) $1189.774

F⎛ ⎞= + =⎜ ⎟⎝ ⎠

$1189.77 – $1025 = $164.77

4. ( )120.08

12

1($25,000) $11,263.09

1

⎡ ⎤⎢ ⎥ =⎢ ⎥+⎣ ⎦

5. ( )180.06

12

.0612

1 1($250) $72,704.68

⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

6. ( ) 48.06

12

.0612

1 1($500) $21,290.16

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

7. ( )

.042

10.042

($12,000) $1095.921 1

⎡ ⎤⎢ ⎥ =⎢ ⎥+ −⎣ ⎦

8. ( )

.0912

120.0912

($200,000) $2533.521 1

−

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

9. 48

.061 ($300) $381.15

12⎛ ⎞+ =⎜ ⎟⎝ ⎠

( )48.0612

.0612

1 1($50) $2704.89

⎡ ⎤+ −⎢ ⎥ =⎢ ⎥⎣ ⎦

$381.15 + $2704.89 = $3086.04

10. .03

($2499.93) $37.502

= interest

$350 – $37.50 = $312.50 paid on principal $2499.93 – $312.50 = $2187.43

11. a. ( )

.0912

60.0912

($8000) $166.071 1

−

⎡ ⎤⎢ ⎥ =⎢ ⎥− +⎣ ⎦

b. .09

($8000) $6012

=

c. .09

1 ($8000) $166.07 $7893.9312

⎛ ⎞+ − =⎜ ⎟⎝ ⎠



10-29

d. ( ) 12.09

12

.0912

1 1($166.07) $1899.00

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

12. ( ) 100.06

4

.064

1 1($3500) $180,686.46

−⎡ ⎤− +⎢ ⎥ =⎢ ⎥⎣ ⎦

$180,686.46 + $45,000 = $225,686.46

13. a. ( )11.06

1

.061

1 1($5000) $74,858.21F

⎡ ⎤+ −⎢ ⎥= =⎢ ⎥⎣ ⎦

This money then earns interest compounded annually for 29 years and grows to 29($74,858.21)(1.06) $405,610.84Balance = =

b. ( )29.06

1

.061

1 1($5000) $368,198.99F

⎡ ⎤+ −⎢ ⎥= =⎢ ⎥⎣ ⎦

14. (1 ) 4000(1 .10 3)

12 12(3)

$144.44

P rtR

t

+ + ⋅= =

=

15. P = 250,000 – 2500 = 247,500 Using the Excel function 12*RATE(180, –2510.31, 247500, 0) gives .09 or 9%.

16. a. It will decrease because the cost of the points will be spread over a longer period of time.

b. Increasing the number of points will increase the effective cost of the loan, therefore, increasing its interest rate.

17. If there are no points, or the mortgage is not terminated early.

18. a. P = $500,000 and 0.078

0.006512

i = =

(0.0065)($500,000) $3250Payment iP= = =

b. ( )

.07812

12 20.07812

($500,000) $4120.181 1

Payment − ×

⎡ ⎤⎢ ⎥= =⎢ ⎥− +⎣ ⎦

Chapter 10 Project

1. total 2 1(1 )(1 ) 1

(1.04)(1.03) 1 .0712 7.12%

i i i= + + −= − = =

2. total 2 1(1 )(1 ) 1

(1.03)(1.04) 1 .0712 7.12%

i i i= + + −= − = =

The answer is the same.

3. B0 = $100,000 B1 = (1.03)100,000 – 10,000 = $93,000 B2 = (1.04)93,000 – 10,000 = $86,720



10-30

4. B0 = $100,000 B1 = (1.04)100,000 – 10,000 = $94,000 B2 = (1.03)94,000 – 10,000 = $86,820 The amount is greater because the larger interest rate is being applied to the larger balance.

5. (100% 4%) (1 .04)

(.96)

P P

P

⋅ − = ⋅ −= ⋅

6. (100% %) (1 )100

rP r P⋅ − = ⋅ −

7. P(100% + r%)(100% + s%) = P(1 + r/100)(1 + s/100) = P(1 + s/100)(1 + r/100)

8. a. P · (1 + .04)(1 – .04) = P · .9984 ≠ P therefore false

b. P(1 + .04)(1 – .03) = P(1 – .03)(1 + .04), true

9. 9(1.08) 2P P⋅ ≈

10. 6(1.12) 2P P⋅ ≈

12. If a quantity doubles and then triples, it will have increased by a multiple of 6.

13. If a quantity increases by a multiple of n and then increases by a multiple of p, it will have increased by a multiple of n ? p.

14. 10 2 5

239/8

72 167 239

(1.08) 9.966 10

N N N= + = + =

≈ ≈

15. 1.5 3 2 114 72 42N N N= − = − =

r = 7, 1.5 / 6 yearsN r =


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