8 Consumer Math - croninmath [licensed for non...

8 Consumer Math

Exercise Set 8-1 1. Percent means per one hundred.

2. To change a percent to a decimal, drop the percent sign and move the decimal point two places to the left; to change a percent to a fraction put the percent number over 100 and reduce.

3. To turn a decimal into a percent move the decimal two places to the right and add the % sign; to change a fraction into a percent, divide numerator by denominator to put it in decimal form, then change the decimal to a percent.

4. When you take a percent “of” something you need to multiply.

5. First find the amount of increase or decrease, then divide by the original amount to find the percent increase or decrease.

6. It is not possible to have more than 100% of a quantity, since percent means per one hundred; you cannot have more than a hundred per one hundred. It is however possible for something to increase more than 100%, for example, “This year’s snowfall is 150% of last year’s snowfall.”

7. 0.63 = 63%

8. 0.87 = 87%

9. 0.025 = 2.5%

10. 0.0872 = 8.72%

11. 1.56 = 156%

12. 3.875 = 387.5%

13. 1 0.2 20%5

= =

14. 5 0.625 62.5%8

= =

15. 2 0.666 66.6%3

= =

16. 1 0.166 16.6%6

= =

17. 11 1.25 125%4

= =

18. 32 2.375 237.5%8

= =

19. 18% = 18.% = 0.18

20. 23% = 23.% = 0.23

21. 6% = 6.% = 0.06

22. 2% = 2.% = 0.02

23. 62.5% = 0.625

24. 75.6% = 0.756

25. 320% = 320.% = 3.2

26. 275% = 275.% = 2.75

27. 266 % 66.6% 0.63

= =

28. 120 % 20.3% 0.2033

= =

29. 24 624%

100 25= =

30. 36 936%

100 25= =

31. 236 9236% 2100 25

= =

32. 520 1520% 5100 5

= =

411

33. 121 1%

2 100 200= =

34. 251

2 2121 25 112 %2 100 100 200 8

= = = =

35. 502

3 3162 50 116 %3 100 100 300 6

= = = =

36. 251

6 641 25 14 %6 100 100 600 24

= = ⋅ = =

37. Find 5% of 299.99. 0.05 × 299.99 = 15 rounded 299.99 + 15 = 314.99 The sales tax is $15 and the total cost is $314.99.

38. Find 7% of 59.95. 0.07 × 59.95 = 4.20 rounded 59.95 + 4.20 = 64.15 The sales tax is $4.20 and the total cost is $64.15.

39. Find 6% of 249.99. 0.06 × 249.99 = 15.00 rounded 249.99 + 15.00 = 264.99 The sales tax is $15.00 and the total cost is $264.99.

40. Find 4.5% of 19.99. 0.045 × 19.99 = 0.90 rounded 19.99 + 0.90 = 20.89 The sales tax is $0.90 and the total cost is $20.89.

41. Find the decrease. 999.99 − 399.99 = 600

600Percent decrease 0.60 60%999.99

= ≈ =

The reduction in price is 60%.

42. Find the decrease. 109.99 − 99.99 = 10

10Percent decrease 0.09 9%109.99

= ≈ =


43. Original price is 249.99 + 80.00 = 329.99 80.00Percent decrease 0.24 24%329.99

= ≈ =


44. Original price = 179.99 + 20.00 = 199.99 20.00Percent decrease 0.10 10%

199.99= ≈ =


45. ___ is 40% of $159.99 x = 0.4 × 159.99 x ≈ 64.00 159.99 − 64.00 = 95.99 The sale price is $95.99.

___ is 20% of $95.99 x = 0.2 × 95.99 x ≈ 19.20 95.99 − 19.20 = 76.80 The sale price with an additional 20% discount is $76.80.

46. ___ is 25% of $24.99

x = 0.25 × 24.99 x ≈ 6.25 24.99 − 6.25 = 18.74 The sale price is $18.74.

___ is 10% of $18.74 x = 0.1 × 18.74 x ≈ 1.87 18.74 − 1.87 = 16.87 The sale price with an additional 10 percent discount is $16.87.

The cost for 4 towels is 4(16.87) = $67.47 47. ___ is 7% of $1799.99

x = 0.07 × 1799.99 x ≈ 126.00 The clerk receives $126.

48. ___ is 12% of $2,499.99 x = 0.12 × 2,499.99 x ≈ 300.00 The commission is $300.

49. $75.36 is 6% of ____ 75.36 0.0675.36 0.060.06 0.061,256

xx

x

= ×

=

=

The item he sold cost $1,256.

50. $96.00 is 6% of ____ 96.00 0.0696.00 0.060.06 0.061,600

xx

x

= ×

=

=

The item cost $1,600.

412

51. The percent decrease from the 95-96 academic year to the 02-03 year is:

71,261 86,087 14,827 17.2%86,087 86,087

− −= = −

The percent increase from the 02-03 academic year to the 08-09 year is:

120,488 71,261 49,227 69.1%

71,261 71,261−

= =

The percent increase from 2003 to 2009 was 69.1 – 17.2 = 51.9% greater than the decrease from 96 to 03.

52. The percent decrease for chefs/head cooks is:

99,800 100,600 800 0.8%99,800 99,800

− −= ≈ −

The percent decrease for food service managers is:

310,000 320,600 10,600 3.31%320,600 320,600

− −= ≈ −

The percent decrease is 3.3 – 0.8 = 2.5% higher for food service managers.

53. $200.00 is 25% of ____ 200 0.25200 0.250.25 0.25

800

xx

x

=

=

=

The original cost of the laptop was $800.

54. Find the increase $54,747 − $50,480 = $4,267. 4,267Percent increase 0.085 8.5%

50,480= ≈ =

The increase in pay is 8.5%.

55. Find the increase from 2000 to 2005: 207.9 − 109.5 = 98.4 million.

98.4Percent increase 89.9%109.5

= ≈

The increase in subscribers is 89.9%. Find the increase from 2005 to 2010: 300.5 − 207.9 = 92.6 million.


= ≈

The increase in subscribers is 44.5%. Even though the actual changes were similar,

the percent change was far less with a larger beginning number.

56. Find the increase from 2001 to 2009 in Asia: 4.1 − 3.8 = 0.3 million.


= ≈

The increase in HIV patients is 7.9%. Find the increase from 2001 to 2009 in North

America: 1.5 − 1.2 = 0.3 million.

0.3Percent increase 25%1.2

= ≈

The increase in HIV patients is 25%. Even though the actual changes were the same,

the percent change was a lot more for the smaller initial number.

57. Commissions for 2000 were 55 − 19 = $36

billion

19Percent increase 52.8%36

= ≈

The increase in commissions from 2000 to 2007 was 52.8%.

Change in commissions from 2007 to 2009 44.3 − 55 = –$10.7 billion

10.7Percent increase 19.5%55

−= ≈ −

The decrease in commissions from 2007 to 2009 was –19.5%.

The average percent change per year from

2000 to 2007 = 52.8 7.5%7

≈ .


2007 to 2009 = 19.5 9.8%2

−≈ − .

58. Change in cost from 2001 to 2007

12,797 − 8,653 = $4,144 billion

4,144Percent increase 47.9%8,653

= ≈

The increase in cost from 2001 to 2007 was 47.9%.

Change in cost from 2007 to 2010 15, 014 − 12,797 = $2,217 billion

2,217Percent increase 17.3%

12,797= ≈

The increase in cost from 2007 to 2010 was 17.3%. The average percent change per year from

2001 to 2007 = 47.9 8%6

≈ .


2007 to 2010 = 17.3 5.8%3

≈ .

413

59. If you used the coupon that offered $20 off you entire purchase, you would pay 129 – 20 = $109.

If you used the coupon that offered 25% off your entire purchase, you would pay 129 – 0.25(129) = $96.75. You should use the 25% off coupon and save an extra 109 – 96.75 = $12.25.

60. The total cost of your items is 49(2) + 39 + 95

= $232. If you used the coupon that offered 15% off your entire purchase, you would pay 232 – 0.15(232) = $197.20.

If you use the 40% off one item coupon, and you’ve been paying attention at all during this section, then you would apply the coupon to the sweater and the new cost would be 95 – 0.40(95) = $57. The total cost of your items would be 49(2) + 39 + 57 = $194.

You would save 197.20 – 194 = $3.20 by using the 40% off one item coupon.

61. a) 2.4% × x = 11,145 0.024 × x = 11,145 x = 11,145 ÷ 0.024 = 464,375 b) The number of green jobs will increase by

26%. 0.26 11,145 2,898× ≈ There will be 11,145 2,898 13, 043+ = green jobs in 2012. c) 4.3% × 464,375 = 0.043 × 464,375 =

19,968.13 464,375 + 19,968.13 = 484,343 x% × 484,343 = 14,043

14,043 0.29 2.9%484,343

x = ≈ =

62. a) 523,000 – 27% × 523,000 = 381,790 523,000 – 381,790 = 141,210 fewer permits b) 28.8% 375,000

0.288 375,0000.712 375,000

526,685

x xx x

xx

− =− =

=

526,685 – 375,000 = 151,685 fewer homes were started in 2011.

c) x × 66% = 298,000 x × 0.66 = 298,000 x ≈ 451,515 There will be 526,685 – 451,515 = 75,170

fewer homes than in 2010.

63. Author’s house:

280,000 205,000 75,000 26.8%

280,000 280,000−

= ≈

Average house price in town:

150,000 120,000 30,000 20%

150,000 150,000−

= ≈

64.

Year Author Town

07 – 08 255,000 280,000

280,00025,000 8.9%

280,000

−

−= ≈ −

143,000 150,000

150,0007,000 4.7%

150,000

−

−= = −

08 – 09 255,000 255,000

280,0000 0%

255,000

−

= =

130,000 143,000143,000

13,000 9.1%143,000

−

−= = −

09 – 10 240,000 255,000

280,00015,000 5.9%

255,000

−

−= = −

136,000 130,000

130,0006,000 4.6%

130,000

−

= =

10 – 11 205,000 240,000

280,00035,000 13.7%

255,000

−

−= = −

120,000 136,000

136,00016,000 11.8%

136,000

−

−= = −

65. The highest value on the chart for the author’s

home is $305,000. The current value is $205,000

305,000 205,000 1.06305,000 1.06205,000

305,000log log1.06205,000305,000log log1.06205,000305,000log205,000

log1.066.82

t

t

t

t

t

t

= ⋅

=

= = ⋅ =

≈

It would take almost 7 years for the author’s home.

The highest value on the chart for the town average is $150,000.

The current value is $120,000

414

150,000 120,000 1.06150,000 1.06120,000

150,000log log1.06120,000150,000log log1.06120,000150,000log120,000

log1.063.82

t

t

t

t

t

t

= ⋅

=

= = ⋅ =

≈

It would take almost 4 years for the average home.

66. Sum of changes for author’s home is

( )( 8.9) 0 ( 5.9) ( 13.7) 28.5%− + + − + − = − Sum of changes for average home is ( 4.7) ( 9.1) 4.6 ( 11.8) 21%− + − + + − = − The percent decrease for the author’s home

was –6.82% for four years. ( )4 6.82% 27.3%× − ≈ which is fairly close to the sum of changes. For the average price of all homes, it was

3.82%− for four years. ( )4 3.82% 15.3%× − ≈ which is not very close to the sum of changes. 67. No; after taking 30% off the price, the cost is

now 70% of its original price, B, or 0.70B. The coupon is for 20% off this price, or the cost is 80% of the price of 0.70B. 0.80(0.70B) = 0.56B The total discount is 1 − 0.56 = 0.44 or 44%.

68. No; suppose the stock was worth $100 at the beginning.

Next day: Find 30% of $100.00. P = 0.3 × 100 P = 30 The stock is worth $100 − $30 or $70 the next day. Next week: Find 30% of $70. P = 0.3 × 70 P = 21 After both fluctuations, the stock is worth $70 + $21 or $91 per share, which means you do not break even.

69. The friend will receive 50% off the discounted price of the TV not the original price, so they will receive an additional 0.5(0.5 ) 0.25P P=off so their total discount will be 75%, not 100%.

70. Percent decrease in grill price =90 60 30 33.3%.

90 90−

= =

Percent increase in rides = 30 50%.60

=

The discount on the grill is actually more than advertised. Allowing for rounding, neither is lying.

71. There’s no comparison: 20% more than what? 72. The difference in ounces is 64 – 56 = 8

ounces. x% × 64 = 8 x = 12.5% 73. $4.00 ÷ 56 = $0.071 per ounce 64 × $0.071 = $4.57

74. 4.57 4 0.57 14.25%4 4

−= =

75. The difference in ounces is 56 – 48 = 8

ounces. x% × 56 = 8 x = 14.3% size reduction $4.00 ÷ 48 = $0.083 per ounce 56 × $0.083 = $4.67 for a 56 ounce container

4.67 4 0.67 16.75%4 4

−= = is the effective

percent increase. 76. The difference in ounces is 64 – 48 = 16

ounces. x% × 64 = 16 x = 25% size reduction $4.00 ÷ 48 = $0.083 per ounce 64 × $0.083 = $5.33 for a 64 ounce container

5.33 4 1.33 33.25%4 4

−= = is the effective

percent increase. The sum of the percent reductions is larger

than the overall, but the sum of the percent increase in price is smaller than the overall.

415

Exercise Set 8-2

416

1. Interest is a fee charged for the use of money and is usually calculated as a percent of the amount borrowed or invested. Simple interest is calculated using the formula I = Prt.

2. Principal refers to the amount borrowed (or

invested) and future value refers to the amount paid back (or accumulated).

3. The term of a loan is the amount of time

before the loan is repaid. 4. The rate of a loan or savings account is

typically described as a percent of the amount borrowed or invested.

5. The Bankers Rule calculates interest based on

a 360 day year instead of 365 days; it helps lenders.

6. A discounted loan is a loan where the interest

is paid up front by the borrower. 7. (12,000)(0.06)(2) 1,440I Prt= = =

The simple interest is $1,440.

8. 61.5 (6,150)(0.0025)

4

I Prtt

t

===

The time is 4 years.

9. (154,625)(0.0475)(30) 220,340.63I Prt= = = The simple interest is $220,340.63.

10. 72 (600)(0.04)

3

I Prtt

t

===


11. 1,927.2 (7,300)( )(6)

0.044

I Prtr

r

===

The simple interest rate is 4.4%.

12. 45 (200)( )(3)

0.075

I Prtr

r

===


13. 354.6 (0.09)(4)

985

I PrtPP

===

The principal is $985.

14. 65,625 (0.15)(7)62,500

I PrtPP

===

The principal is $62,500.

15. 464 ( )(.0116)(5)

8,000

I PrtP

P

===

The principle is $8,000.

16. 375 (1,250)(0.05)

6

I Prtt

t

===


17. (15,620)(0.018)(4.75)1,333.80

I Prt===

The simple interest is $1,333.80.

18. 31 (420)( )(2.5)

0.0295

I Prtr

r

==≈


19. ( )(1,975)(0.072) 3.5

497.7

I Prt===

The simple interest is $497.70.

20. 156 (325)( )(8)0.06

I Prtr

r

===

The simple interest rate is 6%.

21. 141.75 (700)(0.0675)

3

I Prtt

t

===


22. 1,372 (0.056)(3.5)7,000

I PrtPP

===


23. (1 )800(1 0.04(5))800(1 0.2)800(1.2)960

A P rtA

= += += +==

The future value is $960.

417

24. (1 )15,000(1 0.09(7))15,000(1 0.63)15,000(1.63)24,450

A P rtA

= += += +==

The future value is $24,450.

25. (1 )960(1 0.0228(6.5))960(1 0.1482)960(1.1482)1,102.27

A P rtA

= += += +==

The future value is $1,102.27

26. (1 )1,350(1 0.0338(4))1,350(1 0.1352)1,350(1.1352)1,532.52

A P rtA

= += += +==

The future value is $1,532.52.

27. 302,200(0.043)360

7.88

I Prt=

=

=

The interest is $7.88.

28. 135550(0.0175)360

3.61

I Prt=

=

=


29. 451,750(0.023)

3605.03

I Prt=

=

=


30. 225660(0.056)360

23.10

I Prt=

=

=


31. a) 3,000(0.08)(3)720

I PrtI

===

The discount is $720. b) The amount of money received is

$3,000 $720 $2,280− = c) True interest:

720 2,280 (3)

0.10510.5%

I Prtr

r

==≈=

The true interest rate is 10.5%.

32. a) 1,750(0.045)(6)472.5

I PrtI

===

The discount is $472.50. b) The amount of money received is

$1,750 $472.50 $1,277.50− = c) True interest:

472.50 1,277.5 (6)

0.0626.2%

I Prtr

r

==≈=

The true interest rate is 6.2%. 33. a)

6,000(0.075)(4.5)2,025

I PrtI

===

The discount is $2,025. b) The amount of money received is

$6,000 $2,025 $3,975− = c) True interest:

2,025 3,975 (4.5)

0.11311.3%

I Prtr

r

==≈=

The true interest rate is 11.3% 34. a)

33,000(0.036)(7)8,316

I PrtI

===

The discount is $8,316. b) The amount of money received is

$33,000 $8,316 $24,864− =

418

c) True interest:

8,316 24,864 (7)0.0484.8%

I Prtr

r

==≈=

The true interest rate is 4.8%. 35. $8,000

$4,046.406

4,046.40 8000 (6)0.8438.43%

PItI Prt

rr

=======

The interest rate is 8.43%

36. $15,000$18,00012

18,000 15,000 (12)0.110%

PItI Prt

rr

=======

The interest rate is 10%.

37. $150,0000.50.12

(1 )150,000 (1 (0.12)(0.5))150,000 (1.06)

$141,509.43

AtrA P rt

PP

P

==== += +=≈

The principal was $141,509.43.

38. 0.09$1,3503

1,350 (0.09)(3)$5,000

rItI Prt

PP

======

The principle was $5,000. 39.

1, 282.5 (4,500)(0.095)3

I Prtt

t

===

The term is 3 years.

40. 441.15 (8,650)(0.068)

0.75

I Prtt

t

===

The term is 0.75 year or 9 months.

41. 3,170.5 (9,325)(0.08)

4.25

I Prtt

t

==≈

The term is 4.25 years or 51 months.

42. 216 (0.08)(2)

1,350

I PrtPP

===

The principal was $1,350.

43. (4,300)(0.0975)(5) 2,096.25I Prt= = = The interest was $2096.25.

44. (816)(0.045)(3) 110.16I Prt= = = The interest was $110.16.

45. $6,20030.06

(1 )6,200(1 (0.06)(3))$7,316

PtrA P rtAA

==== += +=

46. $12,00060.09

(1 )12,000(1 (0.09)(6))$18,480

PtrA P rtAA

==== += +=

47. (1 )1,829(1 0.11(2))1,829(1 0.22)1,829(1.22)2,231.38

A P rtA

= += += +==

The monthly payment for 24 months is $2,231.38 ÷ 24 = $92.97.

419

48. (1 )1,800(1 0.129(1.5))1,800(1 0.1935)1,800(1.1935)2,148.30

A P rtA

= += += +==

The monthly payment for 18 months is $2,148.30 ÷ 18 = $119.35.

49. a) Answers may vary.

b) (1 )6,300(1 0.06(10))6,300(1 0.6)6,300(1.6)10,080

A P rtA

= += += +==

The monthly payment for the 10 year loan is $10,080 ÷ 120 = $84.

(1 )6,300(1 0.10(5))6,300(1 0.5)6,300(1.5)9,450

A P rtA

= += += +==

The monthly payment for the 5 year loan is $9,450 ÷ 60 = $157.50.

c) The 5 year loan has 10,080 – 9,450 = $630 less interest.

50. The amount she borrows is 530 + 0.25(530) = $662.50.

(1 )662.5(1 0.0629(1.5))662.5(1 0.09435)662.5(1.09435)725.01

A P rtA

= += += +=≈

The monthly payment for 18 months is $725.01 ÷ 18 = $40.28.

51. The total of her payments is 36(34.20) = $1,231.20.

(1 )1,231.20 884.29(1 (3))1,231.20 884.29 2,652.87

346.91 2,652.870.131

A P rtr

rr

r

= += += +=≈

The interest rate is 13.1%.

52. The total of the payments is 18(28.80) = $518.40.

(1 )518.40 412.23(1 (1.5))518.40 412.23 618.35106.17 618.35

0.172

A P rtr

rr

r

= += += +=≈


53. The total cost of the carpet without tax is 312(1.49) = $464.88.

The total cost with tax is 464.88 + 0.065(464.88) = $495.10.

The total of the payments is 24(27.41) = $657.84.

(1 )657.84 495.10(1 (2))657.85 495.10 992.80162.74 992.80

0.164

A P rtr

rr

r

= += += +=≈


54. The total cost of the fence without tax is 194(8.49) = $1,647.06.

The total cost with tax is 1,647.06 + 0.076(1,647.06) = $1,772.24.

The total of the payments is 30(72.12) = $2,163.60.

(1 )2,163.60 1,772.24(1 (2.5))2,163.60 1,772.24 4,430.60

391.36 4,430.600.088

A P rtr

rr

r

= += += +=≈


420

55. $60090 days0.04

90600(0.04)360

$6

PtrI Prt

I

I

====

=

=

56. $950120 days0.0675

120950(0.0675)360

$21.38

PtrI Prt

I

I

====

=

=

57. Bankers rule:

$2,200730 days0.072

7302,200(0.072)360

$321.20

PtrI Prt

I

I

====

=

=

Simple interest:

( )2,200 0.072 (2)$316.80

I PrtII

=

=

=

The interest paid would be $316.80. You would pay $4.40 more using the Banker’s rule.

58. 5 years:

Bankers rule:

$2,2001,825 days0.072

1,8252,200(0.072)360

$803.00

PtrI Prt

I

I

====

=

=

Simple interest:

( )2,200 0.072 (5)$792.00

I PrtII

=

=

=

The interest paid would be $792. You would pay $11 more using the Banker’s rule on a five year loan.

10 years:

Bankers rule:

$2,2003,650 days0.072

3,6502,200(0.072)360

$1,606

PtrI Prt

I

I

====

=

=

Simple interest:

( )2,200 0.072 (10)$1,584.00

I PrtII

=

=

=

The interest paid would be $1,584. You would pay $22 more using the Banker’s rule.

It seems that the extra interest keeps getting larger as the term increases.

59. a) 18,000(0.05)(6)5,400

I PrtI

===

The discount was $5,400. b) The amount of money received was

18,000 5,400 $12,600− = c) True interest:

5, 400 12,600 (6)

0.0717.1%

I Prtr

r

==≈=

60. a) Pr20,000(0.09)(3)5,400

I tI

===

The discount was $5,400. b) The amount of money received was

20,000 5,400 $14,600− =

421

c) True interest:

5, 400 14,600 (3)0.12312.3%

I Prtr

r

==≈=

61. Personal loan: (10,000)(0.09)(6) 5,400I Prt= = =

Auto loan: 60(10,000)(0.08) 4,00012

I Prt = = =

The personal loan has a higher interest amount. This is because it has both a higher interest rate and a longer term.

62. 98,000 (1,000,000)( )(20)0.0049

I Prtr

r

===

The rate is 0.49%. The rate seems unreasonably low.

63. New truck loan: (25,000)(0.18)(10) 45,000I Prt= = =

Repair loan: (18,000)(0.125)(8) 18,000I Prt= = =

The repair loan of $18,000 at 112 %2

for 8

years is less expensive.

64. Bank loan: 48(7,800)(0.095) 2,96412

I Prt = = =

Company loan: 54(7,800)(0.085) 2,983.512

I Prt = = =

The bank loan of $7,800 at 9.5% for 48 months is less expensive.

65. a)

( )10,000 0.065 (5)$3,000

I PrtII

=

=

=

You would pay $3,000 in interest.

b)

End of

Year

Future Value

1 10,000(1 0.06(1)) $10,600A = + = 2 10,600(1 0.06(1)) $11,236A = + = 3 11,236(1 0.06(1)) $11,910.16A = + =

4 11,910.16(1 0.06(1)) $12,624.77A = + =

5 12,624.77(1 0.06(1)) $13,382.26A = + = c) You would pay 13,382.26 – 13,000 ≈ $383

more by having the loan broken up this way.

d)

End of

Year

Future Value

2.5 10,000(1 0.06(2.5)) $11,500A = + = 5 11,500(1 0.06(2.5)) $13,225A = + =

You would pay 13,225 – 13,000 = $225 more by having the loan broken up this way.

66. a) Option 1: 0.04$5,0004

5,000(0.04)(4)$800

rPtI PrtII

======

Option 2:

End of

Year

Future Value

1 5,000(0.035)(1) 175I = = 2 5,175(0.035)(1) 181.13I = = 3 5,356.13(0.035)(1) 187.46I = = 4 5,543.59(0.035)(1) 194.02I = =

The final value is: $5,737.61.

Option 1 is worth 5,800 – 5,737.61 ≈ $62 more than Option 1

b) It won’t affect the result, other than all final amounts will be 10 times as big. So, option 1 is still better and is better by $624.

422

67. If you forget to change the percentage to decimal form, then the interest is WAY too high for the situation.

68. (a) Answers may vary.

(b) 1st loan: 0.5P

2nd loan: 0.5P

(c) 1st loan: monthly payment, 120 months

0.5

0.0125120

P PP

+=

2nd loan: monthly payment, 60 months

0.5

0.02560

P PP

+≈

(d) For the 10-year loan, the payments are half as much but the total paid is the same for both loans.

69. The first loan has a future value of:

(1 )(1 (0.05)(6)) (1.3)

A P rtA P P

= += + =

The second loan has a future value of: (1 )(1 (0.06)(5)) (1.3)

A P rtA P P

= += + =

They are the same regardless of principal.

70. The longer the term, the less money received and the higher the true interest rate. So it is worse for the borrower if the term is longer.

Exercise Set 8-3 1. Simple interest is earned (or charged) on the

total amount once; with compound interest the interest is deposited into the account at the end of the compounding period and begins earning interest as well.

2. If interest is compounded quarterly, it means it is compounded 4 times a year; compounded monthly is 12 times a year.

3. The effective rate of an investment is the simple interest rate that would yield the same future value.

4. An annuity is an investment in which a deposit is made every compounding period and interest does not change during the term of the investment. Interest is earned on the previous compounding period’s balance at the end of the compounding period.

5. In an investment earning compound interest, the entire amount of your principle begins earning interest immediately, which means that the interest earned is going to be higher. In an annuity, the total amount of your principle is not “in the account” until after the very last investment period. Therefore, you have less money in the account at any given time to earn interest on. However, not everyone has a large chunk of change lying around to make an investment. Most of us need to save a small amount each month.

6. The interest rate, r, is the annual interest rate. If it were not divided by n when compounding more than one time per year, it would be like calculating your yearly interest more than once a year. Sounds great, but banks don’t operate that way.

7. 1(10)

1

0.04825 11

1,221.20

ntrA Pn

= +

= +

≈

1, 221.20 825 396.2I = − =

423

The interest is $396.20 and the future value is $1,221.20.

8. 211

1

0.0061,495 12

1,596.84

ntrA Pn

⋅

= +

= +

≈

I = 1,596.84 − 1,495 = 101.84 The interest is $101.84 and the future value is $1,596.84.

9. 4(3)

1

0.0425560 14

635.72

ntrA Pn

= +

= +

≈

I = 635.72 − 560 = 75.72 The interest is $75.72 and the future value is $635.72.

10. 365(1)

1

0.09750 1365

820.62

ntrA Pn

= +

= +

≈


11. 365(4.5)

A 1

0.025320 1365

358.10

ntrPn

= +

= +

=


12. 52(32)

1

0.0419115,000 152

439,304.25

ntrA Pn

= +

= +

≈

I = 439,304.25 − 115,000 = 324,304.25 The interest is $324,304.25 and the future value is $439,304.25.

13. 1(15)

430, 0.07, 1, 15

0.07430 1 11

$10,805.480.071

R r n t

A

= = = =

+ − = =

14. 2(3)

8,000, 0.0161, 2, 3

0.01618,000 1 12

$48,976.430.01612

R r n t

A

= = = =

+ − = =

15. 4(6)

2,750, 0.0034, 4, 6

0.00342,750 1 14

$66,649.190.00344

R r n t

A

= = = =

+ − = =

16. 4(8.5)

3,500, 0.0173, 4, 8.5

0.01733,500 1 14

0.01734

$127,897.39

R r n t

A

= = = =

+ − =

=

17. 7,200, 0.031, 4, 3A r n t= = = =

4(3)

4(3)

4(3)

0.0311 14

7,2000.031

40.03155.8 1 1

4

55.8

0.0311 14

$574.85

R

R

R

R

+ − =

= + −

= + −

=

18. 27,500, 0.053, 12, 8A r n t= = = =

424

12(8)

12(8)

12(8)

0.0531 112

27,5000.053

120.053121.46 1 1

12

121.46

0.0531 112

$230.63

R

R

R

R

+ − =

= + −

= + −

=

19. 1,100,000, 0.072, 12, 28A r n t= = = =

12(28)

12(28)

12(28)

0.0721 112

1,100,0000.072

120.0726,600 1 1

12

6,600

0.0721 112

$1,021.18

R

R

R

R

+ − =

= + −

= + −

=

20. 2, 400,000, 0.0925, 52, 32A r n t= = = =

52(32)

52(32)

52(32)

0.09251 152

2,400,0000.0925

520.09254,269.23 1 1

52

4,269.23

0.09251 152

$233.97

R

R

R

R

+ − =

= + −

= + −

=

21. 4

0.06, 4

0.061 1 0.0614 6.14%4

r n

E

= =

= + − ≈ =

22. 2

0.10, 2

0.101 1 0.1025 10.25%2

r n

E

= =

= + − = =

23. 4

0.065, 4

0.0651 1 0.0666 6.66%4

r n

E

= =

= + − ≈ =

24. 2

0.0955, 2

0.09551 1 0.0978 9.78%2

r n

E

= =

= + − ≈ =

25. Determine the effective rate for each investment:

2

4

Investment 1: 0.045, 2

0.0451 1 0.0455 4.55%2


0.04251 1 0.0432 4.32%4

r n

E

r n

E

= =

= + − ≈ =

= =

= + − ≈ =

Since it has the higher effective rate, 4.5% compounded semiannually is the better investment.


12

2


0.071 1 0.0723 7.23%12


0.0721 1 0.0733 7.33%2

r n

E

r n

E

= =

= + − ≈ =

= =

= + − ≈ =


425


365

4


0.031 1 0.0305 3.05%365


0.0311 1 0.0314 3.14%4

r n

E

r n

E

= =

= + − ≈ =

= =

= + − ≈ =

Since it has the higher effective rate, 3.1% compounded quarterly is the better investment.


2

365


0.05741 1 0.0582 5.82%2


0.0561 1 0.0576 5.76%365

r n

E

r n

E

= =

= + − ≈ =

= =

= + − ≈ =


29. 4(10)

1

0.095,000 14

12,175.94

ntrA Pn

= +

= +

≈

They will have $12,175.94 in 10 years.

30. 2(18)

1

0.0820,000 12

82,078.65

ntrA Pn

= +

= +

≈

There will be $82,078.65 in 18 years.

31. The money will be invested for 50 – 25 = 25 years.

2(25)

1

0.0760,000 12

335,095.61

ntrA Pn

= +

= +

≈

She will have $335,095.61 at age 50.

32. 4(5)

1

0.07510,000 14

14,499.48

ntrA Pn

= +

= +

≈

In 5 years, $14,499.48 will be available.

33. 4(6)

4(6)

1

0.03912,000 14

12,000

0.039149,507.11

ntrA Pn

P

P

P

= + = +

= +

≈

They need to invest $9,507.11 right now in order to have $12,000 in 6 years.

34. 4(6)

4(6)

1

0.052522,000 14

22,000

0.03914

$16,088.29

ntrA Pn

P

P

P

= + = +

= +

≈

She needs to invest $16,088.29 right now in order to have $22,000 in 6 years.

35. 52(5)

52(5)

1

0.0234,500 152

4,500

0.023152$4,011.25

ntrA Pn

P

P

P

= + = +

= +

≈

You need to invest $4,011.25 right now in order to have $4,500 in 5 years.

426

36. 12(18)

12(18)

1

0.06280,000 112

80,000

0.062112$26,282.41

ntrA Pn

P

P

P

= + = +

= +

≈

They need to invest $26,282.41 right now in order to have $80,000 in 18 years.

37.

( )

( )

( )

2

2

2

0.072100,000 73,000 12

100,000 .072173,000 2

100log log 1.03673

100log 2 log 1.03673

100log73

2log 1.0364.45

t

t

t

t

t

t

= +

= +

= =

=

≈

The investment would be worth $100,000 in about 4.5 years.

38.

365

365

365

0.07140,000 17,000 1365

40,000 0.071113,000 365

40 0.071log log 113 36540 0.071log 365 113 365

40log13

0.071365log 136512.05

t

t

t

t

t

t

= +

= +

= + = +

=

+

≈

The investment would be worth $40,000 in about 12.05 years.

39.

( )( ) ( )( ) ( )

( )( )

0.0682 11

2 1.068

log 2 log 1.068log 2 log 1.068

log 2log 1.068

10.54

t

t

t

P P

t

t

t

= +

=

==

=

≈

( )

( )( )

365

365

365

0.0682 1365

0.0682 1365

0.068log 2 log 13650.068log 2 365 1365

log 20.068365log 136510.19

t

t

t

P P

t

t

t

= +

= +

= + = +

= +

≈

If the interest is compounded annually, it will take 10.54 years for the investment to double. If the interest is compounded daily, it will take 10.19 years for the investment to double.

40.

( )

( )

( )

12

12

12

0.182,000 700 112

2,000 0.181700 1220log log 1.015720log 12 1.0157

20log7

12log 1.0155.88

t

t

t

t

t

t

= +

= +

= =

=

≈

It would take 5.88 years to reach $2,000. That is almost triple the original balance. I think there is a lesson to be learned here. . . .

427

41. R = 2,250, n = 2, t = 4, 0.075 0.0375

2rn

= =

2(4)

1 1

(1 0.0375) 12,2500.0375

20,548.25

ntrRn

Arn

+ − =

+ −=

≈

The future value of the annuity is $20,548.25.

42. R = 600, n = 4, t = 3, 0.08 0.02

4rn

= =

4(3)

1 1

(1 0.02) 16000.02

8,047.25

ntrRn

Arn

+ − =

+ −=

≈


43. R = 200, n = 4, t = 20, 0.05 0.0125

4rn

= =

4(20)

1 1

(1 0.0125) 12000.0125

27,223.76

ntrRn

Arn

+ − =

+ −=

≈


44. R = 200, n = 2, t = 2, 0.09 0.045

2rn

= =

2(2)

1 1

(1 0.045) 12000.045

855.64

ntrRn

Arn

+ − =

+ −=

≈

The Washingtons can spend $855.64 for the vacation.

45. a) R = 4,000, n = 1, t = 5, 0.105 0.105

1rn

= =

1(5)

1 1

(1 0.105) 14,0000.105

24,664.64

ntrRn

Arn

+ − =

+ −=

≈

The annuity will be worth $24,664.64 in 5 years.

b) 1(4)

1

0.10520,000 11

$29,818.04

ntrA Pn

A

A

= +

= +

=

The investment would be worth 29,818.04 – 24,664.64 = $5,153.40 more than the annuity.

46. a) R = 160, n = 4, t = 4, 0.04 0.01

4rn

= =

4(4)

1 1

(1 0.01) 11600.01

2,761.26

ntrRn

Arn

+ − =

+ −=

≈


b) The value of the deposits is $160 × 4 × 4 = $2,560.

4(3)

1

0.042,560 14

$2,884.67

ntrA Pn

A

A

= +

= +

=

The investment would be worth 2,884.67 – 2,761.26 = $123.41 more than the annuity.

428

47. a) R = 2,000, n = 2, t =10, 0.08 0.04

2rn

= =

2(10)

1 1

(1 0.04) 12,0000.04

59,556.16

ntrRn

Arn

+ − =

+ −=

≈

They will have saved $59,556.16 in 10 years.

b) The total value of the deposits is $2,000 × 2 × 10 = $40,000. The total interest earned would be 59,556.16 –

40,000 = $19,556.16.

c)

( )

( )

2(9)

18

18

1

0.0859,556.16 12

59,556.16 1.0459,556.16

1.0429,398.60

ntrA Pn

P

P

P

P

= +

= +

=

=

=

They would need to invest $29,398.60 at the beginning of the first year in order to get the same return.

48. a) R = 800, n = 1, t = 3, 0.06 0.06

1rn

= =

1(3)

1 1

(1 0.06) 18000.06

2,546.88

ntrRn

Arn

+ − =

+ −=

=


b) The value of the deposits is $800 × 3 = $2,400. The total interest earned would be 2,546.88 –

2,400 = $146.88.

c) 1(2)

1(2)

1

0.062,546.88 11

2,546.88

0.0611

2,266.71

ntrA Pn

P

P

P

= + = +

= +

=

They would need to invest $2,266.71 at the beginning of the first year in order to get the same return.

49. A = 150,000, n = 12, t = 5, 0.0712

rn

=

12(5)

1 1

0.07150,00012

0.071 112

2,095.18

nt

rAnR

rn

R

=

+ −

=

+ −

=

They need to make a regular monthly payment of $2,095.18 to reach their goal.

50. A = 10,000, n = 52, t = 4, 0.065 0.00125

52rn

= =

( )

( )52(4)

1 1

10,000 0.00125

1 0.00125 142.13

nt

rAnR

rn

R

=

+ −

=+ −

=

You will need to make a regular weekly payment of $42.13 to reach your goal.

429

51. A = 4,000,000, n = 2, t = 10, 0.11 0.055

2rn

= =

( )( )2(10)

1 1

4,000,000 0.055

1 0.055 1114,717.32

nt

rAnR

rn

R

=

+ −

=+ −

≈

They will need to make a regular semiannual payment of $114,717.32 to reach their goal.

52. The period is 50 – 25 = 25 years.

A = 2,000,000, n = 12, t = 25, 0.074

12rn

=

12(25)

1 1

0.0742,0000,00012

0.0741 112

2,316.64

nt

rAnR

rn

R

=

+ −

=

+ −

≈

She will need to make a regular monthly payment of $2,316.64 to reach her goal.

53. A = 12,000, n = 12, t = 6, 0.03912

rn

=

12(6)

1 1

0.03912,00012

0.0391 112

148.20

nt

rAnR

rn

R

=

+ −

=

+ −

=

They need to make a regular monthly payment of $148.20 to reach their goal. Over 6 years (72 months), that would be $10,670.40. With a one-time investment (found in Exercise 33), they would need to invest $9,507.11. So the annuity would cost them 10,670.40 – 9,507.11 = $1,163.29 more.

54. A = 22,000, n = 4, t = 6, 0.0525 0.013125

4rn

= =

( )

( )4(6)

1 1

22,000 0.013125

1 0.013125 1785.81

nt

rAnR

rn

R

=

+ −

=+ −

=

She would need to make a regular quarterly payment of $785.81 to reach her goal. Over 6 years (24 quarters), that would be $18,859.44. With a one-time investment (found in Exercise 34), she would need to invest $16,088.29. So the annuity would cost them 18,859.44 – 16,088.29 = $2,771.15 more.

55. Compound interest:

365(6)

1

0.0455,000 1365

$6,549.71

ntrA Pn

A

A

= +

= +

=

Simple interest:

(1 )

5,000(1 (0.045)(6))$6,350

A P rtA

= += +=

Your ridiculous choice would cost you 6,549.71 – 6,350 = $199.71, and you shouldn’t be applying for any jobs on Wall Street anytime soon.

430

56. Simple interest:

(1 )

10,000(1 (0.12)(10))$22,000

A P rtA

= += +=

Compound interest:

12(10)

1

0.0610,000 112

$18,193.97

ntrA Pn

A

A

= +

= +

=

Annuity: 0.08 0.024

rn

= =

4(10)

1 1

(1 0.02) 12500.02

$15,100.50

ntrRn

A rn

+ − =

+ −=

≈

The simple interest would be the best choice.


4(3)

1

0.0414,000 14

$4,520.71

ntrA Pn

A

A

= +

= +

=

Annuity: 0.054 0.01354

rn

= = ,

4,000 333.334 3

R = =⋅

4(3)

1 1

(1 0.0135) 1333.330.0135

$4,310.74

ntrRn

Arn

+ − =

+ −=

≈

The certificate of deposit would be the better choice.


12(4)

1

0.03830,000 112

$34,916.42

ntrA Pn

A

A

= +

= +

=

Annuity:0.059

12rn

= , 30,00012 4

R =⋅

12(4)

1 1

.059625 1 112

.05912

$33,742.67

ntrRn

Arn

A

+ − =

+ − =

≈

The account paying 3.8% compounded monthly would be the better choice.

431

59.

12( )

12

12

12

12

1 1

0.04980 1 112

15,000.04912

0.04961.25 80 1 112

0.0490.765625 1 112

0.0491.765625 112

0.049log(1.765625) log 112

nt

t

t

t

t

t

rRn

Arn

+ − =

+ − =

= + −

= + − = +

= +

0.049log(1.765625) 12 log 112

log(1.765625)0.04912log 1

1211.6

t

t

t

= +

= +

≈

It would take Bill and Ted 11.6 years to reach their goal. I can only imagine the excellent adventure they are saving up for.

60.

( )

( )

( )

( )

( )

12( )

12

12

12

12

1 1

0.072520 1 112

300,0000.072

121,800 520 1.006 11,800 1.006 1520

1,800 1 1.006520

1,800log 1 log 1.006520

1,800log 1 12 log 1.006520

nt

t

t

t

t

t

rRn

Arn

t

+ − =

+ − =

= − = −

+ =

+ = + =

( )

1,800log 1520

12log 1.00620.8

t

t

+ =

≈

It would take about 20.8 years.

61. a) If nur

= and 1rn u

= and n ur= , then:

11 1nt urtrA P P

n u = + = +

Using the power to the power rule of

exponents, 1 11 1

rturt uP P

u u

+ = +

b) If n keeps getting bigger and bigger and r

stays the same, then nr

is going to keep getting

bigger and bigger. Think of it as a pile of cash that is split amongst 5 people. If the pile of cash keeps getting bigger, then the piece of the pile that the five people get keeps getting bigger.

432

c)

u 11u

u +

50 2.692 100 2.705 500 2.716

1,000 2.717 1,500 2.718 2,000 2.718 2,500 2.718

The formula keeps getting closer and closer to 2.718.

62. rtA Pe=

63. Compounded Continuously:

0.04(20)

5,000, 0.04, 20

5,000$11,127.70

rtP r tA PeA eA

= = ===≈

Compounded Annually:

1(20)

5,000 1

0.045,000 11

$10,955.62

ntrAn

= + = +

≈

64.

t 6% simple

(1 )A P rt= + P=50,000, r = 0.06

6% continuous rtA Pe=

P = 50,000, r = 0.06 5 65,000 67,493 10 80,000 91,106 15 95,000 122,980 20 110,000 166,006 25 125,000 224,084 30 140,000 302,482

65. After the first two years there will be:

2(2)

1

0.045,000 12

5,412.16

ntrA Pn

= +

= +

≈

The principal for the last three years is $5,412.16 + 2,000 = $7,412.16. Amount over the next three years is:

2(3)

1

0.047,412.16 12

8,347.30

ntrA Pn

= +

= +

≈

There is $8,347.30 in the account at the end of 5 years.

66. After the first two years there will be:

4(2)

1

0.033,500 14

3,715.60

ntrA Pn

= +

= +

≈

The principal for the last three years is $3,715.6 − 800 = $2,915.60. Amount over the next 4 years is:

4(4)

1

0.032,915.60 14

3,285.86

ntrA Pn

= +

= +

≈


433

67. 0.02

0.0054

r

n= =

After the first two years there will be:

( )4(2)

1

900 1 0.005936.64

ntrA Pn

= +

= +≈

The principal for the next year is $936.64 + 400 = $1,336.64. Amount over the

next year is:

( )4(1)

1

1,336.64 1 0.0051,363.57

ntrA Pn

= +

= +≈

The principal for the next three years is $1,363.57 − 200 = $1,163.57. The amount over the next three years is:

( )4(3)

1

1,163.57 1 0.0051,235.34

ntrA Pn

= +

= +≈


68. 5 years:

Simple interest: 0.12 5 0.6I Prt P P= = ⋅ ⋅ =

Compound interest: 5t = , 0.075r = , 365 24 8, 760n = ⋅ =

8,760 5

1

0.07518,760

1.45

ntrA Pn

P

P

⋅

= +

= +

≈

7.5% interest compounded hourly would draw more interest over 5 years.

20 years:

Simple interest: 0.12 20 2.4I Prt P P= = ⋅ ⋅ =

Compound interest: 20t = , 0.075r = , 365 24 8, 760n = ⋅ =

8,760 20

1

0.07518,760

4.48

ntrA Pn

P

P

⋅

= +

= +

≈

7.5% interest compounded hourly would still draw more interest.

69. Simple Interest = Compound Interest

12(10)

120

120

120

1

0.05(1 (10)) 112

0.051 10 112

0.0510 1 1120.051 112

100.0647

ntrA Pn

P x P

x

x

x

x

= + + = +

+ = + = + − + − =

=

An interest rate of 6.47% is necessary to make the simple interest investment equal to the compound interest investment.

70. Answers may vary.

71. a) $40,488.76

b) $38,902.85; When you round, you end up with 40,448.76 – 38,902.85 = $1,585.91 less than if you don’t round.

72. a) $99,534.63

b) 84,832.42; When you round, you end up with 99,534.63 – 84,832.42 = $14,702.21 less than if you don’t round.

434

Exercise Set 8-4 1. When a person borrows money to purchase

something and makes periodic payments it’s called an installment loan.

2. A finance charge is a fee charged by a lender for borrowing money.

3. The purchase price is what the item cost, the amount financed is how much was borrowed. The amount financed is the purchase amount minus any down payment.

4. Closed-ended credit is paid off over a specific period of time; open-ended credit does not have to be paid off by any specific date.

5. The annual percentage rate is the true interest charged in a year. It is typically different from the stated interest rate because interest can be calculated using different methods, resulting in different true rates.

6. When interest is computed by the unpaid balance method, the consumer pays only on the balance remaining at the end of the billing cycle. The average daily balance takes into consideration the average balance over the billing cycle and charges based on the average rather than the unpaid balance. There are advantages and disadvantages to both methods.

7. The exact minimum payment formula varies by credit card company, but the federal government has mandated that the minimum payment covers both the interest for the month AND part of the balance.

8. It’s the same as the difference between buying a house and renting a house. Whether you rent, lease or buy, you are going to be making regular monthly payments, but with renting and leasing, you are not going to own the car or house at the end of your lease period. If you buy a house or a car, it’s all yours once the loan is paid off.

9. Down payment: (given) $60. Amount Financed: (purchase price minus

down payment) 460 – 60 = $400. Finance charge: (total payments minus amount

financed) $42 × 10 – 400 = 420 – 400 = $20 Installment Price: (purchase price plus finance

charge) $460 + $20 = $480 10. Down payment: (15% of purchase price)

0.15($1,720) = $258. Amount Financed: (purchase price minus

down payment) $1,720 − $258 = $1,462. Finance charge: (4% of amount financed)

0.04($1,462) = $58.48 Installment Price: (purchase price plus finance

charge) $1,720 + $58.48 = $1,778.48 Monthly payment: (amount financed plus

finance charge divided by number of payments) ($1,462 + $58.48) ÷ 18

= $1,520.48 ÷ 18 = $84.47 11. Down payment: (15% of purchase price)

0.15($375) = $56.25. Amount Financed: (purchase price minus

down payment) $375 − $56.25 = $318.75. Finance charge: (total payments minus amount

financed) $27.25 × 12 − $318.75 = $327 − $318.75 = $8.25 Installment Price: (total monthly payments

plus down payment) $327 + $56.25 = $383.25 12. Down payment: (given) $95. Amount Financed: (purchase price minus

down payment) $845 − $95 = $750. Finance charge: (total payments minus amount

financed) $128.75 × 6 − $750 = $772.5 − $750 = $22.50

Installment Price: (total monthly payments plus down payment) $772.50 + $95 = $867.50

13. Down payment: (20% of purchase price)

0.20(925) = $185. Amount Financed: (purchase price minus

down payment) $925 − $185 = $740. Finance charge: (5% of amount financed)

0.05($740) = $37 Installment Price: (purchase price plus finance

charge) $925 + $37 = $962 Monthly payment: (amount financed plus

finance charge divided by number of payments) ($740 + $37) ÷ 18 = $777 ÷ 18

= $43.17 14. Down payment: (given) $75.

435

Amount Financed: (purchase price minus down payment) $550 − $75 = $475.

Finance charge: (11% of amount financed) 0.11($475) = $52.25

Installment Price: (purchase price plus finance charge) $550 + $52.25 = $602.25

Monthly payment: (amount financed plus finance charge divided by number of payments) ($475 + $52.25) ÷ 12

= $527.25 ÷ 12 = $43.94 15. Amount Financed = Cost of car – down

payment: 14,295.41 – 2,600 = $11,695.41. Total installment price = total of payments +

down payment: (279.40)(48) + 2,600 = $16,011.20.

Finance Charge = installment price – cost of car: 16,011.20 – 14,295.41 = $1,715.79.

16. Amount Financed = Cost of car – down


down payment: (312.15)(60) + 6,300 = $25,029.

Finance Charge = installment price – cost of car: 25,029 – 22,152.37 = $2,876.63.

17. Cost of car = 19,500 + 0.065(19,500) + 375 =

$21,142.50 Amount Financed = Cost of car – down


down payment: (390.25)(60) + 3,500 = $26,915.

Finance Charge = installment price – cost of car: 26,915 – 21,142.50 = $5,772.50.

18. Cost of car = 6,052 + 0.08(6,052) + 260 =

$6,796.16 Amount Financed = Cost of car – down

payment: 6,796.16 – 1,500 = $5296.16. Total installment price = total of payments +

down payment: (180.10)(36) + 1,500 = $7,983.60.

Finance Charge = installment price – cost of car: 7,983.60 – 6,796.16 = $1,187.44.

19. Cost of car = 31,600 – 12,400 + 0.055(31,600 – 12,400) + 410 = $20,666.

Amount Financed = $20,666. Total installment price = total of payments +

trade-in: (471.22)(48) + 12,400 = $35,018.56. Finance Charge = total of payments – amount

financed: 22,618.56 – 20,666 = $1,952.56. 20. Cost of car = 20,600 – 3,600 + 0.045(20,600 –

3,600) + 304 = $18,069. Amount Financed = $18,069 Total installment price = total of payments +

trade-in: (371.48)(60) + 3,600 = $25,888.80. Finance Charge = total of payments – amount

financed: 22,288.80 – 18,069 = $4,219.80. 21. Step 1: Find the finance charge per $100

borrowed. Amount financed = 39,905 – 15,000 = 24,905 Finance charge = total payments – amount

financed: $614(48) − $24,905 = $29,472 – $24,905 = $4,567

Finance Charge $100Amount Financed

$4,567 $100$24,905$18.34

×

= ×

=

Step 2: Find the row in Table 8-1 with 48 payments. Notice that $18.31 is closest to the value from Step 1. Step 3: Move up the table and find the APR at the top of the corresponding column. The APR is 8.5%.

22. Step 1: Find the finance charge per $100

borrowed. Amount financed = 14,400 – 4,000 = 10,000 Finance charge = total payments – amount

financed: $319(36) − $10,400 = $11,484 – 10,400 = $1,084


$1,084 $100$10,400$10.42

×

= ×

=

Step 2: Find the row in Table 8-1 with 36 payments. Notice that $10.34 is closest to the value from Step 1.

436

Step 3: Move up the table and find the APR at the top of the corresponding column. The APR is 6.5%.


borrowed. Finance charge = total payments – amount

financed: $71.05(12) − $800 = $852.60 − $800 = $52.60


$52.60 $100$800

$6.58

×

= ×

=


Step 3: Move up the table and find the APR

at the top of the corresponding column. The APR is 12%.



financed: $84(12) − $950 = $1,008 – $950 = $58


$58 $100$950$6.11

×

= ×

=



at the top of the corresponding column. The APR is 11%.

25. Step 1: Find the finance charge per $100 borrowed.

Finance charge = total payments – amount financed: $90(24) − $1,960 = $2,160 – 1,960

= $200


$200 $100$1,960$10.20

×

= ×

=



at the top of the corresponding column. The APR is 9.5%.



financed: $389.50(60) − $19,900 = $23,370 – $19,900 = $3,470


$3,470 $100$19,900$17.44

×

= ×

=



at the top of the corresponding column. The APR is 6.5%.

27. k = 48 – 30 = 18, R = $614, APR = 8.5% so h

(from table 8-1) = $6.86

10018($614)($6.86)

100 $6.86$709.50

kRhuh

=+

=+

=

Payoff amount: 19 × $614 − $709.50 = $10,956.50

28. k = 36 – 24 = 12, R = $319, APR = 6.5% so h (from table 8-1) = $3.56

437

10012($319)($3.56)

100 $3.56$131.59

kRhuh

=+

=+

=

Payoff amount: 13 × $319 − $131.59 = $4,015.41

29. k = 12 – 6 = 6, R = $71.05, APR = 12% so h

(from table 8-1) = $3.53

1006($71.05)($3.53)

100 $3.53$14.54

kRhuh

=+

=+

=

Payoff amount: 7 × $71.05 − $14.54 = $482.81

30. k = 12 – 6 = 6, R = $84, APR = 11% so h (from table 8-1) = $3.23

1006($84)($3.23)100 $3.23

$15.77

kRhuh

=+

=+

=

Payoff amount: 7 × $84 − $15.77 = $572.23

31. k = 6, R = $90, APR = 9.5% so h (from table 8-1) = $2.79

1006($90)($2.79)100 $2.79

$14.66

kRhuh

=+

=+

=

Payoff amount: 7 × $90 − $14.66 = $615.34

32. k = 6, R = $389.50, APR = 6.5% so h (from table 8-1) = $1.90

1006($389.50)($1.90)

100 $1.90$43.58

kRhuh

=+

=+

=

Payoff amount: 7 × $389.50 − $43.58 = $2,682.92

33. f = (36 × 141.17) − 4,200 = 882.12 k = 36 − 20 = 16

( 1) 882.12(16)(16 1) 180.13( 1) 36(36 1)

fk kun n

+ += = =

+ +

$180.13 was saved.

34. f = (12 × 13.75) − 150 = 15 k = 12 − 6 = 6

( 1) 15(6)(6 1) 4.04( 1) 12(12 1)

fk kun n

+ += = =

+ +

$4.04 will be saved.

35. f = (18 × 13.28) − 200 = 39.04 k = 18 − 10 = 8

( 1) 39.04(8)(8 1) 8.22( 1) 18(18 1)

fk kun n

+ += = =

+ +

$8.22 is saved.

36. f = (24 × 29.50) − 600 = 108 k = 24 − 18 = 6

( 1) 108(6)(6 1) 7.56( 1) 24(24 1)

fk kun n

+ += = =

+ +

$7.56 was saved.

37. f = (10 × 99.75) − 950 = 47.50 k = 10 − 7 = 3

( 1) 47.5(3)(3 1) 5.18( 1) 10(10 1)

fk kun n

+ += = =

+ +

$5.18 was saved.

38. f = (12 × 292.50) − 3,250 = 260 k = 12 − 8 = 4

( 1) 260(4)(4 1) 33.33( 1) 12(12 1)

fk kun n

+ += = =

+ +

$33.33 was saved.

438

39. a) Lease total: 1000 + 48(199) = $10,552 b) First two years buy: $4000 + 24(340.27) =

$12,166.48 First two years lease: 1000 + 24(199) = $5,776 Difference: $12,166.48 – $5,776 = $6,390.48 c) Answers may vary. 40. a) Lease total: 2,900 + 36(429) = $18,344 b) First three years buy: 9,000 + 36(500.95) =

$27,034.20 First three years lease: 2,900 + 36(429) =

$18,344 Difference: 27,034.20 – 18,344 = $8,690.20 c) Total of payments: 60(500.95) = $30,057 Amount financed: 32,415 – 4,000 – 9,000 +

0.065(32,415 – 4,000 – 9,000) + 395 = $22,136.98

Interest = Total of payments – amount financed

60(500.95) – 22,136.98 = $7,920.02 d) Answers may vary. 41. k = 18, R = $614, APR = 8.5% so h (from

Table 8-1) = $6.86 f = (48 × 614) − (39,905 – 15,000) = 4,567

( 1) 4,567(18)(18 1) $664.08( 1) 48(48 1)

fk kun n

+ += = =

+ +

Unearned interest for rule of 72 = $664.08 Unearned interest for actuarial method =

$709.50 709.50 – 664.08 = $45.42 benefit to the lender.

42. a) k = 6, R = $389.50, APR = 6.5% so h (from Table 8-1) = $1.90

f = (60 × 389.50) − (27,900 – 8,000) = 3,470

( 1) 3,470(6)(6 1) $39.82( 1) 60(60 1)

fk kun n

+ += = =

+ +

Unearned interest for rule of 72 = $39.82 Unearned interest for actuarial method =

$43.58 43.58 – 39.82 = $3.76 benefit to the lender b) It appears that the actuarial method is better

for the borrower.

43. a) x = price of new car Amount financed = x – 2,000 Total installment price = (48)(150) + 2,000 =

9,200

( )( )

2,000 4 0.062( 2000) 150(48)

2,000 4 0.062 124 7,2002,000 0.248 496 7,200

1.248 2,496 7,2001.248 9,696

$7,769.23

x x

x xx

xxx

− + − =

− + − =

− + − =− =

==

The most expensive car she can afford is $7,769.23.

b) With 4,000 down payment:

( )( )

4,000 4 0.062( 4000) 150(48)

4,000 4 0.062 248 7,2004,000 0.248 992 7,200

1.248 4,992 7,2001.248 12,192

$9,769.23

x x

x xx

xxx

− + − =

− + − =

− + − =− =

==

She can afford a car that is $2,000 more.

44. a) With a monthly payment of $180.

( )( )

2,000 4 0.062( 2000) 180(48)

2,000 4 0.062 124 8,6402,000 0.248 496 8,640

1.248 2,496 8,6401.248 11,136

$8,923.08

x x

x xx

xxx

− + − =

− + − =

− + − =− =

==

She could afford a car that is 8,923.08 – 7,769.23 = $1,153.85 more expensive.

b) She would end up paying 30(48) = $1,440 more total for the car.

45. (832.5)(0.02)(1) 16.65I Prt= = =

The finance charge is $16.65. New balance 832.5 675 16.65 400

1,124.15= + + −=

The new balance is $1,124.15.

46. (1,131.63)(0.0175)(1) 19.80I Prt= = = The finance charge is $19.80. New balance 1,131.63 19.80 512.58 750

914.01= + + −=

The new balance is $914.01.

439

47. (2,364.79)(0.0167)(1) 39.49I Prt= = = The finance charge is $39.49. New balance 2,364.79 39.49 1,964.32 1,000

3,368.6= + + −=


48. (678.34)(0.0225)(1) 15.26I Prt= = = The finance charge is $15.26. New balance 678.34 15.26 3,479.03 525

3,647.63= + + −=


49. (986.53)(0.0135)(1) 13.32I Prt= = = The finance charge is $13.32. New balance 986.53 13.32 186.5 775

411.35= + + −=


50. (638.19)(0.015)(1) 9.57I Prt= = = The finance charge is $9.57. New balance 638.19 9.57 317.98 475

490.74= + + −=


51.(a) Date Balance Days Balance × days

9-1 627.75 10 − 1 = 9 5,649.75

9-10

627.75 + 87.95 = 715.70

15 − 10 = 5 3,578.50

9-15

715.70 − 200 = 515.70

27 − 15 = 12

6,188.40

9-27

515.70 + 146.22 = 661.92

30 − 27 + 1

= 4 2,647.68

Total 30 18,064.33

18,064 602.14

30≈

The average daily balance is $602.14.

(b) (602.14)(0.012) ≈ 7.23 The finance charge is $7.23

(c) 661.92 + 7.23 = 669.15 The new balance is $669.15.


3-1 2,162.56 3 − 1 = 2 4,325.12

3-3

2,162.56 − 800 =

1,362.56

10 − 3 = 7 9,537.92

3-10

1,362.56 + 329.27 = 1,691.83

21 − 10 = 11

18,610.13

3-21

1,691.83 − 500 =

1,191.83

29 − 21 = 8 9,534.64

3-29

1,191.83 + 197.26 = 1,389.09

31 − 29 + 1

= 3 4,167.27

Total 31 46,175.08

46,175.08 1,489.52

31≈

The average daily balance is $1,489.52.

(b) (1,489.52)(0.02) = 29.79 The finance charge is $29.79.

(c) 1,389.09 + 29.79 = 1,418.88 The new balance is $1,418.88.


6-1 157.95 5 − 1 = 4 631.80

6-5

157.95 + 287.62 = 445.57

20 − 5 = 15 6,683.55

6-20

445.57 − 100 = 345.57

30 − 20 + 1 = 11

3,801.27

Total 30 11,116.62

11,116.62 370.55

30≈


(b) (370.55)(0.014) ≈ 5.19 The finance charge is $5.19.


440


12-1 1325.65 15 − 1 = 14 18,559.10

12-15

1,325.65 + 287.62 = 1,613.27

16 − 15 = 1 1,613.27

12-16

1,613.27 + 439.16 = 2,052.43

22 − 16 = 6 12,314.58

12-22

2,052.43 − 700 =

1,352.43

31 − 22 + 1 = 10

13,524.30

Total 31 46,011.25

46,011.25 1,484.23

31≈

The average daily balance is $1,484.23.

(b) (1,484.23)(0.02) ≈ 29.68 The finance charge is $29.68.

(c) 1,352.43 + 29.68 = 1,382.11 The new balance is $1,382.11.


7-1 65.00 2 − 1 = 1

65.00

7-2 65 + 720.25 = 785.25

8 − 2 = 6

4,711.50

7-8 785.25 − 500 = 285.25

17 − 8 = 9

2,567.25

7-17 285.25 − 100 = 185.25

28 − 17 = 11

2,037.75

7-28 185.25 + 343.97 = 529.22

31 − 28 + 1

= 4

2,116.88

Total 31 11,498.38

11,498.38 370.92

31≈





9-1 50.00 13 − 1 = 12 600.00

9-13

50 + 260.88 = 310.88

17 − 13 = 4 1,243.52

9-17

310.88 − 100 = 210.88

19 − 17 = 2 421.76

9-19

210.88 + 324.15 = 535.03

30 − 19 + 1 = 12

6,420.36

Total 30 8,685.64

8,685.64 289.52

30≈




57. Month 1: 2% of 11,500 = 230 11,500 – 230 = $11,270 Month 2: 2% of 11,270 = 225.40 11,270 – 225.40 = $11,044.60 Month 3: 2% of 11,044.60 = 220.89 11,044.60– 220.89 = $10,823.71 Month 4: 2% of 10,823.71 = 216.47 10,823.71 – 216.47 = $10,607.23 She will pay 11,500 – 10.607.23 = $892.76 of

her balance. 58. Month 1: 1% of 500 = 5 500 – 5= $495 Month 2: 1% of 495 = 4.95 495 – 4.95 = $490.05 Month 3: 1% of 490.05 = 4.90 490.05– 4.90 = $485.15 Month 4: 1% of 485.15 = 4.85

441

485.15 – 4.85 = $480.30 Month 5: 1% of 480.30 = 4.80 480.30 – 4.80 = $475.50 Month 6: 1% of 475.50 = 4.75 475.50 – 4.75 = $470.75

After 6 months of payments, the balance is $470.75.

59. a) Exercise 51: (627.75)(0.012)(1) $7.53I Prt= = = Unpaid balance: = 627.75 + 7.53 + 234.17 –

200 = $669.45 Exercise 53: (157.95)(0.014)(1) $2.21I Prt= = = Unpaid balance: = 157.95 + 2.21 + 287.62 –

100 = $347.78 Exercise 55: (65)(0.011)(1) $0.72I Prt= = = Unpaid balance: = 65 + 0.72 + 1,064.22 – 600

= $529.94 b) Exercise 51: Unpaid balance method is

669.45 – 669.15 = $0.30 higher. Exercise 53: Unpaid balance method is

350.76– 347.78= $2.98 lower. Exercise 55: Unpaid balance method is 533.30

– 529.94 = $3.36 lower. c) When purchases total a lot more than

payments, unpaid balance is the better method, especially if payments are made late in the month.

60. a) Exercise 52: (2,162.56)(0.02)(1) $43.25I Prt= = = Unpaid balance: = 2,162.56 + 43.25 + 526.53

– 1,300 = $1,432.34 Exercise 54: (1,325.65)(0.02)(1) $26.51I Prt= = = Unpaid balance: = 1,325.65 + 26.51 + 726.78

– 700 = $1,378.94 Exercise 56: (50)(0.019)(1) $0.95I Prt= = = Unpaid balance: = 50 + 0.95 + 585.03 – 100 =

$535.98 b) Exercise 52: Unpaid balance method is

1,432.34 – 1,418.88 = $13.46 higher. Exercise 54: Unpaid balance method is

1,382.11– 1,378.94 = $3.17 lower.

Exercise 56: Unpaid balance method is 540.53 – 535.98= $4.55 lower.

c) The conclusions in Exercise 59(c) are reinforced.

61. (40,000)(0.045)(12) $21,600I Prt= = = . Total of loan and interest: 40,000 + 21,600 =

$61,600. Monthly payment: $61,600 ÷ 96 = $641.67. 62. First year: (10,000)(0.045)(12) $5,400I Prt= = = Second year: (10,000)(0.045)(11) $4,950I Prt= = = Third year: (10,000)(0.045)(10) $4,500I Prt= = = Fourth year: (10,000)(0.045)(9) $4,050I Prt= = = Total interest: 5,400 + 4,950 + 4,500 + 4,050 =

$18,900. Cheryl would pay 21,600 – 18,900 = $2,700

less interest than Jamie.

63. (1 )monthly payment

[1 (0.08)(4)]10012(4)

4,8001 0.32

3,636.36

P rtnt

P

P

P

+=

+=

=+

≈

The principal is $3,636.36. 64. I = Prt = (800)(0.12)(1) = 96,

96monthly interest 812

= =

4-month savings = 4(8) = 32 If computed equally over 12 months, $32 is saved after 8 payments. Rule of 78: k = 12 − 8 = 4

( 1) 96(4)(4 1) 12.31( 1) 12(12 1)

fk kun n

+ += = ≈

+ +

Using the rule of 78, $12.31 is saved. The borrower saves more money when computed equally over 12 months.

65. The calculation ends up being

( )60.99 2,300 $2,154.40× = .

442

66. (1)(6)

1

0.012,300 11

$2,165.40

ntrA Pn

= +

= −

≈

Same as Example 8 with slight rounding error. 67.

( )

1

500 2,300 1 0.01500 0.99

2,300500log log0.99

2,300500log log0.99

2,300500log

2,300log0.99

151.84

nt

t

t

t

rA Pn

t

t

t

= +

= −

=

= = =

≈

151.84 months ≈12.65 years.


Exercise Set 8-5 1. If you have a subsidized loan, the government

picks up the interest on the loan while you are still in school. If it is unsubsidized, then the interest is added to the principle instead.

2. For a federal loan, the interest rate is fixed for the life of the loan and they are regulated by congress.

3. If you don’t pay the interest that accumulates on a loan during college, then it is added to the principle and it and we call it capitalized interest.

4. A student loan defers payments until you actually have an income to pay it (meaning you graduate!).

5. A mortgage is a loan for the purchase of real estate where the lender has a right to seize the property if payments are not made.

6. The total interest paid can be calculated by subtracting the original principal from the total payments made.

7. A mortgage with a shorter term results in less interest paid in total, but it also results in a higher monthly payment.

8. An amortization schedule is a schedule of payments on a mortgage that keeps track of the balance at any given time as well as the amount of each payment that goes toward the principal and interest.

9. Daily interest = balance interest 7,200 0.068 $1.34

365.25 365.25× ×

= =

Interest for 30 day month = 1.34(30) = $40.20 Accrual period = 3(12) + 9 + 6 = 51 months. Total interest = 40.20(51) = $2,050.20 10. Daily interest =

balance interest 11,500 0.068 $2.14365.25 365.25

× ×= =

Interest for 30 day month = 2.14(30) = $64.20 Accrual period = 2(12) + 3 = 27 months. Total interest = 64.20(27) = $1,733.40

443

11. Daily interest = balance interest 21,500 0.062 $3.65

365.25 365.25× ×

= =

Interest for 30 day month = 3.65(30) = $109.50

Accrual period = 4(12) + 3 = 51 months. Total interest = 109.50(51) = $5,584.50 12. Daily interest =


× ×= =

Interest for 30 day month = 2.8(30) = $84 Accrual period = 1(12) + 9 + 6 = 27 months. Total interest = 84(27) = $2,268 13. Monthly payment for Exercise 9: 7, 200 2, 050.20 $9, 250.20P = + =

120

120

12

1 112

0.0689,250.2012

0.0681 112

52.420.4924$106.45

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

14. Monthly payment for Exercise 10: 11, 500 1, 733.40 $13, 233.40P = + =

120

120

12

1 112

0.06813,233.4012

0.0681 112

74.990.4924$152.29

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

15. Monthly payment for Exercise 11: 21, 500 5, 584.50 $27, 084.50P = + =

120

120

12

1 112

0.06227,084.5012

0.0621 112

139.940.4612$303.42

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

16. Monthly payment for problem 12: 16, 500 2, 268 $18, 768P = + =

120

120

12

1 112

0.06218,76812

0.0621 112

96.970.4612$210.25

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

17. Loan 1: Daily interest =


× ×= =

Interest for 30 day month = 2.25(30) = $67.50 Accrual period = 4(12) + 9 + 6 = 63 months. Total interest = 67.50(63) = $4,252.50

Loan 2: Daily interest =


× ×= =


Total interest for both loans = 4,252.50 + 3,053.70 = $7,306.20

444

18. Loan 1: Daily interest =


× ×= =

Interest for 30 day month = 0.60(30) = $18 Accrual period = 4(12) + 4 + 3 = 55 months. Total interest = 18(55) = $990

Loan 2: Daily interest =


× ×= =


Total interest for both loans = 990 + 2,269.20 = $3,259.20

19. In Exercise 17, if the first loan is subsidized

and the second is not, then the total interest for both loans would only be the interest on the second loan which is: $3,053.70.

20. In Exercise 18, if the first loan is subsidized

and the second is not, then the total interest for both loans would only be the interest on the second loan which is: $2,269.60.

21. Monthly payment for Mona: Since both loans

were combined, they can be made at 6.8%. 12,100 14, 000 7, 306.20 33, 406.20P = + + =

120

120

12

1 112

0.06833,406.2012

0.0681 112

189.300.4924$384.44

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

22. Monthly payment for Hui Loan 1 ($3,200 at

6.8%):

120

120

12

1 112

0.0684,19012

0.0681 112

23.740.4924$48.22

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

Monthly payment for Hui Loan 2 ($12,400 at 7.2%):

120

120

12

1 112

0.07214,669.212

0.0721 112

88.020.5122$171.84

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

Total of both loan payments = 48.22 + 171.85 = $220.06.

23. a) Daily interest =


× ×= =

Interest per month = 2.44(30) = $73.20

b) Monthly payments:

120

120

12

1 112

0.06813,10012

0.0681 112

74.230.4924$150.76

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

445

c) Accrual period = 5(12) + 6 = 66 months. Total interest = 73.20(66) = $4,831.20 Payments with capitalizing:

120

120

12

1 112

0.06817,931.2012

0.0681 112

101.610.4924$206.36

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

Payment difference = 206.36(120) – [150.76(120) + 73.20(66)] = $1,840.80

24. a) Daily interest =


× ×= =

Interest per month = 1.57(30) = $47.10

b) Monthly payments:

120

120

12

1 112

0.0629,27512

0.0621 112

47.920.4612$103.90

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

c) Accrual period = 14(12) + 3 = 51 months. Total interest = 47.10(51) = $2,402.10 Payments with capitalizing:

120

120

12

1 112

0.06211,677.1012

0.0621 112

60.330.4612$130.81

rPR

r

R

R

R

−

−

⋅=

− +

⋅=

− +

=

=

Payment difference = 130.81(120) – [103.90(120) + 47.10(51)] = $827.10

25. Subsidized loan:

120

120

12

1 112

0.06820,00012

0.0681 112

$230.16

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

Total cost of the loan is 230.16(120) = $27,619.20.

Unsubsidized loan: Principle at end of deferment:

20,000 0.062 (30)(30) 20,000 $23,055.44

365.25×

+ =

120

120

12

1 112

0.06223,055.4412

0.0621 112

$258.28

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=


Private loan: Principle at end of deferment:

20,000 0.06 (30)(30) 20,000 $22,956.88

365.25×

+ =

180

180

12

1 112

0.0622,956.8812

0.061 112

$193.72

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=


Difference in unsubsidized and subsidized: 30,993.60 – 27,619.20 = $3,374.40

Difference in private and subsidized: 34,869.60 – 27,619.20 = $7,250.40

446

26. Subsidized loan:

120

120

12

1 112

0.0740,00012

0.071 112

$464.43

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=


Unsubsidized loan: Principle at end of deferment:

40,000 0.068 (66)(30) 40,000

365.25$54,744.97

×+

=

120

120

12

1 112

0.06854,744.9712

0.0681 112

$630.01

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=



40,000 0.065 (66)(30) 40,000

365.25$54,094.46

×+

=

168

168

12

1 112

0.06554,094.4612

0.0651 112

$491.23

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=


The difference between the unsubsidized loan and the subsidized loan is 75,601.20– 55,731.60 = $19,869.60.

The difference between the subsidized loan and the private loan is 82,526.64– 55,731.60 = $26,795.04.

27. Since the loans are subsidized, your total

principle to pay back at the end of deferment is: 7,100(4) = $28,400.

120

120

12

1 112

0.06828,40012

0.0681 112

$326.84

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

28. Subsidized loans:

120

120

12

1 112

0.06821,20012

0.0681 112

$243.97

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=


4,000 0.075 (15)(30) 4,000365.25

$4,369.61

×+

=

120

120

12

1 112

0.0754,369.6112

0.0751 112

$51.87

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

Total monthly payment = 51.86 + 243.97 = $295.83.

447

29. (a) 0.15 × $145,000 = $21,750 (b) $145,000 − $21,750 = $123,250

(c) $123,250 7.07 $871.38$1,000

× =

(d) ($871.38 × 12 × 25) − $123,250 = $138,164

30. (a) 0.05 × $182,500 = $9,125

(b) $182,500 − $9,125 = $173,375

(c) $173,375 8.99 $1,558.64$1,000

× =

(d) ($1,558.64 × 12 × 15) − $173,375 = $107,180.20

31. (a) 0.40 × $200,000 = $80,000

(b) $200,000 − $80,000 = $120,000

(c) $120,000 6.00 $720.00$1,000

× =

(d) ($720 × 12 × 30) − $120,000 = $139,200

32. (a) 0.12 × $125,000 = $15,000

(b) $125,000 − $15,000 = $110,000

(c) $110,000 $7.39 $812.90$1,000

× =

(d) ($812.90 × 12 × 25) − $110,000 = $133,870

33. (a) 0.10 × $325,000 = $32,500

(b) $325,000 − $32,500 + $3,200= $295,700

(c) $295,700 $6.58 $1,945.71

$1,000× =

(d) ($1,945.71 × 12 × 40) − $295,700 = $638,240.80

34. (a) 0.22 × $175,000 = $38,500

(b) $175,000 − $38,500 + $2,700 = $139,200

(c) $139,200 $7.46 $1,038.43

$1,000× =

(d) ($1,038.43 × 12 × 20) − $139,200 = $110,023.20

35. (a) 0.30 × $1,200,000 = $360,000

(b) $1,200,000 − $360,000 = $840,000

(c) $840,000 $6.88 $5,779.20$1,000

× =

(d) ($5,779.20 × 12 × 20) − $840,000 = $547,008

36. (a) 0.25 × $550,000 = $137,500

(b) $550,000 − $137,500 = $412,500

(c) $412,500 $5.50 $2,268.75$1,000

× =

(d) ($2,268.75 × 12 × 40) − $412,500 = $676,500

37. Monthly payment:$123,250 8.99 $1,108.02

$1,000× =

($1,108.02 × 12 × 15) − $123,250 = $76,193.60

138,164 – 76,193.60 = $61,970.40 less interest on the 15-year loan.

38. Monthly payment$173,375 6.65 $1,152.94

$1,000× ≈

(d) ($1,152.94 × 12 × 30) − $173,375 = $241,683.40

$241,683.40 - $107,180.20 = $134,503.20 more interest on the 30-year loan.

39. Monthly payment: $295,700 $6.32 $1,868.82

$1,000× =

($1,868.82 × 12 × 30) − $295,700 = $377,075.20

638,240.80 – 377,075.20 = $261,165.60 less than the 40-year loan.

448

40. Monthly payment:

$139,200 $8.44 $1,174.85

$1,000× =

($1,174.85 × 12 × 15) − $139,200 = $72,273

110,023.20 – 72,273 = $37,750.20 less than the 20-year mortgage.

41. Down payment: 0.15 × $232,000 = $34,800 Mortgage amount: $232,000 − $34,800 = $197,200

Monthly payment: r = 0.0675, n = 12, t = 20

240

1 1

0.0675$197,20012

0.06751 112

$1,499.44

nt

rPnRrn

−

−

=

− +

=

− +

≈



300

1 1

0.053$247,50012

0.0531 112

$1,490.45

nt

rPnRrn

−

−

=

− +

=

− +

≈

43 Down payment: 0.12 $163, 000 $19, 560× = Mortgage amount: $163,000 − $19,560 = $143,440 Monthly payment: r = 0.0675, n = 12, t = 18

216

1 1

0.0675$143,44012

0.06751 112

$1,148.90

nt

rPnRrn

−

−

=

− +

=

− +

≈



180

1 1

0.0575$231,20012

0.05751 112

$1,919.91

nt

rPnRrn

−

−

=

− +

=

− +

≈



324

1 1

0.0635$652,50012

0.06351 112

$4,215.22

nt

rPnRrn

−

−

=

− +

=

− +

≈

46. Down payment: 0.05 × $162,000 = $8,100 Mortgage amount: $162,000 − $8,100

449

= $153,900


216

1 1

0.056$153,90012

0.0561 112

$1,132.46

nt

rPnRrn

−

−

=

− +

=

− +

≈

47. Mortgage amount:$327,000


120

1 1

0.067$327,00012

0.0671 112

$3,746.38

nt

rPnRrn

−

−

=

− +

=

− +

≈



180

1 1

0.072$337,50012

0.0721 112

$3,071.41

nt

rPnRrn

−

−

=

− +

=

− +

≈

49. Step 1 1123,250(0.07)

12718.96

I Prt= =

≈

Step 2 871.38 − 718.96 = 152.42

Step 3 123,250 − 152.42 = 123,097.58

Step 4 1123,097.58(0.07) 718.0712

I = ≈

Step 5 871.38 − 718.07 = 153.31

Step 6 123,097.58 − 153.31 = 122,944.27

Step 7 1122,944.27(0.07) 717.1712

I = ≈

Step 8 871.38 − 717.17 = 154.21

Step 9 122,944.27 − 154.21 = 122,790.06

Payment number

Interest

Payment on Principal

Balance of Loan

1 $718.96 $152.42 $123,097.58

2 $718.07 $153.31 $122,944.27

3 $717.17 $154.21 $122,790.06

50. Step 1 1173,375(0.07)

121,011.35

I Prt= =

≈

Step 2 1,558.64 − 1,011.35 = 547.29

Step 3 173,375 − 547.29 = 172,827.71

Step 4 1172,827.71(0.07) 1,008.1612

I = ≈

Step 5 1,558.64 − 1,008.16 = 550.48

Step 6 172,827.71 − 550.48 = 172,277.23

Step 7 1172,277.23(0.07) 1,004.9512

I = ≈

Step 8 1,558.64 − 1,004.95 = 553.69

Step 9 172,277.23 − 553.69 = 171,723.54

Payment number

Interest

Payment on

Principal

Balance of Loan

1 $1,011.35 $547.29 $172,827.71

450

2 $1,008.16 $550.48 $172,277.23

3 $1,004.95 $553.69 $171,723.54

51. Step 1 1840,000(0.055)

12$3,850

I Prt= =

≈

Step 2 5,779.20 − 3,850 = 1,929.20

Step 3 840,000 − 1,929.20 = 838,070.80

Step 4 1838,070.80(0.055)12

3,841.16

I =

≈

Step 5 5,779.20 − 3,841.16 = 1,938.04

Step 6 838,070.80 −1,938.04 = 836,132.76

Step 7 1(836,132.76)(0.055)12

3,832.28

I =

≈

Step 8 5,779.20 − 3,832.28 = 1,946.92

Step 9 836,132.76 −1,946.92 = 834,185.84

Payment number

Interest

Payment on

Principal

Balance of

Loan

1 $3,850 $1,929.20 $838,070.80

2 $3,841.16 $1,938.04 $836,132.76

3 $3,832.28 $1,946.92 $834,185.84

52. Step 1 1412,500(0.06)

122,062.50

I Prt= =

=

Step 2 2,268.75 − 2,062.50 = 206.25

Step 3 412,500 − 206.25 = 412,293.75

Step 4 1412,293.75(0.06) 2,061.4712

I = ≈

Step 5 2,268.75 − 2,061.47 = 207.28

Step 6 412,293.75 − 207.28 = 412,086.47

Step 7 1412,086.47(0.06) 2,060.4312

I = =

Step 8 2,268.75 − 2,060.43 = 208.32

Step 9 412,086.47 − 208.32 = 411,878.15

Payment number

Interest

Payment on

Principal

Balance of Loan

1 $2,062.50 $206.25 $412,293.75

2 $2,061.47 $207.28 $412,086.47

3 $2,060.43 $208.32 $411,878.15

53.

360

360

1 1

0.049121,300

0.0491 112

0.0491,300 1 112

0.04912

244,947

nt

rPnRrn

P

P

P

−

−

−

=

− +

=

− +

− + =

≈Maximum amount is 244,947 + 14,000 – 3,000 = $255,947.

451

54.

180

180

1 1

0.042121,300

0.0421 112

0.0421,300 1 112

0.04212

173,391

nt

rPnRrn

P

P

P

−

−

−

=

− +

=

− +

− + =

≈

Maximum amount is 173,391 + 14,000 – 3,000 = $184,391.

55. Answers may vary. Exact amount for 30-year loan: 1,300(30)(12) – 244,947 = $223,053 Exact amount for 15-year loan: 1,300(15)(12) – 173,391 = $60,609 Difference: 205,947 – 153,891 = $162,444 56. 30-year loan:

360

360

1 1

0.049121,500

0.0491 112

0.0491,500 1 112

0.04912

282,631

nt

rPnRrn

P

P

P

−

−

−

=

− +

=

− +

− + =

≈Maximum amount is 282,631 + 14,000 – 3,000 = $293,631.

The difference between this and the 30-year loan with a maximum payment of $13,00 is 293,631 – 255,947 = $37,684.

15-year loan:

180

180

1 1

0.042121,500

0.0421 112

0.0421,500 1 112

0.04212

200,067

nt

rPnRrn

P

P

P

−

−

−

=

− +

=

− +

− + =

≈Maximum amount is 200,067 + 14,000 – 3,000 = $211,067

The difference between this and the 15-year loan with a maximum payment of $13,00 is 211,067 – 184,391 = $26,676.

57. 25% down payment option:

Down payment: 0.25 × $180,000 = $45,000 Mortgage amont: $180,000 − $45,000 = $135,000

Monthly payment: $135,000 $9.00 $1,215$1,000

× =

Total interest paid:

($1,215 × 12 × 20) - $135,000 = $156,600

10% down payment option: Down payment: 0.10 × $180,000 = $18,000 Mortgage amont: $180,000 − $18,000 = $162,000 Monthly payment: $162,000 $7.70 $1,247.40

$1000× =

Total interest paid: ($1,145 × 12 × 25) - $162,000 = $181,602.

If you can afford the higher payment of the 20-year mortgage at 9% after making the 25% down payment, then that is the better option since the total interest paid is less. However, if you can only manage a 10% down payment, or need lower monthly payments, the 25-year mortgage at 7% would be the better choice.

452

58. Suppose a principal of P is borrowed in both cases. For the 30-year mortgage the monthly payment would be

6.32 0.006321,000

PR P= × ≈

The total interest paid would be 360 2.2752 1.2752R P P P P× − = − =

For the 15-year mortgage the monthly payment would be

10.75 0.010751,000

PR P= × =

The total interest paid would be 180 1.935 0.935R P P P P× − = − =

For the 30-year mortgage the cost is about 128% of P while for the 15-year mortgage, the cost is about 94% of P. So, despite the higher interest rate, the 15-year loan costs less.

59. Total interest in Exercise 23(b) is 150.76(120) + 73.20(66) – 13,100 = 9,822.40. The interest

paid is 9,822.40 75%13,100

≈

Total interest in Exercise 29 is $138,164. The

interest paid is 138,164 112.1%123,250

≈ of the loan

amount.

60. Total interest for Exercise 37 is $76,193.60.

The interest paid is 76,193.60 61.8%123,250

≈ of the

loan amount. It is a higher than the student loan at 10 years in Exercise 59 but much less than the mortgage at 25 years. The percentage of the borrowed amount represented by interest paid goes up dramatically as the length of the loan increases.

Exercise Set 8-6 1. Answers may vary.

2. Stocks are partial ownership in the company, bonds are more like a loan to the company.

3. A mutual fund consists of a large number of small investments in various companies by a group of investors and is usually managed by a professional manager.

4. The P/E ratio is the price to earnings ratio. It means that the price of a share of stock is P/E times the company’s annual earnings per share.

5. A stockbroker is a person who buys and sells stock on behalf of an investor at a stock exchange.

6. Sale price is how much is paid for the stock, proceeds are the difference between the cost of the stock and the stockbroker’s commission, and profit (or loss) is the difference between the proceeds and the selling price.

7. Stocks tend to fall in price when the economy is bad. Also, when the economy is bad, the money you earn on bonds is “worth more,” therefore the cost of them is also higher.

8. If an investor needed cash, he might be willing to sell bonds for less than he paid for them.

9. No, if the company fails, you are out of luck.

10. Pretty simple: people like money. Your potential earnings on stocks exceed your potential earnings on bonds.

11. The 52-week high is $97.25 per share.

12. The 52-week low is $57.50 per share.

13. The company paid $1.23 per share in dividends last year.

14. If you owned 175 shares you would have made 175($1.23) = $215.25 last year.

15. There were 4,626 × 1,000 = 4,626,000 shares traded yesterday.

16. Yesterday’s closing price was $62.06.

17. The annual earnings per share is yesterday’s closing price divided by the

P/E ratio = $62.06 ÷ 7 ≈ $8.87 per share.

18. The total cost is the cost per share times the number of shares purchased, plus1.5% = 480(62.06) + 0.015(480(62.06)) ≈ $30,235.63.

453

19. The investor received 623($1.23) = $766.29.

20. The closing price yesterday was $62.06, which was up $0.77 from the day before. So the closing price the day before yesterday was $62.06 − $0.77 = $61.29

21. The 52-week high was $40.08 per share.

22. The 52-week low was $24.75 per share.

23. The amount of the dividend per share the company paid last year was $0.04.

24. If you owned 357 shares you would have made 357($0.04) = $14.28.

25. There were 345 × 1,000 = 345,000 shares traded yesterday.


27. The annual earnings per share is yesterday’s closing price divided by the

P/E ratio = $29.79 ÷ 20 ≈ $1.49 per share.

28. The total cost is the cost per share times the number of shares purchased, plus 2.6% = 1,247(29.79) + 0.026(1,247(29.79))

≈ $38,113.98.

29. The investor received 1,562($0.10) = $156.20.

30 $0.39 from the day before. The closing price the day before yesterday was

$29.79 − $0.39 = $29.40.

31. The 52-week high was $50.87 per share.

32. The 52-week low was $42.31 per share.

33. The amount of the dividend per share the company paid last year was $0.67.

34. If you owned 682 shares you would have made 682($0.67) = $456.94.

35. There were 9,662 × 1,000 = 9,662,000 shares traded yesterday.


37. The annual earnings per share is yesterday’s closing price divided by the P/E ratio

= $48.12 ÷ 19 ≈ $2.53 per share.

38. The total cost is the cost per share times the number of shares purchased, plus 2%

= 842($52.67) + 0.02 (842($52.67)) ≈ $45,235.10.

39. The investor made 1,225($0.67) = $820.75.

40. Yesterday’s closing price of $48.12 is down $0.10 from the day before. The closing price the day before yesterday was

$48.12 + $0.10 = $48.22.

41. P/E ratio is closing price divided by annual earnings: $21.92 ÷ $0.88 ≈ 24.91.

42. P/E ratio is closing price divided by annual earnings: $6.65 ÷ $0.35 = 19.




= $18.53 ÷ 55 ≈ $0.34 per share.


= $75.66 ÷ 25 ≈ $3.03 per share.


= $25.76 ÷ 13 ≈ $1.98 per share.


= $43.73 ÷ 30 ≈ $1.46 per share.

49. Purchase price: 800($63.25) = $50,600 Selling price: 800($65.28) = $52,224 Broker’s fee for purchase: 0.02($50,600) = $1,012 Broker’s fee for sale: 0.02($52,224) =

$1,044.48 Profit: 52,224 − 50,600 −1,012 −1,044.48 = −432.48 The investor lost $432.48 on the transaction.

454

50. Purchase price: 200($93.75) = $18,750 Selling price: 200($89.50) = $17,900 Broker’s fee for purchase: 0.025($18,750) = $468.75 Broker’s fee for sale: 0.025($17,900) = $447.50 Profit: 17,900 − 18,750 − 468.75 − 447.50 = −1,766.25. The investor lost $1,766.25 on the transaction. 51. Purchase price: 550($51.60) = $28,380 Selling price: 550($49.70) = $27,335 Broker’s fee for purchase: 0.02($28,380) = $567.60 Broker’s fee for sale: 0.015($27,335) = $410.03 Profit: 27,335 − 28,380 − 567.60 − 410.03 = −2,022.63. She lost $2,022.63 on the transaction. 52. Purchase price: 670($73.20) = $49,044

Selling price: 670($82.35) = $55,174.50 Broker’s fee for purchase: 0.025($49,044) = $1,226.10 Broker’s fee for sale: 0.025($55,174.50) = $1,379.36 Profit: 55,174.50 − 49,044 − 1,226.10 − 1,379.36 = 3,525.04. He earned $3,525.04 on the transaction.

53. You: Purchase price: 500($22) = $11,000 Dividends paid: 500($1.70)(3) = $2,550 Selling price: 500($38) = $19,000 Profit: $19,000 + $2,550 – $11,000 = $10,550. Friend: Purchase price: 500($20) = $10,000 Selling price: 500($38) = $19,000 Profit: $19,000 – $10,000 = $9,000. Your stock that paid dividends made you

$10,550 – $9,000 = $1,550 dollars richer than your friend.

54. Purchase price: 600(12.89) + 39.95 =

$7,773.95. Selling price: 600(530.38) – 39.95 =

$318,188.05. Her profit would be $318,188.05 – 7,773.95 =

$310,414.10. The difference in profits is $310,414.10 –

303,974.76 = $6,439.34

55. Number of shares = 20,000 ÷ 1.43 = 13,986 Purchase price: = 13,986(1.43) + 19.99 =

$20,019.97. Selling price: = 13,986(8.41) – 19.99 =

$117,602.27. Profit: 117,602.27 – 20,019.97 = $97,582.30. Profit per day: 97,582.30 ÷ 365 = $267.35. 56. a) Selling price for smart person:

13,986(18.65) – 19.99 = $260,818.91 Profit for smart person: 260,818.91 –

20,019.97 = $240,798.94. Difference in being stupid and smart:

240,798.94 – 97,582.30 = $143,216.64. b) Answers may vary. 57.

1,000(0.049)(5)245

I PrtI

===

Value at maturity: $1,245 58.

5,000(0.057)(10)2,850

I PrtI

===


902,500(0.0619)12

1,160.63

I Prt

I

=

=

=

Value at maturity: $3,660.63 60.

423,750(0.0379)12

497.44

I Prt

I

=

=

=

Value at maturity: $4,247.44 61. $1,245 – $925 = $320 62. $7,850 – $4,000 = $3,850 63. $3,660.63 – $2,800 = $860.63 64. $4,247.44 – $4,190 = $57.44

65. 320 34.6%34.6%; 11.5%925 3

= =

66. 3,850 96.25%96.25%; 27.5%4,000 3.5

= =

455

67. 860.63 30.7%30.7%; 5.9%622,80012

= =

68. 57.44 1.37%1.37%; 1.6%104,19012

= =

69. Investment 1:

365(2)

1

$10,000, 0.051, 365, 2

0.051$10,000 1 $11,073.76365

ntrA Pn

P r n t

A

= +

= = = =

= + ≈

The investment earns $1,073.76.

Investment 2:

Purchase price: 1,400($7.11) = $9,954

Broker’s fee for purchase: 0.0075($9,954) ≈ $74.66

Total price paid: $10,028.66

Dividend for the first year: 1,400($0.48) = $672

Dividend for the second year: 1,400($0.36) = $504

Total dividend paid: $1,176

Sale Price: 1,400($7.95) = $11,130

Broker’s fee for sale: 0.0075($11,130) ≈ $83.48

Net proceeds for sale: $11,046.52

The investment made: $11,046.52 + $1,176 − $10,028.66 = $2,193.86

The second investment would be the better choice.


71. 10,000(0.0645)(12)7,740

I PrtI

===

a)Value at maturity: $17,740 Larry: 11,400 – 10,000 = $1,140 Curly: 14,950 – 11,400 = $3,550 Moe: 17,740 – 14,950 = $2,790 Curly made the most profit at $3,350!

b) Larry: 1,140 11.4%11.4%; 4.1%

3310,00012

= =

Curly: 3,550 31.1%31.1%; 4.4%

11,400 7= =

Moe: 2,790 18.7%18.7%; 8.3%2714,95012

= =

Curly had the greatest percent return at 31.1%. c) Moe had the greatest percent return per year

at 8.3%. 72. Answers may vary. 73. Answers may vary. 74. Answers may vary. 75. Answers may vary. 76. Answers may vary.

456

Review Exercises

1. 7 0.875 87.5%8

= =

2. 54 270.54 54%

100 50= = =

3. 185 37185% 1.85100 20

= = =

4. 6 30.06 6%

100 50= = =

5. 3 5755 5.75 575%4 100

= = =

6. 45.5 9145.5% 0.455100 200

= = =

7. ___is 72% of 96 x = 0.72 × 96 x = 69.12

8. 18 is ___% of 60 18 60

0.3xx

= ×=

The percent is 30%.

9. 275 is 25% of _____ 275 0.25

275 0.250.25 0.25

1,100

xx

x

= ×

=

=

10. ____ is 5% of $19.95 0.05 19.95 0.9975 1.00 x = × = ≈

19.95 + 1 = 20.95 The tax is $1.00 and the total cost is $20.95.

11. 3.60 is 6% of ____ 3.6 (0.06)

3.6 0.060.06 0.06

60

xx

x

= ×

=

=

The cost of the table is $60.

12. 2,275 is 7% of ___ 2,275 (0.07)2,275 0.070.07 0.07

32,500

xx

x

= ×

=

=

The price of the home is $32,500.

13. Amount of increase: 5.9 − 3.4 = 2.5

Percent increase: 2.5 0.735 73.5%3.4

≈ =

The increase in credit card offers in the five year period was 73.5%.

14. Amount of decrease: 3,147 − 2,226 = 921

Percent decrease: 921 0.293 29.3%3,147

≈ =

The percent decrease in adolescents under 18 in state prisons was 29.3%.

15. a) The 10% is taken off the discount price, not the original.

b) Cost of jeans at 30% off: 71.50 – 0.30(71.50) = $50.05.

Cost after additional 10% off: 50.05 – 0.10(50.05) = $45.05.

16. $7,326 + 0.12(7,326) = $8,205.12

17. I = Prt = (4,300)(0.09)(6) = 2,322 The simple interest is $2,322.

18. 1920 (16,000)( )(3)0.04

I Prtr

r

===


19. 262.5 (875)(0.12)( )

2.5

I Prtt

t

===

The time is 2.5 years. 20. I = Prt = (50)(0.06)(1.5) = 4.5

The simple interest is $4.50.

21. 104.65 (230)( )(6.5)

0.07

I Prtr

r

===


22. 63.75 (0.03)(5)

425

I PrtPP

===

The principal is $425.

23. 385 (0.14)(2)

1,375

I PrtPP

===


24. 1,130.4 (785)(0.12)( )

12

I Prtt

t

===


457

25. I = Prt = 6,000(0.06)(5) = 1,800 A = 6,000 + 1,800 = 7,800 The simple interest is $1,800 and the future value is $7,800.

26. I = Prt = 13,450(0.08)(15) = 16,140 A = 13,450 + 16,140 = 29,590 The simple interest is $16,140 and the future value is $29,590.

27. 60.48 (4,320)( )(1)0.014

I Prtr

r

===


28. 36(5,300)(0.11) 1,74912

I Prt = = =

A = 5,300 + 1,749 = 7,049 7,049 195.81

36≈

The monthly payments are $195.81.

29. ( ) 80 $2,300 0.05 $25.56360

I Prt = = ≈

30. ( ) 100 $ 8,750 0.085 $206.60360

I Prt = = ≈

31. Discount: $6,000(0.06)(3) = $1,080 David received $6,000 − $1,080 = $4,920 32. Discount: $9,250(0.12)(4) = $4,440 Marla received $9,250 − $4,440 = $4,810

33. 1 6

1

0.051,775 11

2,378.67

ntrA Pn

⋅

= +

= +

≈

I = 2,378.67 − 1,775 = 603.67 The interest is $603.67 and the future value is $2,378.67.

34. 2100.0421 200 1 303.07

2

ntrA Pn

⋅ = + = + ≈


35. 424120.03041 45 1 50.03

4

ntrA Pn

⋅ = + = + ≈

I = 50.03 − 45 = 5.03

The interest is $5.03 and the future value is $50.03.

36. 741212

1

0.051921,000 112

28,901.18

ntrA Pn

⋅

= +

= +

≈


37. 40.121 1 1 1 0.1255

4

nrEn

= + − = + − ≈

The effective rate is 12.55%.

38. Effective rates:

20.043

1 1 0.0434 4.34%2

+ − ≈ =

3650.0427

1 1 0.0436 4.36%365

+ − ≈ =

4.27% daily is the better investment.

39. 12(3)

12(3)

1

0.037510,000 112

10,000

0.0375112

8,937.54

ntrA Pn

P

P

P

= + = +

= +

≈

You would need to invest $8,937.54 now in order to have $10,000 in 3 years.

40.

( )

( ) ( )( ) ( )

( )( )

4( )

4

4

1

0.04620,000 12,500 14

20,000 1.011512,500

log 1.6 log 1.0115log 1.6 4 log 1.0115

log 1.64log 1.0115

10.3

nt

t

t

t

rA Pn

t

t

t

= +

= +

=

==

=

≈

Charles needs to wait about 10.3 years to get his 20K.

458

41. Compounded semiannually:

( )( )

( )

( )

2( )

2( )

2( )

1

0.0842 12

2 1.042log 2 log 1.042log 2 2 log 1.042

log 22log 1.042

8.42

nt

t

t

t

rA Pn

P P

t

t

t

= + = +

===

=

≈

It would take about 8.42 years to double if the compounding is semiannual.

Compounded daily:

365( )

365( )

365( )

1

0.0842 1365

0.0842 1365

0.084log 2 log 1365

0.084log 2 365 log 1365

log 20.084365log 1365

8.25

nt

t

t

t

rA Pn

P P

t

t

t

= + = +

= +

= +

= +

= +

≈

It would take about 8.25 or 8.42 – 8.25 = 0.17 years less to double if it were compounded daily. 0.17 years ≈ 62 days.

42. n = 2, t = 7, 0.083 0.0415

2rn

= =

( )2(7)

1 1

4,000 (1 0.0415) 1

0.041573,925.08

ntrRn

Arn

+ − =

+ −=

≈


43.

( )( )4(3)

9,000, 4, .03, 3 0.03 0.0075

4

1 1

9,000 0.0075

1 0.0075 1719.56

nt

A n r trn

rAnR

rn

R

= = = =

= =

=

+ −

=+ −

≈

The monthly payment needed is about $719.56.

44. Cost of car = $12,942.49 Amount financed = Cost of car – down


down payment: (261.34)(36) + 4,300 = $13,708.24.

Finance charge = installment price – cost of car: 13,708.24 – 12,942.49 = $765.75.

45. Amount financed = 22,400 – 8,100 +

0.0625(22,400 – 8,100) + 325 = $15,518.75. Cost of car = amount financed + down

payment = 15,518.75 + 8,100 = $23,618.75 Total installment price = total of payments +

down payment: (290.40)(60) + 8,100 = $25,524.

Finance charge = installment price – cost of car: 25,524 – 23,618.75 = $1,905.25.

459

46. Down payment: $750(0.15) = $112.50 Amount financed: $750 − $112.50 = $637.50 Finance charge: $637.50(0.06) = $38.25 Total installment price: $637.50 + $38.25 + $112.50 = $788.25

Monthly payment: $637.50 $38.25 $84.47

8+

=

47. Amount financed: $10,900 − $1,000 = $9,900 Finance charge: $310(36) − $9,900 = $1,260 Finance charge per $100:

$1,260 $100 $12.73$9,900

× =

APR from Table 8-1: 8%


$4,160 $100 $28.69$14,500

× ≈

APR from Table 8-1: 10.5%

49. Down payment: (given) $8,000. Amount Financed: (purchase price minus

down payment) $149,500 – $8,000 = $141,500.

Finance charge: (8.5% of amount financed × 25 years) 0.085($141,500) × 25 = $300,687.50

Monthly payment: (amount financed plus finance charge divided by number of payments) ($141,500 + $300,687.50) ÷ (25 × 12) = $442,187.50 ÷ 300 = $1,473.96.

50. k = 12, R = $310, h = $4.39

10012(310)4.39 $156.44100 4.39

kRhuh

u

=+

= ≈+

The unearned interest is $156.44 and the payoff amount is 13($310) − $156.44= $3,873.56.

51. k = 24, R = $311, h = $11.30

10024(311)11.30 $757.80100 11.30

kRhuh

u

=+

= =+

The unearned interest is $757.80 and the payoff amount is 25($311) − $757.80 = $7,017.20.

52. f = (30 × 61.25) − 1,500 = 337.50 k = 30 − 24 = 6

( 1) 337.5(6)(6 1) 15.24( 1) 30(30 1)

fk kun n

+ += = ≈

+ +

$15.24 is saved.

53. I = Prt = (563.25)(0.0175)(1) ≈ 9.86 The finance charge is $9.86. New balance 563.25 9.86 563.25 350

786.36= + + −=


54. Date Balance Days Balance × days

4-1 5,628.00

10 − 1 = 9

50,652

4-10

5,628 + 2,134.60 = 7,762.60

22 − 10 = 12

93,151.20

4-22

7,762.60 − 900 = 6,862.60

28 − 22 = 6 41,175.60

4-28

6,862.60 + 437.80

= 7,300.40

30 − 28 + 1

= 3 21,901.2

Total 30 206,880

$206,880Average daily balance $6,896

30= =

(6,896)(0.018) ≈ 124.13 The finance charge is $124.13. 7,300.40 + 124.13 = 7,424.53 The new balance is $7,424.53.

460

55. Month 1: 2% of 845.32 = 16.91 845.32 – 16.91 = $828.41 Month 2: 2% of 828.41 = 16.57 828.41 – 16.57 = $811.84 Month 3: 2% of 811.84 = 16.24 811.84 – 16.24 = $795.60 Month 4: 2% of 795.60 = 15.91 795.60 – 15.91 = $779.69 Month 5: 2% of 779.69 = 15.59 779.69 – 15.59 = $764.10 Month 6: 2% of 764.10 = 15.28 764.10 – 15.28 = $748.82 56. a) 30-day month:

4,700 0.068 (30) $26.25365.25

×=

31-day month: 4,700 0.068 (31) $27.13

365.25×

=

b) Payment will begin in December 2013. There are 822 days in which interest will accrue.

4,700 0.068 (822) $719.26

365.25×

=

57. Monthly payment for Latwan’s loan:

120

120

12

1 112

0.0685,419.2612

0.0681 112

$62.37

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

58. Monthly payment if it is subsidized:

120

120

12

1 112

0.0684,70012

0.0681 112

$54.09

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

59. If the loan is not subsidized and he capitalizes interest, Latwan will pay 62.37(120) – 4,700 = $2,784.40 in interest.

If the loan is not subsidized and Latwan does not capitalize interest, his total interest payment would be: 54.09(120) – 4,700 + 719.26 = $2,510.06.

So, he would pay an additional 2,784.40 – 2,510.06 = $274.34 by capitalizing.

60. (a) 0.20 × $145,000 = $29,000

(b) $145,000 − $29,000 = $116,000

(c) $116,000 $8.05 $933.80$1,000

× =

(d) Step 1 1116,000(0.085)

12821.67

I Prt= =

≈

Step 2 933.80 − 821.67 = 112.13 Step 3 116,000 − 112.13 = 115,887.87

Step 4 1115,887.87(0.085)

12820.87

I =

≈

Step 5 933.80 − 820.87 = 112.93 Step 6 115,887.87 − 112.93

= 115,774.94

Payment number Interest

Payment on

Principal

Balance of Loan

1 $821.67 $112.13 $115,887.87

2 $820.87 $112.93 $115,774.94

461

61. Down payment: 0.08 × $252,000 = $20,160

Mortgage amount: $252,000 − $20,160 = $231,840


300

1 1

0.0825$231,84012

0.08251 112

$1,827.94

nt

rPnRrn

−

−

=

− +

=

− +

=

62. $116,000 $9.85 $1,142.60

$1,000× =

Total of payments for 25-year loan is: 933.80(25)(12) = $280,140.

Total of payments for 15-year loan is: 1,142.60(15)(12) = $205,668.

The difference is: 280,140 – 205,668 = $74,472.

63. The 52-week high was $34.82 per share and the 52-week low was $27.09 per share.

64. You received 475($0.72) = $342

65. There were 5,528 × 1,000 = 5,528,000 shares of stock traded yesterday.

66. Yesterday’s closing price of $32.79 was down $0.25 from the day before, so the closing price the day before yesterday was

$32.79 + $0.25 = $33.04.

67. Annual earnings is closing price divided by P/E ratio: $32.79 ÷ 30 ≈ $1.09 per share.

68. Purchase price: 90($86.43) = $7,778.70 Broker’s fee for purchase: 0.02($7,778.70) = $155.57 Selling price: 90($92.27) = $8,304.30 Broker’s fee for sale: 0.02($8,304.30) = $166.09. Profit: 8,304.30 − 7,778.70 − 155.57 − 166.09

= 203.94. She earned $203.94 on her investment.

69. Answers may vary. 70.

( )10,000(0.042) 41,680

I PrtI

=

=

=


( )1,000(0.053) 6318

I PrtI

=

=

=

Value at maturity: $1,318 Original owner: 1,145 – 1,000 = $145 profit Buyer: 1,318 – 1,145 – 0.01(1,145) = $161.55

profit. The buyer makes 161.55 – 145 = $16.55 more

profit. 72. The buyers profit was $161.55, which is

161.55 14%1,145

= return on investment.

The percent return per year is: 14% 4.2%4012

= .

Chapter Test

1. 5 0.3125 31.25%

16= =

2. 0.63 = 63%

3. 28 7 4 728%

100 25 4 25⋅

= = =⋅

4. 16.7% = 0.167

462

5. 32 is ____% of 40

32 4032 4040 400.8

xx

x

= ×

=

=

80% of the people had made at least one purchase that day.

6. ____ is 89.6% of 48. x = 0.896 × 48

x = 43 43 states have a lottery.

7. 68 is 19.6% of ____

68 0.19668 0.196

0.196 0.196347

xx

x

= ×

=

≈

There are 347 total teams.

8. ___ is 8% of $29.95. x = 0.08 × 29.95 x ≈ 2.40 The sales tax is $2.40.

The total cost of the toaster with tax is 29.95 + 2.40 = $32.35

9. $385.20 is 15% ____

385.2 0.15385.2 0.150.15 0.15

2,568

xx

x

= ×

=

=

The salesperson sold $2,568 worth of merchandise.

10. 7Percent increase 0.25 25%28

= = =

The percent increase is 25%.

11. I = Prt I = (1,350)(0.12)(3) = 486 The simple interest is $486.

12. 150 (200)( )(15)0.05

I Prtr

r

===


13. I = Prt = (435)(0.0375)(0.5) ≈ 8.16 A = 435 + 8.16 = 443.16

443.16monthly payment 73.866

= =

The interest is $8.16 and the future value is $443.16. The monthly payment is $73.86.

14. With simple interest, the percentage is always calculated on the principle amount. With compound interest, interest is also computed on previously earned interest.

15. I = Prt = (6,000)(0.072)(4) = $1,728

12 3

1

0.0646,000 112

$7,266.31

ntrA Pn

⋅

= +

= +

≈

Interest earned = 7,266.31 – 6,000 = 1,266.31. You would earn 1,728 – 1,266.31 = $461.69 more with the simple interest.

16. 605,000(0.04) $33.33

360

I Prt

I

=

= ≈

17. Discount: I = Prt = 12,650(0.075)(6) = $5,692.50 Latoya received $12,650 − $5,692.50 = $6,957.50

18.2 40.065 1 500 1 645.79

2

ntrA Pn

⋅ = + = + ≈


19. 4 6

1

0.19,750 14

17,635.08

ntrA Pn

⋅

= +

= +

≈


463

20. I = Prt = 12,000(0.095)(4) = 4,560 A = 12,000 + 4,560 = 16,560

16,560monthly payment 34548

= =

Jayden’s monthly payment is $345.

21. 20.081 1 1 1 0.0816

2

nrEn

= + − = + − =

The effective rate is 8.16%.

22. n = 2; t = 3; 0.045 0.0225

2rn

= =

6

1 1

3,000 (1 0.0225) 1

0.022519,043.39

ntrRn

Arn

+ − =

+ − =

≈


23. Down payment: $935(0.30) = $280.50 Financed amount: $935 − $280.50 = $654.50 Finance charge: ($654.50)(0.10)(1/2) = $32.73 Total Installment price: $654.50 + $32.73 + $280.50 = $967.73 Monthly Payment:


$1,640 $100 $12.62$13,000

× ≈

APR from Table 8-1: 6%. 25. k = 24, R = $305, h = $6.37

10024($305)$6.37 $438.36

100 $6.37

kRhuh

u

=+

= ≈+

The unearned interest is $438.36 and the payoff amount is 25($305) − $438.36 = $7,186.64.

26. f = (24 × 111.85) − 2,200 = 484.40

k = 24 − 20 = 4( 1) 484.4(4)(4 1) 16.15( 1) 24(24 1)

fk kun n

+ += = ≈

+ +

$16.15 is saved.

27. I = Prt = 1,250 (0.016)(1) = 20 The finance charge is $20. New balance 1,250 20 560 800 1,030= + + − = The new balance is $1,030.

28. Date Balance Days Balance × days

5-1 474.00 11 − 1 = 10 4,740

5-11 474 − 300 = 174

20 − 11 = 9 1,566

5-20 174 + 86.50 = 260.50

25 − 20 = 5 1,302.50

5-25 260.50 + 120 = 380.50

31 − 25 + 1 = 7 2,663.50

Total 31 10,272

10,272average daily balance 331.35

31= ≈

(331.35)(0.02) ≈ 6.63 The finance charge is $6.63. 380.50 + 6.63 = 387.13 The new balance is $387.13.

29. a) 30 day month:11,200 0.068 (30) $62.55

365.25×

= .

b) She will owe 11,200 + 54(62.55) = $14,577.70.

c) 120

120

12

1 112

0.06814,577.7012

0.0681 112

$167.76

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

464

30. a) Payments if interest is not capitalized:

120

120

12

1 112

0.06811,20012

0.0681 112

$128.89

rPR

r

R

R

−

−

⋅=

− +

⋅=

− +

=

The difference in payments would be 167.76 – 128.89 = $38.87.

b) If Yasmil does not capitalize interest, her total interest payment would be: 128.89(120) – 11,200 + 62.55(54) = $7,644.50.

If she does capitalize interest, her total interest payment would be: 167.76(120) – 11,200 = $8,931.20.

So, she would pay an additional 8,942 – 7,644.50 = $1,286.70 by capitalizing.

31. (a) 0.05 × $180,000 = $9,000

(b) $180,000 − $9,000 = $171,000

(c) $171,000 $6.00 $1,026$1,000

× =

(d) Step 1 1(171,000)(0.06)

12855

I Prt= =

=

Step 2 1,026 − 855 = 171 Step 3 171,000 − 171 = 170,829

Step 4 1(170,829)(0.06)

12854.15

I =

≈

Step 5 1,026 − 854.15 = 171.85 Step 6 170,829 − 171.85 = 170,657.15

Payment number Interest

Payment on

Principal

Balance of Loan

1 $855 $171 $170,829

2 $854.15 $171.85 $170,657.15

32. 15-year term:

$171,000 $8.44 $1,443.24

$1,000× =

The monthly payment would be 1,443.24 – 1,026 = $417.24 more.

Total paid on 30-year term: 1,026(30)(12) = $369,360.

Total paid on 15-year term: 1,443.24(15)(12) = $259,783.20

Difference: 369,360 – 259,783.20 = $109,576.80.

33. The 52-week high is $36.98 and the 52-week low is $23.17.

34. You receive 300($0.20) = $60 in dividends.

35. Yesterday’s closing price of $27.45 is up $0.80 from the day before. So, the closing price the day before yesterday is

$27.45 − $0.80 = $26.65.

36. Annual earnings is closing price divided by P/E ratio: $27.45 ÷ 12 ≈ $2.29 per share.

37. ( )1,000(0.055) 8

440

I PrtI

=

=

=

Value at maturity: $1,440 Original owner: 1,250 – 1,000 = $250 profit Buyer: 1,440 – 1,250 = $190 profit. The original owner made the better

investment: more profit over a shorter time period.

465

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