math 373 chapter 1 homework - purdue universityjbeckley/q/wd/ma373/s… · · 2014-01-30math 373...

MATH 373 Chapter 1 Homework

January 28, 2014

Non-Interest Theory

1. 2000 + 2005 + … + 3500 =

( )(# )2

2000 3500( )(301) 827,750

2

Firstterm Lasttermterms

2. 4 + 16 + 64 + . . . + 16,384=

First term – Next term after last

1 – ratio

=4 - 48

1- 4= 21,844

3. 1 + 0.9 + 0.92 + . . . 0.912 =

1- .913

1- .9= 7.45813

4. If 5(1 ) 1.1,i calculate

5 10 1001 (1 ) (1 ) ... (1 ) .i i i

Geometric series, ratio is (1+i)5 or 1.1

100 20

21

(1 ) (1.1)

1 (1.1)64.0025

1 1.1

i


January 28, 2014

Section 1.3

5. Andrea borrows 10,000 from Stephen. After 3 years, Andrea repays the loan by paying Stephen

12,400.

a. Calculate the principle in this transaction?

b. How much interest does Andrea pay?

c. Determine (0)A .

d. Determine (3)A .

e. Determine (3)a .

a) Principal is the original amount of the loan, so K=10,000

b) Interest = S(the accumulated value) – K(the principal) = 12,400-10,000 = 2400

c) A(0) = K = 10,000

d) A(3) = accumulated value after 3 years = 12,400

e) a(3) = accumulation function = A(3)/A(0) = 12400/10000 = 1.24

6. Pinar lends 500 to Sarah who repays the loan after 4 years. You are given that 2( ) 1 0.03 0.001a t t t . Calculate the amount that Sarah will have to repay at the end of 4

years.

2(4) 1 (.03)(4) (.001)(4) 1.136

500 (4) 500(1.136) 568

a

a

7. You are given that 2( )a t t . You are also given that 1000 invested today will grow to

1360 at the end of 3 years.

Determine and .

2

2

(0) 1 (0) 1 1

1000 (3) 1360 1000(1 (3) ) 1360 1 9 1.36 .04

a

a


January 28, 2014

8. Xinyi invests 2000 in a fund that has an accumulation function of 2t t . At the end of

two years, Xinyi has 2100. At the end of four years, Xinyi has 2280.

Determine the amount that Xinyi has at the end of eight years.

2(0) 1 (0) (0) 1 1

2000 (2) 2000(1 2 4 ) 2100 1 2 4 1.05 2 4 0.05

2000 (4) 2000(1 4 16 ) 2280 1 4 16 1.14 4 16 0.14

(2 4 .05)* 2

4 16 .14

Add the two equations above to get

a

a

a

2

8 0.04 .005

1 2 4(.005) 1.05 .015

2000 (8) 1 (.015)(8) (.005)(8 ) 2880a


January 28, 2014

9. You are given that 2( ) 1 0.03 0.001a t t t .

Calculate:

a. [2,4]i

b. The amount of interest that Keith will earn from time 2 until time 4 assuming he invests

1000 now

c. [3,4]i

d. 4i

e. The annual effective interest rate during the fourth year.

f. [2.5,3.2]i

(4) (2) 1 .03(4) .001(16) [1 .03(2) .001(4)] 1.136 1.064. .067669

(2) 1 .03(4) .001(16) 1.064

a aa

a

b.1000a(4)-1000a(2) =1000(1.136)-1000(1.064) = 72

[3,4]

(4) (3). . We know from above that a(4)=1.136

(3)

(3) 1 .03(3) .001(9) 1.099

1.136 1.099.03366697

1.099

a ac i

a

a

4 [3,4]d. .03366697i i

4 [3,4]e. .0336669 7The annual effective interest rate during the fou i irth year

2 2

[2.5,3.2]

(3.2) (2.5).

(2.5)

(3.2) 1 .03(3.2) .001(3.2) 1.10624 and (2.5) 1 .03(2.5) .001(3.2) 1.08125

1.10624 1.08125.023112

1.08125

a af

a

a a

i


January 28, 2014

10. You are given that:

a. 1 0.05i

b. 2 0.06i

c. 3 0.04i

Calculate (3)a .

(1) (0) (1) 10.05 (1) 1.05

(0) 1

(2) (1) (2) 1.050.06 (2) 1.113

(1) 1.05

(3) (2) (3) 1.1130.04 (3) 1.15752

(2) 1.113

a a aa

a

a a aa

a

a a aa

a

11. Poppy has a choice of two investments. Investment 1 has an accumulation function of 3( ) 1 0.001a t t . Investment 2 has an accumulation function of

2( ) 1 0.01a t t .

a. Which investment should Poppy use if Poppy will be investing for 5 years? Explain why.

b. Which investment should Poppy use if Poppy will be investing for 15 years? Explain why.

c. If Poppy invests 100 into both investments, at what time will the accumulated value of

both investments be equal?

a. Investment 1 would grow to Investment 2 because 31 0.001(5) 1.125 while

Investment 2 would grow to 21 0.01(5) 1.25 . Chose Investment 2 since it would be

greater.

b. Investment 1 would grow to Investment 2 because 31 0.001(15) 4.375 while

Investment 2 would grow to 21 0.01(15) 3.25 . Chose Investment 1 since it would

be greater.

c. 3 2 3 2100(1 0.001 ) 100(1 0.01 ) 0.001 0.01 10t t t t t


January 28, 2014

Section 1.4

12. Kylie invests 1500 in a fund earning simple interest at a rate of 6.4% per year.

a. Write an expression for ( )a t .

b. Write an expression for ( )A t .

c. Determine the amount of interest that Kylie will earn in the first year.

d. Determine the amount of interest that Kylie will earn in the 10th year.

e. How much money will Kylie have after 12 years?

f. Calculate 5i and 10i .

a. ( ) 1 0.064a t t

b. ( ) 1500(1 0.064 )A t t

c. 1500(1 0.064) 1500 96

d. 1500(1 0.064(10)) 1500(1 0.064(9)) 96

e. 1500(1 0.064(12)) 2652

f.

5

10

0.064.050955

1 ( 1) 1 4(0.064)

0.064.040609

1 9(0.064)

ii

n i

i

13. Andrea borrows 10,000 from Stephen. After 3 years, Andrea repays the loan by paying Stephen

12,400. If Andrea is paying simple interest, determine the simple rate of interest.

10000(1 3 ) 12400 1 3 1.24 3 0.24 .08i i i i


January 28, 2014

14. Jack borrows K to buy a new television set. Jack pays simple interest of 15% on the loan. Jack

repays the loan at the end of 8 months with a payment of 819.50.

Determine K .

8(1 (.15)) 189.50 1.1 819.50 745

12K K K

15. Mai invests 1000 in a fund earning simple interest at a rate of %s .

Mai earns an effective rate of interest of 6.25% during the 9th year of the investment.

Determine the amount the Mai will have after 10 years.

9

(9) (8) 1 9 (1 8 )0.0625 0.0625 0.5 0.125

(8) 1 8

1000(1 .125(10)) 2250

a a s si s s s

a s

16. Yang invests in a fund earning simple interest of 10%. The effective interest rate for the nth year (in) is 5%. Calculate n.

( ) ( 1) 1 .1 (1 0.1( 1))0.05 0.05 0.005( 1) 0.1

( 1) 1 0.1( 1)

0.005( 1) 0.05 11

a n a n n nn

a n n

n n


January 28, 2014

Section 1.5

17. Kylie invests 1500 in a fund earning compound interest at a rate of 6.4% per year.

a. Write an expression for ( )a t .

b. Write an expression for ( )A t .

c. Determine the amount of interest that Kylie will earn in the first year.

d. Determine the amount of interest that Kylie will earn in the 10th year.

e. How much money will Kylie have after 12 years?

f. Calculate 5i and 10i .

a. a(t) = (1.064)t

b. A(t) =1500(1.064)t

c. 1500(1.064)-1500 = 96

d. 1500(1.064)10 -1500(1.064)9 =167.7822

e. 1500(1.064)12 = 3157.84

f. i5 = 6.4%

i10 = 6.4%Annual effective rate is the same for all years for compound interest

18. Jack invests 10,000 in an account earning compound interest at a rate of i . After 8 years, Jack

has 18,034.84.

Determine i .

8 8

1/8

10000(1 ) 18,034.84 (1 ) 1.803484

1 (1.803484) 1.0765 7.65%

i i

i i


January 28, 2014

19. April borrows 2000 at a compound interest rate of 6%. She repays the loan at the end of n

years with a payment of 4265.86.

Determine n .

2000(1.06) 4265.86 (1.06) 2.13293 ln[(1.06) ] ln(2.13293)

ln(2.13293)ln(1.06) ln(2.13293) 13

ln(1.06)

n n n

n n

20. Qiyu invests 1275 in a fund that pays compound interest for 10 years.

The interest rates are:

i. 5% for the first five years;

ii. 4% for the next 3 years; and

iii. 3.5% for the last 2 years.

Determine the amount Qiyu will have after 10 years.

Determine the level annual effective interest rate that is equivalent to the rates that Qiyu

earned.

5 3 2

10 10 1/10

1275(1.05) (1.04) (1.035) 1960.8185

1275(1 ) 1960.82 (1 ) 1.537897 1 (1.537897) 1.04398

4.398%

i i i

i


January 28, 2014

21. Kartik invests money in a bank account earning compound interest at an annual effective

interest rate of 6.25%.

Jim invests money in a bank account earning 12.5% simple interest.

What year will Kartik and Jim earn the same annual effective interest rate?

( ) ( 1)0.0625

(n 1)

1 0.125 (1 0.125( 1))0.0625

1 0.125( 1)

0.0625 0.0078125 0.0078125 0.125 9

a n a n

a

n n

n

n n

22. Ryan invests 25,000 in an account earning simple interest of 5%.

Chelsea invests X in an account earning 2% compounded annually.

In year Y, Chelsea and Ryan earn the same annual effective interest rate.

At the end of year Y, the amount of money in Chelsea’s account is equal to the amount of

money in Ryan’s account.

Determine X.

31

1 0.05 (1 0.05(Y 1))0.02 0.02 0.001 0.001 0.05 31

1 0.05(Y 1)

25000(1 0.05(31)) (1.02) 63750 1.847589 34,504.43

YY Y

X X X


January 28, 2014

23. Alex invests 100 in an account earning simple interest. During the 10th year, the amount of

interest that Alex earns is 20.

Taylor invests X into an account earning compound interest of .i During the 10th year, the

amount of interest that Taylor earns is also 20.

During the 11th year, Alex and Taylor earn the same annual effective interest rate.

Determine X .

Solution: Interest is the same in all years for simple interest.

Or (10) (9) 20 100(1 10 ) 100(1 9 ) 20 100 20 0.20A A s s s s

(11) (10) 1 11(0.2) (1 10(0.2))0.06667

(10) 1 10(0.2)compound

a ai

a

10 9(1.06667) (1.06667) 20 0.11917 20 167.827X X X X

100(1 ) 100 20 1 1.2 20%s s s


January 28, 2014

Section 1.6

24. Yifan borrows 800 from a bank to buy a new computer. The bank charges interest in advance at

a rate of 8%. Yifan repays the loan after one year.

a. Determine the principle for Yifan’s loan.

b. Determine the amount of discount for this loan.

c. Determine the amount of money that Yifan at time 0 to spend on a computer.

d. Determine the amount of money that Yifan will need to pay at time 1 to repay the loan.

a. Principle = amount paid up front = K = 800

b. Discount = K*D = 800(.08) = 64

c. K - K*D = 800 - 64 = 736

d. 800

25. You are given that ( ) (1.07)ta t .

Calculate:

a. [2,3]d

b. 3d

c. 10d

d. [2.5,4]d

a. d[2,3] =a(3)- a(2)

a(3)=

1.073 -1.072

1.073= .065421

b. d3 =a(3)- a(2)

a(3)= .065421

c. d10 =a(10)- a(9)

a(10)= .065421because effective rate is the same for compound interest

d. d[2.5,4] =a(4)- a(2.5)

a(4)=

1.074 -1.072.5

1.074= .096508


January 28, 2014

26. You are given that ( ) 1 0.07a t t .

Calculate:

a. [2,3]d

b. 3d

c. 10d

a. d[2,3] =a(3)- a(2)

a(3)=

1+ .07(3)- [1+ .07(2)]

1+ .07(3)= .057851

b. d3 =a(3)- a(2)

a(3)= .057851

c. d10 =a(10)- a(9)

a(10)=

1+ .07(10)- [1+ .07(9)]

1+ .07(10)= .041176

27. If 5 0.05i , calculate 5d .

d5 =i5

1+ i5=

.05

1.05= .047619

28. If 5 0.05d , calculate 5i .

i5 =d5

1- d5

=.05

.95= .052631


January 28, 2014

Section 1.7

29. You are given that 2( ) 1 0.03 0.001a t t t .

Calculate:

a. (5)v .

b. The amount that you must invest today to have 5000 at the end of 5 years.

a. v(5) =1

a(5)=

1

1+ .03(5)+ .001(5)2=

1

1.175= .85106

b.

x ×a(5) = 5000

1.175x = 5000

x = 4255.32

30. Christine borrows money to buy a car. The loan has a simple interest rate of 6%. At the end of 4

years, Christine repays the loan with a payment of 13,000.

Calculate the amount of the loan.

x(1+ 4(.06)) =13000

x = 10,483.87

31. Wei borrows money to buy a car. The loan has compound interest at a rate of 6%. At the end of

4 years, Wei repays the loan with a payment of 13,000.

Calculate the amount of the loan.

x(1.06)4 = 13000

x = 10,297.22

32. Yue wants to have 5000 at the end of 5 years. If Yue invests in a fund earning simple interest of

8%, determine the amount that Yue must invest today to have 5000 at the end of 5 years.

x(1+ 5(.08)) = 5000

x = 3571.43


January 28, 2014

33. Yue wants to have 5000 at the end of 10 years. Yue invests in a fund earning simple interest of

8%. Yue invests D at the end of 5 years and it grows to 5000 at the end of 10 years.

Determine D .

(5) 1 5(.08) 1.43888.89

5000 (10) 1 10(.08) 5000 1.8

D a DD

a

34. You are given that 1

( )( )

v tt

.

Using this discount function, a payment of 200 to be made at time 10 has a present value of 100

at time 0.

Calculate (20).a

1 1(0) 1 1 1 1

(0)

200 (10) 100

1200( ) 100

(10)

200100 200 100 100 .1

10 1

1 1( ) ( ) .1 1

.1 1 ( )

(20) .1(20) 1 3

v

v

v t a t tt v t

a

35. The present value of 12,345 at the end of 7 years is 8,765 assuming a compound annual interest

rate of i .

Calculate i .

8765(1+ i)7 = 12345

(1+ i)7 = 1.40844

1+ i = 1.050143® i = 5.0143%


January 28, 2014

36. Wang Corporation is building a new factory. Wang expects the factory will generate the

following cash flows:

Time Cash Flow

0 - 500,000

1 - 100,000

2 300,000

3 400,000

4 300,000

5 200,000

a. Calculate the Net Present Value at an annual effective interest rate of 10%.

b. Calculate the Net Present Value at an annual effective interest rate of 20%.

c. Calculate the Internal Rate of Return (IRR).

d. What is the Net Present Value at the IRR. (Note: You should be able to answer this

question without doing any work.)

a.

C0 = -500,000

C1 = -100,000

C2 = 300,000

C3 = 400,000

C4 = 300,000

C5 = 200,000

I = 10

NPVCPT = 286,639.015

b.

C0 = -500,000

C1 = -100,000

C2 = 300,000

C3 = 400,000

C4 = 300,000

C5 = 200,000

I = 20

NPVCPT = 81,532.9218


January 28, 2014

c.

C0 = -500,000

C1 = -100,000

C2 = 300,000

C3 = 400,000

C4 = 300,000

C5 = 200,000

IRRCPT = 25.3778%

d. 0 because IRR is the interest rate at which NPV = 0

37. Fisher Corporation invests X million today to build a factory. The factory is expected to produce

the following profits:

End of Year Profits

1 1 million

2 4 million

3 2 million

4 1 million

At the end of 4 years, the factory will be obsolete and will be closed.

The Net Present Value of this project to Fisher Corporation is 1 million at an interest rate of 8%.

Calculate the Internal Rate of Return on this project.

-X +1(1

1.08)+ 4(

1

1.08)2 + 2(

1

1.08)3 +1(

1

1.08)4

X = 5.67798million

C0 = -5.67798

C1 = 1

C2 = 4

C3 = 2

C4 = 1

IRRCPT = 15.913%


January 28, 2014

Section 1.9

38. Tianyi borrows 25,000. The loan is repaid in 4 years at an annual rate of discount of 9.5%.

Calculate:

a. The amount of cash that Tianyi will receive at time 0.

b. The amount of discount that Tianyi will pay.

c. The equivalent compound annual interest rate.

a. 25000(1- .095)4 =16,770.05

b. 25000 -16770.05 = 8229.95

c. i =d

1- d=

.095

1- .095= 10.497%

39. Mikaila wants to buy a new car. The bank offers her a loan which has a compound discount rate

of 7%. Mikaila needs 12,000 to buy the car and want to repay the loan at the end of 4 years.

How much must Mikaila borrow (repay in 4 years) to be able to have 12,000 now to buy the car?

12000(1- .07)-4 =16,041.665

40. Abigail needs 20,000 to pay her tuition for this year. She has the option of the following two

loans:

a. A three year loan at a compound annual interest rate of 7%;

b. A three year loan at a compound annual discount rate of 6.55%.

Which loan should Abigail accept and why?

i =d

1- d=

.0655

1- .0655= .07009

She should take Loan A because the annual effective interest rate is lower, which is what you’re

looking for when taking out a loan.


January 28, 2014

41. Dias has a loan with a bank where the terms call for him to receive 10,000 today and to repay

15,000 at the end of five years. Determine the annual compound discount rate on this loan.

10000(1- d)-5 = 15000

(1- d)-5 = 1.5

1- d = .922108 ® d = 7.7892%

42. Mengyi invests 24,000 in an account that pays interest at a rate equivalent to a 6% discount

rate. Determine the amount of money that Mingyi will have at the end of 5 years.

i =d

1- d=

.06

1.06= .06383

24000(1.06383)5 = 32,701.82

43. Bobby invests 100 in a bank account earning a simple interest rate of 11%. Will invests 100 in a

bank account which earns an annual effective discount rate of d. At the end of 10 years, Bobby

and Will have the same amount of money in their accounts. Calculate d.

100(1+ .11(10)) = 100(1- d)-10

2.1 = (1- d)-10

.92849 = 1- d® d = 7.1508%

Section 1.10

44. Pawel borrows 3000. Pawel’s loan will charge interest at a rate of 8% compounded quarterly.

The loan is repaid at the end of 30 months.

Calculate the amount that Pawel must pay to repay the loan.

1+.08

4

æ

èçö

ø÷

4

= 1.08243216

3000(1.08243216)2.5years = 3656.98


January 28, 2014

45. Yishi borrows money to buy a new car. The loan has an interest rate of 12% compounded

monthly. At the end of five years, Yishi repays the loan with a payment of 40,000.

Calculate the amount that Yishi borrowed.

X 1+.12

12

æ

èçö

ø÷

12i5

= 40,000

X =40,000

1+.12

12

æ

èçö

ø÷

12i5 = 22,017.95

46. Logan invests 1000 in an account earning a nominal interest rate which is compounded

quarterly. At the end of 6 years, Logan has 2000.

Determine the nominal interest rate compounded quarterly that Logan is earning.

1000 1+i(4 )

4

æ

èçö

ø÷

4i6

= 2000

1+i(4 )

4

æ

èçö

ø÷

24

= 2

1+i(4 )

4= 1.0293022 ®

i(4 )

4= .0293022

i(4 ) = 11.72089%

47. You are given that (12) 0.09i . Calculate the equivalent:

a. Monthly effective interest rate

b. Annual effective interest rate

c. (4)i

d. (2)d

a. i(12)

12=

.09

12= .75%

b. 1+i(12)

12

æ

èçö

ø÷

12

= 1.0075( )12

= 1.0938069 = 1+ i

i = 9.38069%


January 28, 2014

c.

1+i(12)

12

æ

èçö

ø÷

12

= 1+i(4 )

4

æ

èçö

ø÷

4

1.0938069 = 1+i(4 )

4

æ

èçö

ø÷

4

1+i(4 )

4= 1.022669

i(4 ) = 9.067%

d.

1+i(12)

12

æ

èçö

ø÷

12

= 1-d (2)

2

æ

èçö

ø÷

-2

1.0938069 = 1-d (2)

2

æ

èçö

ø÷

-2

1-d (2)

2= .95616

d (2) = .087684

48. Peter borrows 10,000 at an interest rate of 5% compounded monthly. At the end of the loan,

Peter pays 15,474.23.

Calculate the length of Peter’s loan in years.

10,000 1+.05

12

æ

èçö

ø÷

12X

= 15,474.23

1+.05

12

æ

èçö

ø÷

12X

= 1.547423

12X(ln1.0041667) = ln1.547423

12X = 104.9999762

X = 8.75


January 28, 2014

49. Noah invest 6000 today. Noah earns the following interest or discount rates for the next nine

years:

i. Quarterly effective rate of interest of 4% for the first two years;

ii. A nominal rate of interest of 6% compounded monthly during the third through

the fifth year;

iii. (2) 0.08d for the last four years.

Determine the value of Noah’s investment at the end of nine years.

6000 1.04( )2i4

1+.06

12

æ

èçö

ø÷

12i3

1-.08

2

æ

èçö

ø÷

-2i4

= 6000(1.36857)(1.19668)(1.386213) = 13,621.54

50. Emily borrows 20,000 for 8 years at a nominal discount rate of 6% compounded quarterly.

Calculate the amount of discount that Emily will pay.

4 8 4 8(4)

Borrow 20,000 means that this is the amount that is repaid at time 8.

0.06Amount at time 0 =20,000 1 20,000 1 12,330.75

4 4

Amount of discount = Amount at time 8 less the Amount at t

d

ime 0 = 20,000-12,330.75=7669.25

51. Lauren invests 13,000 in an account that earns interest at a rate equivalent to a discount rate of 6% convertible semi-annually. How much will Lauren have after 18 months.

13,000 1-

.06

2

æ

èçö

ø÷

-2i1.5

= 14,243.87


January 28, 2014

52. Bo wants to borrow money to buy a car. Bank A offers Bo a loan with an annual effective interest rate of 8.0%. Bank B offers Bo a loan with an interest rate of 7.8% compounded monthly. Which loan should Bo select and state why. (Provide work to support your answer.)

1+.078

12

æ

èçö

ø÷

12

= 1.08085 -1= 8.085%

Bo should select Bank A because we want the lower effective interest rate, and A’s is 8% while B’s is 8.085%.

53. Megan invests 100,000 in an account that earns a nominal interest rate of 10% compounded

every 4 years.

a. Calculate the amount that Megan will have at the end of 4 years.

b. Calculate the amount that Megan will have at the end of 7 years.

a.

100,000 1+

.10

1/ 4

æ

èçö

ø÷

1

4i4

= 140,000

b.

100,000 1+

.10

1/ 4

æ

èçö

ø÷

1

4i7

= 180,187.25

54. Sam invests 10,000 in an account earning a nominal interest rate of 6% convertible every two

years.

Sam’s brother Ben invests K in an account earning a nominal rate of interest of 6% convertible

monthly.

After 10 years, the amount in Sam’s account is equal to the amount in Ben’s account.

Determine K.

10,000 1+.06

1/ 2

æ

èçö

ø÷

1

2i10

= K 1+.06

12

æ

èçö

ø÷

12i10

17,623.42 = 1.819397K

K = 9686.41


January 28, 2014

Section 1.11

55. You are given that (12) 0.05i . Calculate .

1+.05

12

æ

èçö

ø÷

12

= ed

ln1.0511619 = d lne

d = 4.9896%

56. Luiting invests 17,500 in an account that earns 7.25% interest compounded continuously.

Determine the amount that Luiting will have after 5 years.

17500e.0725(5) = 25,146.05

57. Brendan borrows 1500 today. Brendan must repay the loan using a continuously compounded

interest rate of . He repays 3500 at the end of 10 years.

Determine .

1500e10d = 3500

e10d = 2.33333

10d lne = ln2.33333

10d = ln2.33333

d = 8.473%

Section 1.12

58. You are given that 2( ) 1 0.02a t t . Calculate:

a. 1

b. 5

a. d1 =a '(1)

a(1)®

.04t

1+ .02t 2=

.04(1)

1+ (.02)(1)2= .039216

b. d 5 =a '(5)

a(5)®

.04t

1+ .02t 2=

.04(5)

1+ (.02)(5)2= .13333


January 28, 2014

59. You are given:

2

1( )

(1 0.02 )(1 0.04 )v t

t t

Calculate 6 .

a(t) =1

v(t)® a(t) = (1+ .02t)(1+ .04t 2 ) = 1+ .02t + .04t 2 + .0008t 3

d 6 =a '(6)

a(6)=

.02 + .08t + .0024t 2

1+ .02t + .04t 2 + .0008t 3

=.02 + .08(6)+ .0024(6)2

1+ .02(6)+ .04(6)2 + .0008(6)3=

.5864

2.7328= .214578

60. Under simple interest with an interest rate of 5%, calculate 20 .

a(t) = 1+ .05t

a '(t) = .05

d 20 =a '(20)

a(20)=

.05

1+ .05(20)= .025

61. You are given:

a. Interest is earned using simple interest.

b. (10) 0.4v

Determine 20 .

v(t) =1

a(t)=

1

1+ it

v(10) =1

1+10i= .4 ® i = .15

a(t) = 1+ .15t

d 20 =a '(20)

a(20)=

.15

1+ .15(20)= .0375


January 28, 2014

62. You are given that 0.02t t .

a. Cassie invests 5000 today. Determine the amount that Cassie will have at the end of 4

years.

b. Mark invests 5000 at the end of 6 years. Determine the amount that Mark will have at

the end of 10 years.

a. 5000e.02t dt

0

4

ò= 5000e.01t

2|04

= 5000e.01(16) = 5867.55

b. 5000e.02t dt

6

10

ò= 5000e.01t

2|610

= 5000e.01(100-36) = 9482.40

63. Arturo wants to have 50,000 to make a down payment on a house at the end of 10 years.

Arturo deposits D into an account today that will pay a force of interest of 20.01 0.003t t .

Determine the minimum D so that Arturo meets his goal.

De(.01t+.003t2 )dt

0

10

ò= 50,000

De.005t2+.001t 3|010

® De.5+1 ® De1.5 = 50000

D = 11,156.51

64. Kehara invests 1000 in an account earning 0.02t t . At the end of t years, Kehara’s

investment has doubled.

Calculate t ( t will not be an integer.)

1000e.02t

0

t

ò= 2000

1000e.01t2 |0t

= 2000 ®1000e.01t2 = 2000

e.01t2 = 2® .01t 2 lne = ln2

t =ln2

.01= 8.3255


January 28, 2014

65. You are given:

a. 3( ) 1 0.001a t ct t where c is a constant

b. 10 0.14

Miao invests 1000 at time 0. How much will Miao have at time 10?

2 2

10 3 3

3

'(10) 0.003 0.003(10).14

(10) 1 0.001 1 (10) 0.001(10)

.3.14 0.14(2 10 ) .3 0.28 1.4 0.3 0.05

1 10 1

1000 (10) 1000(1 0.05(10) 0.001(10) ) 2500

a c t c

a ct t c

cc c c c c

c

a

66. Collin can invest 1000 in either of the following accounts:

a. Account A earns compound interest at an annual effective rate of i

b. Account B earns interest equivalent to an annual effective discount rate of 6% for the

first five years. Thereafter, Account B accumulates at a 0.05 0.002t t .

At the end of 10 years Collin would have the same amount in either account.

Calculate i .

10

5

2 105

.05 .00210 5

.05 .001 |10 5

10 .05(10) .001(100) .05(5) .001(25)

10

10

1000(1 ) 1000(1 .06)

1000(1 ) 1000(.94)

1000(1 ) 1362.58

1000(1 ) 1885.85

(1 ) 1.88585

6.5493%

tdt

t t

i e

i e

i e

i

i

i


January 28, 2014

Section 1.13

67. You are given that the nominal interest is 8% and the rate of inflation is 2.5%.

Calculate the inflation adjusted interest rate.

1+ j =1+ i

1+ r=

1.08

1.025= 1.0536585

j = 5.36585%

68. If the real interest rate is 6% and the rate of inflation is 3%, calculate the nominal interest rate.

1.06 =1+ i

1.03®1+ i = 1.0918

i = 9.18%

69. Chang has 100 and could use it to buy 80 songs from itunes. Instead, Chang invests his 100 at an

annual effective interest rate of 11.3% for 4 years. The price of songs on itunes is subject to an

annual effective rate of inflation of %r . At the end of 4 years, Chang can buy 101 songs.

Determine r .

100(1.113)4 = 153.4548635 /101loaves

1.519355 / loaf - now

100 / 80 = 1.25 / loaf - before

1.25(1+ r)4 = 1.519355

(1+ r)4 = 1.215484

r = 5%


January 28, 2014

70. A gallon of gasoline costs 3.00 today. Lewis has enough money to buy 100 gallons today. Instead of buying gasoline, Lewis decides to invest his money at an annual interest rate of 6.6%. If the annual rate of inflation over the next five years is 4.1%, calculate how many gallons of gasoline Lewis will be able to buy at the end of five years.

3(100) = $300

300(1.066)5 = 412.9593

3(1.041)5 = 3.66754

412.9593 / 3.66754 = 112.598

math 373 chapter 1 homework - purdue universityjbeckley/q/wd/ma373/s… · · 2014-01-30math 373...

Documents