Financial Math_topic

FINANCIAL MATHEMATICS

Financial Mathematics

Mathematics Learning Centre

Financial Mathematics FINM-A

Objectives................................................................................................. FINM 1

FINM-B

Simple Interest.......................................................................... ............... FINM 2

FINM-C

Compound Interest ............................................................... .................. FINM 6

FINM-D

Effective Interest Rates................ ........................................................... FINM 8

FINM-E

Equations of Value ......................................................................... ......... FINM 9

FINM-F

Summary.................................................................................................. FINM 10

FINM-G

Review Exercise................................ ....................................................... FINM 11

FINM-H

Appendix

FINM-Y

Index......................................................................................................... FINM 13

FINM-Z

Solutions................................................................................................... FINM 14

FINM-A

Transposing Formula............................. Formula........................................................ ........................... FINM 12

Objectives

• To perform calculations involving simple interest • To perform calculations involving compound interest • To use effective interest rate to compare interest rates • To determine unknown debts using equations of value.

FINM 1

Financial Mathematics

FINM-B

Mathematics Learning Centre

Simple Interest

When money is borrowed for a loan or invested, interest accumulates. The amount of interest depends on: • the amount of money borrowed or invested, the principal; • the interest rate; and • the time. To calculate the simple interest, I , accumulated on the principal, P, over an interval of t  years at an annual interest rate of r , the formula  I = P × r × t   is used. The percentage interest rate must be expressed as a decimal for calculations. The symbol p.a. used with interest rates denotes per annum meaning an annual or yearly interest rate. Interest rates stated as flat rates are used in simple interest calculations. The total amount of money that must be repaid on a loan or the total value of an investment can be called the future value, S . The future value is calculated using S = P + I  . The principal is also called the present value.

Examples FINM-B1 1.

A student borrows $600 to buy a camera. The loan is over two years, and the simple interest rate is 6% per year. How much interest does the student pay? What is the total amount of money repaid? Principal = P = $600 Interest rate = 6% per year ∴ r  = 0.06 t = 2 years

• list known information

 I = P r t  = 600 × 0.06 × 2 = 72

• • • •

write formula substitute values evaluate answer in words

• • • •

write formula substitute values evaluate answer in words

Interest to be paid is $72 S = P + I  = 600 + 72 = 672

Total amount ot be paid is $672 2.

An investor lent $12000 to a business associate for 6 months at an interest rate of 8% per year. Calculate the interest the investor earns and how much in total will be repaid. Principal = P = $12 000 Interest rate = 8% per year ∴ r  = 0.08 t = 6 months =

6 12

• list known information

= 0.5 year

 I = P r t  = 12000 × 0.08 × 0.5 = 480

Interest to be earned is $480 S = P + I  = 12000 + 480 = 12480

Total amount ot be paid is $12 480

• • • •

write formula substitute values evaluate answer in words

• • • •

write formula substitute values evaluate answer in words

FINM 2

Financial Mathematics

Mathematics Learning Centre

Examples FINM-B1 continued 3.

Calculate the simple interest earned when $500 is invested at 7.5% p.a. for 21 months. Principal = P = $500 Interest rate = 7.5% per year t = 21 months =

21 12

∴ r  = 0.075

• list known information

= 1.75 years • • • •

 I = P r t  = 500 × 0.075 × 1.75 = 65.625

Interest to be earned is $65.63

write formula substitute values evaluate answer in words

Exercise FINM-B1 1.

2.

3.

For each of the following situations define principal P, the interest rate r , and the time t . (a)

Barry borrows $15000 in order to buy a second-hand car. The loan is over three years and the interest rate is 12% per year.

(b)

Britney wishes to purchase a boat which requires her to take out a $20000 loan at 15% per annum over five years.

Find the simple interest due and the total amount owing for each of the following loans. (i) $2000 at 8% per year for 3 years. (ii)

$18000 at 12.5% per year for 6 months.

(iii)

$5000 at 1.5% per month for 3 years.

(iv)

$16000 at 8% per year for 36 months.

Find the simple interest due, and the total amount to be repaid for each of the following loans. (i) $3000 at 10% per year for 3 years 6 months. (ii)

$3000 at 8% per year for 18 months.

(iii)

$3000 at 12% per year for 2 years 9 months.

(iv)

$3000 at 8.5% per year for 36 months.

It is possible to use the simple interest formula,  I = P r t   to calculate the principal, P, interest rate, r , or the time, t , required to give the specified amount of simple interest,  I . To find these values it will be necessary to substitute in given quantities and then rearrange the equation to solve for the unknown. Sometimes it is necessary to calculate how much money is required to be invested now, (present value) at a given interest rate, to amount to a specified value at some time in the future (future value). To achieve this the formula needs to be manipulated as follows. S = P + I  S = P+P r t  S = P(1 + rt )  S  ∴P = 1 + rt 

FINM 3

Financial Mathematics

Mathematics Learning Centre

Examples FINM-B2 1.

Calculate the amount of money that would earn $750 simple interest if it was invested at 4.5% p.a. for 5 years. Principal = P = ? • list known information Interest = I  = 750 Interest rate = 4.5% per year ∴ r  = 0.04.5 = 5 years t

= P r t   I 750 = P × 0.045 × 5 750 = 0.225 P 750 ∴P = = 3333.33 0.225

• write formula • substitute values • solve for unknown

$3333.33 would need to be invested. 2.

Calculate the time required for $8300 to earn $3087.60 interest at the flat rate of 6.2% p.a.

• list known information

P = $8300  I  = $3087.60 r  = 0.062 t=?  I = P r t  3087.60 = 8300 × 0.062 × t  3087.60 = 514.6t  3087.60 ∴ t  = =6 514.6

• write formula • substitute values • solve for unknown

It would take 6 years to earn $3087.60 interest. 3.

• answer in words

• answer in words

Betty won $245 000 in lotto. She wants to travel with the interest earned by investing her winnings. At what interest rate must betty invest her money so that she will earn $27 500 every 18 months?

• list known information

P = $245 000  I  = $27 500 t = 18 months =

18 12

= 1.5 years

r=?  I 27500 27500

∴ r 

• write formula • substitute values

= P r t  = 245000 × r ×1.5 = 367500r  =

• solve for unknown

27500 367500

= 0.074829 ≈ 0.075 = 7.5% • answer in words

Interest rate required is 7.5% p.a.

FINM 4

Financial Mathematics

Examples FINM-B2 4.

Mathematics Learning Centre

continued

Brendan wants to have $15000 for a new boat in 10 years time. How much should he invest at 5% per year to save this amount (assuming no withdrawals are made and the interest rate does not change)?

S  = $15000

• list known information

t  = 10 years r  = 0.05

S = P + I  S = P+P r t  S = P(1 + rt )  S  ∴P = 1 + rt  15000

=

1 + 0.05 ×10 15000 = = 10000 1.5

OR

• •

write formula

• •

substitute values

transpose formula

S S 15000 15000 15000 15000

1.5

= P + I  • = P+P r t   = P + P × 0.05 ×10 • = P + 0.5P • = 1.5 P

write formula substitute values solve for unknown

= P = 10000

evaluate

• answer in words

Brendan would need to invest $10 000.

When money is borrowed, it is usually paid back on a regular basis over the term of the loan. The total amount of the loan, S , is divided by the required number of payments to give the amount of the regular repayment.

Example FINM-B3 A student borrows $600 to buy a camera. The loan is over two years, and the simple interest rate is 6% per year. How much will her monthly repayments be? P = $600 Interest rate = 6% per year ∴ r  = 0.06 t = 2 years = 24 months  I = P r t S = P + I  = 600 × 0.06 × 2 = 600 + 72 = 72 = 672

repayments =

Total repaid no. of months

=

• list known information • determine amount to be repaid

S  n

=

672 24

= 28

Her repayments will be $28 per month.

• write formula, substitute values and evaluate • answer in words

Exercise FINM-B2 1.

Find the principal that was invested to earn $3348 interest in 6 years at 6.2% p.a.

2.

How long was $4500 borrowed for at 7.1% p.a. to cost $798.75 interest?

3.

At what interest rate was $875 invested for to earn $73.50 interest in 18 months?

4.

What amount of money should be invested now at 10% per year to amount to $1800 in 2 years’ time? What amount of money should be invested now at 6% per year to amount to a future value of $8000 seven months from now? A $120 000 property is purchased with a deposit of $30000. The balance is repayable over 15 years at 7% per year. Find: (a) the balance of the purchase price

5. 6.

7.

(b)

the interest due

(c)

the total amount repayable

(d)

the monthly repayments

Brian wishes to borrow $15000 for a 6 month holiday in Europe. He has approached two  banks and has the following options. The People's Bank quotes an interest rate of 5% per year with the repayments to be made over 3 years, whilst the Friendly Bank interest rate is 4% per year and the payments are to be made over 4 years. Which loan will ultimately be cheaper for Brian?

FINM 5

Financial Mathematics

FINM-C

Mathematics Learning Centre

Compound Interest

In compound interest, the interest earned by an invested amount of money (or principal) is reinvested so that it too earns interest. That is, at the end of each interest period the interest earned for that period is added to the principal so that the interest also earns interest over the next interest period. In the same way, interest due on a debt at the end of a period is subject to interest in the next period. The accumulated amount or compounded amount , S , has same meaning as future value. The difference between the compounded amount and the original principal is called the compound interest,  I = S − P . For an original principal of P  invested at a periodic interest rate of i  for n periods, the compounded amount (or future value), S , is given by S = P (1 + i )

n

.

The periodic interest rate  can be found by dividing the annual interest rate, r ,  by the number of r   periods in a year   , N, i.e. i =  N 

Example FINM-C1 Clarissa invests $5000 at 6.2% p.a. with interest compounded monthly. What would her investment be worth after five years? What amount of interest has been earned during the five years? P = $5000  N = 12 (12 months in a year) r  = 0.062 p.a. r  0.062 per month ∴i = = 12  N  t = 5 years ∴ n = 5 ×12 = 60 months S = P (1 + i )

• list known information

• write formula • substitute values • evaluate using calculator  5000 ( 1 + 0.062 ÷  12 )  x y  60 = • answer in words

n

60

⎛ 0.062 ⎞ S  = 5000 ⎜ 1 + ⎟ 12 ⎠ ⎝ = $6811.69

The investment will be worth $6811.69. S = P + I  ∴  I = S − P = 6811.69 − 5000 = 1811.69

• write formula • substitute values and evaluate • answer in words

The interest earned was $1811.69.

It is possible to calculate the principal if the period of investment and compounded amount are known. n

By transposing the compound interest formula, S = P (1 + i ) ,  the formula for the principal becomes P=

S 

(1 + i )

n

OR P = S (1 + i )

−n

In a similar way the periodic interest rate can be determined by transposing the compound interest formula for  i. 1

i=

n

⎛ S  ⎞ n − 1 or  ⎜ ⎟ − 1 . P ⎝P⎠ S

FINM-H Appendix −  Transposing Formula, shows the transposition processes.

FINM 6

Financial Mathematics

Mathematics Learning Centre

Examples FINM-C2 1.

What principal should be invested now so that it will $13 000 in 5 years time at 12% per annum compounded quarterly? S = $13 000 • list known information  N  = 4 (4 quarters in a year) r  = 12 p.a. r  0.012 ∴i = = per quarter   12  N  t = 5 years ∴ n = 5 × 4 = 20 quarters P = ?

• Solution can involve transposing the

unknown.

n

S = P (1 + i ) S  −n P= or S (1 + i ) n (1 + i) P=

⎛ 0.12 ⎞ ⎜1 + 4 ⎟ ⎝ ⎠

20

⎛ 0.12 ⎞   or  13000 ⎜ 1 + ⎟ 4 ⎠ ⎝

2.

n

S

= P (1 + i )

13000

⎛ 0.12 ⎞ = P ⎜1 + ⎟ 4 ⎠ ⎝

−20

13000

⎛ 0.12 ⎞ ⎜1 + 4 ⎟ ⎝ ⎠

P = 7197.78

The principal required is $7197.78

to

• OR substituting values and solving for the

formula and then substituting values

13000

compound

20

20

= P = 7197.78

• write answer in words

What annual rate of interest, compounded quarterly, would be required if an initial investment of $3000 is to amount to $3500 after 5 years?

• list known information

P = $3000 S = $3500  N  = 4 (4 quarters in a year) t = 5 years ∴ n = 5 × 4 = 20 quarters r = ?

find i first.

S = P (1 + i )

n

• write formula • transpose formula

1

i=

n

S P

⎛ S  ⎞ n

− 1or ⎜ ⎟ − 1 ⎝P⎠ 1

⎛ 3500 ⎞ ⎛ 3500 ⎞ 20 − 1 or  ⎜ i = 20 ⎜ ⎟ ⎟ −1 ⎝ 3000 ⎠ ⎝ 3000 ⎠ = 0.007737313

• substitute values • evaluate using calculator ( 3000 ÷  3500 )  x y  ( 1 ÷  20 ) =

The periodic rate is 0.774%, so the annual rate would be 4

0.774% = 3.096%

Exercise FINM-C1 1.

Find the compounded amount and compound interest for the following: (a) $600 for 8 years at 8% compounded monthly. (b)

$1000 for 5 years at 12% compounded daily.

(c)

$750 for 12 months at 6% compounded weekly.

(d)

$3000 for 4 years at 9% compounded quarterly.

2.

How much should be invested at 8% compounded semi-annually for six years in order to  provide a compounded amount of $10000?

3.

The day Charlotte was born her father deposited $500 in a bank account. If the account was compounded annually and on Charlotte’s twenty-first birthday there was $1326.65 in the account. What was the annual interest rate?

4.

A investor has a choice of investing $4000 at 10% compounded annually or at 9.75% compounded monthly for one year. Which is the better investment?

FINM 7

Financial Mathematics

FINM-D

Mathematics Learning Centre

Effective Interest Rate

The effective rate of interest is the equivalent rate of simple interest earned over a period of one year for an interest rate which is compounded twice or more over the year. The annual simple interest rate will be greater than the annual compounding interest rate to earn the same amount of interest. The effective annual interest rate, r e, is calculated using:

⎛ ⎝

r e = ⎜ 1 +

 N 

r  ⎞

⎟ −1

 N  ⎠

where r  is the annual interest rate (in decimal form) and  N  is the number of periods per year. The annual interest rate, r , for compounding calculations is often called the nominal rate. Effective interest rates can be used to compare investment opportunities.

Examples FINM-D1 1.

What effective interest rate is equivalent to a nominal rate of 12% compounded (a) monthly and (b) daily? 12

(a)

(b)

⎛ 0.12 ⎞ r e = ⎜1 + ⎟ − 1 ≈ 0.1268 12 ⎠ ⎝ = 12.68% ⎛ ⎝

r e = ⎜1 +

• monthly: 12 months in year  N  = 12

365

0.12 ⎞

⎟

365 ⎠

− 1 ≈ 0.1275

• daily: 365 days in year  N  = 365

= 12.75% 2.

Which is the "best" investment, 6% per annum compounded monthly or 6.20% compounded annually? 12

⎛ 0.06 ⎞ compunding monthly = ⎜1 + ⎟ − 1 ≈ 0.0617 = 6.17% 12 ⎠ ⎝ So, an annual rate of 6.20% is better for investment than 6% per annum compounded monthly. Exercise FINM-D1 1.

Find the compounded amount and the compound interest for the given investment and annual rate. (i) $6000 for 8 years at an effective rate of 8%. (ii)

2.

$750 for 12 months at an effective rate of 10%.

Find the effective rate that corresponds to the given nominal rate. (i)

10% compounded quarterly.

(ii)

12% compounded monthly.

(iii)

8% compounded daily.

(iv)

12% compounded daily.

3.

A $6000 certificate of deposit is purchased for $6000 and is held for 7 years. If the certificate earns an effective rate of 8%, what is it worth at the end of that period?

4.

To what sum will $2000 amount in 8 years if invested at a 6% effective rate for the first 4 years and 6% compounded semi-annually thereafter?

FINM 8

Financial Mathematics

FINM-E

Mathematics Learning Centre

Equations of Value

An amount of money can have different values at different times, for a particular interest rate. The value of money is dependant on its ability to earn interest. Equations of value are used to compare the value of money at various times. The equation S = P (1 + i ) P = S (1 + i )

−n

n

 is used to find the value of money, at various points in the future while

 is used to find the value of money at some point prior to this time. To perform

calculations the value of repayments or debts must be determined at only one point in time. For equations of values the general formula used is repayments = debts (at the same point in time). The value of each repayment or debt should be determined at the point in time when the final repayment is to be paid. To assist with calculations a time line can be drawn showing all repayments and debts.

Examples FINM-E1 1.

If Edna owes $200 in 2 years and a further $300 in 5 years, how much does she need to  pay at present to account for these debts? (Assume 8% compounding quarterly.) i = 8% p.a. =

0.08 4

• list given values

per quarter  

S 1 = 200 n1 = 2 × 4quarters = 8

S 2 = 300 n1 = 5 × 4quarters = 20

• amounts moved left use P = S (1 + i ) repayments = debts = debt1 + debt 2

• draw a time line

−n

• write formula −8

−20

⎛ 0.08 ⎞ ⎛ 0.08 ⎞  x = 200 ⎜1 + + 300 ⎜ 1 + ⎟ ⎟ 4 ⎠ 4 ⎠ ⎝ ⎝ = 170.69 + 201.89 = 372.58

• substitute values • evaluate

Edna needs to pay only $372.58 now to cover her future debts. 2.

A debt of $5000, which was due 5 years from now, is instead going to be repaid by three  payments: $2000 now, $1000 in 2 years time and the third payment after 4 years. What will be the final payment if the assumed interest rate is 6% p.a. compounding half-yearly? i = 6% p.a. =

0.06 2

per quarter  

⎛ ⎝

note: ⎜1 +

0.06 ⎞

⎟ = 1.03

2 ⎠

• draw a time line • move right use S = P (1 + i ) • move left use P = S (1 + i ) repayments repay1 + repay2 + repay3

= debts = debt  

8

4 −2 ⎛ 0.06 ⎞ 2000 ⎜1 + +1000 (1.03 ) + x = 5000 (1.03 ) ⎟ 2 ⎠ ⎝ 2533.54 +1125.51 + x = 4712.98  x = 1053.93

The final repayment will be $1053.93

FINM 9

• write formula • substitute values • evaluate • answer in words

n

−n

Financial Mathematics

Mathematics Learning Centre

Exercise FINM-E1 1.

A debt of $550 due in 4 years and $550 due in 5 years is to be repaid by a single payment now. Find how much the payment is if an interest rate of 10% compounded quarterly is assumed. 2. A debt of $600 due is 3 years and $800 due in 4 years is to be repaid by a single payment 2 years from now. If the interest rate is 8% compounded semiannually, how much is the  payment? 3. A debt of $5000 due in 5 years is to be repaid by a payment of $2000 now and a second  payment at the end of 6 years. How much should the second payment be if the interest rate is 6% compounded quarterly? 4. A debt of $5000 due 5 years from now and $5000 due 10 years from now is to be repaid by a  payment of $2000 in 2 years, a payment of $4000 in 4 years, and a final payment at the end of 6 years. If the interest rate is 7% compounded annually, how much is the final payment?

FINM-F

Summary

To calculate the simple interest, I , accumulated on the principal, P, over an interval of t  years at an annual interest rate of r , the formula  I = P × r × t . The total amount of money that must be repaid on a loan or the total value of an investment can be called the future value, S . and is calculated using S = P + I  . It is possible to use the simple interest formula to calculate the principal, P, interest rate, r , or the time, t , required to give the specified amount of simple interest, I . It is also to determine the principal, P, to S   be invested now to accumulate to some future value, S , using the formula P = . 1 + rt  The total amount of the loan, S , is divided by the required number of payments to give the amount of the regular payment. For an original principal of P  invested at a periodic interest rate of i  for n periods, the compounded amount (or future value), S , is given by S = P (1 + i ) n . The periodic interest rate  can be found by dividing the annual interest rate, r , by the number of periods in a year  , N. It is possible to calculate the principal if the period of investment and compounded amount are known S  −n using: P = OR P = S (1 + i ) . In a similar way the periodic interest rate can be determined n (1 + i) 1

 by i =

n

⎛ S  ⎞ n − 1 or  ⎜ ⎟ − 1 . P ⎝P⎠ S

⎛ ⎝

The effective annual interest rate, r e, is calculated using r e = ⎜1 +

 N 

r  ⎞

⎟ −1 ,

 N  ⎠ interest rate (in decimal form) and  N  is the number of periods per year.

where r   is the annual

Equations of value are used to compare the value of money at various times. The equation n −n S = P (1 + i )  is used to find the value of money at various points in the future, while P = S (1 + i )  is used to find the value of money at some point before this time. For equations of value, the general formula used is repayments = debts (at the same point in time).

FINM 10

Financial Mathematics

FINM-G 1.

2.

Mathematics Learning Centre

Review Exercise

Calculate the simple interest and accumulated amount for each of the following. (i)

$2000 at 8% p.a. for 5 years

(ii)

$35 000 at 5.6% p.a. for 30 months.

Angela bought a computer for $2100. She paid a deposit of $300 and took out a loan for the remainder at 8 12 % p.a. simple interest over 3 years. (a)

How much interest did she pay?

(b)

What was the total cost of buying the computer?

(c)

How much did she have to pay per month?

3.

How much needs to be invested at 11% p.a. simple interest to have an accumulated amount of $7800 in 5 years from now?

4.

Calculate the compounded amount and the compound interest for the following. (i)

$2000 at 8% p.a. for 5 years compounded semi-annually

(ii)

$35 000 at 5.6% p.a. for 30 months compounded monthly.

5.

Find the compound interest on a loan of $600 borrowed at 7.5% compounded quarterly for 3 years.

6.

What annual interest rate, compounded monthly, would be required for an investment of $5000 to amount to $6500 in 4 years?

7.

Leading company A offers rates of 8.2% p.a. compounded quarterly, while company B advertises a rate of 8.1% compounded daily. Which company offers the best rate for a loan?

8.

Two debts, $2000 due in 3 years and $1500 due in 5 years, are to be repaid by $1000 now and another repayment in 4 years. What is the value of the final r epayment if the interest rate is 10% compounded monthly?

FINM 11

Financial Mathematics

FINM-H

Mathematics Learning Centre

Appendix – Transposing Formula n

Solving S = P (1 + i )  for the principal, P.

= P (1 + i )

S S 

(1 + i ) ∴P =

n

P (1 + i )

=

(1 + i ) S 

(1 + i )

n

n

n

n

or P = S (1 + i )

⎡1 −n ⎤ ⎢⎣ x n = x ⎥⎦

−n

n

Solving S = P (1 + i )  for the interest rate, i.

= P (1 + i )

S S 

=

P S  S

n

S P S

n

P

n

OR

= 1+ i

P n

n

= (1 + i )

S

n

P

n

P n

P (1 + i )

= (1 + i )

P

n

⎛ S  ⎞ ⎜ ⎟ ⎝P⎠

−n

⎛ S  ⎞ ⎜ ⎟ ⎝P⎠

−n

(

n

= (1 + i )

)

−n

= 1+ i −n

−1 = 1 + i −1

⎛ S  ⎞ ⎜ P ⎟ − 1 = 1 + i −1 ⎝ ⎠

−1 = i

⎛ S  ⎞ ⎜ P ⎟ −1 = i ⎝ ⎠

−n

n

Solving S = P (1 + i )  for the number of investment periods, n.

= P (1 + i )

S S  P S  P

=

n

P (1 + i )

n

P

= (1 + i )

n

log ( PS ) = log (1 + i )

n

log ( PS ) = n log (1 + i ) log ( SP ) log (1 + i )

=n

OR

n

ln ( S P )

= ln (1 + i )

ln ( S P )

= n ln (1 + i )

ln ( PS  ) ln (1 + i )

=n

Manipulation of logarithmic equations can be found in LOGS - Logarithms Module.

FINM 12

Financial Mathematics

FINM-Y

Mathematics Learning Centre

Index

Topic

Page

Annual Interest Rate ........................................................ ........

FINM 2, 6, 8

Compound Amount, S.............................................................. Compound Interest...................................................................

FINM 6 FINM 6

Effective Annual Interest Rate................................................. Equations of Value...................................................................

FINM 8 FINM 9

Future Value, S ........................................................................

FINM 2

Interest − compound ................................................................ Interest −  simple.......................................................................

FINM 8 FINM 2, 3

 Nominal Rate .................................................................. .........

FINM 8

 p.a. − per annum ............................................................ .......... Periodic Interest Rate............................................................... Present Value, P....................................................................... Principal, P ..............................................................................

FINM 2 FINM 6 FINM 4 FINM 2, 3, 7

Repayments..................................................................... .........

FINM 5, 9

Simple Interest .................................................................... .....

FINM 2, 3

Transposing Formula ...............................................................

FINM 17

FINM 13

SOLUTIONS

Financial Mathematics

FINM-Z

Solutions

FINM -B

Simple Interest

FINM -C

FINM-B1

Simple Interest ..........................................................

FINM 15

FINM-B1

Simple Interest Manipulation ....................................

FINM 16

Compound Interest FINM-C1

FINM -D

Effective Interest Rates .............................................

FINM 19

Equations of Value FINM-E1

FINM -G

FINM 17

Effective Interest Rates FINM-D1

FINM -E

Compound Interest ...................................................

Equations of Value....................................................

FINM 20

Review Exercise FINM-G

Review Exercise........................................................

FINM 14

FINM 21

SOLUTIONS

FINM-B1 1 .

2 .

(a)

(i)

(iii)

3 .

(i)

Financial Mathematics

Simple Interest P = $15 000

(b)

r  = 12% = 0.12

r  = 15% = 0.15

t = 3 years

t = 5 years

P = $2000

r  = 8% = 0.08

(ii)

P = $18 000 r  = 12.5% = 0.125

t = 3 years

t = 6 months = 0.5 years

 I = Prt  = 2000 × 0.08 × 3

 I = Prt  = 18000 × 0.125 × 0.5

= $480 Total Amount = P + I  = 2000 + 480 = $2480

= $1125 Total Amount = P + I  = 18000 + 1125 = $19125

P = $5 000

t = 3 years

(iv)

P = $16 000

r  = 8% = 0.08

r = 1.5% p.m. = 1.5% ×12 p.a

t = 36 months = 3 years

= 0.015 ×12 = 0.18  I = Prt  = 5000 × 0.18 × 3 = $2700 Total Amount = P + I  = 5000 + 2700 = $7700

 I = Prt  = 16000 × 0.08 × 3

P = $3 000

r  = 10% = 0.1

= $3840 Total Amount = P + I  = 16000 + 3840 = $19 840

(ii)

P = $3 000

r  = 8% = 0.08

t = 3 y 6months = 3.5 years

t = 18months = 1.5 years

 I = Prt  = 3000 × 0.1× 3.5

 I = Prt  = 3000 × 0.08 ×1.5

= $1050 Total Amount = P + I  = 3000 + 1050 = $4 050 (iii)

P = $20 000

P = $3 000

= $360 Total Amount = P + I 

= 3000 + 360 = $3 360

r  = 12% = 0.12

(iv)

P = $3 000

r  = 8.5% = 0.085

t = 2 y 9months = 2 129 years = 2.75 y

t = 36 months = 3 years

 I = Prt  = 3000 × 0.12 × 2.75

 I = Prt  = 3000 × 0.085 × 3

= $990

= $765

Total Amount = P + I 

Total Amount = P + I 

= 3000 + 765 = $3 765

= 3000 + 990 = $3 990

FINM 15

SOLUTIONS

Financial Mathematics

FINM-B2 1.

Simple Interest Manipulation P=?

r = 6.2% = 0.062 t = 6 years I =  3348

 I

= Prt

 I 

I

= Prt 

=P

3348

= P × 0.062 × 6

=P

3348

= 0.372P

∴ P = $9 000

3348

rt  3348

0.062 × 6

OR

= P = $9 000 0.372 A principal of $9000 would need to be invested. 2.

P = 4500 r = 7.1% = 0.071 t = ? I = 798.75  

= Prt

 I  I  Pr  798.75

4500 × 0.071

OR

= Prt 

I

=t

798.75 = 4500 × 0.071 × t 

=t

798.75 = 319.5t  798.75

∴ t = 2.5 years

319.5

= t = 2.5 years

The money would be invested for 2.5 years. 3.

P = 875 r = ?

t = 18months = 1.5 years

= Prt

 I  I  Pt  73.50

875 ×1.5

OR

I  = 73.50

= Prt 

I

=r

73.50 = 875 × r × 1.5

=r

73.50 = 1312.5r  73.50

∴ r = 0.056 = 5.6%

1312.5

= r  = 0.056 = 5.6%

The interest rate would be 5.6%. 4.

P=?

r = 10% = 0.1

t = 12 years

Amount =  1800

OR

S

= P (1 + rt )

= P + Prt

1800

= P (1 + 0.1 × 2 )

= P (1 + rt )

1800

= P ×1.2

=P

1800

=P

 Amount = P + I S

S 

(1 + rt ) 1800

(1 + 0.1× 2 )

1.2

=P

∴ P = $1500

$1500 should be invested. 5.

P=?

r = 6% = 0.06

 Amount = P + I S

S 

(1 + rt )

t = 7months = 127 years

Amount =   $8000

S

= P (1 + rt )

= P + Prt

8000

= P (1 + 0.06 × 127 )

= P (1 + rt )

8000

= P ×1.035

=P

8000

=P

8000

(1 + 0.06 × 127 )

OR

1.035

=P

∴ P ≈ $7729.47

$7729.47 should be invested.

FINM 16

SOLUTIONS

FINM-B2

Financial Mathematics

continued

6.

(a)

 Balance = $120 000 − $30 000

= $90 000 (b)

 I = $90 000

r  = 7% = 0.07

t = 15 years = 15 ×12 = 180months  I = Prt 

= 90000 × 0.07 × 15 = $94 500 (c)

Total Repayable = 90 000 + 94 500

= $184 500 (d)

 Monthly Repay =

180 = $1025

P = $15 000

7.

184500

r people = 5% = 0.05

t people = 3 years

r friend = 4% = 0.04

t friend  = 4 years

 I  people = Prt

I friend  = Prt 

= 15000 × 0.05 × 3 = $2250 Total people = 15000 + 2250

= 15000 × 0.04 × 4 = $2400 Total friend  = 15000 + 2400

= $17250

= $17400

The cheaper loan is available from the People's Bank. FINM-C1 1.

Compound Interest (a)

P = $600 S = P (1 + i )

n = 8 ×12 = 96 i =

8% 12

=

0.08 12

n

96

= 600 (1 + 0.08 ≈ $1135.47 12 )  I = S − P

= 1135.47 − 600 = $535.47 (b)

P = $1000 S = P (1 + i )

n = 5 × 365 = 1825

i=

12%

6%

=

365

=

n

1825

= 1000 (1 + 0.12 365 )

≈ $1821.94

 I = S − P

= 1821.94 − 1000 = $821.94 (c)

P = $750 S = P (1 + i )

n=

12 12

× 52 = 52

i=

52

n

52

= 750 (1 + 0.06 ≈ $796.35 52 )  I = S − P

= 796.35 − 750 = $46.35

FINM 17

0.06 52

0.12 365

SOLUTIONS

Financial Mathematics

FINM-C1 1.

continued P = $3000

(d)

n = 4 × 4 = 16

S = P (1 + i )

i=

9%

0.09

=

4

4

n

16

= 3000 (1 + 0.09 ≈ 4282.86 4 )  I = S − P

= 4282.86 − 3000 = $1282.86

2.

P=?

S = $10 000

n = 6 × 2 = 12 i = S = P (1 + i )

P = S (1 + i )

−n

0.08

=

2

2

= 0.04

n

P=

OR

= 10000 (1 + 0.04 )

8%

−12

=

S 

(1 + i )

n

10000 12

(1 + 0.04 )

∴ P ≈ $6245.97 $6245.97 should be invested. 3.

P = $500

S = $1326.65 n = 21 S = P (1 + i )

i =?

n

1

i=

n

S P

−1

OR

⎛ S  ⎞ n i = ⎜ ⎟ −1 ⎝P⎠ 1

=

1326.65 21 600

⎛ 1326.65 ⎞ 21 =⎜ ⎟ −1 ⎝ 600 ⎠ ≈ 0.0385 = 3.85%

−1

≈ 0.0385 = 3.85%

The interest rate was approximately 4.8%. 4.

P = 4000

For 10% compounded annually

For 9.75% compounded monthly

i = 10% = 0.1

i=

S10% = P (1 + i )

n =1 n

9.75% 12

=

0.0975

S9.75% = P (1 + i ) 1

n = 12

12 n

12

= 4000 (1 + 0.1)

= 4000 (1 + 0.0975 12 )

= $4400

= $4407.91

9.75% compounded monthly is the better investment.

FINM 18

SOLUTIONS

FINM-D1 1.

Financial Mathematics

Effective Interest Rates (i)

P = $6000

n=8

S = P (1 + i )

i = 8% = 0.08

n

I = S −P 8

(ii)

= 6000 (1 + 0.08 )

= 11105.58 − 6000

≈ $11105.58

= $5105.58

P = $750

n=

S = P (1 + i )

12 12

=1

n

I = S −P 1

2.

(i)

i = 10% = 0.1

= 750 (1 + 0.1)

= 825 − 750

≈ $825

= $75

r = 10% = 0.1

N  = 4

(iii)

 N 

r = 8% = 0.08  N 

r  ⎞ ⎛ r e = ⎜1 + ⎟ − 1 ⎝  N  ⎠

r  ⎞ ⎛ r e = ⎜1 + ⎟ − 1 ⎝  N  ⎠

4

365

⎛ 0.1 ⎞ = ⎜1 + ⎟ −1 4 ⎠ ⎝ ≈ 0.1038 = 10.38% (ii)

r = 12% = 0.12

⎛ 0.08 ⎞ = ⎜1 + ⎟ −1 365 ⎠ ⎝ ≈ 0.0833 = 8.33%

N  = 12

(iv)

 N 

r = 12% = 0.12 r  ⎞ ⎛ r e = ⎜1 + ⎟ − 1 ⎝  N  ⎠

12

365

⎛ 0.12 ⎞ = ⎜1 + ⎟ −1 12 ⎠ ⎝ ≈ 0.1268 = 12.68% P = $6000

n=7

S = P (1 + i )

⎛ 0.12 ⎞ = ⎜1 + ⎟ −1 365 ⎠ ⎝ ≈ 0.1275 = 12.75%

i = 8% = 0.08

n

= 6000 (1 + 0.08 )

7

≈ 10282.95 The certificate is worth $10 282.95 4.

First 4 years: P = $2000 i = 6% = 0.06 n = 4 S = P (1 + i )

n

4

= 2000 (1 + 0.06 ) ≈ $2524.95 Last 4 years: P = $2524.95 i = S = P (1 + i )

N  = 365

 N 

r  ⎞ ⎛ r e = ⎜1 + ⎟ − 1 ⎝  N  ⎠

3.

N  = 365

6% 2

=

0.06 2

n = 4×2 = 8

n

8

= 2524.95 (1 + 0.06 ≈ $3198.53 2 ) The $2000 will amount to $3198.53.

FINM 19

SOLUTIONS

FINM-E1

Financial Mathematics

Equations of Value

1.

i=

10% 4

=

⎛ 0.1 ⎞ ⎜1 + 4 ⎟ = 1.025 ⎝ ⎠

0.1 4

repayments = debts repay1 = debt1  x = 550

+

( 1.025 )

−16

debt2  

+ 550 ( 1.025 )

= 370.49 + 335.65 = $706.14

The repayment would be $706.14 now. 2.

i=

8% 2

=

⎛ 0.08 ⎞ ⎜1 + 2 ⎟ = 1.04 ⎝ ⎠

0.08 2

repayments = debts repay1 = debt1  x = 600

+

debt2  

−2

− ( 1.04 ) + 800 ( 1.04 )

4

= 554.73 + 683.84 = $1238.57

The repayment would be $1238.57. 3.

i=

6% 4

=

⎛ 0.06 ⎞ ⎜1 + 4 ⎟ = 1.015 ⎝ ⎠

0.06 4

repayments

=debts

repay1 + repay2 = debt 1

2000

The second repayment should be $2447.81.

24

( 1.015 ) +  x = 5000 (1.015) 2859.01 +  x = 5306.82  x = $2447.81

4.

i = 7% = 0.07

(1 + 0.07 ) = 1.07 = debts = debt1 +

repayments repay1 + repay2 + repay3 4

2

debt 2  1

2000 (1.07 ) + 4000 (1.07 ) +  x = 5000 (1.07 ) + 5000 (1.07 ) 2621.59 + 4579.60 +  x = 5350.00 + 3814.48  x

= $1963.29

The final repayment will be $1963.29.

FINM 20

−4

4

−20

SOLUTIONS

Financial Mathematics

FINM-G 1.

2.

Review Exercise P = $2000

(i)

r  = 8% = 0.08

r  = 5.6% = 0.056

t = 5 years

t = 30months = 2.5 years

 I = Prt 

 I = Prt 

= 2000 × 0.08 × 5 = $800 Total Amount = P + I  = 2000 + 800 = $2800

= 35000 × 0.056 × 2.5 = $4 900 Total Amount = P + I  = 35000 + 4900 = $39 900

P = 2100 − 300 = $1800

(a)

P = $35000

(ii)

r = 8.5% = 0.085 t = 3 years = 36months

 I = Prt 

= 1800 × 0.085 × 3 = $459 S  = 1800 + 459

(b)

(c)

= $2259

 Repayments =

Total Cost  = 2259 + 300 = $2559

3.

P=?

r = 11% = 0.11

t = 5 years

Amount =  $7800

OR

S

= P (1 + rt )

= P + Prt

7800

= P (1 + 0.11 × 5 )

= P (1 + rt )

7800

= P ×1.55

=P

7800

=P

 Amount = P + I S

S 

(1 + rt ) 7800

(1 + 0.11× 5 )

1.55

=P

∴ P ≈ $5032.26

$5032.26 should be invested. 4.

(i)

P = $2000 S = P (1 + i )

n = 5 × 2 = 10

i=

8% 2

=

n

10

= 2000 (1 + 0.08 ≈ $2960.49 2 )  I = S − P

= 2960.49 − 2000 = $960.49 (ii)

P = $35 000 S = P (1 + i )

n = 30

i=

5.6% 12

=

0.056 12

n

30

= 35000 (1 + 0.056 ≈ $40 246.47 12 )  I = S − P

= 40 246.47 − 35000 = $5 246.47

FINM 21

0.08 2

2259 36

= $62.75

SOLUTIONS

Financial Mathematics

FINM-G 5.

continued P = $600

n = 3 × 4 = 12 i =

S = P (1 + i )

n

7.5% 4

=

0.075 4

12

= 600 (1 + 0.075 ≈ $749.83 4 )  I = S − P

= 749.83 − 600 = $149.83 The compound interest is $149.83. 6.

P = $5000

S = $6500

n = 4 ×12 = 48

S = P (1 + i )

i =?

n

1

i=

n

S P

−1

OR

⎛ S  ⎞ n i = ⎜ ⎟ −1 ⎝P⎠ 1

= 48

6500 5000

⎛ 6500 ⎞ 48 =⎜ ⎟ −1 ⎝ 5000 ⎠

−1

∴ i ≈ 0.00548 = 0.548% p.m. annual rate = 0.548% × 12 = 6.576% 7.

r A = 8.2% = 0.082  N

r ⎞ ⎛ reA = ⎜1 + ⎟ − 1 ⎝  N ⎠ 4

⎛ 0.082 ⎞ = ⎜1 + ⎟ −1 4 ⎠ ⎝ ≈ 0.0846 = 8.46%

NA = 4

rB = 8.1% = 0.081

N B = 365  

N 

r  ⎞ ⎛ r eB = ⎜1 + ⎟ − 1 ⎝ N  ⎠ 365

⎛ 0.081 ⎞ = ⎜1 + ⎟ −1 365 ⎠ ⎝ ≈ 0.0844 = 8.44%

Company B offers the best rate for a loan. 8.

i=

10%

=

0.1

12 12 repayments

=debts

repay1 + repay2

= debt1 +

48

debt 2  12

+ x = 2000 (1 + 0.1 + 1500( 1 + 0.1 1000 (1 + 0.1 12 ) 12 ) 12 )

The final repayment would be $2077.90

FINM 22

1489.35 + x  x

= 2209.43 = $2077.90

+1357.82

−12

SOLUTIONS

Financial Mathematics

FINM 23