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“IN THE NAME OF ALLAH THE MOST
BENEFICIENT, THE MOST MERCIFUL”
1
SCIENTIFIC CALCULATORS
2
Success is a ladder,
on which you cannot
climb
with
both
hands
in
your
pockets.
3
Course Objectives
 In the course the main concentration is on the applications
of mathematics to businesses.
 These may be seen in the form of stated problems from the
text book & other reference material.
 Our focus will be on the procedures to find the solutions to
these problems.
 Time to time some business terminologies will be used in the
course.
4
A WARM
WEL COME
5
PERCENTAGE
6
Percentage is another aspect of fraction, specifically numbers
expressed as fractions of 100. by expressing two or more
numbers in the same terms – as parts of 100 – we can compare
them directly. Thus 100 is the denominator of the fraction, and
the number is said be so many hundredths or so many percent
of the total. Percent is the Latin expression for hundredths.
CHANGING A DECIMAL TO A PERCENT:
Because percent means hundredths, we can change the
expression 0.04 (read as 4 hundredths) to 4% (read as 4
percent). To change a decimal to a percent:
Move the decimal point two places to the right (two places
stands fro hundredths), and place the percent sign to the right
of the number.
CHANGING A PERCENT TO A DECIMAL
To convert a percent to a decimal, move the decimal point two places
to the left and remove the percent sign. Remember that whole
number percents between 1% and 99% are written in the first two
decimal places.
CHANGING A PERCENT TO A FRACTION
1.
2.
3.
4.
5.
7
Change the percent to the decimal equivalent.
Read the value of the decimal as “ – hundredths ,” and write this
value as a fraction.
Reduce the fraction to lowest terms.
The only exception to this procedure is with a percent containing a
fraction. Because percent means hundredths (1/100), take the
percent containing a fraction and multiply by 1/100.
Reduce the resulting fraction to the lowest terms.
Examples
1.
2.
3.
4.
5.
6.
7.
4% = 4 hundredths (% sign means hundredths) = 0.04
85% = 85.% = 0.85
46.7% = 46.7% = 0.467
243.2% = 243.2% = 2.432
0.8% = 00.8% = 0.008
All the above examples tell how to convert a percentage to
the decimals.
94 47
94% = 0.94 = 94 hundredths = 100  50
150% = 1.50 = one and 50 hundredths
50
1
1
1
100
2
8.
25.5% = 0.255 = 255 thousands

8
255
51

1000 200
PROJECT 1 PERCENTS
A. Express each of the following as percent.
1. 0.89 =
4. 0.0037=
2. 6.93=
5. 0.045=
3. 0.0007=
6. 0.508=
B. Express each of the following as a decimal.
7. 35%=
9. 0.03%=
8. 72.6%=
10. 125.04%=
C. Express each of the following as common fraction in the
lowest terms.
11. 36%=
14. 0.035%=
12. 224%=
15. 0.0062%=
13. 6.3%=
9
APPLICATIONS OF PERCENTS
Percents are always expressed in some variation of the form
R B = P
where R is the rate
B is the base
P is the percentage
The base is the number or quantity taken as a whole, the total, or
100%. It may be total students in KARDAN, gross or net sales for
the year, the population of a country in a particular year, etc.
The rate is the number of hundredths of the total (the base) under
consideration. Without a base, the rate is meaningless. If one says
that 50 students represent 10%, the question immediately arises, 10%
of what?
The percentage is the product of the rate times
the base. It is easy to compute anyone of
the three parts of the formula if you know the
other two parts.
10
Examples
1.
2.
What is 35% of $756?
P=R B

P = 0.35 756
(of means multiply)
P = $264.60
What percent of $428 is $64.20.
RB = P
R  $428 = 64.20
R  428 64.20

428
428
Divide by 428
64.20
R
 0.15  15%
428
[Note: when finding the rate remember that the final answer must be
expressed in percent from.]
11
Examples (cont…)
3.
12.5% of what is 625?
R B = P
0.125  B = 625
0.125B 625

Divide by 0.125
0.125 0.125
625
B
 5,000
0.125
There are many uses of this percentage formula. One is finding the
rate of increase or decrease (some times call percent of change).
When computing rate of change always use the original , or
beginning, quantity as the base.
12
Examples (cont…)
The formula R  B = P may be restated as
Rate x Base = Amount of increase or decrease
4.
The price of a pound of ground beef was $1.00 last year and
$1.50 this year. What was the rate of increase?
R  B = Amount of increase
Amount of increase = 1.50 – 1.00 = 0.50
R 1.00 = 0.50
R 1.00 0.50

Divide by 1.00
1.00
1.00
1
R   50%
2
13
Examples (cont…)
5.
A portable radio bought for $24.50 is sold for $32.
What percent of the cost is profit?
R  24.50 = (32.00 – 24.50)
R  24.50 = 7.50
R  24.50 7.50

24.50
24.50
R  30.60%
14
Profit is the amount
of increase
Divide by 24.50
PROJECT 2
PERCENTS: APPLICATIONS
1. What is 45% of $850?
2. Rida bought a calculator for $75. She made a down
payment of 20% of the price. (a) How much was the
down payment? (b) what percent of the price was the
unpaid balance?
3. What percent of $180 is $15?
4. $31.25 is what percent of $375?
5. If 4% of a number is $3.56, what is the number?
15
6. 17% of what number is 20.23?
PROJECT 2 (cont…)
7. $525 is what percent more than $500?
8. $500 is what percent less than $525?
9. There were 500 freshman last year and 496 this year in a
small Midwestern college. Find the percentage decrease.
10. Sarah Gold received a dividend of $450, which is 5% of
her investment. What is the size of the investment?
11. Last year’s taxes on a house were $1,550. this year’ taxes
are $1,250. What percent are this year’s taxes of last
year’s taxes?
16
PROJECT 2 (cont…)
12. Five out of 25 students in Professor Ford’s class received an”A”
grade. What percent of the class did not receive “A” grade?
13. In 2003, Dunkin’ Donuts Company had $300,000 in doughnuts
sales. In 2004, sales were up 40%. What are Dunkin’ Donuts sales
for 2004?
14. The price of an Apple computer dropped from $1,600 to $1,200.
What was the percent decrease?
15. In 1982, a ticket to the Boston Celtics cost $ 14. In 2003, a ticket
cost $50. What is the percent increase to the nearest hundredth
percent?
17
PROJECT 2 (cont…)
16. At a local McDonald’s, a survey showed that out of 6,000 customers
eating lunch, 1,500 ordered diet coke with their meal. What percent of
customers ordered Diet Coke?
17. Out of 6,000 college students surveyed, 600 responded that they do not eat
breakfast. What percent of the students do not eat breakfast?
18. What percent of the college students in problem 17 eat breakfast?
19. Borders bookstore ordered 80 Marketing books but received 60 books.
What percent of the order was missing?
20. The sales tax rate is 6%. If Jim bought a new Buick and paid a sales tax of
$1,920. what was the cost of the Buick before the tax?
18
PROJECT 3
1.
PERCENTS APPLICATIONS
John bought a new car and made a down payment of $1,080. after
the payment, he owes 85% of the purchase price. What is the price
of the car?
2.
Salma bought a piece of jewelry for her friend for $28.00, which
includes a federal tax of 10% and a state sales tax of 2%. What is
the price excluding the taxes?
3.
Nancy needs 352.8 inches of cloth to complete her project of home
economics class. The cloth will shrink 2% after washing. How
long should the piece be before washing?
19
PROJECT 3 (cont…)
4.
Fancy furniture company sold a bedroom set for $960, for a loss of
$163.20 on the original price. What was the percent of loss on the
original purchase price?
5.
Jim Hill works for a salary of $285 per week plus a commission of
6% of his sales. What were his sales during a week in which he
earned $407.85?
6.
Handy Office Supply bought a manual typewriter for $220 from
the factory and sold it to customer for $420. What percent of the
selling price is the cost?
20
PROJECT 3 (cont…)
7.
The Sewing Center Shop cut the price of a dress fabric from $6 a
yard to $4. what percent price cut did they advertise?
8.
Sam’s Suit Shop is selling for one day a man’s three-piece suit for
$232.50, a loss of 7% on the cost from the manufacturer. What was
the cost of each suit?
9.
Mountain Peak Community College decided to cut the size of the
entering freshman class by 15% because of the 15% budget cut.
Last year’s class had 1,180 students. How large is this year’s class?
21
PROJECT 3 (cont…)
10. The value of the inventory of Johnny’s Corner Variety Store
increases from $90,000 to $97,200 during one-year period. What
percent rise in value did this represent?
11. In Kalamazoo, the number of building permits issued during April
dropped from 720 last year to 576 this year. What percent decrease
in permits did this represent?
12. The cost of a table was increased by 15% of itself to get the selling
price. What was the cost of the table if the selling price was $161?
22
CHAPTER NO 9
Mathematics of
Finance 1
23
Profit:
 The difference between the selling price and the cost
price of an item is called the profit or mark-up. Thus if
“S” is the selling price and “C” is the cost price the
mark-up can be calculated as:
P=S–C
 The mark-up is expressed as the percentage of the cost
price or the selling price.
 Often the mark-up is based on the cost.
 The selling price can be determined by adding the
mark-up to the cost.
24
Continue…
 If “C” and “S” represents the cost price and the sale price
respectively and “r” is the percentage mark-up, then:
S = C + Cr = C(1+r)
Example:
If the Cost price of an item is $ 2,400 and the
mark-up on cost is 23%. Find the sale price.
Solution:
Using the formula
S = C(1+r)
Putting the values
S = 2,400(1+0.23) = 2,952
25
When mark-up is stated as a percentage on sale
If “d” is the percentage mark-up on sales. Then
d.S would represent the disount or mark-down.
Now
Profit = Sales price – Cost price
d.S = S – C
Or
C = S - d.S
C = S(1- d)
Example:
After a mark-up of 30% on sales a watch sells for $ 225.
i)
What is its cost price?
ii)
What is the %age mark-up on sales if the cost price of watch would have
been $ 153?
26
Simple Interest and Present Value
o Interest is a fee which is paid for having the use
of money.
o The amount of money that is lent or invested
is called principal.
o Interest is usually paid in proportion to the principal
and the period of time over which the money is used.
o The interest rate specifies the rate at which
interest accumulates.
o The interest rate is typically stated as a
percentage of the principal per period of time.
27
Continue…
o The interest paid only on the principal is
called simple interest.
o Simple interest is usually associated with
loans or investments which are short-term in
nature.
Computation
Simple interest = (principal ) * ( interest rate) * (number
of time period)
Or
I = Prt
Where I = simple interest
P = principal
( also denoted by PV )
r = interest rate ( also denoted by i )
t = number of time period of loan
( denoted by n also )
28
Continue…
The total amount “A” to be repaid is the principal
plus the accumulated interest, or
A = FV = P + I
=P+Prt
= P(1 + r t) or FV=PV(1 + r t )
If the future value “A” is known, the present value of
an amount “P” at simple interest “r” can be written
as :
A
FV
P

(1  r t ) (1  r t )
29
Important
Note that in computing the interest it is
customary to consider a 360-day year instead
of a 365-day year. Thus 30 days will be
30
1

considered as 360of 12an ordinary year and so
on. The interest thus obtained is called
“ordinary interest” but if it is based on 365
days, it is called the “exact interest”.
30
Application of the formula
Example:
A credit union has issued a 3-year loan of $5,000. Simple
interest is charged at the rate of 10% per year. The
principal plus interest is to be paid at the end of third year.
Compute the interest for the 3-year period. What amount
will be repaid at the end of the third year?
Solution:
On board
31
Simple Discount
If “A” is the amount to be paid at maturity after
a time “t” at the simple interest rate of “r
percent” per annum. Then the simple discount
“D” on maturity value “A” in time “t” is given by
D  Art
To get the present value “P”, we must subtract
this from “A”.
P  A  Art
P  A(1  rt )
32
Discounting Negotiable Instrument
A written promise to pay money at a certain
specific date is called Negotiable Instrument.
They are of two types,
 Non interest bearing
 Interest bearing
The basic principles of discounting a bill of
exchange or short term note at a bank or at any
other party are the same as those of obtaining a
loan from a bank which deducts interest in
advance.
33
Discounting non-Interest-Bearing Note
Example:
After Khalid accepted a bill for $ 4,500. Hanif discounted it at
National Bank Karachi on April 15. The maturity date of the bill was
May 15. How much did Hanif receive if the bill was discounted at
8%?
Solution:
Period of discount = 30 days or t = 1/12 year.
A = 4,500 & r = 8%
the discounted value
P = A(1-rt) = $ 4,470
34
Discounting Interest-Bearing Note
We follow the following two steps in discounting an interestbearing note.
i. Find the maturity value from the face value of the note after
adding the interest which would have been earned up to the
maturity date at the given rate.
ii. Find the proceeds by discounting the maturity value obtained in
step (i) at the discounting rate.
35
Example
Mohsin had a note for $ 15,000 with an interest rate of
6%. The note was dated January 12, 1983 and the
maturity date was 90 days after date. On January 27,
1983 he took the note to his bank which discounted it at
a rate of 7%. How much did he receive?
Solution:
Step-1:
Find maturity value FV
Step-2:
Discount the maturity value at 7%
36
Equivalent Values of Different Debts
and their Payments
Sometimes a situation arises when a single debt or a set of
debts are to be paid on different dates by means of a single
payment or a set of payments. To satisfy both the creditor
and the debtor, the values of payments should be
equivalent to the values of the original debts on a certain
date called the comparison date.
Due to interest, a sum of money has different values at
different times. Therefore a comparison date should first be
chosen to equate the sum of the values of the original debts
with the values of the desired payments on the same date.
This process will bring the different debts and the
subsequent payments on the same footing.
37
Example:
A man owes $ 800, $ 1,000, and $ 200 due in 30
days, 60 days and 90 days respectively. If the
rate of interest is 8%, Find the amount of the
single payment, if the payment is made:
1.after 75 days from now
2.after 110 days from now
3.after 30 days from now
38
Getting More Involved
Solving
PRACTICE SET 9, p # 110-111
&
PROBLEM SET 9,
p # 112-113
39
CHAPTER 10
MATHEMATICS
OF
FINANCE-II
40
Compound Interest
 Compounding involves the calculation of interest periodically




41
over the life of the loan (or investment).
After each calculation the interest is added to the principal.
Future calculations are on the adjusted principal (old principal
plus interest).
Compound interest is the interest on the principal plus the
interest of prior periods.
Future value, or compound amount, is the final amount of the
loan or investment at the end the last period.
Lets see
How $1 will grow if it is calculated for 4 years at 8% annually?
1.6
1.4
1.2
1
After 2
After 1 periods
Present period $1
$ 1.1667
value
$ 1.08
$ 1.00
After 3 After 4
periods periods
$1.2597 $1.3605
0.8
0.6
0.4
0.2
0
0
42
1
2
3
4
Some term to understand
 Compounded annually: Interest calculated on the





43
balance once a year.
Compounded semiannually: Interest calculated on
the balance every 6 months or every ½ yeas
Compounded quarterly: Interest calculated on the
balance every 3 months or every ¼ yeas
Compounded monthly: Interest calculated on the
balance each month.
Compounded daily: Interest calculated on the
balance each day.
Number of periods: Number of years multiplied by
number of times the interest is compounded per year.
For Example
If we compound $1 for 4 years at 8% annually, semiannually,
or quarterly, the following periods will result:
Annually:
4 years * 1 = 4 periods
Semiannually:
4 years * 2 = 8 periods
Quarterly:
4 years * 4 = 16 periods
 Rate for each period: Annual interest rate divided by
the number of times the interest is compounded per year.
Compounding changes the interest rate for annual,
semiannual, and quarterly periods as follows:
44
Continue…
Annually:
Semiannually:
Quarterly:
8% / 1 = 8%
8% / 2 = 4%
8% / 4 = 2%
Note:
Both the number of periods (4) and
the
interest rate (8%) for the annual
example do
not change. The rate and periods (not years) will
always change unless the interest is compounded
yearly.
45
Compound Amount Formula
Let
P = Principal
i = interest rate per compounding period
n = number of compounding periods
(number of periods in which the principal
earned interest)
S = compound amount
The compound amount after one period is
S = P + iP
S = P(1 + i)
46
has
Continue…
compound
amount
after two periods
=
compound amount
after one period
interest earned
+ during the second
period
S  P1  i   i[ P (1  i )]
S  P (1  i )(1  i )
S  P (1  i )
2
Similarly, the compound amount after three periods is given by:
S  P(1  i )
47
3
Continue…
Thus we have the following definition:
If an amount of money “P” earns interest compounded at a
rate of “i” percent per period, it will grow after “n” periods
to the compound amount “S”, where
This equation is often referred as n
the compound interest
S  P (1  i )
formula.
The compound interest is given by:
Compound interest = S - P
48
Examples
1. Find out the compound amount and the
compound interest at the end of three years on
a sum of $ 20,000 borrowed at 6%
compounded annually.
2. If $ 3,000 are invested at 6% interest
compounded semi-annually, what would it
amount to at the end of 8 years?
49
Effective Interest Rates
 Interest rates are typically stated as the
annual percentages. The stated annual rate
is usually referred to as the nominal rate.
 When interest is compounded semiannually,
quarterly, and monthly, the interest earned
during a year is greater than if compounded
annually.
 When compounding is done more frequently
than annually, an effective annual
interest rate can be determined.
50
Continue…
Definition:
The effective interest rate is the interest rate
compounded annually which is equivalent to a
nominal rate compounded more frequently than
annually. The two rates would be considered
equivalent if both will result in the same compound
amount.
Example:
The nominal rate of 8% compounded quarterly is
equivalent to the effective rate of interest 8.2432%.
51
Formula for finding the Effective
rate of interest
e  1  i   1
f
Where
e = the effective rate
i = interest rate per conversion period
f = number of conversion periods
in one year (frequency)
52
Formula for Finding the Equivalent
Rates of Interest
# of periods for
nominal rate
(frequency)
f1


r1 
r2 
1    1  
f1 
f2 


f2
# of periods for
the given rate
(frequency)
Given interest rate
# of times the compounding is
required per year (for given rate)
Unknown nominal interest rate
53
# of times the compounding is
required per year (for nominal rate)
Example
At what nominal rate compounded
quarterly
will
a
principal
accumulate to the same amount as
at 8% compounded semiannually?
SOLUTION ON BOARD
54
Depreciation by Reducing Balance Method
If ‘C’ is the original cost of a machinery and ‘T’ is the trade in or
scrap value of the machinery after ‘n’ years of useful life and ‘r’
is the percentage rate of depreciation on the reduced balance
each year.
Then,
Depreciation for 1st year = Cr
Residual value after 1st year = C – Cr = C(1-r)
Depreciation for the 2nd year = C(1-r)r
Residual value after 2nd year = C(1-r) – C(1-r)r = C(1-r)2
Continuing in this way we get:
Residual value after ‘n’ years = C(1-r)n
55
Continued….
Since ‘n’ is the useful life of the machinery when its
residual value is ‘T’, therefore,
C (1  r ) n  T
T
n
(1  r ) 
C
1 r 
n
T
C
r  1
or
56
n
T
C
T 
r 1  

C 
1
n
Getting More Involved
Discussion
on
Practice Set 10-A,
(P# 121/122)
57
PRESENT VALUE
(AT COMPOUND INTEREST)

The compound amount formula is given by:
S  P 1  i 
n
 A slight rearrangement of the formula gives:
S
n
P
 S 1  i 
n
1  i 
This formula is called the Present value or discounting formula.
 The factor
or
is
1
n
1  i 
called the present
value
1 
i  factor or the discounting factor.
n
58
Examples
Find the present value of $ 4,814.07 due at the end of 8
years if money is worth 6% compounded semi-annually.
ii. What sum of money invested at 6% compounded
annually will amount to $ 500 in 4 years?
iii. Find the present value of $ 4,958.54 due at the end of
8½ years if money is worth 6% compounded semiannually.
i.
59
Application on Discounting Interest bearing and
Non-interest bearing Notes
Example:
A non-interest bearing note of $ 3,000 is due in 5 years from
now. If the note is discounted now at 6% compounded semiannually, what will be the proceeds and the compound
discount?
(SOL ON BOARD)
60
Example: (interest bearing note)
An interest bearing note of $ 5,000 dated
January 1,1980 at 6% compounded
quarterly for 10 years was discounted on
January 1, 1984. what were the proceeds
and the compound discount if the note
was discounted at 8% compounded semiannually?
61
Examples
1. Mr. Ahmad owes Mr. Bashir $ 5,000 in three years
and $ 10,000 in 5½ years. How much should Mr.
Ahmad pay at the end of 4 years which may be
acceptable to Mr. Bashir if money is worth 8%
compounded semi-annually?
2. Mr.Zahir owes to Mr. Mohmood $ 4,000 due in 2
years and $ 5,000 due in 4 years. If he agree to pay $
4,500 now, how much he will have to pay to Mr.
Mahmood three years from now to settle his two
debts, if money is worth 6% compounded semiannually?
62
Continued…
Discussion on
Practice Set 10-B, P#(127-128)
&
Problem Set 10,
P# (128-129)
63
CHAPTER # 11
MATHEMATICS
OF FINANCE – III
ANNUITIES
64
DEFINITIONS
An annuity is a series of periodic payments
(usually equal in amounts).
 The payments are made at regular intervals of
time such as annually, semi-annually, quarterly
or monthly.
 Examples of annuities include
 Regular deposits to a savings account
 Monthly car
 Mortgage
 Insurance payments
 Periodic payments to a person from a retirement fund
65
Continue…
If the payments are made at the end of the
payment periods the annuity is called an
“ordinary annuity”
If the payments are made at the beginning of
each interval the annuity is called “annuity due”
The time between two successive payment dates
is called “payment period”.
The time between the beginning of the 1st
payment & the end of the last payment period is
called the “term of the annuity”
66
Sum of an annuity
Let
R= payment per period
i= interest rate per period
n= number of annuity payments
(also number of periods)
S= sum (future value) of the annuity
after n periods (payments)
The sum of the annuity is given by:
 1  i n  1
S  R

i


67
Example
1.
Find the amount of an annuity of $ 500 payable at
the end of each year for 10 years, if the interest rate
is 6% compounded annually.
Finding R when S is Known
2.
68
A father at the time of birth of his daughter decides
to deposit a certain amount at the end of each year in
the form of an annuity. He wants that the sum of $
20,000 should be made available for meeting the
expenses of his daughter’s marriage which he expects
to be solemnized just after her 18th birthday. If the
payments accumulate at 8% compounded annually,
how much should he start depositing annually?
Use of the annuity table
 In table 5 at book page 309 values of the sum
of an ordinary annuity of Re.1 are given. We
use the symbol sni read as “s angle n at i ” is
used for the factor (1  i)  1 .
n
i
 To search an entry in the table we consult the
table against i and n, and then multiply by the
payment per period.
 The use of the table will be discussed in the
solution of problems.
69
Finding n when S is known
Example:
How many semi-annual payments of $ 100 each at an
account in the form of an ordinary annuity will
accumulate $ 3,000 if the interest rate is 8% semiannually?
70
Present Value of an Ordinary
Annuity
 The present value of an annuity is an amount of money
today which is equivalent to a series of equal payments
in the future.
 For example
If a loan has been made, we may be interested in
determining the series of payments (annuity)
necessary to repay the loan with interest.
71
Finding the Present Value of an
Annuity
Let
R = amount of an annuity
i = Interest rate per compounding
periods
n = number of annuity payments (also,
the number of compounding)
P = Present Value of the annuity
Then,
n
1  (1  i )
P  R
i

72



Using table
 Table – 6 (book page 311) gives the
present value of an ordinary
annuity of Rupee 1 per period. The
factor 1 (1i i)  is denoted by the


symbol ani ( a angle n at i ).
Thus the general formula for the
present value of an annuity
becomes:
n
73
1  (1  i)  n 
P  R ani  R 

i


Examples
1. A loan of $ 94 is to be paid back in monthly
installments in one year the first one starting
after one month of the starting of the loan. If
the interest is charged at the rate of 24%
monthly on the unpaid principal, what will be
the amount of the monthly installment?
1. Find the present value of an annuity of $ 600
payable at the end of each year for 15 years
if the interest rate
is 5% compounded
annually.
74
Finding R when P is known
 Mr. Ahmad borrows $ 21,000 from a bank to
build a house with the condition that he would
pay back the loan in the semi-annual equal
installments in four years with interest rate at
6% compounded semi-annually. If the first
payment is to start at the end of first six monthly
period, what would be amount of each
installment?
75
Finding n when P is known
 Mr. Ashraf wants to deposit his savings of $
50,000 in a bank which offers 8% interest
compounded semi-annually so as to withdraw
$2,500 at the end of each six months from the
date of deposit. How many withdrawals will he
or his heir (in case of his death) be able to make
before the entire amount is exhausted?
76
Annuity Due
 When the periodic payments of an annuity starts at the
beginning of an interval rather than at the end of
interval the annuity is called an annuity due. Its term
begins on the date of the first payment and ends on
one interval after the last payment is made.
 The annuity due has a payment at the beginning of
each interest period but none at the end of the term.
Therefore the formula for calculating an annuity due is
given as:
S (due)  R sn 1i R
 R ( sn 1i 1)
or
77
 (1  i ) n 1 1 
S (due)  R 
1
i


Present Value of an annuity due
 For the present value of an annuity due, we find out the
present value of (n-1) periods of an ordinary annuity and
then add the 1st payment which has the same present value.
The formula for calculating the present value of an annuity
due is given by:
P(due)  R an 1i  R
 R (an 1i  1)
or
78
 1  (1  i)  n 1

P(due)  R 
1
i


Examples
1. If $ 250 are deposited at the beginning of
each quarter in a fund which earns interest at
the rate of 8% compounded quarterly what
will it amount to after the end of the year?
2. Mrs. Ahmad bought a sewing machine by
paying $ 50 each month for 10 months,
beginning from now. If money is worth 12%
compounded monthly, what was the selling
price of the machine on cash payment basis?
79
PERPETUITY
 An annuity whose payments starts on certain date and
continues indefinitely is called perpetuity. As the
payments continues for ever, it is impossible to compute
the amount of the perpetuity but its present value can be
determined easily.
 The formula for calculating the present value of the
perpetuity is given as:
R
P
i
where R is the size of periodic payment and i is the
interest rate per period.
80
Examples
81
1.
Pakistan Manufacturing Co. is expecting to pay $ 4.80
every 6 months on the share of its stocks. What is the
present value of a share if money is worth 8%
compounded semi-annually?
2.
Find the present value of Karachi Toy Company share
which is expected to earn $ 5.60 quarterly, if money is
worth 8% compounded quarterly.
Discussion on Selected
Exercises from
Practice sets 11-A (p# 139),
11-B(p#157-159) &
Problem set 11 (p# 160-162)
82
Chapter 16
Matrix Algebra
83
Matrix
 A matrix is a rectangular array of numbers enclosed in
brackets or in bold face parenthesis. Matrices are
represented by capital letters such as A, B, C, X, and Y etc.
 Examples of matrices are:
1 4
A

8 0
 67 
C  
9
84
5 1
B   3 3 
 5 4
Continued...
 A matrix is described by 1st stating its number of rows
and then its number of columns. This description of a
matrix is known as order of a matrix.
 In the above examples matrix A has the order 2 x 2 (2
by 2), B has the order 3 x 2 (3 by 2), and the order of
C is 2 x 1 (2 by 1).
 Generally if there are m rows in a matrix and n
columns, the order of the matrix would be m x n and
we may call it as m x n matrix.
 If m = n, the matrix is called a square matrix.
85
General Form of m x n Matrix
 Generally an m x n matrix may be in the form given below:
 a11

 a21
 a31




 am1
86
a12
a22
a32
a13
a23
a33
am 2
am3
a1n 

a2 n 
a3n 




amn 
Operation with Matrices
Addition / Subtraction
(i)

Matrices of the same order can be added/Subtracted.

While adding/subtracting two matrices of the same order we
add/subtract their corresponding elements.
Given two matrices:
 a11 a12 a13 


A   a21 a22 a23 
a a a 
 31 32 33 
&
 b11

B   b21
b
 31
b12
b22
b32
b13 

b23 
b33 
then
 a11  b11 a12  b12 a13  b13 


A  B   a21  b21 a22  b22 a23  b23 
a b a b a b 
 31 31 32 32 33 33 
87
Continued...
(ii) Multiplication
a)
Multiplication of a matrix by a real number (Scalar)


When a matrix is multiplied by a real number, each element of
the matrix is multiplied by that real number.
The product obtained is a matrix of the same order.
Example
then
88
Let
6
9
1


3.D  3.  0 1 2 
1
5
3


3* 6
 3*1

  3* 0 3* ( 1)
 3*1
3* 5

3

 0
3

18
3
15
27 

6 
9 

3* 9 

3* 2 
3* 3 

1 6 9


D   0 1 2 
1 5 3


b)
Multiplication of a matrix by another matrix






89
Multiplication of two matrices is only possible if the number of
columns in the first matrix is equal to the number of rows in the
second matrix.
If this condition is not satisfied multiplication will not be
possible.
If the order of the first matrix is m x n and the order of the
second matrix is n x p multiplication will be possible and the
order of the resultant matrix will be m x p.
To obtain any element in the product matrix, 1st determine the
row and column (in which the element lies) in the product
matrix.
Multiply the row of the first matrix with that column of the
second matrix, this value will give us that element.
Further the method is explained in the following examples.
Examples
 If possible multiply the following matrices.
 2 5 
A

7 8 
with
C  1 5 6 
with
1

2

M
5

9
90
3
4
3
7
5
6
1
6
7

8
7

2
 3
B 
 4
 0 2 8


D   9 3 5
 4 1 1


0 8 1
with N   3 4 2 
4 6 5


Continued…
In the given matrices
3 9 
A 
,

 2 4
5
D  2
3
5 6
B
,

1 7 
1
7,
6
4 6 3
,
E  

1 5 8
perform the following operations.
(a) A + B
(b)
C - D
(c) A . B
(d)
A . E
(e) E . C
(f)
C . F
91
1
C  3
8
2
9 
4
2 4 7 

F  

5 3 1
Example
A bakery makes three types of bread using the ingredients listed in the given table in convenient
units per loaf of bread.
these ingredients can be put in the matrix
form as:
Ingredients Required
Type of
bread
A
B
C
D
E
I
II
III
3
1
2
2
1
1
1
1
2
1
1
1
0
1
1
 3 2 1 1 0
1 1 1 1 1 


2 1 2 1 1
If an order is placed for 60 loaves of type I, 75 loaves of type II, and 50 loaves of type III which
can be shown in the matrix form as
60
1.
2.
92
75 50
Find the number of units of each ingredient required by the bakery to fill the order.
If per unit costs to the bakery of the ingredients A, B, C, D, and E are given by the
matrix given below, then find the cost for each type of bread $0.10 
$0.08 


$0.06 


$
0
.
05


$0.07 
DETERMINANT
A determinant is a rectangular arrangement of numbers in rows and columns
in two vertical lines. It is written in a manner similar to its associated form
of square matrix except that the bracket of the matrix is replaced by two
vertical lines.
For Example: The determinant of the matrix
is represented as
1
A
7
5
3
1 5
A 

7
3


and the determinant of the matrix
 2 7 9
M   1 0 3
 5 4 6
93
2
is M  1
5
7
0
4
9
3
6
How to evaluate the determinant?
The value of a determinant can be evaluated by two methods
i.
By cross multiplication
ii.
By finding the minors
Evaluation of Determinant by Cross Multiplication
Consider a 2 x 2 square matrix
 a11
A 
a21
a12 

a22 
The determinant of A is given by:
A
a11
a12
a21
a22
s1
Secondary Diagonal
94
p1
Primary Diagonal
Continue…
The determinant of a 3 x 3 matrix has 3 primary diagonals p1, p2,
p3 and 3 secondary diagonals s1, s2, s3
The 1st & 2nd columns are copied
b11
B  b21
b31
s1
s
b12
b22
b32
s
b13
b23
b33
b11
b21
b31
p
b12
b22
b32
p
p
3 is found out as:1
2
3
The numerical value of2 a determinant
a) Multiply the elements on each primary diagonal and add their products
b) Multiply the elements on each secondary diagonal and add their products
c) Subtract the results.
95
Exercises
Find the value of the following determinants:
(a)
(c )
( e)
(g)
( h)
96
2
8
4
9
(b)
9
0
0
4
2
2
3
4
7
5
8
6
9
8
6
0
(d )
3
0
1
0
3
2
4
2
2
5
1
1
0
3
4
(f)
4
5
9
7
5
8
3
5
2
2
0
2
2
1
1
2
0
Example
The National Center for Higher Education Management System uses matrices
as model to study college management. The elements of an important matrix in
this model, the induced course-load matrix, are the average numbers of units
taken in each field by students classified according to majors.
3.5
English 3.6

Mathematics 2.5
Biology 1.6

Chemistry 0.2

Accounting 0.4
Economics 1.5

Physical Ed. 1.5
History
97
Total 15
3.8
4.5
1.8
1.5
1.5
1.6
1.0
1.5
2.6
1.5
2.5
2.0
1.8
0.1
6.0
1.4
1.9
5.0
1.0
0.0
0.0
1.5
1.5
0.0
1.5
1.5
0.0
1.5
1.5
3.5
3.0
1.5
2.8
3.2 
2.5

2.0
1.2 

0.3
1.5 

1.5 
15
15
15
15
15
From Past
 ALGEBRAIC EXPRESSION
 EQUATION
(Statement which indicates that two algebraic
expressions are equal)
 SYSTEM OF LINEAR EQUATIONS
(We have discussed only the systems consisting of
two equations and two variables and the elimination
method for finding the solution of these systems).
 In this chapter we will do the same but this time with
a different method known as the Cramer’s Rule.
98
Cramer’s Rule
 Given a system of linear equations of the form
AX=B
Where A is an (n x n) square matrix of coefficients, X is the
column matrix of the variables, and B is the column matrix
which contains the values from the right sides of the equations
in the system.
 Cramer’s Rule provides a method of solving the system by
using determinants. To solve for the value of the jth variable,
form the matrix Aj by replacing the jth column of A with
column vector B.
 If we denote the determinant of Aj by  j the value of the jth
variable is determined as:
j
xj 

99
Continue…
  0 , the given system of equations has a unique solution.
 If   0 , the computation of x j is undefined. If   0 and
1   2     n  0, the system has infinitely
many solutions.
 If   0 , and any  j  0 , then the system has no solution.
 Further explanation of the method is given in the following
example.
Examples:
Using Cramer’s Rule, determine the solution to the given
system .
 If
3x1  2 x2  80
2 x1  4 x2  80
100
Exercises
 Determine the solution to the given systems using Cramer’s
Rule. If the system has no solution, or infinitely many
solutions, so state.
1.


2y
6y


 13
0
x

2y

4
2x

4y

18
x
5. 3 x
3x



y
2y
2y



z
4z
z
3.
101
3x
4x
x
6. 2 x
3x



2y
4y
6y



3z
z
9z
2.
4.






1
5
2
24
12
16
5x
3x


4y
5y


8
47
4x

8y

4
5x

3y

34
Exercises
7.
9.
Continue…
x

2y

 13
 4x

8y

52
x

2y

4z

3
2x
 3x


5y
6y


z
12 z


2
9
8.
 3x

4y

10
9x

12 y

30
Problem:
A store has three different mixes of nuts in sixteen kilogram bags marked A, B,
and C. The contents of each type of bag are given in the following table
Bags
Ground Nuts
Almonds
Apricots
A
8Kg
5Kg
3Kg
B
6Kg
4Kg
6Kg
C
10Kg
2Kg
4Kg
An order is received for a mixture of 132 Kg of ground nuts, 54 Kg of almonds
and 70 Kg of apricots. How can the store fill this order.
102