Business Accounting
Business Accounting
BFC 1300; 1302 / RBA1002; 10x2
Study Unit 4:
Simple and Compound Interest
&
Rounding and Conversion
IMPORTANT TO NOTE
ALWAYS WRITE OUT YOUR FORMULA!!!!
Study Unit 4 –Simple and Compound & Rounding and Conversion Notes
Page 1
Business Accounting
.
Simple and Compound Interest
Every business at some time will borrow or invest money. If you wish to borrow money you
must pay for the use of that money (interest expense). If you invest money you will earn
income on the money invested (interest income), because you are giving up the right to use
that money for a certain period.
After studying these chapters, you should be able to:
4.1
-
Understand the difference between simple and compound interest
-
Calculate the different simple and compound interest formulae
Simple interest
Simple interest is used when a loan or investment is repaid in one lump sum and is usually
used for short-term loans or investments. The formula used to calculate simple interest is:
i
=
Pxrxt
The price paid for using the money is called interest ( i ) in Rand
The capital you borrowed is the principal (P) or present value (Pv)
The time period (t) or (n) is always expressed in terms of a year
The sum to be repaid (equal to the principal/present value plus interest) is called the amount
(A) or future value (Fv).
The interest rate is ( r).
Example 1
Find the simple interest on R600 for 4 years at 2% per annum.
i
=
Pxrxt
where: P or Pv
=
R600
=
R600 x 0.02 x 4
r
=
0.02 or 2%
=
R48
t or n
=
4 years
Note: Time (t or n) is always expressed in terms of a year.
Study Unit 4 – Simple and Compound Interest Notes
Page 2
Business Accounting
Example 2
Find the simple interest on R10 000 invested for 9 months at 15% p.a..
i
=
Pxrxt
=
R10 000 x 15 % x
=
R1 125
where: P or Pv
=
R10 000
r
=
0.15 or 15%
t or n
=
9 months
9
12
(12 months is 1 year)
Note: Time (t or n) is always expressed in terms of a year, therefore
9
12
Example 3
Find the simple interest on R350.60 for 82 days at 5% p.a.
i
=
Pxrxt
=
R350.60 x 5 % x
=
R3.94
where: P or Pv
=
R350.60
r
=
0.05 or 5%
t or n
=
82 days
82
365
(365 days is 1 year)
Note: Time (t or n) is always expressed in terms of a year, therefore
Do not round off t or n in a sum. Thus
Study Unit 4 – Simple and Compound Interest Notes
82
365
82
stays that, you do not use 0.22.
365
Page 3
Business Accounting
t or n (time period)
Denominator (time periods in a year)
Days
365
Weeks
52
Months
12
Quarterly
4
Biannually
2
Annually
1
4.1.1 Calculating interest over a number of days
Note: The first day is not counted, but the last day is counted.
Example 1
Find the number of days from 2nd March to 3rd July
March
29 days (31 – 2)
April
30 days
May
31 days
June
30 days
July
3 days
Total
123 days
t =
123
365
Study Unit 4 – Simple and Compound Interest Notes
Page 4
Business Accounting
Example 2
Loan R15 000 from the bank from 7 April to 28 September at 20% simple interest per
annum. Calculate the interest.
April
………………..
May
………………..
June
………………..
July
………………..
August
………………..
September
………………..
Total
………………..
i
Note:
=
Pxrxt
=
R15 000 x ………. x ……….
=
………………..
In a leap year February has 29 days and then there are 366 days in the year. A leap
year occurs every four years. (The year 2000 was a leap year therefore the next leap
year would occur in 2004, then 2008, then 2012 etc.)
Days in each month:
January
February
March
April
May
June
July
August
September
October
November
December
31
28
31
30
31
30
31
31
30
31
30
31
Study Unit 4 – Simple and Compound Interest Notes
Page 5
Business Accounting
4.1.2 Finding the principal (P) / present value (Pv)
i
=
Px r x t
or
i
=
Pv x r x n
By using the rules for changing the subject of the formula:
Make P or Pv the subject of the formula (Calculate principal/present value)
P
=
i
rxt
or
Pv
=
………………..
Example 1
An investment is made at the bank for 2 years at a simple interest rate of 20% per annum.
The interest received is R2 000. How much money was originally invested?
i
=
R2 000
r
=
20%
t
=
2
P
=
?
i
=
P rt
P
=
i
rxt
P
=
R2 000
0.20 x 2
P
=
R5 000
(to find the principal, make P the subject of the formula)
Study Unit 4 – Simple and Compound Interest Notes
Page 6
Business Accounting
4.1.3 Finding the principal (P)/present value (Pv), amount and (A)/future value (Fv),
The principal (P)/present value (Pv) is the original amount borrowed or invested.
The amount (A)/future value (Fv) is equal to the principal (P)/present value (Pv) plus
interest.
Example 1
Find the principal that will amount to R628.50 in 5 years, at 6.5% per annum simple interest.
A
=
P+i
but
i
=
Pxrxt
Therefore:
A
=
P + (P x r x t),
or
A
=
P(1 + {r x t}),
or
A
=
P(1 + rt)
By using the rules for changing the subject of the formula:
Make P the subject of the formula:
P
=
A
1  rt
P
=
R628.50
1  0.065 x 5
P
=
R474.34
A
Study Unit 4 – Simple and Compound Interest Notes
=
P(1 + rt)
Page 7
Business Accounting
Example 2
Invest R10 000 at 15% simple interest per annum. What will the value of the investment be
after 2 years?
A = P+i
A = P + (P x r x t)
A = R10 000 + (R10 000 x 0.15 x 2)
A = R13 000
Example 3
Invest R10 000 at 15% simple interest per annum. What will the value of the investment be
after 9 months?
………………………………….
………………………………….
………………………………….
………………………………….
Example 4
Invest R10 000 at 15% simple interest per annum. What will the value of the investment be
after 200 days?
………………………………….
………………………………….
………………………………….
………………………………….
Study Unit 4 – Simple and Compound Interest Notes
Page 8
Business Accounting
4.2
Compound interest
4.2.1 Basic concepts
When money is invested at simple interest rate, the investor draws the interest due to him at
the end of each year. Therefore, the interest for the following year is again calculated on the
same capital. Some investors, however, prefer to re-invest that interest so that the principal
for the following year may be greater. This means: interest is added to the initial investment
(or loan) and interest for the following period is calculated on the initial amount plus interest
earned in the previous period. The interest has been compounded.
When money is borrowed for longer periods the interest is compounded, meaning interest is
calculated more than once during the term of the loan/investment and this interest is added
to the original principal. The new amount then becomes the principal for the next calculation
of interest.
With this method you pay/earn interest on interest.
Example 1
Find the compound interest on R150.40 for 2 years at 6% per annum.
Principal/Present value at beginning of 1st year
Interest
=
R150.40 x
6
100
Principal/Present value at beginning of 2nd year
Interest
=
R159.42 x
6
100
R150.40
R 9.02
R159.42
R 9.57
Amount/Future value (at end of 2 years)
R168.99
Less: Principal/Present value
(R150.40)
Compound interest
R 18.59
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Therefore,
compound interest is equal to amount (A)
-
principal (P)
compound interest is equal to future value (Fv)
-
present value (Pv)
or,
Example 2
Find the compound interest on R5 000 for 3 years at 15% per annum.
Principal/Present value at beginning of 1st year
Interest
=
…………………….
Principal/Present value at beginning of 2nd year
Interest
=
…………………….
Principal/Present value at beginning of 3rd year
Interest
=
…………………….
Amount/Future value (at end 3 years)
Less: Principal/Present value
Compound interest
Study Unit 4 – Simple and Compound Interest Notes
R5 000.00
R.………..
R…………
R…………
R…………
R…………
R…………
(R5 000.00)
R…………
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Business Accounting
4.2.2 Compound interest formula
Compound interest for longer periods can be calculated by using a formula.
To deduce a formula for calculating compound interest:
Say R 1 is invested at 15 % per annum compound interest.
Interest on R 1 at 5 % per annum for 1 year
=
Pxrxt
=
R1 x 0.05 x 1
=
R 0.05
Amount (A) / Present value (Pv) at the end of 1 year =
or
R 1 + R 0.05
=
R 1.05
=
R 1 x 1.05
Amount (A) / Present value (Pv) at the end of 2 years =
=
(R1 x 1.05) x 1.05
R1 (1.05) 2
Formula:
A = P(1 + r)t
and
Compound interest = A – P
and
Compound interest = Fv – Pv
or
Fv = Pv(1 + r)n
Study Unit 4 – Simple and Compound Interest Notes
Page 11
Business Accounting
Recalculate examples 1 and 2 from 5.2.1 using the formula
Example 1
Find the amount (future value) and compound interest on R150.40 for 2 years at 6% per
annum.
A
=
P(1 + r)t
A
=
R150.40(1 + 0.06)2
A
=
R168.99
Compound interest = A – P
= R168.99 – R150.40
= R18.59
Example 2
Find the amount (future value) and compound interest on R5 000 for 3 years at 15% per
annum.
A
=
…………………………
=
…………………………
=
…………………………
Compound interest
=
…………………………
=
…………………………
=
…………………………
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Example 3
Find the amount (future value) and compound interest on R800 for 3 years at 13% per
annum.
A
=
…………………………
=
…………………………
=
…………………………
Compound interest
=
…………………………
=
…………………………
=
…………………………
4.2.3 Finding the principal/(present value)
To find the amount/future value of any principal/present value invested at 5% for 1 year the
principal/present value is multiplied by 1.05. Therefore, if the amount/future value is known
and we are asked to find the principal/present value, the amount/future value must be
decreased in the same ratio as the principal/present value was increased previously:
To increase principal/present value to amount/future value: multiply by 1.05
AND
To decrease amount/future value to principal/present value: divide by 1.05, therefore,
1
1.05
For 1 year the ratio is
1
1
; for two tears the ratio is
and for t years the ratio is
1.05
1.05 2
Study Unit 4 – Simple and Compound Interest Notes
Page 13
Business Accounting
1
1.05 t
1
1.05 t
P
=
A x
P
=
A
1.05 t
P
=
A
(1  r) t
By changing the subject of the formula A = P(1 + r)t, the principal/present value can
also be calculated.
Make P the subject of the formula A = P(1 + r)t:
P
=
A
(1  r) t
Example 1
Find the value a man has to invest now at 5% per annum compound interest for it to
produce a future value of R3 500 in 8 years.
P
=
P
=
P
=
A
(1  r) t
R3 500
(1  0.05) 8
R2 368.94
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Example 2
Jasper wants to buy a car for R22 350 in five years time. What sum must be invested now at
5% per annum compound interest?
………………………………….
………………………………….
………………………………….
4.3. Rounding and Conversion Tables
4.3.1 Rounding
Each digit in a number has a place value. The place value of the digit is determined by its
6
4
5
2
.
3
6
Thousandths
Hundredths
DECIMAL
POINT
Tenths
Ones
Tens
7
Hundreds
8
Thousands
Ten
thousands
9
Hundred
thousands
Millions
location within the number. The placed value chart is given:
8
Numbers to the left of the decimal point are whole numbers. Numbers to the right of the
decimal point are decimals. Numbers to the right of the decimal point have values of tenths
of a whole number, hundredths of a whole number, thousandths of a whole number and so
on. Let’s determine the place value of each digit in the following number: 9 876 452.368
Study Unit 4 – Simple and Compound Interest Notes
Page 15
Business Accounting
Digit
Place in chart
Place value
9
Millions
1 000 000 x 9
= 9 000 000
8
Hundred thousands
100 000 x 8
= 800 000
7
Ten thousands
10 000 x 7
= 70 000
6
Thousands
1 000 x 6
= 6 000
4
Hundreds
100 x 4
= 400
5
Tens
10 x 5
= 50
2
Ones
1x2
=2
3
Tenths
3 x 0.10
= 0.3 or
6
Hundredths
6 x 0.01
= 0.06 or
8
Thousandths
8 x 0.001
= 0.008 or
3
10
6
100
8
1000
In order to round whole numbers, the following three steps must be followed:
Step 1:
Determine the place value that is required to be rounded
Step 2:
Look at the digit immediately to the right of the desired rounding place.
Determine whether this digit is less than 5, or, 5 or more.
Step 3:
a) If the digit to the right of the desired rounding place is less than 5, then
leave the digit as it is and replace the succeeding digits with zeros.
b) If the digit to the right of the desired place is 5 or more, then add 1 to the
digit in the desired rounding place and replace succeeding digits with zeros.
Study Unit 4 – Simple and Compound Interest Notes
Page 16
Business Accounting
Example 1: Round 149 to the nearest ten:
Step 1 The place value to be rounded is the 4, because 4 is in the
tens’ place.
149
Step 2 The digit to the right of the desired rounding place is 9, that is more
than 5.
149
Step 3 Add 1 to the digit in the desired rounding place and replace
succeeding digits with zeros
150
Example 2: Round 134 to the nearest ten:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………..
Example 3: Round 3 248 to the nearest hundred:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Example 4: Round 2 152 to the nearest hundred:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
The same process is followed when rounding decimals:
Example 1: Round 19.23781 to two decimal places:
Step 1 The place value to be rounded is the 3, because 3 is in the
hundredth’s place.
19.23781
Step 2 The digit to the right of the desired rounding place is 7, that
is more than 5.
19.23781
Step 3 Add 1 to the digit in the desired rounding place and remove
all the succeeding digits after the hundredth’s place
19.24
Example 2: Round 3.87296 to two decimal places:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
4. 3.2 Conversion Tables
Length
1 kilometre (km)
=
1 000 metres (m)
1 metre (m)
=
100 centimetres (cm)
1 metre (m)
=
1 000 millimetres (mm)
1 centimetre (cm)
=
10 millimetres (mm)
1 litre (𝑙)
=
1 000 millilitres (ml)
1 cup
=
250 ml
1 teaspoon
=
5 ml
1 tonne (t)
=
1 000 kilogram (kg)
1 kilogram (kg)
=
1 000 grams (g)
1 gram (g)
=
1 000 milligrams (mg)
1 pound
=
454 g
Volume and Capacity
Mass
Example 1
Convert the following length 5 km to ? m
5 km X 1 000 (1 km = 1 000 m)
= 5 000 m
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Example 2
Convert the following length 5 m to ? km
…………………………………………………………………………………………………..
…………………………………………………………………………………………………..
Example 3
Convert the following volume and capacity 60 𝑙 to ? ml.
60 𝑙 X 1 000 ml (1 𝑙 = 1 000 ml)
= 60 000 ml
Example 4
Convert the following volume and capacity 60 ml to ? 𝑙.
…………………………………………………………………………………………………..
…………………………………………………………………………………………………..
Example 5
Convert the following mass 5 kg to ? g.
5 kg X 1 000 g (1 kg = 1 000 g)
= 5 000 g
Study Unit 4 – Simple and Compound Interest Notes
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Business Accounting
Example 6
Convert the following mass 5 g to ? kg.
…………………………………………………………………………………………………..
…………………………………………………………………………………………………..
Study Unit 4 – Simple and Compound Interest Notes
Page 21