Business math CH 1

1
Business Math
Chapter 11:
Simple Interest and
Simple Discount

2
Cleaves/Hobbs: Business Math, 7e Copyright 2005 by Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved
11.1 The Simple Interest Formula
 Find simple interest by using the simple interest
formula.
 Find the maturity of a loan.
 Convert months to a fractional or decimal part of
the year.
 Find the principal, rate or time using the simple
interest formula.

3
Key Terms
 Interest: an amount paid or earned for the use
of money.
 Simple interest: interest earned when a loan or
investment is repaid in a lump sum.
 Principal: the amount of money borrowed or
invested.
 Rate: the percent of the principal paid as
interest per time period.
 Time: the number of days, months or years that
the money is borrowed or invested.

4
11.1.1 The Simple Interest Formula
 The interest formula shows how interest, rate,
and time are related and gives us a way of
finding one of these values if the other three
values are known.
I = P x R x T

5
Find the simple interest using
the simple interest formula

6
Identify the principal,
rate and time.
P= R x B
 The interest is a percentage.
 Principal is the amount borrowed or invested.
 Rate of interest is a percent for a given time
period, usually one year.
 Time must be expressed in the same unit of time
as the rate. (i.e. one year)

7
Find the interest paid on a loan.
 Principal = (P) $1,500.
 Interest rate = 9% (or 0.09)
 Time = 1 year
 Interest = P x R x T
 Interest = 1,500 x 0.09 x 1
 Interest = $135
 The interest on the loan is $135.

8
Try these examples.
 Find the interest on a 2-year loan of $4,000 at a
6% rate.
 $480
 Find the interest earned on a 3-year investment
of $5,000 at 4.5% interest.
 $675

9
11.1.2 Find the maturity value
of a loan.
 Maturity value: the total amount of money due
by the end of a loan period; the amount of the
loan and interest.
 If the principal and the interest are known, add
them.
 MV = principal + PRT
 MV = P(1+RT)

10
Look at this example.
 Marcus Logan can purchase furniture on a
2-year simple interest loan at 9% interest per
year.
 What is the maturity value for a $2,500 loan?
 MV = P (1 + RT) Substitute known values.
 MV = $2,500 ( 1 + 0.09 x 2)
(See next slide)

11
What is the maturity value?
 MV = $2,500 ( 1 + 0.09 x 2)
 MV = $2,500 (1 + 0.18)
 MV = $2,500 (1.18)
 MV = $2,950
 Marcus will pay $2,950 at the end of two years.

12
Try these examples.
 Terry Williams is going to borrow $4,000 at 7.5%
interest. What is the maturity value of the loan
after three years?
 $4,900
 Jim Sherman will invest $3,000 at 8% for 5
years. What is the maturity value of the
investment?
 $4,200

13
11.1.3 Convert months to a
fractional or decimal part of a year.
 Write the number of months as the numerator
of a fraction.
 Write 12 as the denominator of the fraction.
 Reduce the fraction to lowest terms if using the
fractional equivalent.
 Divide the numerator by the denominator to get
the decimal equivalent of the fraction.

14
Convert the following to fractional
or decimal part of a year.
 Convert 9 months and 15 months, respectively,
to years, expressing both as fractions and
decimals.
 9/12 = ¾ = 0.75
 9 months = ¾ or 0.75 of a year
 15/12 = 1 3/12 = 1 ¼ = 1.25
 15 months = 1 ¼ or 1.25 of a year.

15
 To save money, Stan Wright invested $2,500 for
45 months at 3 ½ % simple interest. How much
interest did he earn?
 45 months = 45/12 = 3.75
 I = P x R x T
 I = $2,500 x 0.035 x 3.75
 I = $328.13
 Stan will earn $328.13

16
Try these examples.
 Akiko is saving a little extra money to pay for her
car insurance next year. If she invests $1,000
for 18 months at 4%, how much interest can she
earn?
 $60
 Habib is going to borrow $2,000 for 42 months at
7% . What will the amount of interest owed be?
 $490

17
11.1.4 Find the principal, rate or time
using the simple interest formula.

18
Find the principal using the simple
interest formula.
 P = I / RT
 Judy paid $108 in interest on a loan that she had
for 6 months. The interest rate was 12%. How
much was the principal?
 Substitute the known values and solve.
 P = 108/ 0.12 x 0.5
 P = $1,800

19
 R = I / PT
 Sam wants to borrow $1,500 for 15 months and
will have to pay $225 in interest. What is the
rate he is being charged?
 R = 225/ $1,500 x 1.25
 R = .12 or 12%
 The rate Sam will pay is 12%.
Find the rate using the simple
interest formula.

20
 T = I / RP
 Shelby borrowed $10,000 at 8% and paid $1,600
in interest. What was the length of the loan?
 T = $1,600/0.08 x $10,000
 T = 2
 The length of the loan was two years.
Find the time using the simple
interest formula.

21
11.2 Ordinary and Exact
Time and Interest
 Find ordinary and exact time.
 Find the due date.
 Find the interest using the ordinary and exact
interest rates.
 Find simple interest using a table.

22
11.2.1 Find ordinary and
exact time.
 Ordinary time: time that is based on counting
30 days in each month.
 Exact time: time that is based on counting the
exact number of days in a time period.

23
Examples
 The ordinary time from July 12 to September
12 is 60 days.
 To find the exact time from July 12 to
September 12, add the following:
Days in July (31 -12 =) 19
Days in August 31
Days in September +12
62 days

24
Sequential Numbers for
Dates of the Year
 Find the exact time of a loan using the sequential
numbers table.
(Table 11-1 in the text)
 If the beginning and due dates of the loan fall
within the same year, subtract the beginning
date’s sequential number from the due date’s
sequential number.
 Ex.: From May 15 to October 15
 288-135 = 153 days is the exact time.

25
Beginning and due dates in
different years.
 Subtract the beginning date’s sequential number
from 365.
 Add the due date’s sequential number to the
result from the previous step.
 If February 29 falls between the two dates, add
1. (Is it a leap year?)

26
 Find the exact time from May 15 on Year 1 to
March 15 in Year 2.
 365 – 135 = 230
 230 + 74 = 304 days
 The exact time is 304 days.
 Note: If Year 2 is a leap year, the exact time is
305 days.

27
Try this example.
 A loan made on September 5 is due July 5 of the
following year.
 Find: a) ordinary time
b) exact time in a non-leap year
c) exact time in a leap year.
 Ordinary time = 300 days
 Exact time (non-leap year) = 303 days
 Exact time (leap year) = 304 days

28
11.2.3 Find the ordinary interest rate per day
and the exact interest rate per day.
 Ordinary interest: a rate per day that
assumes 360 days per year.
 Exact interest: a rate per day that
assumes 365 days per year.
 Banker’s rule: calculating interest on a
loan based on ordinary interest and
exact time which yields a slightly higher
amount of interest.

29
Find the ordinary interest per day.
 For ordinary interest rate per day, divide the
annual interest rate by 360.
Ordinary interest rate per day =
Interest rate per year
360

30
Find the exact interest per day.
 For exact interest rate per day, divide the annual
interest rate by 365.
Exact interest rate per day =
Interest rate per year
365

31
Use ordinary time to find the
ordinary interest on a loan.
 A loan of $500 at 7% annual interest rate. The
loan was made on March 15 and due on May
15. (Principal = $500) I = P x R x T
 Length of loan (ordinary time) = 60 days
 Rate = 0.07/360 (ordinary interest)
 Interest = $500 x 0.07/360 x 60
 Interest = $5.83

32
Find the ordinary interest using
exact time for the previous loan.
 Length of loan (exact time) = 61 days
 Rate = 0.07/360 (ordinary interest)
 Interest = $500 x 0.07/360 x 61

33
Find the exact interest using exact
time for the previous loan.
 Length of loan (exact time) = 61 days
 Rate = 0.07/365 (exact interest)
 Interest = $500 x 0.07/365 x 61

34
11.2.4 Find simple interest
using a table.
1. Identify the amount of money that the table
uses as the principal. (Usually $1, $100 or
$1000)
2. Divide the loan principal by the table principal.
3. Select the days row corresponding to the time
period (in days) of the loan.
(continue on next slide)

35
Find simple interest
using a table.
4.Select the annual rate column corresponding to
the annual interest rate of the loan.
5.Locate the value in the cell where the two
intersect.
6. Multiply the quotient from step 2 by the value
from step 5.

36
 Find the exact interest on a loan of $6,500 at
7.5% annually for 45 days.
 Use Table 11-2 in your text to locate the interest
for $100. Move across the 45-days row to the
7.5% column.
 The number is 0.924658.
 Divide $6,500 by $100 ( = 65)
 Multiply 0.924658 x 65 = $860.11
 The exact interest is $860.11

37
Try these examples.
 Find the exact interest on a $5,000 loan for 30
days at 8%.
 $32.88
 Find the exact interest on a $1,800 loan for 20
days at 8.5%.
 $8.38

38
11.3.1 Find the bank discount and proceeds
for a simple discount note.
 For the bank discount, use:
Bank discount = face value x disc. rate x time
[I = P x R x T]
 For the proceeds, use:
Proceeds = face value – bank discount
A = P - I

39
A promissory note

40
11.3.2 Find the third party discount and
proceeds for a third party discount note.
 For the bank discount, use:
Third party discount = maturity value of the
original note x discount rate x discount period.
 For the proceeds, use:
Proceeds = maturity value of original note –
third-party discount
A = P - I

41
 Mihoc Trailer Sales made a note of $10,000
with Darcy Mihoc,owner, at 9% simple interest
based on exact interest and exact time. The
note is made on August 12 and due November
10. Since Mihoc Trailer Sales needs cash, the
note is taken to a third party on September 5.
 The third-party agrees to accept the note with
a 13% annual discount using the banker’s rule.
 Find the proceeds of the note.

42
Mihoc Trailer Sales
 To find the proceeds, we find the maturity value
of the original note, then the third-party
discount.
 Exact time is 90 days (314-224)
 Exact interest rate is .09/365
 MV = P(1+ RT)
 MV = $10,000 ( 1 + 0.09/365 x 90)
 MV = $10.221.92

43
Find the proceeds of the note.
 Exact time of the discount period is 66 days.
(314 - 248) period between Sept. 5 and Nov. 10.
 Ordinary discount rate is 0.13/ 360.
 Third party discount = I = PRT
 Third party discount = $10,221.92 ( 0.13/360)(66)
 Third party discount = $243.62
 Proceeds = A = P – I
 Proceeds = $10,221.92 - $243.62 = $9,978.30

Business math CH 1

More Related Content

What's hot

Similar to Business math CH 1

Recently uploaded

Business math CH 1