Chapter 8 | Simple Interest and Applications
8.1 |
259
Introduction
Interest is a fee paid by borrowers to lenders for using money temporarily. Interest is an expense
when we borrow money and an income when we invest money.
For example, when we invest money, the financial institution uses our money and therefore, pays us
interest for the use of the money for the time period it has been invested. Similarly, when we borrow
money from a financial institution, we pay interest to them for the use of the money for the time
period borrowed, unless otherwise specified. In both cases, when money is returned after a period
of time, interest is added to the original amount borrowed. Therefore, as time goes by, the value
of money increases by the amount of interest earned for that period. This is called time value of
money: money grows with time when it is invested or borrowed at a
particular interest rate.
Interest is a fee
paid by borrowers
to lenders for using
money temporarily.
In simple interest calculations, interest is calculated only on the
initial amount invested or borrowed (not on any interest earned
during the past periods). For this reason, simple interest is usually
used for short term investments or loans.
For long term investments or loans, compound interest is used, where
interest is calculated on the amount borrowed or invested in addition
to the interest earned during the previous period. You will learn about
compound interest in Chapter 9.
8.2 |
Exhibit 8.1: Simple Interest
Calculating Amount of Simple Interest
In simple interest calculations, the amount of interest, for a period of time, is calculated as a
percent of the amount invested or borrowed. This percent is referred to as the rate of interest (or
interest rate), which is always stated for a unit period of time; i.e., r% per annum (r% p.a. or r%
per year).
In simple interest
calculations, interest
is calculated only
on the principal and
not on the interest
earned.
For example,
Interest for $1000 at 10% p.a. for 1 year = 1000 # 0.10 # 1 = $100.00
Interest for $1000 at 10% p.a. for 3 years = 1000 # 0.10 # 3 = $300.00
Similarly,
Interest for $P at r% p.a. for t years Therefore,
=P#r#t
Amount of Interest = Principal # Rate # Time Period
Therefore, the amount of interest is expressed by the following formula:
Formula 8.2
Amount of Interest
I = Prt
In calculations,
'r' is used
as a decimal
or fractional
equivalent of the
percent rate.
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Chapter 8 | Simple Interest and Applications
Notation
I = Amount of interest. It is the amount (in dollars) charged or earned for borrowing or investing
the principal amount for a period of time.
P = Principal. It is the amount of money borrowed or invested at the beginning of a period.
r = Simple interest rate. It is the rate at which the principal is borrowed or invested. It is
expressed as a percent for a given period of time, usually per year, unless otherwise specified. If
the interest rate is r%, it is understood that r% is an annual interest rate (r% p.a.).
t = Time period. It is the period of time for which the principal amount is borrowed or invested
during which interest is charged or earned.
Interest can be calculated
either using a 365-day year
(called the exact interest
method) or a 360-day year
(called the ordinary interest
method). We follow the
exact interest method in
this textbook.
Example 8.2(a)
Note:
I
Since r # t refers to the interest percent of the principal b r # t = P l , the units for
'r' and 't' should match.
For example,
Usually, the unit of 't' is
converted to match the
unit of 'r' using 1 year = 12
months or 1 year = 365 days.
■■ If 'r' is per annum (r% p.a.), then 't' should be in years.
■■ If 'r' is per month (r% p.m.), then 't' should be in months.
Calculating 'I' When 't' is Expressed in Years
Sabrina borrowed $4250 from her friend for 2 years. If she was charged a simple interest rate of
6% p.a. how much interest would she have to pay on the loan?
Solution
P = $4250,
r = 6% p.a. = 0.06 p.a. t = 2 years (unit of 't' and
Using Formula 8.2,
I = Prt
Substituting values,
I = (4250)(0.06)(2)
= $510.00
Therefore, she would have to pay interest of $510.00.
Example 8.2(b)
Calculating 'I' When 't' is Expressed in Months
unit of 'r' are the same)
Calculate the amount of interest earned from an investment of $1750 for 9 months at 4 12 % p.a.
Solution
P = $1750, r = 4 12 % p.a.,
t = 9 months
Method 1: Match the unit of 't' to that of 'r'
Method 2: Match the unit of 'r' to that of 't'
r = 4 % p.a. = 0.045 p.a.
t = 9 months = 9 years = 0.75 years
12
t = 9 months
Using Formula 8.2, I = Prt
Using Formula 8.2,
I = Prt
Substituting values, 0.045
I = (1750)` 12 j (9)
= $59.06
1
2
Substituting values, I = (1750)(0.045)(0.75)
= $59.06
Therefore, interest of $59.06 will be earned.
r = 4 12 % p.a. = 0.045 p.m.
12
Chapter 8 | Simple Interest and Applications
Example 8.2(c)
261
Calculating 'I' When 't' is Expressed in Days
Calculate the amount of interest charged on a loan of $3200 at 6% p.a. for 125 days.
Matching the unit of ‘t‘ to that of ‘r‘
Solution
P = $3200,
r = 6% p.a. = 0.06 p.a.
t = 125 days =a
Using Formula 8.2,
I = Prt
Substituting values,
I = (3200)(0.06) a
125
k years
365
125
k
365
= $65.75
Therefore, interest of $65.75 will be charged.
Example 8.2(d)
Calculating 'I' When 't' is Expressed in Days and 'r' is Expressed Per Month
Calculate the amount of interest earned on an investment of $2275 earning interest at 0.75%
p.m. for 90 days.
Solution
Method 1: Match the unit of ‘t’ to that of ‘r’
Method 2: Match the unit of ‘r’ to that of ‘t’
P = $2275, r = 0.75% p.m. = 0.0075 p.m
P = $2275, r = 0.75% p.m. = 0.0075 # 12 p.a. = 0.09 p.a
t = 90 days =
90
# 12 months
365
Using I = Prt
t = 90 days =
90
years
365
Using I = Prt
90
I = (2275)(0.0075)` 365 # 12j
90
I = (2275)(0.09) a 365 k
= $50.486301...
= $50.486301...
= $50.49
= $50.49
Therefore, interest of $50.49 is earned.
Example 8.2(e)
Calculating 'I' When 't' is Expressed in Years and 'r' is Expressed Per Month
Calculate the amount of interest earned on an investment of $5680 at 0.25% p.m. for 1 12 years.
Solution
Method 1: Match the unit of ‘t’ to that of ‘r’
Method 2: Match the unit of ‘r’ to that of ‘t’
P = $5680, r = 0.25% p.m. = 0.0025 p.m.
P = $5680, r = 0.25% p.m. = 0.0025 # 12 p.a. = 0.03 p.a
Using I = Prt
t = 1 12 years = 3 years
2
Using I = Prt
= (5680)(0.0025)(18)
= (5680)(0.03)S 3 X
= $255.60
= $255.60
1
t = 1 12 years = 1 2 # 12 months = 18 months
Therefore, interest of $255.60 is earned.
2
262
Chapter 8 | Simple Interest and Applications
Determining the Time Period in Days Between Dates
Since simple interest is mostly used for loans or investments with short time periods, it is necessary
to calculate the time period in days, unless it is given in months or years. The time period in days
is then converted to its equivalent in years to use as 't' in the simple interest formula.
The following months have 31 days: January, March, May, July, August, October, and December
The following months have 30 days: April, June, September, and November
There are 28 days in February, except in leap years, when there are 29 days.
Leap years occur every 4 years. For example, 1996, 2000, 2004, 2008, 2012, 2016, etc. are leap years.
However, century years can be leap years only if they are divisible by 400. For example, 1800,
1900, 2100, 2200, and 2300 are not leap years; however, 2000 and 2400 are leap years.
An Easy Way to Remember the Number of Days
in Each Month
Make a fist with one hand, hiding your thumb.
Using your other hand, start counting January, February,
March, April, May, June, July, on the knuckle and the space in
between the knuckles, as shown in blue on the left.
When you reach the end, go back to the first knuckle and
continue counting August, September, October, November,
and December, as shown in green.
All months that fall on a knuckle have 31 days and the others
have 30 days, except for February.
Exhibit 8.2: Days in Each Month
*February usually has 28 days, except in leap years where it has 29 days.
Described below are 3 methods to determine the number of days for a time period between two
dates:
Method 1: Determining the Number of Days Using Calendar Days in Each Month
In calculating the
number of days
in a time period,
you must either
exclude the first
day or the last day
in the day count.
While counting the number of days in a time period, we include either the first day or the last day
of the investment or loan, but not both days.
Financial institutions generally include the first day of the period and exclude the last day. This is
done because they base their financial calculations on the end of the day's closing balance. On the first
day, as you would either borrow or invest money, the closing balance would not be zero. However,
on the last day of the period, since you would either pay-back the loan or withdraw the investment
making the closing balance zero, financial institutions would not include the last day in calculations.
For example, the number of days between January 25 and February 05 are:
January 25 to January 31: 7 days (First day included)
February 01 to February 05: 4 days (Last day excluded)
Therefore, there are 11 days between January 25 and February 05.
In the following two examples, we have illustrated calculations that :
■■include the first day and exclude the last day, as done by financial institutions
■■exclude the first day and include the last day, which can also be done to determine the
number of days.
Chapter 8 | Simple Interest and Applications
Example 8.2(f)
Calculating the Number of Days in a Time Period During a Non-Leap Year
Calculate the number of days in the time period from October 18, 2013 to April 05, 2014.
Excluding the first day and including the last day
Including the first day and excluding the last day
Solution
October: October:
14 days
(First day excluded, counting from October 19 to
October 31 = 13 days or 31 - 18 = 13 days)
November: 30 days
November: 30 days
December: 31 days
December: 31 days
January: 31 days
January: 31 days
February: 28 days
February: 28 days
(Year 2014 is not a leap year)
(Year 2014 is not a leap year)
March: 31 days
March: 31 days
April: 4 days
April: 5 days
(Last day included, counting from
April 01 to April 05 = 5 days)
(Last day excluded, counting from
April 01 to April 04 = 4 days)
_______________________________
Total number of days = 169 days
169
Example 8.2(g)
13 days
(First day included, counting from October 18 to
October 31 = 14 days)
t = 365 years
_______________________________
Total number of days = 169 days
169
t = 365 years
Calculating the Number of Days in a Time Period During a Leap Year
Calculate the number of days in the time period from January 15, 2016 to March 20, 2016.
Solution
Including the first day and excluding the last day
January: 17 days
Excluding the first day and including the last day
January: 16 days
(First day included, counting from
January 15 to January 31 = 17 days)
(First day excluded, counting from January16 to
January 31 = 16 days or 31 - 15 = 16)
February: 29 days
February: 19 days
March: (Year 2016 is a leap year)
March: (Last day excluded, counting from
March 01 to March 19 = 19 days)
_______________________________
Total number of days = 65 days
29 days
(Year 2016 is a leap year)
20 days
(Last day included, counting from
March 01 to March 20 = 20 days)
_______________________________
Total number of days = 65 days
65
years
t = 365
65
t = 365 years
Method 2: Determining the Number of Days Using Days-Table
In this method, we use a table that has the days of the year serially numbered from 1 to 365 as
shown in Table 8.2. The number of days between two dates that are in the same calendar year is
determined by calculating the difference between the serial numbers of those dates in the table.
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Chapter 8 | Simple Interest and Applications
Table 8.2
Table to Determine the Number of Days in a Time Period
Day of
month
Jan
Feb Mar Apr May
Jun
Jul
Aug Sep Oct Nov Dec
Day of
month
1
1
32
60
91
121
152
182
213
244
274
305
335
1
2
2
33
61
92
122
153
183
214
245
275
306
336
2
3
3
34
62
93
123
154
184
215
246
276
307
337
3
4
4
35
63
94
124
155
185
216
247
277
308
338
4
5
5
36
64
95
125
156
186
217
248
278
309
339
5
6
6
37
65
96
126
157
187
218
249
279
310
340
6
7
7
38
66
97
127
158
188
219
250
280
311
341
7
8
8
39
67
98
128
159
189
220
251
281
312
342
8
9
9
40
68
99
129
160
190
221
252
282
313
343
9
10
10
41
69
100
130
161
191
222
253
283
314
344
10
11
11
42
70
101
131
162
192
223
254
284
315
345
11
12
12
43
71
102
132
163
193
224
255
285
316
346
12
13
13
44
72
103
133
164
194
225
256
286
317
347
13
14
14
45
73
104
134
165
195
226
257
287
318
348
14
15
15
46
74
105
135
166
196
227
258
288
319
349
15
16
16
47
75
106
136
167
197
228
259
289
320
350
16
17
17
48
76
107
137
168
198
229
260
290
321
351
17
18
18
49
77
108
138
169
199
230
261
291
322
352
18
19
19
50
78
109
139
170
200
231
262
292
323
353
19
20
20
51
79
110
140
171
201
232
263
293
324
354
20
21
21
52
80
111
141
172
202
233
264
294
325
355
21
22
22
53
81
112
142
173
203
234
265
295
326
356
22
23
23
54
82
113
143
174
204
235
266
296
327
357
23
24
24
55
83
114
144
175
205
236
267
297
328
358
24
25
25
56
84
115
145
176
206
237
268
298
329
359
25
26
26
57
85
116
146
177
207
238
269
299
330
360
26
27
27
58
86
117
147
178
208
239
270
300
331
361
27
28
28
59
87
118
148
179
209
240
271
301
332
362
28
29
29
88
119
149
180
210
241
272
302
333
363
29
30
30
89
120
150
181
211
242
273
303
334
364
30
31
31
90
151
212
243
304
365
31
Note: For leap years, February 29 becomes day number 60. Therefore, you will need to add one day
to any date that includes February 29 in a leap year.
Example 8.2(h)
Calculating the Number of Days Within a Single Calendar Year
Calculate the number of days in the time period from February 18, 2014 to May 17, 2014.
Solution
Therefore, the number of days in the time period = 88 days (t =
88
years)
365
Chapter 8 | Simple Interest and Applications
Example 8.2(i)
265
Calculating the Number of Days in a Time Period During a Non-Leap Year When the Time
Period is Over Two Calendar Years
Calculate the number of days in the time period from October 18, 2013 to April 05, 2014.
Solution
169
Therefore, the number of days in the time period = 74 + 95 = 169 days (t = 365 years)
Example 8.2(j)
Calculating the Number of Days in a Time Period During a Leap Year
Calculate the number of days in the time period from January 15, 2012 to March 20, 2013.
Solution
Therefore, the number of days in the time period = 351 + 79 = 430 days (t =
430
years)
365
Method 3: Determining the Number of Days Using a Financial Calculator
You can use a financial calculator, such as the Texas Instruments BA II Plus, to determine the
number of days in a time period as illustrated below:
The following example will illustrate the use of the calculator to determine the number of days.
Pressing the 2ND button then DATE (press 1 to access DATE) will take you to the date
worksheet to determine the number of days between two dates.
2
Date 1 is the earlier date or the first date you will enter. Dates are shown in mm-dd-yyyy
format and entered as mm-dd-yy (you will understand this better while working through an
example).
3
This is to toggle between different items in the date worksheet.
4
Date 2 is the later date or the second date you will enter, If you enter the dates in a reverse
order you will get a negative value for the number of days.
1
5
6
Days between dates will give you the number of days between the first date entered and the
second date entered. It includes the first day and excludes the last day in the calculation.
It shows you that the setting for your date worksheet takes the actual number of days into
consideration: i.e., it includes adjustments for leap years. If it is set to ‘360’, change it by
pressing 2ND then SET to get ACT.
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Chapter 8 | Simple Interest and Applications
Example 8.2(k)
Calculating the Number of Days in a Time Period Using a Financial Calculator
Calculate the number of days in the time period from October 18, 2013 to April 05, 2014.
Solution
1
Enter Oct 18, 2013 as 10.1813 then press ENTER
2
Use the down arrow to go to DT2 and enter April 05, 2014 as 4.0514
then press ENTER.
3
Use the down arrow to go to DBD and press CPT to compute the number of
days.
Therefore, the number of days in the time period is 169.
4
8.2 |
Exercises Answers to the odd-numbered problems are available at the end of the textbook
For the following problems, express the answers rounded to two decimal places, wherever applicable.
1. Calculate the number of days in the following time periods and express it in terms of years:
a. January 01, 2015 to February 19, 2015
b. February 26, 2014 to December 02, 2014
c. November 23, 2014 to April 04, 2015
d. August 25, 2015 to September 06, 2016
2. Calculate the number of days in the following time periods and express it in terms of years:
a. August 18, 2015 to September 19, 2015
b. January 23, 2014 to October 06, 2014
c. July 18, 2014 to July 18, 2015
d. October 19, 2015 to November 06, 2016
3. Calculate the amount of interest earned on the following investments:
Principal
Rate
Time period
a.
$200
10% p.a.
200 days
b.
$2200
4% p.a.
April 15, 2014 to November 01, 2014
c.
$5605
1.2% p.m.
January 24, 2014 to February 15, 2015
d.
$150
0.8% p.m.
1 year and 9 months
Chapter 8 | Simple Interest and Applications
4. Calculate the amount of interest earned on the following investments:
Principal
Rate
Time period
a.
$20,000
5% p.a.
250 days
b.
$200
12% p.a.
November 06, 2014 to February 28, 2015
c.
$800
1.10% p.m.
March 15, 2014 to August 25, 2015
d.
$500
0.75% p.m.
2 years and 3 months
5. What was the amount of interest charged on a loan of $12,500 received on January 18, 2014 and settled on April
24, 2014 if the interest rate on the loan was 3.25% p.a.?
6. Saira invested $3450 into a savings account that was earning 2.5% p.a. If she invested this amount on May 25,
2014, how much interest did she make by November 23, 2014?
7. Salma earned 2.2% p.a. on a short-term investment of $5000. Calculate the interest amount earned on this
investment if she deposited the money on November 18, 2014 and withdrew it on February 23, 2015.
8. On December 25, 2014, Amy borrowed $4750 from Cynthia at 5.5% p.a. If she repaid the amount on April 27, 2015,
calculate the amount of interest charged on the loan.
9. Owen lent Hua $5000 at 5% p.a. simple interest for 1 year and 3 months. Calculate the amount of interest charged
at the end of the term.
10. Madison won a prize amount of $1500 and she deposited the entire amount into a savings account at a local bank.
How much interest would she have earned after 18 months if interest on the savings account was 3% p.a.?
11. Nathan obtained a loan of $750 to purchase a bike. How much interest would Nathan have to pay at the end of 11
months if the loan was at an interest rate of 5.5% p.a.?
12. Ernie wanted to purchase a motorbike but was short $4350. He approached his friend who agreed to give him the
amount as long as he paid back the money in 4 months at an interest rate of 6% p.a. How much interest would Ernie
have to pay his friend at the end of 4 months?
13. Brenda withdrew $8000 from her business and deposited this amount into a savings account on January 02, 2015.
If the account provided her with an interest rate of 4% p.a., calculate the amount of interest she would earn by
March 15, 2016.
14. Kumar's website development business was doing well and he wanted to lend the surplus earnings he made to his
friend who was in need of money. He lent $5000 at 6% p.a. to his friend on February 7, 2015. Calculate how much
interest his friend had to pay Kumar on July 11, 2016.
15. Benjamin borrowed $2500 from a money lender. If the money lender charges interest at 12.5% p.a., calculate the
amount of interest Benjamin had to pay the lender at the end of 20 days.
16. Zack wanted to propose to his girlfriend and decided to give her a diamond ring. The ring he liked was selling for
$3200 but he only had $1500 with him. His friend, Ethan, agreed to loan him the remaining amount at an interest rate
of 12% p.a. If Zack returned the money in 40 days, calculate the amount of interest he had to pay Ethan for the loan.
17. Aidan borrowed $5000 from a lender at 1.5% p.m. If he repaid the amount in 45 days, calculate the amount of
interest he had to pay.
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Chapter 8 | Simple Interest and Applications
18. Maya had a savings account with a bank in Canada that was providing her with a very low interest rate of 0.05%
p.m. If she deposited $10,000 in her savings account, calculate how much interest she earned after 60 days.
19. Which rate is more attractive for a lender of a loan: 9.8% p.a. or 0.85% p.m.? Explain your answer.
20. Lian's friend offered to lend her $1000 at an interest rate of 6.5% p.a. However, her sister offered her the same
amount at 0.52% p.m. Who was charging her lesser interest for the loan? Explain your answer.
21. Angelo took a loan of $2500 from a bank at 4.6% p.a. for 18 months and gave it to his friend Alisa as a loan at a
rate of 0.5% p.m. At the end of 18 months, how much interest should Alisa have paid Angelo? Of this amount, how
much interest should Angelo have paid the bank?
22. Aba invested $5200 for 210 days at 0.4% p.m. Tabitha invested the same amount for the same period but at 4.3%
p.a. Who earned more and by how much?
23. Jada invested a total of $7500. Out of this, $3500 was invested at 4.6% p.a. and the remainder at 0.4% p.m.
Calculate the total interest earned from both investments after 8 months.
24. Babiya borrowed a total of $3250 to pay for college fees. $2000 of this was from a bank at 6.25% p.a. and the
remainder from a credit union at 0.9% p.m. Calculate the total interest she should have paid after 9 months for
both the loans.
25. Yara took a loan of $2600 for 180 days at 11% p.a. How much more or less money would be required to pay off the
loan if the interest rate is 1.2% p.m. instead of 11% p.a.?
26. Afsoon invested $9500 for 320 days at 8% p.a. How much more or less interest would he have earned on the
investment if his money was growing at an interest rate of 0.75% p.m. instead of 8% p.a.?
8.3 |
27.
Calculating
Principal, Interest Rate, Time,
and Maturity Value
In simple interest calculations, depending on the information provided, the simple interest Formula 8.2
can be rearranged to solve for the unknown variable, as shown below:
Formula 8.2,
I = Prt
Therefore,
Prt = I
Dividing both sides by,
(i) rt, we have,
(ii) Pt, we have,
(iii) Pr, we have,
= I
PP =
rt
rr == I
Pt
tt == I
Pr