The document discusses compound interest, providing examples and formulas. It defines compound interest as interest earned periodically added to the principal to earn further interest. Formulas are provided for compound interest calculated annually, semiannually, quarterly, monthly, and daily. An example calculates the accumulated amount from $1000 invested at 8% compounded annually, semiannually, quarterly, monthly, and daily over 3 years. A quiz question asks the reader to calculate the value of a $3700 investment earning 2.5% interest compounded quarterly over 10 years.

1. COMPOUND INTEREST
CONTEMPORARY MATHEMATICS FOR
COLLEGE
FINISHSTART

2. COMPOUND INTEREST
ACTIVITY
ABSTRACTION
APPLICATION
ANALYSIS

3. ACTIVITY
4 PICS
1 WORD

4. Y M E U T
R A O L F
F O R M U L A
Y E T

5. Y I E X S
M A N R O
I X
M N O
Y E A R S

6. A L Y N L
R N E U A
L Y
R E
A N N U A L

7. R E S N P
G T A I C
R A T E
S N P
G I C

8. M L E I S
Y E D K A
D A I L Y
M E S
E K

9. S V A M E
O L I C D
S V
O
D E C I M A L

10. T O U S I
M P W R E
O U S
P W R
T I M E

11. M Y E N I
O T L A H
Y E I
L A
M O N T H

12. E N D O F A C T I V I T Y
4 PICS
1 WORD
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13. ANALYSIS
HOW LONG DOES IT TAKE TO DOUBLE YOUR MONEY?
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14. ABSTRACTION

15. COMPOUND INTEREST
 FREQUENTLY, INTEREST EARNED IS PERIODICALLY ADDED TO THE
PRINCIPAL AND THEREAFTER EARNS INTEREST ITSELF IN THE SAME RATE.
 COMPOUND INTEREST USES THE SAME INFORMATION AS SIMPLE
INTEREST, BET WAS IS NEW IS THE FREQUENCY OF COMPOUNDING N.
 N=1 ANNUAL, N=2 SEMI-ANNUAL, N=4 QUARTERLY, N=12 MONTHLY,
N=52 WEEKLY, N=365 DAILY
http://www.shsu.edu/ldg005/data/mth199/chapter4.ppt / http://www.asu.edu/courses/mat142ej/finance/powerpoints/compound_interest_dalesandro.ppt

16. COMPOUND INTEREST
 SUPPOSE ₱1000 IS DEPOSITED IN A BANK FOR A TERM
OF 3 YEARS, EARNING INTEREST AT THE RATE OF 8%
PER YEAR COMPOUNDED ANNUALLY.
http://www.shsu.edu/ldg005/data/mth199/cha
𝐴1 = 𝑃(1 + 𝑟𝑡)
=1000[1+0.08(1)]
=1000(1.08)
=1080
𝐴2 = 𝑃(1 + 𝑟𝑡) = 𝐴1(1 + 𝑟𝑡)
= 1000 1 + 0.08 1 1 + 0.08 1
= 1000(1 + 0.08)2 = 1000(1.08)2
≈ 166.40 𝑜𝑟 ₱166.40.

17. http://www.shsu.edu/ldg005/data/mth199/cha
• NOTE THAT THE ACCUMULATED AMOUNTS AT
THE END OF EACH YEAR HAVE THE
FOLLOWING FORM.
𝐴3 = 𝑃 1 + 𝑟𝑡 = 𝐴2 1 + 𝑟𝑡
= 1000[1 + 0.08(1)]2 1 + 0.08 1
= 1000(1 + 0.08)3
= 1000(1.08)3
≈ 1259.71 𝑜𝑟 ₱1259.71.
𝐴1 = 1000(1.08) 𝐴1 = 𝑃 1 + 𝑟
𝐴2 = 1000(1.08)2 or 𝐴2 = 𝑃(1 + 𝑟)2
𝐴3 = 1000(1.08)3
𝐴3 = 𝑃(1 + 𝑟)3

18. THE FORMULA (SIMPLE INTEREST)
• This formula was derived under the assumption that interest
was compounded annually
http://www.shsu.edu/ldg005/data/mth199/ch
𝐴 = 𝑃(1 + 𝑟) 𝑡

19. COMPOUND INTEREST
• TO FIND A GENERAL FORMULA FOR THE ACCUMULATED AMMOUNT, WE
APPLY
REPEATEDLY WITH THE INTEREST RATE i = r/m.
• WE SEE THAT THE ACCUMULATED AMMOUNT AT THE END OF EACH
PERIOD IS AS FOLLOWS:
FIRST PERIOD: 𝐴1 = 𝑃 1 + 𝑖
SECOND PERIOD: 𝐴2 = 𝐴1 1 + 𝑖 = 𝑃 1 + 𝑖 1 + 𝑖 = 𝑃 1 + 𝑖 2
THIRD PERIOD: 𝐴3 = 𝐴2 1 + 𝑖 = [𝑃(1 + 𝑖)2](1+i) = P(1=i)3
-
-
-
nth PERIOD: 𝐴 𝑛 = 𝐴 𝑛−1 1 + 𝑖 = [𝑃 1 + 𝑖) 𝑛−1 1 + 𝑖 = 𝑃(1 + 𝑖) 𝑛
http://www.shsu.edu/ldg005/data/mth199/cha
𝐴 = 𝑃(1 + 𝑟) 𝑡

20. amount at
the end
Principal
(amount
at start)
annual interest
rate
(as a decimal)
mt
m
r
PA 





 1 time
(in years)
number of times per
year that interest in
compounded
COMPOUND INTEREST FORMULA
http://www.mathxtc.com/Downloads/NumberAlg/files/Simple%20Compound%20

21. EXAMPLE
• FIND THE ACCUMULATED AMOUNT AFTER 3 YEARS IF ₱1000 IS
INVESTED AT 8% PER YEAR COMPOUNDED
A. ANNUALLY
B. SEMIANNUALLY
C. QUARTERLY
D. MONTHLY
E. DAILY
http://www.shsu.edu/ldg005/data/mth199/cha

22. EXAMPLE
• FIND THE ACCUMULATED AMOUNT AFTER 3 YEARS IF ₱1000 IS
INVESTED AT 8% PER YEAR COMPOUNDED
A. ANNUALLY
B. SEMIANNUALLY
C. QUARTERLY
D. MONTHLY
E. DAILY
http://www.shsu.edu/ldg005/data/mth199/cha
HOME

23. SOLUTION
B. Semiannually,
Here, P = 1000, r = 0.08, and m = 2.
Thus, 𝑖 =
0.08
2
and n = (3)(2) = 6, so
http://www.shsu.edu/ldg005/data/mth199/cha
𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
= 1000 1 +
0.08
2
6
= 1000(1.04)6
≈ 1265.32
or ₱1265.32
BACK

24. SOLUTION
A. Annually,
Here, P=1000, r=0.08, and m=1.
Thus, i = r = 0.08 and n=3, SO
http://www.shsu.edu/ldg005/data/mth199/cha
𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
= 1000 1 +
0.08
1
3
= 1000(1.08)3
≈ 1259.71
𝑜𝑟 ₱1257.71.
BACK

25. SOLUTION
C. Quarterly,
Here, P = 1000, r = 0.08, and m = 4.
Thus, 𝑖 =
0.08
4
and n = (3)(4) = 12, so
𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
= 1000 1 +
0.08
4
12
= 1000(1.02)12
≈ 1268.24
𝑜𝑟 ₱1268.24
http://www.shsu.edu/ldg005/data/mth199/cha
BACK

26. SOLUTION
E. Daily,
Here, P = 1000, r = 0.08, and m = 365.
Thus, 𝑖 =
0.08
365
and n = (3)(365) = 1095, so
𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
= 1000 1 +
0.08
365
1095
= 1000(1.0002192)1095
≈ 1271.22
𝑜𝑟 ₱1271.22
http://www.shsu.edu/ldg005/data/mth199/cha
NEXT

27. SOLUTION
D. Monthly,
Here, P = 1000, r = 0.08, and m = 12.
Thus, 𝑖 =
0.08
12
and n = (3)(12) = 36, so
𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
= 1000 1 +
0.08
12
36
= 1000(1.0066667)36
≈ 1270.24
𝑜𝑟 ₱1270.24.
http://www.shsu.edu/ldg005/data/mth199/cha
BACK

28. QUIZ BEE

29. QUIZ BEE
A. Yvette invested ₱3700 in a savings account that
pays 2.5% interest compounded quarterly. Find
the value of the investments in 10 years.
 Use the compound interest formula.
 Substitute.
 Simplify.
 Add the parentheses.
 Find (1.00625)40 and round.
 Multiply and round to the nearest cent.
 After 10 years, what will be the value of the
investment?

30. 𝐴 = 𝑃 1 +
𝑟
𝑚
𝑛
ANSWER
BACK

31. ANSWER
= 3700(1 +
0.025
4
)4(10)
BACK

32. ANSWER
= 3700(1 + 0.00625)40
BACK

33. ANSWER
= 3700(1.00625)40
BACK

34. ANSWER
= 3700(1.28303)
BACK

35. ANSWER
= 4,747.20
BACK

36. ANSWER
= ₱4,747.20
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38. COMPOUND INTEREST GROUP
GE-2 RONALD BURGOS
WYETH CALANTOC
GE-6 ROCHELLE KAYE VILLENA
CARLA YANGO
GE-7 MERWIN ANDREI SIMON
NICOLE JANE RINGOR