Mathematics+of+Money.ppt_simple and compound interest

It is a discussion about simple and compound interest, it has specific examples and clear solutions on how was the answers obtained.

1. loan
Simple Interest
Compound Interest
Installment Buying
Loans in Daily Life

2. What is an Interest and how does it
work?
Interest is the “rent” that a borrower pays a lender
to use the lender’s money.

3. Simple Interest
Interest is the amount of money charged for
borrowing or using money. When you deposit
money into a savings account, you are paid
interest. Simple interest is one type of fee paid
for the use of money.
I = P

r
 t
Simple interest
is money paid
only on the
principal.
Principal is the amount of
money borrowed or invested.
Rate of interest is the
percent charged or
earned.
Time in years that
the money is
borrowed or invested

4. Simple Interest
To buy a car, Jessica borrowed $15,000 for 3
years at an annual simple interest rate of 9%.
How much interest will she pay if she pays the
entire loan off at the end of the third year?
What is the total amount that she will repay?
Additional Example 1: Finding Interest and Total
Payment on a Loan
First, find the interest she will pay.
I = P

r
 t Use the formula.
I = 15,000

0.09

3 Substitute. Use 0.09 for 9%.
I = 4050 Solve for I.

5. Simple Interest
Additional Example 1 Continued
Jessica will pay $4050 in interest.
P + I = A principal + interest = amount
15,000 + 4050 = A Substitute.
19,050 = A Solve for A.
You can find the total amount A to be repaid on a
loan by adding the principal P to the interest I.
Jessica will repay a total of $19,050 on her loan.

6. Simple Interest
To buy a laptop computer, Elaine borrowed
$2,000 for 3 years at an annual simple interest
rate of 5%. How much interest will she pay if
she pays the entire loan off at the end of the
third year? What is the total amount that she
will repay?
Check It Out: Example 1
First, find the interest she will pay.
I = P

r
 t Use the formula.
I = 2,000

0.05

3 Substitute. Use 0.05 for 5%.
I = 300 Solve for I.

7. Simple Interest
Check It Out: Example 1 Continued
Elaine will pay $300 in interest.
P + I = A principal + interest = amount
2000 + 300 = A Substitute.
2300 = A Solve for A.
You can find the total amount A to be repaid on a
loan by adding the principal P to the interest I.
Elaine will repay a total of $2300 on her loan.

8. Simple Interest
Additional Example 2: Determining the Amount of
Investment Time
I = P

r

t Use the formula.
450 = 6,000
 0.03

t Substitute values into
the equation.
2.5 = t Solve for t.
Nancy invested $6000 in a bond at a yearly
rate of 3%. She earned $450 in interest. How
long was the money invested?
450 = 180t
The money was invested for 2.5 years, or 2 years
and 6 months.

9. Simple Interest
Check It Out: Example 2
I = P

r

t Use the formula.
200 = 4,000
 0.02

t Substitute values into
the equation.
2.5 = t Solve for t.
TJ invested $4000 in a bond at a yearly rate of
2%. He earned $200 in interest. How long was
the money invested?
200 = 80t
The money was invested for 2.5 years, or 2
years and 6 months.

10. Simple Interest
Mr. Johnson borrowed $8000 for 4 years to
make home improvements. If he repaid a total
of $10,320, at what interest rate did he
borrow the money?
Additional Example 3: Finding the Rate of Interest
P + I = A Use the formula.
8000 + I = 10,320 Substitute.
I = 10,320 – 8000 = 2320 Subtract 8000
from both sides.
He paid $2320 in interest. Use the amount of
interest to find the interest rate.

11. Simple Interest
Additional Example 3 Continued
2320 = 32,000

r Simplify.
I = P

r

t Use the formula.
2320 = 8000

r

4 Substitute.
2320
32,000
= r Divide both sides by
32,000.
0.0725 = r
Mr. Johnson borrowed the money at an annual rate
of 7.25%, or 7 %.
1
4

12. Simple Interest
Mr. Mogi borrowed $9000 for 10 years to
make home improvements. If he repaid a total
of $20,000 at what interest rate did he borrow
the money?
Check It Out: Example 4
P + I = A Use the formula.
9000 + I = 20,000 Substitute.
I = 20,000 – 9000 = 11,000 Subtract 9000
from both sides.
He paid $11,000 in interest. Use the amount of
interest to find the interest rate.

13. Simple Interest

14. Simple Interest

15. Simple Interest

16. Remember…
Simple interest is a type of interest that is
paid only on the original amount deposit
and not on past interest paid.

17. Want to Be a Millionaire?
You Can!
If you leave your money to grow for a long time,
€ 100 can turn into a million euros. No, seriously.
How?
Through compounding.

18. Does anyone have any interest in
interest?
Very few banks today pay interest
based on the simple interest
formula. Instead, they pay interest
by using a principle called
compounding.
The difference between simple and
compound interest is this: Simple
interest grows slowly, compounding
speeds up the process.

19. How it works.
Simple interest is interest on the
principle amount.
Compound interest is when your
principle and any earned interest
both earn interest.

20. Consider this example: You begin with
$100 invested at 10% annual interest.
After Simple Interest Compound
Interest
1 year 110 110
2 years 120 121
3 years 130 133
4 years 140 146
5 years 150 161
10 years 200 259
20 years 300 672
50 years 600 11,739

21. Compound Interest Wins!!
From this example, it is easy to see
that if you are saving money, you
would prefer compound interest.

22. Time is on Your Side
The longer you save, the greater the
effect of compound interest.
But also…
The longer you borrow, the quicker
your debts grow.

23. Compounding Periods
When calculating compound interest, the
number of calculating periods makes a
significant difference.
The basic rule is that the higher the
number of calculating periods, the greater
the amount of compound interest.

24. A bank may pay interest as follows:
 Semi-annually: twice a year or every 180 days
Quarterly: 4 times a year or every 90 days
Monthly: 12 times a year or every 30 days
Daily: 360 times a year

25. Calculate compound interest using this
formula:
A—Total amount
p —principle
r —interest rate
n —number of compounding periods
t —time in years
nt
n
r
pA 





1

26. Example: $100 is invested at 10%
interest compounded yearly for 6 years
177.16

27. $250 invested at 6.5% for 8 years
compounded monthly.
419.92

28. Example……
$500 invested at 12% for 10 years
compounded yearly.

29. Answer……
Problem:
$500 invested at
12% for 10 years
compounded
yearly.
Answer:

93.1552
12.1500
1
12.
1500
1
10
101

















A
A
A
n
r
PA
nt

30. Example……
$1000 at 7.25% for 9 years
compounded monthly.

31. Answer……
Problem:
$1000 at 7.25%
for 9 years
compounded
monthly.
Answer:
57.1916
12
0725.
11000
1
)912(
















A
A
n
r
PA
nt

32. Try these:
1.$750 at 6.5% for 5 years compounded
annually
2.$25,000 at 8% for 3 years compounded
annually
3.$680 at 5.5% for 1.5 years compounded
monthly
4.$1500 at 4.5% for 2 years compounded
monthly