Chapter 11 Consumer Math
Page 1 of 9
Chapter 11
Loans
Spring 2016 1/9/16
- Simple interest loans
- Loan ammortization schedule
(home mortgage)
- Compound interest loans
Chapter 11 Consumer Math
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In this chapter, ...
Don’t
round numbers inside the problem
....round conventionally at answer only…learn to use your calculator’s
memory
Simple Interest Loan
- A lender is willing to loan money to a borrower. We usually think of a bank as the lender, but if
you have a savings account at a bank, YOU are the lender. The bank is borrowing your money.
- The principal, (p), is the amount of money lent….the amount of money that changes hands AT
THE START of the loan. It’s also called present value.
- Interest (i) is the money a borrower pays for the use of the lender’s money. One type of interest
is called simple interest. Interest rate is the percentage of a sum of money charged for its
use. Interest rate is expressed as a percent. Percent must be converted to a decimal for calculator
operations.
Review High School Math:
5.25% = .0525
12.3333% = .123333
.15 = 15%
.074 = 7.4%
- A loan takes place over a period of time. Time (t) is the number of days, months, or years for
which the money will be lent. Time is expressed as the same period as the interest rate…… t
and r must be in same time frame. For example, if the interest rate is 2% per month, the time
must be expressed in months, if interest rate is 2% annual, time must be in years.
The formula used to find simple interest is
i = prt
- The amount to be repaid (A)…the amount of money that changes hands AT THE END (also
called future value) of the loan is equal to the principal plus the interest:
A=p+i
A = p + prt
A = p(1 + rt)
same as
Chapter 11 Consumer Math
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Sherry needs to borrow $6200 to replace her air conditioner. She obtains a 30-month loan with
an annual simple interest rate of 5.75%.
Determine the interest on the loan
p = $6200
r = 5.75% = 0.0575
t = 30/12 = 2.5 years (to convert mos. to years, divide months by 12)
i = prt
= 6200(0.0575)(2.5)
= $891.25
Determine how much must be paid at end of loan in 30 months
p = $6200
i = $891.25
A=p+i
A = 6200 + 891.25 = $7091.25
Another way to do it with same ranswer:
A = p(1 + rt)
= 6200(1 + .0575(2.5))
= 6200(1 + .14375)
= 6200(1.14375)
= $7,091.25
Sharlene needs $11,000 in 42 months from now to pay off a student loan. How much should she
put into a bank account today so that she will have $11,000 in 42 months. The annual interest
rate is 4.72%
A = $11,000
p=?
r = 4.72% = .0472
t = 42 months = 42/12 = 3.5 years
A = p(1 + rt)
11,000 = p(1 + .0472(3.5))
11,000 = p(1 + 0.1652)
11,000 = p(1.1652)
round to cents
11000 = p = $9,440.43941 = $9,440.44
1.1652
Chapter 11 Consumer Math
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Loan Ammortization Schedule
Purchase home, borrow $211,650.00 mortgage
Payment = $1408.11
monthly for 30 years
(t
=
1/12
year)
(30 years = 360 months)
Payment
r
=
7.0%
(
=
.07)
Coupon
Book
Bank gives you a book, tear out a page each month,
and send your payment
Loan ammortization schedule is the payment book
Borrowed P = $211,650.00
Payment
number
Interest
for this
month
Principal
Paydown
Balance you
owe
$211,650.00
1 = $1,408.11
$1,234.63
$173.49
$211,476.51
2 =$1408.11
3 =$1408.11
4 =$1408.11
$1,233.61
$1232.60
$174.50
$175.52
$211,302.01
$211,126.49
358 = $1408.11
359 = $1408.11
360 = $1408.11
$0.00
Interest for Payment 1 = Prt = $211,650.00 (.07)(1/12) = $1234.63
Principal Paydown = $1408.11 - $1234.63 = $173.49
Balance you owe = $211,650.00 - $173.49 = $211,476.51
Interest for Payment 2 = Prt = $211,476.51 (.07)(1/12) = $1233.61
Principal Paydown = $1408.11 - $1233.61 = $174.50
Balance you owe = $211,476.51 - $174.50 = $211,302.01
Interest for Payment 3 = Prt = $211,302.01 (.07)(1/12) = $1232.60
Principal Paydown = $1408.11 - $1232.60 = $175.52
Balance you owe = $211,302.01 - $175.52 = $211,126.49
In real life,
do these calculations
360 times for 360
months of payments
Chapter 11 Consumer Math
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SOLUTIONS
Practice problems page 28
Review High School Math:
9.) Change 11/20 to %: 11/20 = .55
= 55% move decimal 2 places to right
11.) Change .007654 to %: .007654
= .7654% move decimal 2 places to right
17.) Change 5.15% to decimal: 5.15% = .0515 move decimal 2 places to left
Simple interest:
81.) Given: p = $465, r = 2.75% = .0275, t = 1.25 years
calculate interest = i = prt
= 465(.0275)(1.25)
= 15.984375 = $15.98 rounded to cents
95.) Given: p = $1650.00, t = 6.5 years, i = $343.20
calculate interest rate r: i = prt
$343.20 = $1650.00(r)(6.5)
r = 343.20 = 0.032 = 3.2%
1650(6.5)
98.) Karen borrowed $3500 for 6 months (must be years). The bank charged 7½% (=.075)
annual interest.
a.) Calculate the interest paid:
i = prt = 3500(.075)(6/12) (t must be years)
= 3500(.075)(.5)
i = $131.25
b.) Determine the amount she repaid:
A = p + i = $3500 + $131.25
A = $3631.25
….another way with same result: A = p(1 + rt)
= 3500(1 + .075(6/12))
= 3500(1 + .0375)
= 3500(1.0375)
= $3631.25
Chapter 11 Consumer Math
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Practice problem (not in text, text is beyond scope of Math 102)
Given: monthly payment = $804.62
mortgage loan : $100,000
interest rate = 9% (.09)
Payment
number
Interest
for this
month
Principal
Paydown
Construct 1st 3 lines
of
loan amortization schedule
Balance you
owe
$100,000.00
1 = $804.62
$750.00
$54.62
$99,945.38
2 = $804.62
3 = $804.62
$749.59
$749.18
$55.03
$55.44
$99,890.35
$99,834.91
Interest for Payment 1 = Prt = $100,000.00 (.09)(1/12) = $750.00
Principal Paydown = $804.62 - $750.00 =
$54.62
Balance you owe = $100,000.00 - $54.62 = $99,945.38
Interest for Payment 2 = Prt = $99,945.38 (.09)(1/12) = $749.59
Principal Paydown = $804.62 - $749.59 = $55.03
Balance you owe = $99,945.38 - $55.03 = $99,890.35
Interest for Payment 3 = Prt = $99,890.35 (.09)(1/12) = $749.18
Principal Paydown = $804.62 - $749.18 = $55.44
Balance you owe = $99,890.35 - $55.44 = $99,834.91
Chapter 11 Consumer Math
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Compound Interest Loan (hint - look for the word “compound” written in the
problem)
Future Value - money that changes hands at end of loan = A
A = p(1 + r/n) nt
where A is the amount that accumulates in the account (future value), p is the principal (present
value), r is the annual interest rate as a decimal number, n is the number of compounding periods
per year, and t is the time in years.
typical values for n: monthly compounding
daily compounding
weekly compounding
quarterly compounding
semi-annual compounding
n = 12
n = 360
n = 52
n=4
n=2
Kathy Mowers invested $3000 in a savings account with an interest rate of 1.8%
compounded monthly. If Kathy makes no other deposits into this account, determine the amount in the savings account after 2 years.
A = p(1 + r/n)nt
p = $3000.
r = 1.8% = 0.018
interest is compounded monthly, n = 12.
t =2
A = p(1 + r/n) nt
A = 3000(1 + .018/12) 12(2)
A = 3000(1 + .0015) 24
don’t round here
24
A = 3000(1.0015)
A = 3000(1.036627885) = 3109.883655 round to $3,109.88
--------------------------Calculate the interest on $650 at 8% compounded semiannually (n = 2) for 3 years
p = $650
n =2
r = 8% = 0.08
t = 3 years
A = p(1 + r/n) nt
A = 650(1 + .08/2) 2(3)
A = 650(1 + .04) 6
A = 650(1.04) 6
A = 650(1.265319018) = 822.457362 = $822.46
A=p+i
$822.46 = $650 + i
$172.46 = i
Chapter 11 Consumer Math
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Present Value – (Money that changes hands at start of loan) What amount of money must you
deposit in an account today to have a certain amount of money in the future?...deposit today at a
given rate of interest so that it will accumulate to $30,000 for college costs in 6 years in an
account that has an annual rate of 2.5% (0.025) compounded quarterly (n = 4)
p
=
A
(1 + r/n) nt
same formula A = p(1 + r/n) nt rearranged
p = invested now is called the present value
A = amount needed later = $30,000 future value
r = 2.5% = .025 annual interest rate as a decimal number
n=4
t = 6 = time in years.
p =
A
=
300000
=
30000
=
(1 + r/n) nt
(1 + .025/4) 4(6)
(1.00625) 4(6)
p
=
30000
1.1612922018
=
don’t round here
30000
(1.00625) 24
25833.29562 = $25,833.30 round to cents
Chapter 11 Consumer Math
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Practice problems page 369
143.) Calculate accumulated amount A (future value) and interest earned i for a compound
interest loan $7,000 for 3 years at 5.5% compounded monthly. Round to nearest cent.
p = $7,000
t = 3 years
r = 5.5% = .055
n = 12
A = p(1 + r/n) nt = 7000(1 + .055/12) 12(3)
= 7000(1.004583333) 36
don’t round here
= 7000(1.178948602)
= 8252.640216
= $8,252.64 this is where you round to nearest cent
Interest earned = A - p = $8,252.64 - $7,000 = $1252.64
149.) The desired accumulated amount is $100,000 after 4 years invested in an account with 4%
interest compounded quarterly. Calculate p (present value).
A = $100,000
t=4
r = 4% = .04
n=4
p
=
=
note this
A
= 100000
= 100000
= 100000
(1 + r/n) nt
(1 + .04/4) 4(4) (1 + .01) 16
1.172578645
85282.12622
rounded to
don’t round
$85,282.13
162.) Jean opened a Kids Club savings account for her daughter at the Royal Credit Union. She
deposited $500, and the account pays 0.25% interest compounded daily. If Jean made no
further deposits or withdrawals to the account, what is the amount in the account 18 years
later (future value)?
p = $500.00
r = 0.25% = .0025
t = 18
n = 360
360 seems strange but it’s convention
A = p(1 + r/n) nt = 500(1 + .0025/360) 360(18)
500(1 + (
)) 360(18)
= 500(1.00000694444444444) 6480
= 500(1.046027666)
= $523.0138332
= $523.01 round to penny
.0025/360 = 6.94444444444(10)-6
= .00000694444444444
don’t round here
let your calculator store numbers