1. Contemporary MathContemporary Math
11.2 Personal Loans and Simple11.2 Personal Loans and Simple
InterestInterest

2. Simple Interest FormulaSimple Interest Formula
 I = prtI = prt
 I(nterest) is the money the borrower paysI(nterest) is the money the borrower pays
for the use of the lender’s money.for the use of the lender’s money.
 P(rincipal) is the amount of moneyP(rincipal) is the amount of money
borrowed.borrowed.
 R(ate) is the percentage rateR(ate) is the percentage rate
 T(ime) is the amount of time the money isT(ime) is the amount of time the money is
borrowed.borrowed.

3. Simple InterestSimple Interest
 Most common is Ordinary InterestMost common is Ordinary Interest
 30 days in a month30 days in a month
 12 months in year12 months in year
 360 days in a year360 days in a year
 Rate and time must be expressed in theRate and time must be expressed in the
same units.same units.

4. Find the simple interest – assume rate is annual.Find the simple interest – assume rate is annual.
1) p=$420 r=9% t=1 year1) p=$420 r=9% t=1 year
I=(420)(.09)(1) =I=(420)(.09)(1) = $37.80$37.80
2) p=$365.45 r=11.5% t=8 months2) p=$365.45 r=11.5% t=8 months
I=(365.45)(.115)(8/12) =I=(365.45)(.115)(8/12) = $28.02$28.02
3) p=$4372.80 r=5.25% t=60 days3) p=$4372.80 r=5.25% t=60 days
I=(4372.80)(.0525)(60/360) =I=(4372.80)(.0525)(60/360) = $38.26$38.26
4) p=$12,752 r=0.055% per day t=90 days4) p=$12,752 r=0.055% per day t=90 days
I=(12,752)(.00055)(90) =I=(12,752)(.00055)(90) = $631.22$631.22

5. Use the simple interest formula toUse the simple interest formula to
find the missing value.find the missing value.
5) p = $8000, r = 6%, t = ?, i = $64.005) p = $8000, r = 6%, t = ?, i = $64.00
64=(8000)(.06)(t)64=(8000)(.06)(t)
t=64/(8000 x .06)t=64/(8000 x .06)
t=.133 x 360 = 48 dayst=.133 x 360 = 48 days
6) p = $?, r = 10%, t = 4 months, i = $9.506) p = $?, r = 10%, t = 4 months, i = $9.50
9.5=(p)(.1)(4/12)9.5=(p)(.1)(4/12)
p=9.5/(.1 x 4/12)p=9.5/(.1 x 4/12)
p=$285.00p=$285.00

6. 7) Ward borrowed $1500 from his bank for 60 days7) Ward borrowed $1500 from his bank for 60 days
at a simple interest rate of 10 ½ %.at a simple interest rate of 10 ½ %.
 How much did he pay for the use of the money?How much did he pay for the use of the money?
 i=(1500)(.105)(60/360)i=(1500)(.105)(60/360)
 i=$26.25i=$26.25
 What is the amount he paid to the bank on the date ofWhat is the amount he paid to the bank on the date of
maturity?maturity?
 1500+26.25 =1500+26.25 = $1526.25$1526.25

7. Discount NotesDiscount Notes
 WATCH OUT!WATCH OUT! In the previous loans theIn the previous loans the
principal plus interest were paid on the date ofprincipal plus interest were paid on the date of
maturity. In amaturity. In a discount notediscount note the bank takesthe bank takes
the interest off the top of the loan on the daythe interest off the top of the loan on the day
they give it. This interest is called thethey give it. This interest is called the bankbank
discountdiscount. On the maturity date the original. On the maturity date the original
principal is paid back.principal is paid back.
 The actual rate of interest isThe actual rate of interest is HIGHERHIGHER thanthan
stated.stated.

8. 8) Carol borrowed $3650 from the bank for 88) Carol borrowed $3650 from the bank for 8
months. The bankmonths. The bank discounteddiscounted the loan at 12%.the loan at 12%.
 How much interest did Carol pay the bank forHow much interest did Carol pay the bank for
the use of its money?the use of its money?
i=(3650)(.12)(8/12) =i=(3650)(.12)(8/12) = $292$292
 How much did she receive from the bank?How much did she receive from the bank?
3650-292 = $33583650-292 = $3358
 What was the actual rate of interest she paid?What was the actual rate of interest she paid?
292 = (3358)(r)(8/12)292 = (3358)(r)(8/12)
r=292/(3358 x 8/12)r=292/(3358 x 8/12)
r=13.04%r=13.04%

9. Early and Partial PaymentsEarly and Partial Payments
 TheThe United State’s RuleUnited State’s Rule allows you to pay a partialallows you to pay a partial
payment before the due date of your loan. When youpayment before the due date of your loan. When you
make a partial payment the interest is calculated for themake a partial payment the interest is calculated for the
number of days since your last payment (or thenumber of days since your last payment (or the
origination date) and your payment first pays for theorigination date) and your payment first pays for the
interest and second pays off your principal. When youinterest and second pays off your principal. When you
make your next partial payment the interest is calculatedmake your next partial payment the interest is calculated
using this new principal amount.using this new principal amount.
 TheThe Banker’s RuleBanker’s Rule is used when applying the Unitedis used when applying the United
State’s Rule. One year is equal to 360 days but any partState’s Rule. One year is equal to 360 days but any part
of a year is the exact number of days.of a year is the exact number of days.
 Use Table 11.1 in your textbook.Use Table 11.1 in your textbook.

10. 9. Find the exact time from the first date to the9. Find the exact time from the first date to the
second date.second date.
 a) April 4 to October 11a) April 4 to October 11
 284-94 =284-94 = 190 days190 days
 b) May 19 to September 17b) May 19 to September 17
260-139 =260-139 = 121 days121 days

11. 10. Determine the due date using10. Determine the due date using
the exact time.the exact time.
 a) 90 days after April 18a) 90 days after April 18
 108+90 = 198108+90 = 198thth
day isday is July 17July 17..
 b) 120 days after June 8b) 120 days after June 8
 159+120 = 279159+120 = 279thth
day isday is October 6October 6..

12. 11. Use the U.S. Rule to determine the balance11. Use the U.S. Rule to determine the balance
due on the note at the date of maturity.due on the note at the date of maturity.
 P = $2500P = $2500
 R = 8%R = 8%
 Start Date = July 1Start Date = July 1
 Due Date = October 15Due Date = October 15
 Partial payment = $300Partial payment = $300
on August 1on August 1
 i=(2500)(.08)(31/360)i=(2500)(.08)(31/360)
 i=17.22i=17.22
 2500+17.22-300=2217.222500+17.22-300=2217.22
 i=(2217.22)(.08)(75/360)i=(2217.22)(.08)(75/360)
 i=36.95i=36.95
 2217.22+36.952217.22+36.95=$2254.17=$2254.17
Due on the date of maturity.Due on the date of maturity.

13. 12. Use the U.S. Rule to determine the balance12. Use the U.S. Rule to determine the balance
due on the note at the date of maturity.due on the note at the date of maturity.
 P = $2400P = $2400
 R = 7%R = 7%
 Start Date = April 10Start Date = April 10
 Due Date = July 7Due Date = July 7
 Partial payment = $500Partial payment = $500
on May 5on May 5
 i=(2400)(.07)(25/360)i=(2400)(.07)(25/360)
 i=11.67i=11.67
 2400+11.67-500=1911.672400+11.67-500=1911.67
 i=(1911.67)(.07)(63/360)i=(1911.67)(.07)(63/360)
 i=23.42i=23.42
 1911.67+23.421911.67+23.42=$1935.09=$1935.09
Due on the date of maturity.Due on the date of maturity.

14.  13. On March 1, the Zwick Balloon Company Signed a13. On March 1, the Zwick Balloon Company Signed a
$6500 note with the simple interest of 10 ½ % for 180$6500 note with the simple interest of 10 ½ % for 180
days. The company made payments of $1750 on May 1days. The company made payments of $1750 on May 1
and $2350 on July 1. How much will the company owe onand $2350 on July 1. How much will the company owe on
the date of maturity?the date of maturity?
 i=(6500)(.105)(61/360)=115.65i=(6500)(.105)(61/360)=115.65
 6500+115.65-1750=4865.656500+115.65-1750=4865.65
 i=(4865.65)(.105)(61/360)=86.57i=(4865.65)(.105)(61/360)=86.57
 4865.65+86.57-2350=2602.224865.65+86.57-2350=2602.22
 i=(2602.22)(.105)(58/360)=44.02i=(2602.22)(.105)(58/360)=44.02
 2602.22+44.022602.22+44.02=$2646.24 due on August 28=$2646.24 due on August 28thth
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