Simple interest: I = PRT, where I = interest, P = principal, R = interest rate, and T = number of periods
Once simple interest is calculated, the amount to which the investment has grown can be found using A = P + I
ex. $4000 is invested earning 5.6% p.a. simple interest; how much interest will be earned in 3.5 years? I = PRT → I =
4000 x 0.056 x 3.5 = $784. what amount will the investment grow to in 3.5 years? A = P + I → A = 4000 + 784 = $4784
Hire purchase is paying off an item over time, where higher interest is charged when the terms are not followed.
Pay deposit – interest is charged on balance, or deferred payment – if paid late interest is charged on whole amount.
ex. a fridge is advertised for $1200 cash or on terms at 10% deposit and $57 a month for 2 years; how much more will
it cost than if bought for cash? deposit + 24 monthly repayments = 10% x $1200 + $57 x 24 = $1488 → difference =
$1488 − $1200 = $288. what is the flat rate of interest charged per annum? P = 1200 – 120 = $1080 → R = I / PT →
R = x 2 = 13%
Compound interest: A = P(I + R)t compounded annually, or A = P(I + R / n)tn compounded n times a year
ex. $3000 is invested at 5% p.a. compounded yearly; calculate the investment at the end of 2 years. I = PRT → I =
3000 x 0.05 x 1 = $150 → A = P + I → 3000 + 150 = $3150 → I = PRT → I = 3150 x 0.05 x 1 = $157.50 → A = P + I →
A = 3150 + 157.50 = $3307.50. calculate the compound interest earned after 2 years. interest = final investment –
initial principal = 3307.50 – 3000 = $307.05. ex. $65000 is invested at 7.5% p.a. compounded yearly; write a
recurrence relation describing the value after one year. tn+1 = rtn, where t0 = a → tn+1 = (1 + 0.075)tn, where t0 = 65000
Effective interest rate: e = (1 + i / n)n – 1, where e = effective yearly interest rate, i = nominal yearly interest rate, and
n = number of compounding periods in a year, ex. what is the effective interest rate for an investment earning 7% p.a.
compounding monthly? e = (1 + i / n)n – 1 → e = (1 + 0.)12 – 1 = 7.23%
Reducing balance loans: An = An-1(1 + r), where A0 = initial amount, and r = compounding rate as a decimal.
ex. 40000 is borrowed on 15th March, 2014. The bank lends this amount with monthly interest at 11.2% p.a., with
repayments of $873.69 a month for 5 years; how much is owed until 15th May? 40000(1 + 0.) – 873.69 =
$39499.64 → 39499.64(1 + 0.) – 873.69 = $38994.62.
Amount still owing after the nth payment: An = (1 + r / 100)An-1 – monthly repayment. ex. $25000 was borrowed with
monthly repayments of $500 and interest at 9.5% p.a.; what is the recurrence relation? An = (1 + 9. x 100)An-1 –
500, where A0 = 25000.
Using the ClassPad
$140000 is borrowed at 7.5% p.a. compound interest adjusted monthly; find the monthly repayment if the loan is to be
repaid after 15 years. N = 180, I = 7.5, PV = 140000, PMT = ?, FV = 0, P/Y = 12, C/Y = 12 → PMT = -1297.82. find the
total amount repaid on the loan. 180 x 1297.82 = $233608. find the amount of interest paid on the loan. amount repaid
– amount borrowed = 233608 – 140000 = $93608.
ex. 100000 is borrowed at 8% p.a. compounding monthly to be repaid over 10 years with repayments of $1213.28 per
month; after 5 years, the interest rate is reduced to 7.5% p.a. find the new repayment required for the load to be paid
off. N = 60, I = 8, PV = 100000, PMT = -123.28, FV = ?, P/Y = 12, C/Y = 12 → FV = -59836.57 → N = 60, I = 7.5, PV =
59836.57, PMT = ?, FV = 0, P/Y = 12, C/Y = 12 → PMT = -1199 = $1199. how much is saved due to the rate cut? total
repaid with old amount: 60 x 1213.28 = $72796.80, total repaid with new amount: 60 x 1199 = $71940; interest saved
= 72796 – 71940 = $856.80
Annuities are a type of investment in which withdrawals are made at regular intervals, the formula being An+1 = rAn + d,
where A1 = a, ex. an inheritance of 150000 is received and invested at 7.5% p.a., with interest paid at the end of each
year and a withdrawal of $12000; what is the value of the investment at the start of the tenth year? An+1 = 1.075An –
12000, where A1 = 150000. find value on ClassPad. PV is negative.
Perpetuities have regular payments out that never run out, where the deposited amount remains in the account and
each year only the interest is paid out. Formula is Q = PR, where Q = value of regular payment per period, P = amount
invested, and R = interest rate as a decimal. ex. $500000 is set up for a perpetuity, investing the money in bonds that
return 3.5% p.a. compounded annually; how much is received? Q = PR / 100 → Q = 500000 x 0.035 = $17500.
Once simple interest is calculated, the amount to which the investment has grown can be found using A = P + I
ex. $4000 is invested earning 5.6% p.a. simple interest; how much interest will be earned in 3.5 years? I = PRT → I =
4000 x 0.056 x 3.5 = $784. what amount will the investment grow to in 3.5 years? A = P + I → A = 4000 + 784 = $4784
Hire purchase is paying off an item over time, where higher interest is charged when the terms are not followed.
Pay deposit – interest is charged on balance, or deferred payment – if paid late interest is charged on whole amount.
ex. a fridge is advertised for $1200 cash or on terms at 10% deposit and $57 a month for 2 years; how much more will
it cost than if bought for cash? deposit + 24 monthly repayments = 10% x $1200 + $57 x 24 = $1488 → difference =
$1488 − $1200 = $288. what is the flat rate of interest charged per annum? P = 1200 – 120 = $1080 → R = I / PT →
R = x 2 = 13%
Compound interest: A = P(I + R)t compounded annually, or A = P(I + R / n)tn compounded n times a year
ex. $3000 is invested at 5% p.a. compounded yearly; calculate the investment at the end of 2 years. I = PRT → I =
3000 x 0.05 x 1 = $150 → A = P + I → 3000 + 150 = $3150 → I = PRT → I = 3150 x 0.05 x 1 = $157.50 → A = P + I →
A = 3150 + 157.50 = $3307.50. calculate the compound interest earned after 2 years. interest = final investment –
initial principal = 3307.50 – 3000 = $307.05. ex. $65000 is invested at 7.5% p.a. compounded yearly; write a
recurrence relation describing the value after one year. tn+1 = rtn, where t0 = a → tn+1 = (1 + 0.075)tn, where t0 = 65000
Effective interest rate: e = (1 + i / n)n – 1, where e = effective yearly interest rate, i = nominal yearly interest rate, and
n = number of compounding periods in a year, ex. what is the effective interest rate for an investment earning 7% p.a.
compounding monthly? e = (1 + i / n)n – 1 → e = (1 + 0.)12 – 1 = 7.23%
Reducing balance loans: An = An-1(1 + r), where A0 = initial amount, and r = compounding rate as a decimal.
ex. 40000 is borrowed on 15th March, 2014. The bank lends this amount with monthly interest at 11.2% p.a., with
repayments of $873.69 a month for 5 years; how much is owed until 15th May? 40000(1 + 0.) – 873.69 =
$39499.64 → 39499.64(1 + 0.) – 873.69 = $38994.62.
Amount still owing after the nth payment: An = (1 + r / 100)An-1 – monthly repayment. ex. $25000 was borrowed with
monthly repayments of $500 and interest at 9.5% p.a.; what is the recurrence relation? An = (1 + 9. x 100)An-1 –
500, where A0 = 25000.
Using the ClassPad
$140000 is borrowed at 7.5% p.a. compound interest adjusted monthly; find the monthly repayment if the loan is to be
repaid after 15 years. N = 180, I = 7.5, PV = 140000, PMT = ?, FV = 0, P/Y = 12, C/Y = 12 → PMT = -1297.82. find the
total amount repaid on the loan. 180 x 1297.82 = $233608. find the amount of interest paid on the loan. amount repaid
– amount borrowed = 233608 – 140000 = $93608.
ex. 100000 is borrowed at 8% p.a. compounding monthly to be repaid over 10 years with repayments of $1213.28 per
month; after 5 years, the interest rate is reduced to 7.5% p.a. find the new repayment required for the load to be paid
off. N = 60, I = 8, PV = 100000, PMT = -123.28, FV = ?, P/Y = 12, C/Y = 12 → FV = -59836.57 → N = 60, I = 7.5, PV =
59836.57, PMT = ?, FV = 0, P/Y = 12, C/Y = 12 → PMT = -1199 = $1199. how much is saved due to the rate cut? total
repaid with old amount: 60 x 1213.28 = $72796.80, total repaid with new amount: 60 x 1199 = $71940; interest saved
= 72796 – 71940 = $856.80
Annuities are a type of investment in which withdrawals are made at regular intervals, the formula being An+1 = rAn + d,
where A1 = a, ex. an inheritance of 150000 is received and invested at 7.5% p.a., with interest paid at the end of each
year and a withdrawal of $12000; what is the value of the investment at the start of the tenth year? An+1 = 1.075An –
12000, where A1 = 150000. find value on ClassPad. PV is negative.
Perpetuities have regular payments out that never run out, where the deposited amount remains in the account and
each year only the interest is paid out. Formula is Q = PR, where Q = value of regular payment per period, P = amount
invested, and R = interest rate as a decimal. ex. $500000 is set up for a perpetuity, investing the money in bonds that
return 3.5% p.a. compounded annually; how much is received? Q = PR / 100 → Q = 500000 x 0.035 = $17500.
