Reducible Loans 2

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Year 11

Consumer Maths

Reducible Loans

Reducible Loans 2

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Question 1 of 5

Daniela borrowed

$ 29000

$$29000$ to purchase a car. The interest rate she will be paying is

8.5 %

$8.5%$ per annum and her monthly repayments are

$ 715

$$715$ . The table below sets out her monthly repayment schedule for the first four months. Find the missing values

x, y

$x, y$ and

z

$z$ .

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$x$ $x$	$205.42$ $205.42$	$29205.42$ $29205.42$	$28490.42$ $28490.42$
$2$ $2$	$28490.42$ $28490.42$	$y$ $y$	$28692.23$ $28692.23$	$27977.22$ $27977.22$
$3$ $3$	$27977.22$ $27977.22$	$198.17$ $198.17$	$28175.39$ $28175.39$	$27460.39$ $27460.39$
$4$ $4$	$27460.39$ $27460.39$	$194.51$ $194.51$	$27654.90$ $27654.90$	$z$ $z$

Round off decimals to

2

$2$ places

$x = $$ $x=$$ (29000, 29,000)

$y = $$ $y=$$ (201.81)

$z = $$ $z=$$

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A Reducible Loan is a loan where the interest is paid on the balance owing, or the remaining amount of the loan that is yet to be paid.

Reducible Loan Table

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
	Remaining loan amount	Interest amount for the month	Principal $+$ $+$ Interest	Principal $+$ $+$ Interest $-$ $-$ Repayment

Finding $x$ $x$

First, summarise the data given in the problem.

Principal (

P

$P$ )

= $ 29000

$=$29000$

Interest (

I

$I$ )

= 8.5 %

$=8.5%$ per annum

Monthly Repayment

= $ 715

$=$715$

In the table, $x$ $x$ is the principal amount in the $1$ $1$ st month.

Since it has just been the start of the loan, that means:

x = $ 29000

$x=$29000$

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$29000$ $29000$	$205.42$ $205.42$	$29205.42$ $29205.42$	$28490.42$ $28490.42$
$2$ $2$	$28490.42$ $28490.42$	$y$ $y$	$28692.23$ $28692.23$	$27977.22$ $27977.22$
$3$ $3$	$27977.22$ $27977.22$	$198.17$ $198.17$	$28175.39$ $28175.39$	$27460.39$ $27460.39$
$4$ $4$	$27460.39$ $27460.39$	$194.51$ $194.51$	$27654.90$ $27654.90$	$z$ $z$

Finding $y$ $y$

From the table, $y$ $y$ is the interest amount in the $2$ $2$ nd month.

Solve for the monthly interest rate by dividing

I

$I$ by

12

$12$ .

Interest (

I

$I$ )

= 8.5 %

$=8.5%$ per annum/year

	$8.5 % \div 12$ $8.5%-:12$

$=$ $=$	$\frac{8.5}{100} \div 12$ $8.5/100-:12$	Convert from percentage to decimal

$=$ $=$	$0.0070833$ $0.0070833$

Multiply the monthly interest rate to the

2

$2$ nd month’s principal amount

2

$2$ nd month Principal Amount

= $ 28490.42

$=$28490.42$

$y$ $y$	$=$ $=$	$0.0070833 \times 28490.42$ $0.0070833xx28490.42$
	$=$ $=$	$$ 201.81$ $$201.81$

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$29000$ $29000$	$205.42$ $205.42$	$29205.42$ $29205.42$	$28490.42$ $28490.42$
$2$ $2$	$28490.42$ $28490.42$	$201.81$ $201.81$	$28692.23$ $28692.23$	$27977.22$ $27977.22$
$3$ $3$	$27977.22$ $27977.22$	$198.17$ $198.17$	$28175.39$ $28175.39$	$27460.39$ $27460.39$
$4$ $4$	$27460.39$ $27460.39$	$194.51$ $194.51$	$27654.90$ $27654.90$	$z$ $z$

Finding $z$ $z$

From the table, $z$ $z$ is the

P + I - R

$P+I-R$ for the $4$ $4$ th month.

P

$P$ and

I

$I$ (

4

$4$ th month)

= $ 27654.90

$=$27654.90$

R

$R$ (Monthly Repayment)

= $ 715

$=$715$

$P + I - R$ $P+I-R$	$=$ $=$	$27654.90 - 715$ $27654.90-715$
$z$ $z$	$=$ $=$	$$ 26939.90$ $$26939.90$

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$29000$ $29000$	$205.42$ $205.42$	$29205.42$ $29205.42$	$28490.42$ $28490.42$
$2$ $2$	$28490.42$ $28490.42$	$201.81$ $201.81$	$28692.23$ $28692.23$	$27977.22$ $27977.22$
$3$ $3$	$27977.22$ $27977.22$	$198.17$ $198.17$	$28175.39$ $28175.39$	$27460.39$ $27460.39$
$4$ $4$	$27460.39$ $27460.39$	$194.51$ $194.51$	$27654.90$ $27654.90$	$26939.90$ $26939.90$

x = $ 29000

$x=$29000$

y = $ 201.81

$y=$201.81$

z = $ 26939.90

$z=$26939.90$

Question 2 of 5

2. Question
Daniela borrowed $$ 29000$ $$29000$ to purchase a car. The interest rate she will be paying is $8.5 %$ $8.5%$ per annum and her monthly repayments are $$ 715$ $$715$ . If she repaid the loan in $4$ $4$ years, how much did she pay in total?
- $$$ $$$ (34320, 34,320)
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A Reducible Loan is a loan where the interest is paid on the balance owing, or the remaining amount of the loan that is yet to be paid.

To find the total repayments, multiply the monthly repayment to $12$ $12$ and to the loan duration (in years).

Monthly Repayment $= $ 715$ $=$715$

Loan Duration $= 4$ $=4$ years

$$ 715 \times 12 \times 4$ $$715xx12xx4$ $=$ $=$ $$ 34320$ $$34 320$

$$ 34320$ $$34 320$

Question 3 of 5

Yani borrowed

$ 280000

$$280 000$ to buy an apartment. The interest rate for the loan is

4.25 %

$4.25%$ per annum and the monthly repayment is

$ 1517

$$1517$ . Complete the table below which sets out his monthly repayment schedule for the first four months.

Round off decimals to

2

$2$ places

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$280000$ $280 000$	$991.67$ $991.67$	$280 991.67$ $280 991.67$	$279 474.67$ $279 474.67$
$2$ $2$	$279 474.67$ $279 474.67$	$989.81$ $989.81$	$280 464.48$ $280 464.48$	$278 947.48$ $278 947.48$
$3$ $3$	$278 947.48$ $278 947.48$	$987.94$ $987.94$	$279 935.42$ $279 935.42$	(278418.42)
$4$ $4$	(278418.42)	(986.07)	(279404.49)	(277887.49)

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A Reducible Loan is a loan where the interest is paid on the balance owing, or the remaining amount of the loan that is yet to be paid.

Reducible Loan Table

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
	Remaining loan amount	Interest amount for the month	Principal $+$ $+$ Interest	Principal $+$ $+$ Interest $-$ $-$ Repayment

First, summarise the data given in the problem.

Principal (

P

$P$ )

= $ 280000

$=$280 000$

Interest (

I

$I$ )

= 4.25 %

$=4.25%$ per annum

Monthly Repayment

= $ 1517

$=$1517$

The first value missing in the table is the

P + I - R

$P+I-R$ in the $3$ $3$ rd month.

Simply subtract the monthly repayment from the

P

$P$ and

I

$I$ on that month

P

$P$ and

I

$I$ (

3

$3$ rd month)

= $ 279 935.42

$=$279 935.42$

Monthly Repayment

= $ 1517

$=$1517$

$ 279 935.42 - $ 1517

$$279 935.42-$1517$

=

$=$

$ 278 418.42

$$278 418.42$

Since

P + I - R

$P+I-R$ becomes the next month’s new balance, copy this value to the principal amount for the $4$ $4$ th month.

Month	Principal ( $P$ $P$ )	Interest ( $I$ $I$ )	$P$ $P$ and $I$ $I$	$P + I - R$ $P+I-R$
$1$ $1$	$280000$ $280 000$	$991.67$ $991.67$	$280 991.67$ $280 991.67$	$279 474.67$ $279 474.67$
$2$ $2$	$279 474.67$ $279 474.67$	$989.81$ $989.81$	$280 464.48$ $280 464.48$	$278 947.48$ $278 947.48$
$3$ $3$	$278 947.48$ $278 947.48$	$987.94$	$279 935.42$	$278 418.42$
$4$	$278 418.42$

The next missing value is the interest amount in the $4$ th month.

Solve for the monthly interest rate by dividing

$I$ by

$12$ .

Interest (

$I$ )

$=4.25%$ per annum/year

	$4.25%-:12$

$=$	$4.25/100-:12$	Convert from percentage to decimal

$=$	$0.003541666dot6$

Multiply the monthly interest rate to the

$4$ th month’s principal amount

$4$ th month Principal Amount

$=$278 418.42$

$0.003541666dot6xx$278 418.42$

$=$

$$986.07$

Month	Principal ( $P$ )	Interest ( $I$ )	$P$ and $I$	$P+I-R$
$1$	$280 000$	$991.67$	$280 991.67$	$279 474.67$
$2$	$279 474.67$	$989.81$	$280 464.48$	$278 947.48$
$3$	$278 947.48$	$987.94$	$279 935.42$	$278 418.42$
$4$	$278 418.42$	$986.07$

Next, add the

$P$ and

$I$ for the $4$ th month.

$P$ (

$4$ th month)

$=$278 418.42$

$I$ (

$4$ th month)

$=$986.07$

$$278 418.42+$986.07$

$=$

$$279 404.49$

Month	Principal ( $P$ )	Interest ( $I$ )	$P$ and $I$	$P+I-R$
$1$	$280 000$	$991.67$	$280 991.67$	$279 474.67$
$2$	$279 474.67$	$989.81$	$280 464.48$	$278 947.48$
$3$	$278 947.48$	$987.94$	$279 935.42$	$278 418.42$
$4$	$278 418.42$	$986.07$	$279 404.49$

Lastly, subtract the monthly repayment from the

$P$ and

$I$ of the

$4$ th month.

$P$ and

$I$ (

$4$ th month)

$=$279 404.49$

Monthly Repayment

$=$1517$

$$279 404.49-$1517$

$=$

$$277 887.49$

Month	Principal ( $P$ )	Interest ( $I$ )	$P$ and $I$	$P+I-R$
$1$	$280 000$	$991.67$	$280 991.67$	$279 474.67$
$2$	$279 474.67$	$989.81$	$280 464.48$	$278 947.48$
$3$	$278 947.48$	$987.94$	$279 935.42$	$278 418.42$
$4$	$278 418.42$	$986.07$	$279 404.49$	$277 887.49$

Question 4 of 5

Yani borrowed

$$280 000$ to buy an apartment. The interest rate for the loan is

$4.25%$ per annum and the monthly repayment is

$$1517$ . What is the total amount of interest Yani pays for the first

$4$ months?

Month	Principal ( $P$ )	Interest ( $I$ )	$P$ and $I$	$P+I-R$
$1$	$280 000$	$991.67$	$280 991.67$	$279 474.67$
$2$	$279 474.67$	$989.81$	$280 464.48$	$278 947.48$
$3$	$278 947.48$	$987.94$	$279 935.42$	$278 418.42$
$4$	$278 418.42$	$986.07$	$279 404.49$	$277 887.49$

Round off decimals to

$2$ places

$$$ (3955.49)

Question 5 of 5

5. Question
Roberta borrows $$410 000$ as a reducible loan, in order to purchase a home. She has monthly repayments of $$2397$ . The annual interest rate is $5%$ and the loan period is $25$ years.

$(i)$ Find the total repayments.

$(ii)$ Find the total interest paid.
- $(i) $$ (719100, 719,100)
  
  $(ii) $$ (309100, 309,100)
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A Reducible Loan is a loan where the interest is paid on the balance owing, or the remaining amount of the loan that is yet to be paid.

$(i)$ Find the total repayments.

To find the total repayments, multiply the monthly repayment to $12$ and to the loan duration (in years).

Monthly Repayment $=$2397$

Loan Duration $=25$ years

$$2397xx12xx25$ $=$ $$719 100$

$(ii)$ Find the total interest paid.

Simply subtract the principal amount from the total repayments.

Total Repayment $=$719 100$

Principal ( $P$ ) $=$410 000$

$$719 100-$410 000$ $=$ $$309 100$

$(i) $719 100$

$(ii) $309 100$

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