Arithmetic and geometric progressions: DHS 2024 Promo Q6

Dunman High School2024 PromoPromo P2 Q6●●○ Standard7 marks

What this question tests

APGP (Loan Modelling).

Question

. Jane decides to purchase an apartment. To finance her apartment, she intends to apply for a bank housing loan of $847 500 which charges a monthly interest rate of $0.25\%$ . Interest is charged on the outstanding loan on the first day of each month, starting from the second month. A fixed monthly repayment of $ $M$ is done on the last day of each month throughout the loan tenure, starting from the first month.

Find an expression for the loan balance at the end of the second month. Hence show that the calculated loan balance at the end of the $n$ th loan month is given by $847500(1.0025)^{n-1} + 400M\bigl(1 - 1.0025^{\,n}\bigr).$ [3]
TDSR, also known as total debt servicing ratio, refers to the portion of a borrower’s gross monthly income that goes towards the total monthly debt. For a bank to approve any loan, the borrower’s TDSR should be less than or equal to $55\%$ . The formula for calculating TDSR is as follows: $\frac{\text{Borrower's total monthly debt}}{\text{Borrower's gross monthly income}} \times 100.$ If Jane intends to repay the loan in $25$ years, determine whether the bank will approve Jane’s loan given that her minimum gross monthly income is $7 300. [3]
State one assumption that you have used in your calculation for part (b). [1]

Show full worked solution▾

(a)

$n$	Loan balance at beginning of month (after interest)	Loan balance at end of month (after repayment)
$1$	$847500$	$847500-M$
$2$	$(847500-M)(1.0025)=847500(1.0025)-1.0025M$	$847500(1.0025)-1.0025M-M$
$3$	$847500(1.0025)^2-1.0025^2M-1.0025M$	$847500(1.0025)^2-M(1.0025^2+1.0025+1)$
$\vdots$	$\vdots$	$\vdots$
$n$		$847500(1.0025)^{n-1}-M(1.0025^{n-1}+\ldots+1.0025+1)$

Loan at the end of second month $\begin{aligned} &=847500(1.0025)-1.0025M-M \end{aligned}$ Loan at the end of nth month $\begin{aligned} &=847500(1.0025)^{n-1}-M(1.0025^{n-1}+\ldots+1.0025+1)\\ &=847500(1.0025)^{n-1}-M\left(\frac{1-1.0025^n}{1-1.0025}\right)\\ &=847500(1.0025)^{n-1}+400M(1-1.0025^n) \end{aligned}$

(b)

For loan to be repaid in 25 years, $\begin{aligned} 847500(1.0025)^{300-1}+400M(1-1.0025^{300})&=0 \end{aligned}$ Monthly repayment $=M=4008.9185=\$4008.92$ $\frac{4008.92}{7300}\times100=54.916\leq55$ Thus the bank will approve Jane’s loan.

(c)

Assume that the housing loan is the only debt that Jane has.

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