<p>Wei is saving for a property that she intends to buy in the future. Wei needs to save a minimum of $\$50 \,\, 000$. She saves regularly in an account which offers no interest. She makes an initial deposit on 31 January 2021 of $\$a$. Each subsequent month, she deposits $\$50$ more than she deposited in the previous month. Her final deposit is made on 31 December 2023.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find, to the nearest cent, the smallest value of $a$ so that she saves at least $\$50\,\,000$.<span class="inline-marks">[2]</span></div></li></ol>
<p>Wei has arranged to purchase a property on 1 January 2024 with the aid of a loan of $\$400\,\,000$. The terms of the loan are that interest of $0.1\%$ is added to the amount owing at the start of every month, with the first interest amount added on 1 January 2024. Wei makes a monthly repayment of $\$x$ at the end of every month, with the first repayment on 31 January 2024.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Show that the amount, in dollars, that Wei owes at the end of $n$ months is</span></div><div class="list-item-paragraph" style="text-align: center">$400\,\,000\times {{1.001}^{n}}-1000x\left( {{1.001}^{n}}-1 \right)$.<span class="inline-marks">[3]</span></div></li><li><ol class="i-list"><li><div class="list-item-paragraph">Suppose that the loan is repaid in $360$ monthly repayments. Find the value of the monthly repayment to the nearest cent.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Use your answer to part <strong>(c)(i)</strong> to calculate the total interest paid on the loan in this case.<span class="inline-marks">[1]</span></div></li></ol></li><li><ol class="i-list"><li><div class="list-item-paragraph">Now suppose Wei decides to repay the loan at the rate of $\$1600$ per month for&nbsp;$k$ months, plus a final monthly repayment of $\$y$, where&nbsp;$y&lt;1600$. Find $k$ and $y$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Hence find the total saving to Wei by making the repayments in part <strong>(d)(i)</strong> rather than the repayments in part <strong>(c)(i)</strong>, giving your answer to $4$ significant figures.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

Wei is saving for a property that she intends to buy in the future. Wei needs to save a minimum of $\$50 \,\, 000$. She saves regularly in an account which offers no interest. She makes an initial deposit on 31 January 2021 of $\$a$. Each subsequent month, she deposits $\$50$ more than she deposited in the previous month. Her final deposit is made on 31 December 2023.

Find, to the nearest cent, the smallest value of $a$ so that she saves at least $\$50\,\,000$.

Wei has arranged to purchase a property on 1 January 2024 with the aid of a loan of $\$400\,\,000$. The terms of the loan are that interest of $0.1\%$ is added to the amount owing at the start of every month, with the first interest amount added on 1 January 2024. Wei makes a monthly repayment of $\$x$ at the end of every month, with the first repayment on 31 January 2024.

Show that the amount, in dollars, that Wei owes at the end of $n$ months is

$400\,\,000\times {{1.001}^{n}}-1000x\left( {{1.001}^{n}}-1 \right)$.

Suppose that the loan is repaid in $360$ monthly repayments. Find the value of the monthly repayment to the nearest cent.

Use your answer to part (c)(i) to calculate the total interest paid on the loan in this case.

Now suppose Wei decides to repay the loan at the rate of $\$1600$ per month for $k$ months, plus a final monthly repayment of $\$y$, where $y<1600$. Find $k$ and $y$.

Hence find the total saving to Wei by making the repayments in part (d)(i) rather than the repayments in part (c)(i), giving your answer to $4$ significant figures.

APGP

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2023 P1 Q11

Deposits $a on 31st of each month, earning $50 interest each month.

- n=1 (31 Jan 2021): $a
- n=2 (31 Feb 2021): $(a+50) + a = $2a + 50
- After n months: AP with first term a, common difference 50

After 36 months (31 Dec 2023), total ≥ $50000:
$S_{36} \ge 50000$
$\frac{36}{2}[2a + (36-1)(50)] \ge 50000$
$18[2a + 1750] \ge 50000$
$2a + 1750 \ge \frac{50000}{18}$
$2a \ge \frac{2500}{9} - 1750$
$a \ge 513.\overline{8}$

Smallest value of $a = \boxed{\$513.89}$ (nearest cent)

Loan: $400000, monthly interest 0.1%, monthly payment $x

**Tracking the loan balance:**
- End of Jan 2024: $400000(1.001) - x$
- End of Feb 2024: $400000(1.001)^2 - [1.001x + x]$
- End of Mar 2024: $400000(1.001)^3 - [1.001^2x + 1.001x + x]$

At end of n-th month:
$= 400000(1.001)^n - S_n$
where $S_n = x + 1.001x + ... + 1.001^{n-1}x$ is a GP with $a = x$, $r = 1.001$

$= 400000(1.001)^n - \frac{x(1.001^n - 1)}{1.001 - 1}$
$= 400000(1.001)^n - 1000x(1.001^n - 1)$ (shown)

For $n = 360$ months, amount owed ≤ 0:
$400000(1.001)^{360} - 1000x(1.001^{360} - 1) \le 0$
$400000(1.001)^{360} \le 1000x(1.001^{360} - 1)$
$x \ge \frac{400000(1.001)^{360}}{1000(1.001^{360} - 1)}$
$x \ge 1323.634...$
$\boxed{x = \$1323.64}$ (nearest cent)

Interest = Total amount paid - Original price
$= 360 \times \$1323.64 - \$400000$
$= \$476510.40 - \$400000$
$= \boxed{\$76510.40}$

Let amount owed at end of k months be $A_k$:
$A_k = 400000(1.001)^k - 1000(1600)(1.001^k - 1)$

From GC:
- k = 286: \$2917.30
- k = 287: \$1320.22
- k = 288: -\$278.46

So $k = 287$, and final payment = $1320.22 \times 1.001 = \boxed{\$1321.54}$

Total savings = Amount paid in (b)(i) - Amount paid in (c)(i)
$= 360 \times \$1323.64 - [287 \times \$1600 + \$1321.54]$
$= \$476510.40 - [\$459200 + \$1321.54]$
$= \$476510.40 - \$460521.54$
$= \$15988.86$
$\approx \boxed{\$15990}$ (4 s.f.)

N2023 P1 Q11

N2023 P1 Q11

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