Monthly deposits and the first crossing day · source reference

Printed question

Chloe plans to invest $k$ dollars on day 1 of each month from 1 January 2021. The bank offers:

Plan A: Each investment earns a fixed 5-dollar bonus at the end of every month it has been in the account. The bonus is added to the account. Accumulated bonuses earn no further bonuses.

(a) For Plan A, find $k$ if all investments including bonuses are worth 22200 dollars at the end of 31 December 2022. [3]

Plan B: Interest is 0.2% monthly, so on the last day of each month the amount increases by 0.2%.

(b) Assume Plan B for the rest.

(i) Find in terms of $k$ the interest on the first $k$ invested on 1 January 2021 at the end of 31 December 2022. [1]

(ii) Find minimum $k$ (nearest integer) for total at least 22200 dollars at that date. [3]

(iii) Let $k=900$. Find the month the total first exceeds 30000 dollars. Explain whether it happens on the first or last day of the month. [5]

Printed working

(a) Total value at 24 months:

$$24k+5+5(2)+\cdots+5(24)=24k+\frac{24}2(5+120)=24k+1500.$$

$24k+1500=22200\Rightarrow k=862.50$.

(b)(i) Interest $=1.002^{24}k-k=k(1.002^{24}-1)$.

(ii) Total at end of 24 months

$$1.002k+1.002^2k+\cdots+1.002^{24}k=\frac{1.002k(1.002^{24}-1)}{1.002-1}=24.60930k.$$

$24.60930k\ge22200\Rightarrow k\ge902.097$. Least integer $k=903$.

(iii) Total at end of $n$ months for $k=900$:

$$1.002(900)+1.002^2(900)+\cdots+1.002^n(900)=\frac{1.002(900)(1.002^n-1)}{1.002-1}=450900(1.002^n-1).$$

Consider $450900(1.002^n-1)\ge30000$.

At $n=32$, amount 29770.40 <30000 (source rounded display). At $n=33$, 30731.68 >30000. On the first day of month 33, $29770.3396+900=30670.3396>30000$. Thus first exceeds 30000 on 1 September 2023.

Source tables and calculator displays

Plan A:

End of	Amount invested	Bonus in account
Jan 2021	$k$	5
Feb 2021	$2k$	$5+2(5)$
…	…	…
Dec 2022	$24k$	$5+2(5)+\cdots+24(5)$

Plan B:

Month	Amount at start	Amount at end
Jan 2021	$k$	$1.002k$
Feb 2021	$k+1.002k$	$1.002k+1.002^2k$
Mar 2021	$k+1.002k+1.002^2k$	$1.002k+1.002^2k+1.002^3k$
…	…	…
Dec 2022	$k+1.002k+\cdots+1.002^{23}k$	$1.002k+1.002^2k+\cdots+1.002^{24}k$

Printed answer key

(a) 862.50; (b)(i) $k(1.002^{24}-1)$; (ii) 903; (iii) 1 September 2023

Mathematical review and qualifications

(a) $24k+5(1+2+\cdots+24)=24k+1500=22200$, so $k=862.50$.

(b)(i) The first deposit earns $k(1.002^{24}-1)$. (ii) Total is $k\sum_{j=1}^{24}1.002^j=\frac{1.002k(1.002^{24}-1)}{0.002}$. This requires $k\ge902.0979\ldots$, hence minimum integer 903.

(iii) For $k=900$, month-end balance after $n$ months is $450900(1.002^n-1)$. Month 32 (August 2023) ends with about 29770.34, below 30000. The next deposit takes it to 30670.34 immediately on 1 September 2023. Therefore the first crossing occurs on the first day of September, before interest is added.

A month-end formula alone can give the right month but the wrong day; check the deposit at the start of that month.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2020/TMJC/Prelim/I/10] Question PDF pages 5, 6; solution PDF pages 25, 26. The original files remain private QA records outside this native content.