Arithmetic & Geometric Progressions: Financial: AP savings vs GP compound interest — TJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
Mabel and Janice decided to start a 5-year savings plan beginning in January 2026.
Mabel saves using a piggy bank. At the start of January 2026, she deposits \$101. Each subsequent month, she increases her deposit by \$1 — so she deposits \$101 in January, \$102 in February, \$103 in March, and so on, until \$112 in December. At the start of each new year, she resets her monthly deposit to \$101 in January and repeats the same pattern through December. She continues this routine from 2026 to 2030, inclusive.
Janice, on the other hand, deposits \$100 at the start of every month into a bank account that earns 0.3% interest per month, with interest calculated and added into the account at the end of each month.
- Show that Janice will have more money in her savings account than Mabel has in her piggy bank at the end of December 2030.
- Find the month and year when Janice's savings first exceed Mabel's savings.
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(a)
Mabel: Amount saved in one year \(= \$(101+102+\cdots+112) = \dfrac{12}{2}(101+112) = \$1278\).
Amount saved in 5 years \(= 1278\times 5 = \$6390\).
Janice: She deposits \$100 at the start of each month at 0.3% monthly interest. The amount at the end of month \(n\) follows the pattern:
| Month | Start of month | End of month |
|---|---|---|
| 1 | \(100\) | \(100\times 1.003\) |
| 2 | \(100 + 100\times1.003\) | \(100\times1.003^2 + 100\times1.003\) |
| \(n\) | \(\cdots\) | \(100\times1.003\cdot\dfrac{1.003^n-1}{1.003-1}\) |
At Dec 2030, \(n = 60\). Amount Janice saves in 5 years: \[ = \frac{100\times1.003\,(1.003^{60}-1)}{1.003-1} = \$6582.85\text{ (2 d.p.)} \]
(b) At end of 4th year (\(n=48\)): Janice has \(\dfrac{100\times1.003(1.003^{48}-1)}{0.003} = \$5169.97 > \$5112\) (Mabel).
At end of 3rd year (\(n=36\)): Janice has \(\$3806.97 < \$3834\) (Mabel).
So Janice overtakes Mabel during the 4th year (i.e. 2029). Let it be the \(k\)-th month of 2029.
After \(36+k\) months total, Janice's amount \(>\) Mabel's amount: \[ \frac{100\times1.003\,(1.003^{36+k}-1)}{0.003} > 3834 + \frac{k}{2}\bigl[2(101) + (k-1)\bigr] \]
Using GC, smallest \(k = 4\).