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Worked solution

Track beginning-of-month deposits and monthly interest

A student deposits $10 on 1 January 2009 into an account paying 2%2\% compound interest on the last day of each month, then another $10 on the first day of each later month.

(a) How much interest has the original $10 earned after two years?

(b) What is the total balance after two years?

(c) After how many complete months does the total first exceed $2000?

[2008/I/10(ii)(modified)]

CJC · Discussion · Question 18

Answer

(a) $6.08 interest. (b) $310.30 balance. (c) 81 complete months.

(a) Follow the original deposit for 24 interest credits

The first deposit has value 10(1.02)n10(1.02)^n at the end of month nn. After 2424 months the interest, excluding its principal, is 10(1.02)24−10≈6.0810(1.02)^{24}-10\approx6.08 dollars (2 decimal places).

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(b) Sum deposits with their correct interest durations

At each month end the new deposit has already earned one month’s interest. The balances start as B1=10(1.02)B_1=10(1.02), B2=10(1.02)2+10(1.02)B_2=10(1.02)^2+10(1.02) and B3=10(1.02)3+10(1.02)2+10(1.02)B_3=10(1.02)^3+10(1.02)^2+10(1.02). In general Bn=10∑j=1n1.02j=10(1.02)(1.02n−1)/0.02B_n=10\sum_{j=1}^n1.02^j=10(1.02)(1.02^n-1)/0.02. Hence B24≈310.30B_{24}\approx310.30 dollars (2 decimal places).

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(c) Solve the strict threshold and check the adjacent months

Bn>2000B_n>2000 gives 1.02(1.02n−1)>41.02(1.02^n-1)>4, so n>ln⁡(1+4/1.02)/ln⁡1.02≈80.47n>\ln(1+4/1.02)/\ln1.02\approx80.47. The first complete month is 8181; B80<2000B_{80}<2000 and B81>2000B_{81}>2000.

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Common mistakes

The final deposit’s exponent is 1, not 0, when measuring the end-of-month balance.

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