Two Savings Offers with a Fee and a Rounding Trap
Maya has $6,000 to deposit for five years. Bank A pays 4.2% per year compounded quarterly and charges no fees. Bank B pays 4.5% per year compounded monthly but deducts a one-off $40 account fee at opening. A classmate argues that B is better because 4.5% is 'almost the same' as 4.2% and monthly compounding adds extra interest. Maya wants to know which bank leaves her with more money after exactly five years, and she is also curious how long B would need to overtake A if both balances were left untouched. The bank quotes nominal annual rates, and neither account allows partial years for the comparison she wants. Maya's calculator is set to five decimal places, and she tends to round the monthly rate before substituting, which her teacher warns can shift the final answer by a few dollars.
1. Calculate the balance in each bank after exactly five years, treating the $40 fee as deducted at opening. State which balance is larger and by how much. Use unrounded rates in your calculation.
Read the worked answer
Bank A has a quarterly rate of 0.042/4 = 0.0105 and 5×4 = 20 quarters.
A = 6000(1.0105)^20 ≈ 6000×1.232328 = $7,393.97.
Bank B has opening principal 6000−40 = $5,960 and monthly rate 0.045/12 = 0.00375, with 5×12 = 60 months.
B = 5960(1.00375)^60 ≈ 5960×1.251793 = $7,460.69.
Bank B is larger by about $7,460.69−$7,393.97 = $66.72. If the monthly rate is rounded to 0.004 before substitution, the B balance changes by several dollars, so the unrounded rate is important.
2. Find the smallest whole number of years n for which Bank B's balance would exceed Bank A's, still with the $40 fee at opening. Set up an inequality, solve it using logarithms, and explain why the final answer must be rounded up rather than to the nearest integer.
Read the worked answer
After n years, Bank A has 6000(1.0105)^(4n) and Bank B has 5960(1.00375)^(12n). We need
5960(1.00375)^(12n) > 6000(1.0105)^(4n).
Divide by 6000 and take natural logs:
ln(5960/6000) + 12n ln(1.00375) > 4n ln(1.0105).
Using ln(5960/6000) ≈ −0.006688, ln(1.00375) ≈ 0.00374298, and ln(1.0105) ≈ 0.01044526:
−0.006688 + 12n(0.00374298) > 4n(0.01044526) −0.006688 + 0.0449158n > 0.0417810n 0.0031348n > 0.006688 n > 2.133.
The balances are equal at about n = 2.13 years. For whole-year comparisons, n = 2 fails the inequality, while n = 3 works. Rounding up is required because the inequality must hold for the whole year; rounding to the nearest integer would give n = 2, which is too early.
3. Suppose Bank B instead deducts the $40 fee at the end of the five years. Recalculate the five-year B balance and the difference from Bank A. Does the ranking change? Explain why the timing of the fee matters.
Read the worked answer
With the fee at the end, the whole $6,000 earns interest for 60 months:
B_end = 6000(1.00375)^60 − 40 ≈ 6000×1.251793 − 40 = 7510.76−40 = $7,470.76.
Bank A is still $7,393.97, so B is ahead by about $7,470.76−$7,393.97 = $76.79. The ranking does not change, but B is about $10.07 better off than in part (a). The timing matters because a fee deducted at opening reduces the principal that earns compound interest for the whole five years, while a fee at the end does not.