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You have 1000 that you want to deposit in a savings account. Bank A computes the amount value of your investment using the amount function $A_1(t) = 1+0.049t$ whereas Bank B uses the amount function $A_2(t) = > (1.4)^{12t}$. Where should you put your money?

Correct answer : B

I solved just by comparing effective interest rate for first period, $i_1 = \frac{A(1)-A(0)}{A(0)}$ but couldn't either one of them become larger depending on which period you choose? What is the proper way to solve this problem?

MinYoung Kim
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  • Is $A_2(t)$ really equal to $(1.4)^{12t}$? That growth seems way too fast. – Minus One-Twelfth Jul 03 '20 at 10:12
  • If $t$ is in years the first looks like 4.9% pa, based only on the original amount. But the second would be 40% per month compounding, which looks like a mistake. I would find it hard to trust the second option - if it looks too good to be true it probably is! – Peter Jul 03 '20 at 10:38
  • yup to the 12th power. – MinYoung Kim Jul 03 '20 at 18:28

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