You deposit $1000 in each of two savings accounts. The interest for the
accounts is paid according to the two options described in Question 1. (Annual interest rate of 8.05% compounded monthly or annual interest rate of 8% compounded continuously)
How long would it take for the balance in one of the accounts to exceed
the balance in the other account by $100? By $100,000?

Question

27-12-2023
Mathematics

Answered

You deposit $1000 in each of two savings accounts. The interest for the
accounts is paid according to the two options described in Question 1. (Annual interest rate of 8.05% compounded monthly or annual interest rate of 8% compounded continuously)
How long would it take for the balance in one of the accounts to exceed
the balance in the other account by $100? By $100,000?

Answer :

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sqdancefan sqdancefan · Answer 1 · 2023-12-27T17:41:55-05:00

Answer:

32.4 years
104.0 years

Step-by-step explanation:

Two accounts start with a balance of $1000. One earns 8.05% interest compounded monthly; the other earns 8% compounded continuously. You want to know how long before the account balances differ by $100 and by $100,000.

Balance

The balance in the 8.05% account is given by ...

A = 1000(1 +0.0805/12)^(12t)

This can be written in terms of equivalent continuously compounded interest as ...

A = 1000e^(at) . . . . . . . where a=12·ln(1 +0.0805/12) ≈ 0.0802312

The balance in the 8% account is given by ...

A = 1000e^(0.08t)

Difference

The difference between the two account balances is ...

1000e^(at) -1000e^(.08t) = d

Dividing by the second term gives ...

e^((a-0.08)t) -1 = d/1000e^(-.08t)

And subtracting the term on the right gives a function of time that we can set to zero:

f(t) = 0

f(t) = e^((a-0.08)t) -1 -d/1000e^(-.08t)

Solution

This equation has no algebraic solution. The attached graph shows graphical solutions for d=100 and d=100,000.

$100 difference — 32.4 years

$100,000 difference — 104.0 years