Question

Suppose you have two empty accounts,A1 andA2. Each pays 8% APR. You put $100,000intoA1 now. One year from now you start making annual deposits intoA2 of $10,000 at thebeginning of each year (thus, the first payment made at the beginning of one year from nowmeans that the first payment is actually made at the end of the first year, and so on). Afterhow many full years will the amount inA2 become greater than that inA1? Comment onthe solution same problem if the annual payments intoA2 are $7,000 instead of $10,000.

Answer 1

After "26 years"  the amount in A2 become greater than that in A1.

According to the question,

The future value of funds in A1:

= $100000(1.08)^n$

here, n = Number of years

The future value of A2 will be:

= $\frac{10000}{0.08}[1.08^n-1]$

= $125000[1.08^n-1]$

now,

→ $125000[1.08^n -1] > 100000(1.08)^n$

→        $25000(1.08^n) > 125000$

→                    $1.08^n > 5$

By taking "log", we get

→                          $n> \frac{ln \ 5}{ln \ 1.08} = 20.9$ or, $21 \ years$ (n "Number of years")  

Each year, with $7000 in A2

→ $\frac{7000}{0.08} [1.08^n -1]> 100000(1.08^n)$

→         $1.08^n -1 > 1.142857 (1.08^n)$

→               $1.08^n> \frac{1}{0.142857} = 7$

By taking "log", we get

→                     $n > \frac{ln \ 7}{ln \ 1.08} = 25.28$ or, $26 \ years$ (n "Number of years")

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