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Alex787 [66]
3 years ago
11

You have $36,948.61 in a brokerage account, and you plan to deposit an additional $3,000 at the end of every future year until y

our account totals $280,000. you expect to earn 11% annually on the account. how many years will it take to reach your goal? round your answer to two decimal places at the end of the calculations.
Mathematics
1 answer:
Luba_88 [7]3 years ago
8 0
Current amount in account
P=36948.61

Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n

Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11

We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000

We can solve this by trial and error.


The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000.  It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.

So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.

So it takes 15.00 years to reach the goal of 280000 years.
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