compounding interest

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August 25th 2010, 02:30 PM
donnagirl

compounding interest

Mike deposits x dollars in a bank to open an account that earns 10% interest compounded annually. All interest earned is immediately deposited into the account. Exactly two years later Donna deposits x dollars in the same bank to open an identical account. After the initial deposit, Mike and Donna make no additional deposits/withdrawals. Assume that when her account is 10 years old, Donna has $10000 in it. How much money is in Mike's account at this time?

I have no idea how they get 12,100 for this problem.
August 25th 2010, 02:57 PM
pickslides

Have you seen this formula?

$\displaystyle A = P e^{rt}$
August 25th 2010, 03:06 PM
skeeter

Donna's account after 10 yrs ...

$10000 = x(1.1)^{10}<br />$

Mike's account after 12 years, let $A$ = amount in his account at that time ...

$A = x(1.1)^{12}$

divide the last equation by the first ...

$\displaystyle \frac{A}{10000} = \frac{x(1.1)^{12}}{x(1.1)^{10}}$

$\displaystyle \frac{A}{10000} = (1.1)^2$

$A = 10000(1.1)^2 = 12100$
August 25th 2010, 03:36 PM
thegreenmathdr

For your interest problem

The answer is not too difficult if you go back to the basic formula for interest: A=P(1+r)^t. She had her money in for 10 years at 10% so it would be 10,000= x(1+.1)^10 therefore 10'000/(1.1^10)=x. Now you go back to the guy and he has had his money in for 12 years. Therefore A=(x~3855.43)(1.1)^12 which equals your answer. I left my TI-89 in the car but I did it the hard way on a cheap calculator with no functions to check it out and I was within a few cents of your answer which would be due to rounding on the cheap tools I had available to me.

I hope this helped.