# Math

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Lee Holmes deposited \$16,600 in a new savings account at 9% interest compounded semiannually. At the beginning of year 4, Lee deposits an additional \$41,600 at 9% interest compounded semiannually.

At the end of year 6, what is the balance in Lee’s account?
I have tried and I got \$44691.78 but that is incorrect. I solved each part separately then added the what he deposited. PLEASE HELP ME!!
Math - Steve, Wednesday, October 12, 2011 at 1:58pm
It's not quite clear from the language, but it appears that after 3 years, an additional deposit was made, and then the current balance is left to draw 9% interest for two more years. If that's the case, then

to start: 16600
at the end of year 3, he has 16600*1.045^6 = 21617.52

After two more years, 63217.52*1.045^4 = 75388.07

Math - Steve, Wednesday, October 12, 2011 at 2:00pm
Oops. That just takes us 5 years.

63217.52*1.045^6 = 82325.66

Math - Reiny, Wednesday, October 12, 2011 at 4:50pm
Make a time graph to see how the periods work

amount = 16600(1.045)^12 + 41600(1.045)^6
= 82 325.65

Math - Lydia, Saturday, October 15, 2011 at 9:08am
Thannk you both of you but both answers are incorrect. I can input the answer to see if it is right and both are wrong. We are all missing something. If anyone else wants to try please do!!!

• Math -

I stand by my solution where I said"

"Math - Reiny, Wednesday, October 12, 2011 at 4:50pm
Make a time graph to see how the periods work

amount = 16600(1.045)^12 + 41600(1.045)^6
= 82 325.65 "

rate = .045 per half-year

Balance now: 16600
balance after 1 yr = 16600(1.045)^2 = 18127.62
balance after 2 yrs. = 18127.62(1.045)^2 = 19795.81
balance after 3yrs = 19795.81(1.045)^2 = 21.617.52
at that point, beginning of year 4, an additional 41600 in added
balance after 3 yrs = 63217.52
balance after 4 yrs = 63217.52(1.045)^2 = 69035.11
balance after 5 yrs = 69035.11(1.045)^2 = 75388.07
balance after 6 yrs = 75388.07(1.045)^2 = 82325.65

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