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Posted by **Lydia** on Saturday, October 15, 2011 at 9:17am.

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

Add 41600 = 63217.52

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 -
**Reiny**, Saturday, October 15, 2011 at 10:06amI 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 "

BTW, Steve had the same answer.

Proof my answer is correct:

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**

PLease accept the inescapable conclusion that this answer is correct

**Answer this Question**

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