June 2nd, 2016, 12:40 PM  #1 
Senior Member Joined: May 2008 Posts: 299 Thanks: 81  Tricky compound interest problem Beer soaked ramblings follow. Got stumped by the following problem for about a day. I be sure my friends Sir Denis and Sir Jeff will appreciate it even more than I did. Hated it. Not that it's a bad problem but simply because I couldn't solve it sober. No solution needed. Just for enjoyment. On March 1, 1993, Mr. H deposited \$4,200 into an account that used a 4% annual effective interest rate when the balance was under \$5,000 and a 5.5% annual effective interest rate when the balance is at least $5,000. Mr. H withdrew \$1,000 on March 1, 1999. If there were no other deposits or withdrawals, find Mr. H’s account balance on March 1, 2003. Ans. \$5,257.17 P.S. How come the backslashes are getting displayed? 
June 2nd, 2016, 06:32 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991 
2 questions: 1: is the interest credited once per year, on Mar 1st? 2: is the "5000 rule" in effect after credit of interest, or is the account "looked at" daily to calculate if the account balance plus accrued interest is at least 5000? 
June 2nd, 2016, 07:07 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549  
June 3rd, 2016, 11:53 AM  #4  
Senior Member Joined: May 2008 Posts: 299 Thanks: 81  Beer soaked ramblings follow. Quote:
Yes.  
June 3rd, 2016, 01:54 PM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991  Quote:
4200(1.04)^5 = 5109.94 : Mar1/1998 5109.94(1.055)  1000 = 4390.98 : Mar1/1999 4398.98(1.04)^4 = 5136.84 : Mar1/2003 So the bonus 5.5 rate paid only for year period Mar1/98 to Mar199 Given answer 5257.17 hereby branded wrong  
March 21st, 2017, 08:52 AM  #6 
Senior Member Joined: May 2008 Posts: 299 Thanks: 81  Beer soaked reckonings follow. What is {4,200(1.04)^x*(1.055)^[(19991993)x]1,000}(1.04)^y*(1.055)^[(20031999)y] where x is derived from 4,200(1.04)^x=5,000 and y is derived from {4,200(1.04)^x*(1.055)^[(19991993)x]1,000}(1.04)^y=5,000 
March 21st, 2017, 10:44 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991  
March 21st, 2017, 12:16 PM  #8 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8 
First step is to calculate how long it is going to take for the balance to reach 5000 at a 4% annually compounded rate With a financial calculator it is 4.445 years which will lead you to about August 10 1997 ( I may be off by a day or so). i chose a financial calculator to save time. Formulas here are not an issue. the math is easy Second step is to calculate the future value on March 1 1999 from August 10th 1997 at an annual compounded rate of 5.5%. That is going to be equal to 5,432.64 Third step is to subtract the 1,000 withdrawal which leads to a balance of 4,432.64 fourth step is to calculate how long it is going to take for the balance to reach 5,000 again at a rate of 4% since the balance is below 5,000. The answer is about 3.07 years Final step is to calculate the future value with the remaining 0.93 years and drumrolll answer is 5,255.27 Answer is off because of rounding but Cheers Last edited by dthiaw; March 21st, 2017 at 12:31 PM. 
March 21st, 2017, 12:28 PM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991  Quote:
 
March 21st, 2017, 12:32 PM  #10 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  

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