April 6th, 2014, 03:36 PM  #1 
Newbie Joined: Apr 2014 From: USA Posts: 10 Thanks: 8  How does the algebra problem fit with the word problem?
I'm having a lot of trouble with an algebra question, or rather, with my lack of logic in understanding how it should be solved. I'm just going to quote the whole bit and then try to explain how I'm thinking it through by showing how I'm solving it. I hope that this is in the right forum. I'm sorry if it's not. The problem is: Jared places a small inheritance of 2475 in a certificate of deposit that earns 6% interest compounded quarterly. The total in the CD after 10 years is given by the expression 2475 (1+0.06/4)^4(10). I can already tell you the answer is 4489.70, according to the back of the book and online algebra calculators and I understand how to solve it, so that's fine. My problem is that I don't understand how the algebra problem fits with the word problem. If I had looked at the word problem without looking at the expression, I would have assumed that what I needed to do is take 6% of 2475, which is 148.50, multiply that by 4 to represent the quarterly interest, which would give me 594 and then multiply that number by 10 to represent the 10 years that the CD is collecting interest, which is 5940. Then, to actually show the total in the CD I'd add the original 2475 to 5940. So I guess what I'm saying is that if I were to come up with an expression on my own for the word problem, it would look more like 2475(0.06 times 40) + 2475 (I'm not too familiar with algebraic expressions, so I don't know if I did that right.) I would just like someone to explain to me in exactly what way I'm wrong. I know I am wrong, but it's just that my way seems logical to me even though it can't be. Thank you so much! 
April 6th, 2014, 04:09 PM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
The interest is added to the principle every quarter year so the principle increases every quarter year. You then earn 6% interest on the new principle. After the first quarter you have $2475 + 0.06(2475) = 2623.50$ After the second quarter you have $2623.50 + 0.06(2623.50) = 2780.91 $ After the third quarter you have $2780.91 + 0.06(2780.91) = 2947.76 $ And so on , get the idea? The formula used in your book does the same thing ... but much quicker. 
April 6th, 2014, 04:24 PM  #3 
Newbie Joined: Apr 2014 From: USA Posts: 10 Thanks: 8 
Thank you, agentredlum! That makes a lot more sense.

April 6th, 2014, 05:42 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,301 Thanks: 1023 
In Bank statement format, this is what it'll look like: Code: QTR INTEREST BALANCE 0 .00 2475.00 1 37.12 2512.12 2 ***37.68 2549.80 .... 39 65.37 4423.35 40 66.35 4489.70 : agrees with what you show using formula The 6% is the annual rate, paid quarterly at .06/4 = .015 (1.5 %) Comes out to simply multiplying the account balances by 6% (.06) then DIVIDING by 4 to make it 1/4 of year. If you want to be my Banker and MULTIPLY by 4 instead, I'll move all my money to your Bank Hope that helps you see WHY this is used: 2475 (1+0.06/4)^4(10). Can be simplified to: 2475 (1.015)^40, which can be "read" as: "future value of 2475 at 1.5% per quarter, for 40 quarters" 
April 6th, 2014, 09:31 PM  #5 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Whoops!! I made a big mistake! Now i can't edit it because it seems time expired. My arithmetic is wrong , sorry about that , I'm giving you 6% interest compounded every quarter , which is clearly insane! Replace 0.06 with 0.015 in my arithmetic to correct my error please. My good friend Denis is correct. ... must be losing my mind ... Last edited by agentredlum; April 6th, 2014 at 09:34 PM. 
April 6th, 2014, 09:56 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,301 Thanks: 1023  
April 7th, 2014, 06:04 AM  #7 
Newbie Joined: Apr 2014 From: USA Posts: 10 Thanks: 8 
It's no problem agentredlum! I still appreciate your answer =). Thank you Denis! The bank statement format is really interesting to see. And don't worry, as soon as I open my own bank you'll be the first to know. First ten customers get a million dollars free! With 6% interest compounded quarterly, of course. I do have another question about this problem. It's probably really obvious, but, you know, not to me. I understand how to get the answer to the problem and I understand what most numbers in the problem mean. The 2475 is the initial investment, the 0.06/4 is the 6% and the four quarters you divide it by, etc. What I don't understand is the 1, as in 1+0.06/4. Is it just there because it makes solving the problem work? I'm not suggesting it shouldn't be there, obviously it does make the problem work, but every other number seems to actually show up in the word problem, and I'm not seeing where the 1 comes in. I really don't know if that makes sense at all, but thanks in advance to anyone who answers it. 
April 7th, 2014, 06:36 AM  #8  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote:
Deposit $1000 with 6%, gives ($1000)(1 + 0.06) If you remove the 1, then after a year, you have only $60 in your account.  
April 7th, 2014, 06:53 AM  #9 
Newbie Joined: Apr 2014 From: USA Posts: 10 Thanks: 8 
Thanks Pero. Then it's just a placeholder for the 2475? I mean, the money you multiply the interest by? Yeah, you'd probably be better off not banking with me. Although, if you were one of the first ten you'd still get the one million. One million + 60 is not so bad. 
April 7th, 2014, 11:05 AM  #10 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
The 1 comes from factoring the expanded simplified version. LET $A = $ Amount in the account after each compounding period $P = $ Principal investment (the money you give to the bank at the start) $ i = $ interest per year $ n = $ number of compounding periods per year. $k = $ the total number of compounding for the duration of your investment with the bank After the 1st compounding period , $k = 1 $ $$A = P + \frac{i}{n}P$$ Factor $$A = P(1 + \frac{i}{n})$$ After the 2nd compounding period , $k = 2 $ $$A = P(1 + \frac{i}{n}) + \frac{i}{n} P(1 + \frac{i}{n})$$ Factor $$A = P(1 + \frac{i}{n})[1 + \frac{i}{n}]$$ $$A = P(1 + \frac{i}{n})^2 $$ Note that $k = 2$ has shown up as the exponent After the 3rd compounding period , $k = 3$ $$A = P(1 + \frac{i}{n})^2 + \frac{i}{n} P(1 + \frac{i}{n})^2 $$ Factor $$A = P(1 + \frac{i}{n})^2 [1 + \frac{i}{n}] $$ $$A = P(1 + \frac{i}{n})^3 $$ Note that $k = 3$ has shown up as the exponent . . . After the kth compounding period , k = k $$A = P(1 + \frac{i}{n})^k $$ We expect k to show up as the exponent. In your book the author(s) represent $k = 4(10)$ which is $$ \frac{number \ of \ compoundings}{year} \cdot (number \ of \ years) $$ 

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