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"Investment" Word Problems (page 1 of 2) Investment problems usually involve simple annual interest (as opposed to compounded interest), using the interest formula I = Prt, where I stands for the interest on the original investment, P stands for the original investment (called "principal"), r is the interest rate, and t is time. (Note that, for annual, that is to say, yearly, interest, the time t must be in years. That is, if they give you a time of, say, nine months, convert this to 9/12 = 3/4 = 0.75 years. Otherwise, you'll get the wrong answer. The same requirement holds true for other time units. If it's a monthly interest rate from the local loan shark, time must be in terms of months; etc. Note also that these problems are not terribly realistic; in "real life", interest is pretty much always compounded somehow, and investments are not generally all for just one year. But you'll get to more "practical" stuff later; this is just warm-up, to prepare you for later.) In all cases of these problems, you will want to substitute all known information into the "I = Prt" equation, and solve for whatever is left. For simple problems, this involves something like this:
In this case, P = $1000, r = 0.06 (remember to convert from a percent to a number by moving that decimal point!), and t = 2. Substituting, I get: I = (1000)(0.06)(2) = 120 I get $120 in interest. Another example would be: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
For this problem, I have P = $500, I = $650 – 500 = $150, and t = 3. Substituting, I get: 150 = (500)(r)(3)
Remember to convert this decimal to a percentage. I was getting 10% interest. The hard part comes when the problems involve multiple investments. But there is a trick to these that makes them do-able.
The problem here comes from the fact that I'm splitting that $50,000 in principal into two smaller amounts. Here's how to handle this:
How do I fill in for those question marks? I'll start with the principal P. Let's say that I put "x" dollars into Fund X, and "y" dollars into Fund Y. Then x + y = 50,000. This doesn't help much, since I only know how to solve equations in one variable. But then I notice that, if x + y = 50,000, then y = $50,000 – x. THIS TECHNIQUE IS IMPORTANT! The amount in Fund Y is (the total) less (what we've already accounted for) in Fund X, or 50,000 – x. You will need this technique, this "how much is left" construction, in the future!
Now I will show you why I set up the table like this. By organizing the columns according to the interest formula, I can now multiply across and fill in the interest column.
Since the interest from Fund X and the interest from Fund Y will add up to $4,500, I can add down the interest column, and get my equation: 0.06x
+ 0.14(50,000 – x) = 4,500
Then y = 50,000 – 31,250 = 18,750. I should put $31,250 into Fund X, and $18,750 into Fund Y. Note that the answer did not involve "neat" values like "$10,000" or "$35,000". You should understand that this means that you cannot always expect to be able to use "guess-n-check" to find your answers. You really do need to know how to do these problems. Top | 1 | 2 | Return to Index Next >>
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