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"Investment" Word Problems

"Investment" Word Problems

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 amount of the original investment (called the "principal"), r is the interest rate (expressed in decimal form), and t is the time.

For annual interest, the time t must be in years. If they give you a time of, say, nine months, you must first convert this to 9/12 = 3/4 = 0.75 years. Otherwise, you'll get the wrong answer. The time units must match the interest-rate units. If you got a loan from your friendly neighborhood loan shark, where the interest rate is monthly, rather than yearly, then your time must be measured in terms of months.

Investment word problems are not generally terribly realistic; in "real life", interest is pretty much always compounded somehow, and investments are not generally all for whole numbers of years. 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 then solve for whatever is left.

* You put $1000 into an investment yielding 6% annual interest; you left the money in for two years. How much interest do you get at the end of those two years?

In this case, P = $1000, r = 0.06 (because I have to convert the percent to decimal form), and the time is t = 2. Substituting, I get:

I = (1000)(0.06)(2) = 120

I will get $120 in interest.

Another example would be: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

* You invested $500 and received $650 after three years. What had been the interest rate?

For this exercise, I first need to find the amount of the interest. Since interest is added to the principal, and since P = $500, then I = $650 – 500 = $150. The time is t = 3. Substituting all of these values into the simple-interest formula, I get:

150 = (500)(r)(3)
150 = 1500r
150/1500 = r = 0.10

Of course, I need to remember to convert this decimal to a percentage.

I was getting 10% interest.

The hard part comes when the exercises involve multiple investments. But there is a trick to these that makes them fairly easy to handle.

* You have $50,000 to invest, and two funds that you'd like to invest in. The You-Risk-It Fund (Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. Because of college financial-aid implications, you don't think you can afford to earn more than $4,500 in interest income this year. How much should you put in each fund?"

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:

I P r t
Fund X ? ? 0.06 1
Fund Y ? ? 0.14 1
total 4,500 50,000 --- ---

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 I can solve x + y = 50,000 to get 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, so make sure you understand it now.

I P r t
Fund X ? x 0.06 1
Fund Y ? 50,000 – x 0.14 1
total 4,500 50,000 --- ---

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 (right to left) and fill in the "interest" column.
I P r t
Fund X 0.06x x 0.06 1
Fund Y 0.14(50,000 – x) 50,000 – x 0.14 1
total 4,500 50,000 --- ---

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 set this sum equal to the given total interest:

0.06x + 0.14(50,000 – x) = 4,500
0.06x + 7,000 – 0.14x = 4,500
7,000 – 0.08x = 4,500
–0.08x = –2,500
x = 31,250

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 exercises.



"Investment" Word Problems: Examples

If you set up your investment word problems so everything is labeled and well-organized, they should all work out fairly easily. Just take your time and do things in an orderly fashion. I've done the set-up (but not the complete solutions) for a few more examples:

* An investment of $3,000 is made at an annual simple interest rate of 5%. How much additional money must be invested at an annual simple interest rate of 9% so that the total annual interest earned is 7.5% of the total investment?

I P r t
first (3,000)(0.05) = 150 3,000 0.05 1
additional 0.09 x x 0.09 1
total (3,000 + x)(0.075) 3,000 + x 0.075 1

First I fill in the P, r, and t columns with the given values.

Then I multiply across the rows (from the right to the left) in order to fill in the I column.

Then add down the I column to get the equation 150 + 0.09 x = (3,000 + x)(0.075).

To find the solution, I would solve for the value of x.

* A total of $6,000 is invested into two simple interest accounts. The annual simple interest rate on one account is 9%; on the second account, the annual simple interest rate is 6%. How much should be invested in each account so that both accounts earn the same amount of annual interest?

I P r t
9% account 0.09x x 0.09 1
6% account (6,000 – x)(0.06) 6,000 – x 0.06 1
total --- 6,000 --- ---

In this problem, I don't actually need the "total" row at all.

First I'll fill in the P, r, and t columns, and multiply to the left to fill in the I column.

From the interest column, I then get the equation 0.09x = ($6,000 – x)(0.06), because the yields are required to be equal.

Then I'd solve for the value of x, and back-solve to find the value invested in the 6% account.

(This exercise's set-up used that "how much is left" construction, mentioned earlier.)

* An investor deposited an amount of money into a high-yield mutual fund that returns a 9% annual simple interest rate. A second deposit, $2,500 more than the first, was placed in a certificate of deposit that returns a 5% annual simple interest rate. The total interest earned on both investments for one year was $475. How much money was deposited in the mutual fund?

The amount invested in the CD is defined in terms of the amount invested in the mutual fund,so I will let "x" be the amount invested in the mutual fund.

I P r t
mutual fund 0.09x x 0.09 1
cert. of deposit (x + 2,500)(0.05) x + $2,500 0.05 1
total 475 2x + $2,500 --- ---

In this problem, I don't actually need the "total" for the "rate" or "time" columns.

First I'll fill in the P, r, and t columns, multiplying to the left to fill in the I column.

Then I'll add down the I column to get the equation 0.09x + (x + 2,500)(0.05) = 475.

Then I'd solve for the value of x. Copyright © Elizabeth Stapel 1999-2010 All Rights Reserved

* The manager of a mutual fund placed 30% of the fund's available cash in a 6% simple interest account, 25% in 8% corporate bonds, and the remainder in a money market fund that earns 7.5% annual simple interest. The total annual interest from the investments was $35,875. What was the total amount invested?

For this problem, I'll let "x" stand for the total amount invested.

I P r t
6% account (0.30x)(0.06) = 0.018x 0.30x 0.06 1
8% bonds (0.25x)(0.08) = 0.02x 0.25x 0.08 1
7.5% fund (0.45x)(0.075) = 0.03375x 0.45x 0.075 1
total $35,875 x --- ---

Once 30% and 25% was accounted for in the 6% and 8% accounts, then there is 100% – 30% – 25% = 45% left for the third account. I can use this information to fill in the "principal" column.

Then I'll fill out the "rate" and "time" columns, and multiply to the left to fill in the "interest" column.

From the interest column, I get the equation 0.018x + 0.02x + 0.03375x = 35,875.

Then I'd solve for the value of x.