Heading towards semester exams, some students face questions like this one. The tutor works this example.

Consider the following problem:

Sue has $2200 to invest for one year.  She has a choice two investments:  investment A pays 4%, while investment B pays 5.4%.  She wants to earn $100 interest for the year. How much should she invest in each choice?

Solution:

Let x=amount of money she puts in A (at 4%)

Let y=amount of money she puts in B (at 5.4%)

Now, we know that

x + y = 2200 (equation 1)

To get the percent of a number, you change the percent into a decimal (by dividing by 100), then multiply.  For instance, 29% of $246=0.29($246)=$71.34.

Therefore, 4% of x is 0.04x and 5.4% of y is 0.054y.  We have

0.04x + 0.054y = 100 (equation 2)

We can solve this system of two equations using substitution:  notice that equation 1 implies

y = 2200 – x

We can substitute 2200 – x, instead of y, in equation 2:

0.04x + 0.054(2200 – x) = 100

Now we simplify and solve:

0.04x + 118.8 – 0.054x = 100

Subtract 118.8 from both sides:

0.04x – 0.054x = -18.8

Simplify the left:

-0.014x = -18.8

Divide both sides by -0.014:

x = -18.8/-0.014 = 1342.86

Now, looking back at equation 1

x + y = 2200

we see that

1342.86 + y = 2200

Subtracting 1342.86 from both sides gives

y = 857.14

Looking back at our let statements, we see that

x = amount invested at 4%

y = amount invested at 5.4%

We now know that she should invest $1342.86 at 4% and $857.14 at 5.4%.

Check:  1342.86(0.04) + 857.14(0.054) = 53.71 + 46.29 = 100

From a high school math point of view, this question is settled.

Questions arise about the investment aspect of this problem.  However, they belong to a different discipline.  We might explore some of those questions in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.