Expert: Josh - 11/12/2008

Question
QUESTION: Hello:

An investor invested $30,000 for one year at simple interest; one portion was invested a 5% and the remainder at 2%.  How much money did he invest at each rate if the total return was at the rate of 4% or $1,200?

Answers: $20,000 at 5% and $10,000 at 2%

5% X $20,000 + 2% X $10,000 = $1,200

My question is why does the above have only the following solutions, that is, $20,000 and $10,00?

For example: 4% X a + 2% X b = 26, a and b can represent numerous numbers or fractions or whole numbers and fractions.

I thank you for your reply.

ANSWER: Hi Kenneth,

I am not sure how you pulled the numbers and the relationship of "4% X a + 2% X b = 26" with the original question.

The problem has a unique solution, because we have to find (two amounts) X and Y that satisfy 0.05*X+0.02Y=0.04*(X+Y) ....[Equation 1] subject to the constraint X+Y=30000 ....[Equation 2].

We have 2 equations for 2 unknowns. Since the two equations are linearly independent, there can only be one set of solutions.

From [2], Y=30000-X.
Substituting this in [1],
5X + 2*(30000-X)=4*(X+30000-X)
3X + 60000 = 120000
Thus, we get X=20000, Y=10000.

---------- FOLLOW-UP ----------

QUESTION: Hello Josh:

I'll explain my first question as follows:

(1) 2% X 5 = 0.1 + 5% X 10 = 0.5 = 0.1 + 0.5 = 0.6

(2) 2% X 8 = 0.16 + 5% X 8.8 = 0.16 + 0.44 = 0.6

The percentages remain the same, but the factors are different in the above calculations, but the product remains the same, 0.6.

In the investor question, the calculation is similar.

5% X $20,000 + 2% X $10,000 = $1,200

I think that I know the reason why the above investor calculation only has one solution. The factors of $20,000 and $10,000 must equal $30,000.

In the other calculations, the factors 5 and 10 equal 15 and 8 and 8.8 equal 16.8.
If in (1) the two factors had to equal 15, then only 5 and 10 can be used.

Is this correct?

I thank you for your follow-up reply.  

Answer
Regarding your explanation, they make sense if you delete "2% x 5" and "2% x 8" from (1) and (2) respectively.

This equation has the form a*X+b*Y=c, where a, b and c represent (known) constant values. Without further constraints, we can pick infinitely many solution pairs for the unknown quantities X and Y. In particularly, Y only has to satisfy Y=(c-aX)/b for a given value of X.

We will very likely (but not always) obtain a unique solution for X and Y if there is another condition imposed on X and Y, such as X+Y=d.

In linear algebra, this can be proven to be the case (can obtain unique pair of solution for X and Y) if the two equations are linearly independent. Intuitively, this means the condition X+Y=d must tell us something we don't already know from the first equation a*X+b*Y=c.