Word problem

Question

An executive invests twenty three thousand, some at seven percent and some at six percent annual interest rates. If he receives an annual return of one thousand five hundred and fifty thousand, how much is invested at each rate?

Answer 1

So this question is testing you on whether you know how to set this up in terms of two variables: we'll call them x and y

 

The first line states that x amount is invested at 7% (or 0.07) and y amount is invested at 6% (or 0.06) annual interest. And the total amount invested is $23,000.

We can convert this information into this equation. (x + y = $23,000) because we know that both amounts must equal twenty three thousand.

Our second equation is: (0.07x + 0.06y = $1,550)

 

For solving a set of two equations, you can use either the substitution method or the addition method. 

Here, it would be easier to do substitution.

If you solve for x, 

x= $23,000 - y

substitution this into the second equation and solve for x.

 

0.07 (23,000 - y)  + 0.06y = 1,550

 

Distribute

1,610 - 0.07 y + 0.06 y = 1,550

Combine like variables

1,610 - 0.01y = 1,550

Isolate the y variable by subtracting 1,610 from both sides.

-0.01y = -60

divide by - 0.01 on both sides

y = -60/-0.01 = $6,000

 

If you plug y's value back into the original equation x + y = $23,000, you will see that x equals the rest of the amount, which is 23,000 - 6,000, which is $17,000. 

 

Thus, you can define how much was invested, and at which rates.