Pat invested a total of256, how much did Pat invest at 10 percent and how much at 8 percent

Pat invested a total of
256, how much did Pat invest at 10 percent and how much at 8 percent?

The Correct Answer and Explanation is:

Answer:

Let the amount Pat invested at 10% be x dollars.
Then, the amount invested at 8% is (256 – x) dollars.

Assume the interest earned from both investments is the same.

Then:

Interest from 10% investment = Interest from 8% investment
0.10x=0.08(256−x)0.10x = 0.08(256 – x)0.10x=0.08(256−x)

Solve the equation:0.10x=0.08(256−x)0.10x = 0.08(256 – x)0.10x=0.08(256−x)0.10x=20.48−0.08×0.10x = 20.48 – 0.08×0.10x=20.48−0.08×0.10x+0.08x=20.480.10x + 0.08x = 20.480.10x+0.08x=20.480.18x=20.480.18x = 20.480.18x=20.48x=20.480.18=113.78 (rounded to two decimal places)x = \frac{20.48}{0.18} = 113.78 \ (\text{rounded to two decimal places})x=0.1820.48​=113.78 (rounded to two decimal places)

Therefore:

  • Pat invested $113.78 at 10%
  • Pat invested $142.22 at 8% (since 256 − 113.78 = 142.22)

Explanation

To solve a problem involving two parts of an investment at different interest rates, the key is to set up a variable to represent one part of the investment and express the other part in terms of that same variable. In this case, the total amount Pat invested is $256. Suppose the amount invested at 10% interest is x dollars. Then the remaining amount, which is (256 – x) dollars, must have been invested at 8% interest.

The problem assumes that the interest earned from both investments is equal. To express this, we use the formula for simple interest:
Interest = Principal × Rate.

The interest from the 10% investment is 0.10 × x, and the interest from the 8% investment is 0.08 × (256 − x). Setting these equal gives a linear equation, which can be solved by isolating the variable. After solving, it turns out Pat invested $113.78 at 10% and the rest, $142.22, at 8%.

This method of breaking a total into parts using a variable, then setting up an equation based on equal outcomes (like equal interest), is a common strategy in algebra. It allows for a logical way to handle mixture and investment problems and to find how much money was allocated to each interest rate, even when the total and conditions are all that’s known.

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