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Two investments totaling $36,000$⁢36,000 produce an annual income of $2715$⁢2715.  One investment yields 6%6% per year, while the other yields 9%9% per year. How much is invested at each rate?

Question

Two investments totaling 36,00036,000⁢36,000 produce an annual income of 27152715⁢2715.  One investment yields 6%6% per year, while the other yields 9%9% per year. How much is invested at each rate?

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Solution

Let's denote the amount invested at 6% as x and the amount invested at 9% as y.

From the problem, we know two things:

  1. The total amount invested is $36,000. So, we can write this as an equation: x + y = 36,000.

  2. The total annual income from both investments is $2715. The income from the first investment is 6% of x and the income from the second investment is 9% of y. We can write this as another equation: 0.06x + 0.09y = 2715.

Now we have a system of two equations, and we can solve it step by step:

Step 1: Multiply the first equation by 0.06 to make the coefficients of x in both equations the same:

0.06x + 0.06y = 2160.

Step 2: Subtract the new equation from the second equation:

0.03y = 555.

Step 3: Solve for y by dividing both sides by 0.03:

y = 18500.

Step 4: Substitute y = 18500 into the first equation:

x + 18500 = 36000,

x = 17500.

So, 17,500isinvestedat617,500 is invested at 6% and 18,500 is invested at 9%.

This problem has been solved

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