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An investment club placed a portion of its funds in a 9% annual simple interest account and the remainder in an 8% annual simple interest account. the amount of interest earned for one year was $810. if the amounts placed in each account had been reversed, the interest earned would have been $805. how much was invested in each account?

a. amount at 9%
b. $ amount at 8%

User Egaga
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1 Answer

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Final answer:

The investment club placed $5,000 at 9% and $4,500 at 8%. This was determined by solving a system of linear equations representing the total interest earned from each account and with amounts reversed in each account.

Step-by-step explanation:

The question relates to solving a system of linear equations derived from the information about the investments made by an investment club in two different simple interest accounts.

Let's denote the amount invested at 9% as x and the amount invested at 8% as y.

The total interest earned for one year is $810, which gives us the first equation:

0.09x + 0.08y = 810

If the amounts placed in each account had been reversed, the interest earned would have been $805. This gives us the second equation:

0.08x + 0.09y = 805

Solving these two equations simultaneously will give the amounts placed in each account at 9% and 8% respectively.

Let's multiply the first equation by 9 and the second equation by 8 to eliminate the decimals and simplify:

0.81x + 0.72y = 7290

0.64x + 0.72y = 6440

Subtracting the second equation from the first yields:

0.17x = 850

Divide both sides by 0.17 to find x:

x = 5000

Substitute x back into one of the original equations to solve for y:

0.09(5000) + 0.08y = 810

450 + 0.08y = 810

0.08y = 810 - 450

0.08y = 360

Divide both sides by 0.08 to find y:

y = 4500

Therefore, the investment club placed $5,000 at 9% and $4,500 at 8%.

User Plamen Nikolov
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