Question

A total of $24,000 was invested for one

year, part at 9% and the rest at 12%.
At the end of the year, $2,610 was
collected in interest. How much was
invested at each rate?

Answer 1

A total of $9,000 was invested at 9% and $15,000 was invested at 12%.

Step-by-step explanation:

Let's assume that the amount invested at 9% is x, and the amount invested at 12% is $24,000 - x.

Now, we can set up the equation: 0.09x + 0.12($24,000 - x) = $2,610

Simplifying, we get: 0.09x + $2,880 - 0.12x = $2,610

Combining like terms, we have: 0.03x = $2,610 - $2,880

Solving for x, we get: x = ($2,610 - $2,880) / 0.03

Plugging in the values, we find that x = $9,000

Therefore, $9,000 was invested at 9% and $15,000 was invested at 12%.

Answer 2

$9000 at 9% and $15000 at 12%

Step-by-step explanation:

x = amount at 9%; y = amount at 12%

x + y = 24000

0.09x + 0.12y = 2610

0.09x + 0.09y = 2160

0.09x + 0.12y = 2610

0.03y = 450

y = 15,000

x + y = 24000

x + 15000 = 24000

x = 9000