Question

$13,883 is invested, part at 8% and the rest at 6%. If the interest earned from the amount invested at 8% exceeds the interest earned from the amount invested at 6% by $625.26, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

An amount of $10,423.14 was invested at an 8% interest rate, and $3,459.86 was invested at a 6% interest rate to satisfy the given conditions of total investment and difference in interest earned.

Step-by-step explanation:

To solve the problem, we'll let x be the amount invested at 8% and y be the amount invested at 6%. According to the conditions given, we have a system of two equations:

• x + y = $13,883
• 0.08x - 0.06y = $625.26

We can solve the first equation for y by subtracting x from both sides, yielding:

• y = $13,883 - x

We can then substitute this expression for y into the second equation:

• 0.08x - 0.06($13,883 - x) = $625.26

After simplifying, we get:

• 0.08x - $833.98 + 0.06x = $625.26
• 0.14x = $1,459.24
• x = $10,423.14

Now, using the first equation again with the value for x, we find y:

• y = $13,883 - $10,423.14
• y = $3,459.86

Therefore, $10,423.14 was invested at 8% and $3,459.86 was invested at 6%.