Question

$13,168is invested, part at 13%and the rest at

5%. If the interest earned from the amount invested at
13%exceeds the interest earned from the amount invested at
5% by $1438.24, how much is invested at each rate?

Answer 1

To find out how much was invested at each rate, set up two equations reflecting the total investment and the difference in interest earned and solve for the two unknowns. The result shows that $11659.11 was invested at 13% and $1508.89 was invested at 5%.

Step-by-step explanation:

To solve this problem, we need to set up two equations based on the information given. Let's call the amount invested at 13% interest x, and the amount invested at 5% interest y. The total investment is $13168, so we have our first equation:

x + y = 13168

The second equation is based on the interest earned. The interest from the amount invested at 13% exceeds the interest from 5% by $1438.24:

0.13x - 0.05y = 1438.24

Now, we solve this system of equations using either substitution or elimination method. Let's use the substitution method. From the first equation:y = 13168 - x. Substitute y in the second equation:

0.13x - 0.05(13168 - x) = 1438.24

Simplify this equation to find the value of x:

0.13x - 0.05 * 13168 + 0.05x = 1438.24
0.18x = 1438.24 + 0.05 * 13168
x = (1438.24 + 658.4) / 0.18
x = $11659.11

Now, calculate y:

y = 13168 - x
y = 13168 - 11659.11
y = $1508.89

Therefore, $11659.11 was invested at 13% interest, and $1508.89 was invested at 5% interest.