Question

$18,389 is invested, part at 10 % and the rest at 8 % . If the interest earned from the amount invested at 10 % exceeds the interest earned from the amount invested at 8% by $621.02, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

Approximately $11,623.00 is invested at 10% interest, and approximately $6,766.00 is invested at 8% interest.

Step-by-step explanation:

Let's assume that the amount invested at 10% interest is x, and the amount invested at 8% interest is 18,389 - x.

The interest earned from the amount invested at 10% is 0.1x, and the interest earned from the amount invested at 8% is 0.08(18,389 - x).

According to the given information, the interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 8% by $621.02:

0.1x - 0.08(18,389 - x) = 621.02

Simplifying the equation, we get:

0.1x - 0.08(18,389) + 0.08x = 621.02

0.1x - 1,471.12 + 0.08x = 621.02

0.18x = 2,092.14

x = 2,092.14 / 0.18

x ≈ 11,623.00

Therefore, approximately $11,623.00 is invested at 10% interest, and approximately $6,766.00 is invested at 8% interest.

Answer 2

Let x be the amount invested in 10% interest. So, the amount invested in 8% interest will be 18389-x.

Express the formula for interset.

I=Prt.

Here, P is the principal amount, r is the rate of interest and t is the time.

Calculate the interest earned by investing x amount.

`I_1=x*0.1`

Calculate the interest earned by investing 18389-x amount.

`I_2=(18389-x)*0.8`

Given that, I1-I2=621.02

Therefore,

`\begin{gathered} \text{0}.1x-((18389-x)*0.8)=621.02 \\ 0.1x-14711.2+0.8x=621.02 \\ 0.9x=15332.22 \\ x=17035.8 \end{gathered}`

Therefore, the invesment under 10% is 17035.80 dollars

Then, the invesment under 8% is 1353.80 dollars.