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

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.

$I_1=x\times0.1$

Calculate the interest earned by investing 18389-x amount.

$I_2=(18389-x)\times0.8$

Given that, I1-I2=621.02

Therefore,

$\begin{gathered} \text{0}.1x-((18389-x)\times0.8)=621.02 \\ 0.1x-14711.2+0.8x=621.02 \\ 0.9x=15332.22 \\ x=17035.8 \end{gathered}$

Therefore, the invesment under 10% is 17035.80 dollars

Then, the invesment under 8% is 1353.80 dollars.