Respuesta :
We have to find how much is invested at each rate.
We then can construct a system of 2 equations with 2 unknowns with the information given.
Let's call x to the amount invested at 11% and y the amount invested at 10%.
We know that the total amount invested is $25,000, so we have:
[tex]x+y=25000[/tex]We also know that the interest earned from x exceeds the interest earned by y by $242.60.
The interest earned by x is equal to the interest rate times the amount, so it is 0.11*x.
The interest earned by y, in the same way, is equal to 0.10*y.
Then, we can write:
[tex]0.11x=0.10y+242.60[/tex]We can define y in function of x as:
[tex]x+y=25000\longrightarrow y=25000-x[/tex]and replace it in the second equation:
[tex]\begin{gathered} 0.11x=0.10y+242.60 \\ 0.11x=0.10(25000-x)+242.60 \end{gathered}[/tex]Then we can solve for x as:
[tex]\begin{gathered} 0.11x=0.10(25000-x)+242.60 \\ 0.11x=0.10\cdot25000-0.10x+242.60 \\ 0.11x+0.10x=2500+242.60 \\ 0.21x=2742.60 \\ x=\frac{2742.60}{0.21} \\ x=13060 \end{gathered}[/tex]and y is:
[tex]y=25000-x=25000-13060=11940[/tex]We can check the interest earned by the two investment as:
[tex]\begin{gathered} I_x=0.11\cdot x=0.11\cdot13060=1436.6 \\ I_y=0.10\cdot y=0.10\cdot11940=1194 \\ I_x-I_y=1436.60-1194=242.60 \end{gathered}[/tex]Answer:
There are $13,060 invested at 11% and $11,940 invested at 10%.
