$2,637 is invested, part at 13% and the rest at 11%. If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 11% by $65.61, how much is invested at each rate? (Round to two decimal places if necessary.)


Respuesta :

x = total amount invested.

we know 2637 was invested on one part, and the rest on the other, well, if the total amount is "x", then the rest is simply just the slack from 2637 and "x" or namely x - 2637.

[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{13\% of 2637}}{\left( \cfrac{13}{100} \right)2637}~\hfill \stackrel{\textit{11\% of x-2637}}{\left( \cfrac{11}{100} \right)(x-2637)}[/tex]

we also know that, if we subtract the 11% amount from the 13% amount, we'd end up with 65.61, so then

[tex]\bf \stackrel{\textit{13\% of 2637}}{\left( \cfrac{13}{100} \right)2637}~~-~~ \stackrel{\textit{11\% of x-2637}}{\left( \cfrac{11}{100} \right)(x-2637)}~~=~~65.61 \\\\\\ 342.81 - 0.11x+290.07 = 65.61\implies 632.88-0.11x = 65.61 \\\\\\ 632.88 = 65.61+0.11x\implies 567.27 = 0.11x\\\\\\ \cfrac{567.27}{0.11}=x\implies 5157=x \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf \stackrel{\textit{13\% of 2637}}{\left( \cfrac{13}{100} \right)2637}\implies 342.81~\hfill \stackrel{\textit{11\% of 5157-2637}}{\left( \cfrac{11}{100} \right)2520}\implies 277.2[/tex]