Respuesta :
let's recall that I = Prt
so... hmm we're assuming here, it doesn't say there, but this is for 1 year period.
ok... so... let's use two quantities for such interest, say "a" is the Interest for the Principal of 6000, and let's use "b" is the Interest for the Principal of 13000.
[tex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$6000\\ r=rate\to r\%\to& \frac{r}{100}\\ t=years\to &1 \end{cases} \\\\\\ a=6000\left( \frac{r}{100} \right)1\implies \stackrel{\textit{P=6000}}{a=60r}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$13000\\ r=rate\to (r-1)\%\to &\frac{r-1}{100}\\ t=years\to &1 \end{cases} \\\\\\ b=13000\left( \frac{r-1}{100} \right)1\implies \stackrel{\textit{P=13000}}{b=130(r-1)}[/tex]
now, we know that, whatever the interest amount of "a" is, is 220 less than whatever "b" is.
we know what "b" is, b = 130(r-1), and 220 less than that is just b - 220, or 130(r-1) - 220.
[tex]\bf a=60r\qquad \boxed{a=130(r-1) - 220}\implies 130(r-1)-220=60r \\\\\\ 130r-130-220=60r\implies 70r=350\implies r=\cfrac{350}{70}\implies r=5[/tex]
so... hmm we're assuming here, it doesn't say there, but this is for 1 year period.
ok... so... let's use two quantities for such interest, say "a" is the Interest for the Principal of 6000, and let's use "b" is the Interest for the Principal of 13000.
[tex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$6000\\ r=rate\to r\%\to& \frac{r}{100}\\ t=years\to &1 \end{cases} \\\\\\ a=6000\left( \frac{r}{100} \right)1\implies \stackrel{\textit{P=6000}}{a=60r}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$13000\\ r=rate\to (r-1)\%\to &\frac{r-1}{100}\\ t=years\to &1 \end{cases} \\\\\\ b=13000\left( \frac{r-1}{100} \right)1\implies \stackrel{\textit{P=13000}}{b=130(r-1)}[/tex]
now, we know that, whatever the interest amount of "a" is, is 220 less than whatever "b" is.
we know what "b" is, b = 130(r-1), and 220 less than that is just b - 220, or 130(r-1) - 220.
[tex]\bf a=60r\qquad \boxed{a=130(r-1) - 220}\implies 130(r-1)-220=60r \\\\\\ 130r-130-220=60r\implies 70r=350\implies r=\cfrac{350}{70}\implies r=5[/tex]
