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boyakko [2]
3 years ago
15

Cesar invests $8200 in two different accounts. The first account paid 13 %, the second account paid 2 % in interest. At the end

of the first year he had earned $461 in interest. How much was in each account?
Mathematics
1 answer:
Ugo [173]3 years ago
3 0

a = amount invested at 13%

b = amount invested at 2%

we know the total amount invested is 8200, so if the first amount is say "a", the second amount must be the slack left from subtracting "a", namely b = 8200 - a.

we also know 13% of "a" plus 2% of "b" yielded an interest of $461 in a year.

\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 a}}{\left( \cfrac{13}{100}\right)a\implies 0.13a}~\hfill \stackrel{\textit{2\% of b}}{\left( \cfrac{2}{100} \right)b\implies 0.02b} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{sum of the percentages}}{0.13a+0.02b}~~=~~\stackrel{\textit{yield for 1 year}}{461}\implies 0.13a+0.02\left(\boxed{8200-a} \right) = 461 \\\\\\ 0.13a+164-0.02a=461\implies 0.11a+164=461\implies 0.11a=297 \\\\\\ a = \cfrac{297}{0.11}\implies \blacktriangleright a = 2700 \blacktriangleleft ~\hfill b = 8200-2700\implies \blacktriangleright b = 5500 \blacktriangleleft

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