Question

A person plans to invest four times as much in an account at 5.5% interest as in an account at 14% annual interest. How much should the person invest in each account to earn a total of $550 in one year?

Answer 1

Account A earns at a 5.5% interest rate

Account B earns at a 14% interest rate

We want to invest 4 times as much in account A compared to account B

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Let x be the amount invested in account B. So the amount invested in account A is 4x.

Use the simple interest formula, i = P*r*t, to find that

interest earned with account A = 4x*0.055*1 = 0.22x

interest earned with account B = x*0.14*1 = 0.14x

Add these up and set this equal to 550. Solve for x. Then compute 4x.

0.22x+0.14x = 550

0.36x = 550

x = 550/0.36

x = 1527.77777777778

x = 1527.78 round to the nearest penny

4x = 4*1527.78

4x = 6111.12

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Answers:

6111.12 dollars needs to be invested at 5.5%

1527.78 dollars needs to be invested at 14%

Answer 2

x = amount invested at 14%

4x = amount invested at 5.5%, 4 times as "x".

`\bf \begin{array} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hfill \stackrel{\textit{14\% of x}}{\left( \cfrac{14}{100} \right)x}\implies 0.14x~\hfill \stackrel{\textit{5.5\% of 4 times x}}{\left( \cfrac{5.5}{100} \right)4x}\implies 0.22x \\\\[-0.35em] ~\dotfill`

`\bf 0.14x+0.22x~~=~~\stackrel{\textit{total annual yield}}{550}\implies 0.36x=550 \\\\\\ x=\cfrac{550}{0.36}\implies \stackrel{\textit{invested at 14\%}}{x=1527.\overline{7}}~\hfill \stackrel{\textit{invested at 5.5\%}}{4(1527.\overline{7})\approx 6111.11}`