Question

You invested $14,000 in two accounts paying 3% and 6% annual interest, respectively. If the total interest earned for the year was 5570, how much was invested at each rate?

Answer 1

Let x be the amount invested at 3%

Then, (14,000-x) will be the amount invested at 6%.

Interest earned at 3% will be:

`\begin{gathered} I=(PRT)/(100) \\ P\colon\text{Principal} \\ R\colon\text{Rate} \\ T\colon\text{Time} \\ I=(x*3*1)/(100) \\ I=(3x)/(100)=0.03x \end{gathered}`

Interest earned at 6% will be:

`\begin{gathered} I=\frac{\text{PRT}}{100} \\ I=((14,000-x)*6*1)/(100) \\ I=0.06(14,000-x) \end{gathered}`

Total interest earned is given to $570, thus we have:

`\begin{gathered} 0.03x+0.06(14000-x)=570 \\ 0.03x+840-0.06x=570 \\ 0.03x-0.06x=570-840 \\ -0.03x=-270 \\ x=(-270)/(-0.03) \\ x=\text{ \$9,000} \end{gathered}`

Amount invested at 6% will be 14,000 - 9,000 = $5,000.

Hence, the amount invested at 3% and 6% are $9,000 and $5,000 respectively