Question

A total of $7000 is invested: part at 6% and the remainder at 10 %. How much is invested at each rate if the annual Interest is $520?

Answer 1

`\begin{gathered} I=(prt)/(100) \\ \end{gathered}`

total interest = $520

Therefore,

let

x = amount invested in 6%

y = amount invested in 10%

`\begin{gathered} x+y=7000 \\ y=7000-x \end{gathered}`

Interest total

`\begin{gathered} 520=(x*6*1)/(100)+(y*10*1)/(100) \\ 520=0.06x+0.1y \end{gathered}`
`\begin{gathered} 520=0.06x+0.1(7000-x) \\ 520=0.06x+700-0.1x \\ 520-700=0.06x-0.1x \\ -180=-0.04x \\ x=(-180)/(-0.04) \\ x=\text{ \$4500} \end{gathered}`

y = 7000 - 4500 = $2500

$4500 was invested at a 6% rate

$2500 was invested at a 10% rate