Question

the 8000 that he invested in fund a returned a 6% profit.the amount that he invested in fund b returned a 2% profit how much did he invest in fund b if both funds together returned a 3 % profit

Answer 1

Let

`\begin{gathered} x=\text{ the amount invested in fund b} \\ m=\text{ the profit from the fund a investment} \\ n=\text{ the profit from the fund b investment} \end{gathered}`

Therefore,

`(m)/(8000)=(6)/(100)`

Hence,

`m=(6)/(100)*8000=6*80=\text{ \$480}`
`\begin{gathered} \text{ Also} \\ n=0.02x------(1) \\ (m+n)/(8000+x)=0.03 \\ \text{therefore} \\ (n+480)/(8000+x)=0.03 \end{gathered}`

Thus

`n+480=0.03(8000+x)-----------------(2)`

Substituting equation(1) into equation(2), we have

`\begin{gathered} 0.02x+480=0.03(8000+x) \\ 0.02x+480=240+0.03x \\ \text{ Thus} \\ 0.03x-0.02x=480-240 \\ 0.01x=240 \\ x=(240)/(0.01)=2400 \end{gathered}`

Hence, the amount he invested is $2400