Question

A banker invested money in two investments. The first investment returned 4% simple interest. The second investment returned 11% simple interest. If the second investment had $340.00 more money than the first, and the total interest for both investments was $142.40, find the amount invested in each investment.

Answer 1

The rate of simple interest for the first investment, R1=4%.

The rate of simple interest for the second investment, R2=11%.

The total interest for both investments, SI= $142.40.

Let P1 be the amount invested in the first investment and P2 be the amount invested in the second investment.

Since the second investment had $340.00 more money than the first,

we can write

`P2=P1+340`

Now, the total interest for both investments can be expressed as,

`SI=(R1)/(100)* P1+(R2)/(100)* P2`

Put the avlues in the above equation.

`\begin{gathered} 142.4=(4)/(100)* P1+(11)/(100)*(P1+340) \\ 142.4*100=4P1+11*(P1+340) \\ 14240=4P1+11P1+11*340 \\ 14240=15P1+3740_{} \\ 15P1=14240-3740 \\ 15P1=10500 \\ P1=(10500)/(15) \\ P1=700 \end{gathered}`

Now, P2 can be found as,

`\begin{gathered} P2=P1+340 \\ =700+340 \\ =1040 \end{gathered}`

Therefore, the amount invested in the first investment is $700 and the amount invested in the second investment is $1040.