Question

Two investments totaling $56,000 produce an annual income of $1400. One investment yields 4% per year, while the other yields 2% per year. How much is invested at each rate?

Answer 1

Let x represent one of the investments

Since the total of the two investments is $56,000, the other investment will be

`=56000-x`

If x is invested at 4% and the other is invested at 2%, then, it can be expressed as

`\begin{gathered} 4\%\text{ of x}=(4)/(100)* x=0.04x \\ 2\%\text{ of \lparen56000-x\rparen}=(2)/(100)*(56000-x)=0.02(56000-x)=1120-0.02x \end{gathered}`

And, their income is $1400, then, it can be expressed as

`0.04x+(1120-0.02x)=1400`

Solve for x

`\begin{gathered} 0.04x+(1120-0.02x)=1400 \\ 0.04x+1120-0.02x=1400 \\ Collect\text{ like terms} \\ 0.04x-0.02x=1400-1120 \\ 0.02x=280 \\ Divide\text{ both sides by 0.02} \\ (0.02x)/(0.02)=(280)/(0.02) \\ x=\text{\$14000} \end{gathered}`

The other investment will be

`=56000-14000=\text{\$42000}`

Hence,

At a 4% rate, the investment is $14000 and at a 2% rate, the investment is $42000