Question

A total of $6000 is invested: part at 5% and the remainder at 10%. How much is invested at each rate if the annualinterest is $5902

Answer 1

Given:

Principal amount = $6000.

Let x be the amount invested at 5%.

So, 6000-x be the amount invested at 10%.

The annual interest is $590

The equation is written as,

`\begin{gathered} 5\text{ \%x+(6000-x)10 \%=590} \\ (5)/(100)x+(6000-x)(10)/(100)=590 \\ 0.05x+(6000-x)0.1=590 \\ 0.05x+600-0.1x=590 \\ -0.05x=-10 \\ x=(10)/(0.05) \\ x=200 \\ 6000-x=6000-200=5800 \end{gathered}`

Answer: At 5%, the amount invested is $200, and at 10%, the amount invested is $5800.