Answered

$6949 is invested part at 11% and the rest at 8% if the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 8% by $224.60 how much is invested at each rate

Answer :

Let x be the part that is invested at 11% and let y be the part that is invested at 8%.

We have that the total amount invested is $6949, then, we have the following equation:

[tex]x+y=6949[/tex]

next, we have that the amount x exceeds the amount y by 224.60, then, we have the following:

[tex]0.11x-0.08y=224.60[/tex]

if we solve for x the first equation, we have:

[tex]x=6949-y[/tex]

using this expression on the second equation, we get the following:

[tex]\begin{gathered} 0.11(6949-y)=224.60 \\ \Rightarrow764.39-0.11y-0.08y=224.60 \\ \Rightarrow-0.19y=224.60-764.39=-539.79 \\ \Rightarrow y=\frac{-539.79}{-0.19}=2841 \\ y=2841 \end{gathered}[/tex]

now that we know that y = 2841, we can use this value to find the value of x:

[tex]x=6949-y=6949-2841=4108[/tex]

therefore, $4108 was invested at 11% and $2841 was invested at 8%

Other Questions