Respuesta :
Paul needs to invest $10000 at 6% interest and $30000 at 10% interest to make a total of 9% return on his $40000.
Explanation
Suppose, the amount of money at 6% interest is [tex]x[/tex]
As Paul has total $40000 to invest, so the amount of money at 10% interest will be: [tex](40000-x) dollar[/tex]
His intent is to earn 9% interest on his $40000 investment. So, the equation will be......
[tex]0.06x+0.10(40000-x)=0.09*40000\\ \\ 0.06x+4000-0.10x=3600\\ \\ -0.04x=3600-4000=-400\\ \\ x=\frac{-400}{-0.04}=10000[/tex]
So, Paul needs to invest $10000 at 6% interest and ($40000- $10000) or $30000 at 10% interest to make a total of 9% return on his $40000.
Given
Paul has $40000 to invest &he can invest part of his money at 6% and part at 10% interest.
Find out Paul invest in each option to make a total of 9% return on $40000 .
To proof
Paul invest = $ 40000
let us assume that x be the part of money he invested at 6%.
let us assume that (40000 - x) be the part of money he invested at 10 %.
6% written in the simpler form
[tex]= \frac{6}{100}[/tex]
= 0.06
similarly,
10% written in the simpler form = 0 .1
9% written in the simpler form = 0.09
than the equation becomes
0.09 (40000) = (40000 - x ) 0.1 + 0.06x
3600 = 4000 - 0.1x + 0.06x
0.1x - 0.06x = 4000 - 3600
0.04x = 400
x = 10000
$ 10000 be the part of money he invested at 6%.
$30000 be the part of money he invested at 10 %.
Hence proved
