Answer-
$2000 were invested at 5%
Solution-
Let x amount of money was invested at 5% and (6000-x) amount was invested as 3%
We know that,
[tex]i=\dfrac{P\cdot r\cdot t}{100}[/tex]
Putting the values,
[tex]i_1=\dfrac{P_1\cdot r_1\cdot t}{100}[/tex]
[tex]i_1=\dfrac{x\cdot 5\cdot 1}{100}=\dfrac{5x}{100}[/tex]
And
[tex]i_2=\dfrac{P_2\cdot r_2\cdot t}{100}[/tex]
[tex]i_2=\dfrac{(6000-x)\cdot 3\cdot 1}{100}=\dfrac{3(6000-x)}{100}[/tex]
According to the question,
[tex]\Rightarrow \dfrac{5x}{100}+\dfrac{3(6000-x)}{100}=220[/tex]
[tex]\Rightarrow \dfrac{5x+3(6000-x)}{100}=220[/tex]
[tex]\Rightarrow 5x+3(6000-x)=22000[/tex]
[tex]\Rightarrow 5x+18000-3x=22000[/tex]
[tex]\Rightarrow 5x-3x=22000-18000[/tex]
[tex]\Rightarrow 2x=4000[/tex]
[tex]\Rightarrow x=2000[/tex]
Therefore, $2000 were invested at 5%
