Answer:
He invested [tex]\$1000[/tex] in first account and [tex]\$1800[/tex] in the second account.
Step-by-step explanation:
Let the man invest [tex]\$\:x[/tex] in the first account.
It is given that he gets a 2% interest on the first amount.
Let [tex]I_{1}[/tex] represents the interest received after 1 year.
We know that [tex]I=\frac{P\times R\times T}{100}[/tex]
Here, [tex]P=x,\,R=2, T=1[/tex]
So, [tex]I_{1}=\frac{x\times 2\times 1}{100}[/tex]
[tex]I_{1}=\frac{2x}{100}[/tex]
It is given that he invest [tex]\$800[/tex] more at 4% interest on the second amount.
Let [tex]I_{2}[/tex] represents the interest received after 1 year.
We know that [tex]I=\frac{P\times R\times T}{100}[/tex]
In that case, [tex]P=x+800,\,R=2, T=1[/tex]
So, [tex]I_{2}=\frac{(x+800)\times 4\times 1}{100}[/tex]
I_{2}=\frac{4x+3200}{100}
Also, it is given that the total interest he earned is [tex]\$92[/tex]
So, we have [tex]I_{1} +I_{2} =92[/tex]
[tex]\frac{2x}{100}+\frac{4x+3200}{100}=92[/tex]
[tex]\frac{2x+4x+3200}{100}=92[/tex]
[tex]6x+3200=9200[/tex]
[tex]6x=9200-3200[/tex]
[tex]6x=6000[/tex]
[tex]x=1000[/tex]
Hence, he invested [tex]\$1000[/tex] in first account and [tex]\$1000+\$800=\$1800[/tex] in the second account.
