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