Question

A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account?

Answer 1

x*2/100 + (x+800)*4/100 = 92
2x+4x+3200 = 9200
6x = 9200-3200
6x = 6000
x = 1000
So the man invested $ 1000 at 2% interest and $ 1800 at 4% interest.

Answer 2

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=(P* R* T)/(100)`

Here,
`P=x,\,R=2, T=1`

So,
`I_(1)=(x* 2* 1)/(100)`

`I_(1)=(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=(P* R* T)/(100)`

In that case,
`P=x+800,\,R=2, T=1`

So,
`I_(2)=((x+800)* 4* 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`

`(2x)/(100)+(4x+3200)/(100)=92`

`(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.