# A man invests a certain amount of money at 2% and $800 more than that amount in another account at 4% interest. At the end of the one year, he earned$112 in interest. How much money was invested in each account?

Oct 24, 2016

$1,333.33 invested at 2% $2,1333.33 invested at 4%

#### Explanation:

As the time is only 1 year, it does matter whether it is simple or compound interest.

Define each amount of money using a variable.

Let the smaller amount of money (at 2%) be $x$
The larger amount (at 4%) is $x + 800$

The interest on the first amount = $\frac{x \times 2}{100} = \textcolor{red}{\frac{2 x}{100}}$

The interest on the larger amount = $\textcolor{b l u e}{\frac{\left(x + 800\right) \times 4}{100}}$

The total interest is $112 $\textcolor{red}{\frac{2 x}{100}} + \textcolor{b l u e}{\frac{4 \left(x + 800\right)}{100}} = 112$$\frac{2 x + 4 x + 3200}{100} = 112$$\frac{6 x + 3200}{100} = 112 \text{ } \leftarrow \times 100$$6 x + 3200 = 11200$$6 x = 8000$x =$1,333.33

x+ 800= \$2,133.33