Question

Kevin has money in two savings accounts. One rate is 6% and the other is 12%. If he has $600 more in the 12% account and the total interest is $279, how much is invested in each savings account?

Answer 1

• The amount invested at 6% is $1,150.

,

• The amount invested at 12% is $1,750.

Explanation:

Let the amount invested at 6% = x

`\text{Interest earned at 6\%}=0.06x`

Kelvin has $600 more in the 12% account, therefore:

The amount invested at 12% = $(x+600).

`\text{Interest earned at 12\%}=0.12(x+600)`

The total interest is $279.

`0.06x+0.12(x+600)=279`

We solve the equation for x.

`\begin{gathered} 0.06x+0.12x+72=279 \\ 0.18x+72=279 \\ \text{Subtract 72 from both sides} \\ 0.18x+72-72=279-72 \\ 0.18x=207 \\ \text{Divide both sides by 0.18} \\ (0.18x)/(0.18)=(207)/(0.18) \\ x=1150 \end{gathered}`

The amount invested at 6% is $1,150.

The amount invested at 12% is:

`1150+600=\$1750`

The amount invested at 12% is $1,750.