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

Answer:

• 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.

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

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.