Question

You invested $3000 between two accounts paying 6% and 7% annual interest respectively if the total interest earned for the year was $190 how much was invest in each rate?

Answer 1

Answer:

$2000 was invested at 6% and $1000 was invested at 7%

Explanation:

Let:

x= the amount invested at 6% annual interest

y= the amount invested at 7% annual interest

The total amount invested was $3000. So,

x+y =3000

or

y=3000-x ===> Equation 1

Then, the total interest earned for the year was $190. So,

0.06x+0.07y=190 =====> Equation 2

Substitute y=3000-x into 0.06x+0.07y=190.

$\begin{gathered} 0.06x+0.07y=190 \\ 0.06x+0.07(3000-x)=190 \\ \text{Simplify and rearrange} \\ 0.06x+210-0.07x=190 \\ -0.01x+210=190 \\ 0.01x=210-190 \\ 0.01x=20 \\ x=\frac{20}{0.01} \\ \text{Calculate} \\ x=2000 \end{gathered}$

Substitute x=2000 into x+y =3000.

So,

$\begin{gathered} x+y=3000 \\ 2000+y=3000 \\ \text{Simplify and rearrange} \\ y=3000-2000 \\ \text{Calculate} \\ y=1000 \end{gathered}$

Therefore, $2000 was invested at 6% and $1000 was invested at 7%.