Q. 150 - Intermediate Algebra

Question

Hattie had $\$3000$ to invest and wants to earn $10.6\%$ interest per year. She will put some of the money into an account that earns $12\%$ per year and the rest into an account that earns $10\%$ per year. How much money should she put into each account?

Step-by-Step Solution

Answer

Hattie should put $\$900$ into her first account at $12\%$ and put $\$2100$ into her second account at$10\%$.

Step 1. Given information.

Consider the given question,

Total amount of money that Hattie had is $\$3000$.

Interest rate per year is $10.6\%$.

Assume $x$ to be the amount invested in first account and $y$ to be the amount invested in second account.

Step 2. Form the equation.

Consider the given question,

$x+y=3000 ...... \left(i\right)$

Also, $$\begin{gathered} 12\% of x+10\% of y=10.6\% of \left(3000\right) \\ 0.12x+0.1y=0.106\left(3000\right) \\ 0.12x+0.1y=318 ...... \left(ii\right) \end{gathered}$$

From equation (i),

$x=3000-y ...... \left(iii\right)$

Step 3. Substitute the values in equations and solve.

Substitute the values of $x$ in equation (ii),

$$\begin{gathered} 0.12\left(3000-y\right)+0.1y=318 \\ 360-0.12y+0.1y=318 \\ -0.02y=-42 \\ y=\frac{-42}{-0.02} \\ y=2100 \end{gathered}$$

Substitute the value of $y$ in equation (iii),

$$\begin{gathered} x+2100=3000 \\ x=3000-2100 \\ x=900 \end{gathered}$$

Step 4. Check the answers.

Substitute the values in equation (i),

$$\begin{gathered} 900+2100=3000 \\ 3000=3000 \end{gathered}$$

This is true.

Substitute the values in equation (ii),

$$\begin{gathered} 0.12x+0.1y=318 \\ 0.12\left(900\right)+0.1\left(2100\right)=318 \\ 108+210=318 \\ 318=318 \end{gathered}$$

This is also true.

Therefore, Hattie should should put $\$900$ into her first account at $12\%$ and put $\$2100$ into her second account at $10\%$.