Question

Last year, Anna had $30,000 to invest. She invested some of it in an account that paid 5% simple interest, and she invested the rest in an account that paid 7% simple interest per year. After one year, she received a total of $1780 in interest. How much did she invest in each account?

Answer 1

The rule of the simple interest is

$I=\text{PRT}$

P is the amount of investment

R is the interest rate in decimal

T is the time

Let her invested x dollars in the 1st account and y dollars in the second amount

Since her total investment is 30,000 dollars, then

$x+y=30000\rightarrow(1)$

Since the 1st account gives 5%, then

R = 5/100 = 0.05

Since the time is 1 year, then

T = 1

The interest of 1st account is

$\begin{gathered} I_1=x(0.05)(1) \\ I_1=0.05x \end{gathered}$

Since the 2nd account gives 7%, then

R = 7/100 = 0.07

The interest of 2nd account is

$\begin{gathered} I_2=y(0.07)(1) \\ I_2=0.07y \end{gathered}$

Since she took a total interest of 1780 dollars, then

$\begin{gathered} I_1+I_2=1780 \\ 0.05x+0.07y=1780\rightarrow(2) \end{gathered}$

Now, we have a system of equations to solve it to find x and y

Multiply equation (1) by -0.07 to eliminate y

$\begin{gathered} x(-0.07)+y(-0.07)=30000(-0.07) \\ -0.07x-0.07y=-2100\rightarrow(3) \end{gathered}$

Add (2) and (3)

$\begin{gathered} (0.05x-0.07x)+(0.07y-0.07y)=(1780-2100) \\ -0.02x+0=-320 \\ -0.02x=-320 \end{gathered}$

Divide both sides by -0.02 to find x

$\begin{gathered} \frac{-0.02x}{-0.02}=\frac{-320}{-0.02} \\ x=16000 \end{gathered}$

Substitute x in (1) by 16 000 to find y

$16000+y=30000$

Subtract 16000 from both sides

$\begin{gathered} 16000-16000+y=30000-16000 \\ y=14000 \end{gathered}$

Then she invested 16 000 dollars in the 1st account and 14 000 dollars in the 2nd account