Question

Let's assume Sarah's father invests $$12,000$ in interest-bearing accounts, with a portion at $3.2$% and the remainder at $2.6$%. Total investment revenue (interest) for the year is $$369$. At each rate, how much money did Sarah's father put in?

• A.
  $3500,8500$
• B.
  $2500,9500$
• C.
  $4500,7500$
• D.
  $5500,6500$

Answer 1

The correct option is B

$2500,9500$
Let $x=$ amount invested at $3.2$%
$y=$ amount invested at $2.6$%
Step 1: Total amount invested is $$12000$.
$x+y=12000\dots \left(1\right)$

Step 2: Total interest collected per year is $$525$.
So,
$x\times \frac{3.2}{100}+y\times \frac{2.6}{100}=369$

$0.032x+0.026y=369\dots \left(2\right)$

Step 3: Multiply equation $\left(1\right)$ by $0.032$
$0.032x+0.032y=384\dots \left(3\right)$

Step 4: Substract equation $\left(3\right)$ from equation $\left(2\right)$

We get,
$-0.006y=-15$

$y=\frac{15}{0.006}$

$\Rightarrow y=$ $$2500$

Plug $y=2500$ into equation $\left(1\right)$

$x+2500=12000$

$\Rightarrow x=$ $$9500$