Question

Sarah's father invests $\$12,000$ in interest-bearing accounts, some part at $3.2$% and the remaining at $2.6$%. Total investment profit (interest) for the year is $\$369$. At $2.6\%$ rate of interest, how much money did Sarah's father put in?

• A. $\$9500$
• B. $\$8500$
• C. $\$2500$

Answer 1

The correct option is A $\$9500$

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: Substitute value of $x$ in terms of $y$ from eq(1) in eq(2)

$\Rightarrow 0.032(12000-y)+0.026y=369\Rightarrow 32\times 12-0.032y+0.026y=369\Rightarrow 384-0.006y=369\Rightarrow 15=0.006y\Rightarrow y=\frac{15000}{6}=\$2500$

Step 4: put value of $y$ in eq(1)
We get,
$x=12000-y=12000-2500=\$9500$

Hence, Sarah's father put $\$2500$ amount at $3.2\%$ interest and $\$9500$ amount at $2.6\%$ interest.