Q26. - Algebra 1

Question

Ms. Brinkman invested \(30,000; part at 5%, and part at 8%. The total interest on the investment was \)2100 after one year. How much did she invest at 8%?

1. \(10,000
2. \)15,000
3. \(20,000
4. \)25,000

Step-by-Step Solution

Answer

Investment at 8% is $20,000. Option C.

Step 1. Write the formula for Simple interest.

Interest is given by:

$S.I.=\frac{p\times r\times t}{100}$

where it $p$ is the principal, $r$ is the rate and $t$ is the time.

Step 2. Form the equation .

Let interest at $5\%$ be $\$x$.

Therefore, investment at $8\%$ is $\$\left(30000-x\right)$.

Interest on $\$x$ is calculated as:

$$\begin{gathered} S.I.=\frac{p\times r\times t}{100} \\ =\frac{x\times 5\times 1}{100} \\ =\frac{5x}{100} \end{gathered}$$

Interest on $\$\left(30000-x\right)$ is calculated as:

 $$\begin{gathered} S.I.=\frac{p\times r\times t}{100} \\ =\frac{\left(30000-x\right)\times 8\times 1}{100} \\ =\frac{8\left(30000-x\right)}{100} \end{gathered}$$

As per the question,

Total interest on the investment is $\$2100$.

The equation can be written as:

$\frac{5x}{100}+\frac{8\left(30000-x\right)}{100}=2100$

Step 3. Solve the equation.

The equation can be solved as follows:

$$\begin{gathered} \frac{5x}{100}+\frac{8\left(30000-x\right)}{100}=2100 \\ \frac{5x+8\left(30000-x\right)}{100}=2100 \\ \frac{5x+240000-8x}{100}=2100 \\ \frac{-3x+240000}{100}=2100 \\ \frac{-3x+240000}{100}\times 100=2100\times 100 \end{gathered}$$

$$\begin{gathered} -3x+240000=210000 \\ -3x+240000-240000=210000-240000 \\ -3x=-3000 \\ \frac{3x}{3}=\frac{30000}{3} \\ x=10000 \end{gathered}$$

Investment at $5\%=\$x=\$10,000.$

Investment at $5\%=\$\left(30,000-x\right)=\$\left(30,000-10,000\right)=\$20,000.$