Question

Kendra invested twice as much money at an annual rate of 7% as she did at an annual rate of 5.5%. How much did she invest at the higher rate if her income from these investments for one year totaled $175.50?

Answer 1

Answer:

$1,679.42

Step-by-step explanation:

Let $x$ be the amount Kendra invested at 5.5%. Since the amount invested at 7% is twice the amount she invested at 5.5%, she invested $2x$ at 7%.

Remember that $I=Prt$ , where:

$I$ is the interest earned

$P$ is the principal or initial investment

$r$ is the interest rate in decimal form

$t$ is the time in years

- For the 5.5% investment:

$P=x$

$r=\frac{5.5}{100} =0.055$

$t=1$

Let's replace the values in our formula, $I=x(0.055)(1)$, which simplified is $I=0.055x$

- For the 7% investment:

$P=2x$

$r=\frac{7.7}{100} =0.077$

$t=1$

$I=2x(0.077)(1)$

$I=0.154x$

Now, we that the the amount of interest earned combining the two investment is $175.50, so:

$0.055x+0.154x=175.50$

$0.209x=175.50$

$x=\frac{175.50}{0.209}$

$x=839.71$

She invested $839.71 at 5.5%, and since she invested twice that amount at 7%, she invested 2*($839.71 ) = $1,679.42 at 7% (the higher rate).