Question

An actor invested some money at 6​% simple​ interest, and ​$48 comma 000 more than three times the amount at 7 %. The total annual interest earned from the investment was ​$37 comma 110. How much did he invest at each​ amount?
He invested ​$__
nothing at 6​% and ​__
nothing at 7​%

Answer 1

Answer:

He invested $125,000 at 6% and $423,000 at 7%.

Step-by-step explanation:

Let $x$ be the amount of money invested at 6%. Since the amount of money invested at 7% is 3 times the amount of money invested at 6% plus $48,000, $3x+48,000$ is the amount invested at 7%.

We know that both investments yield $37,110 in interest, so the sum of the money invested at 6% and the money invested at 7% is $37,110; in other words:

6%$x$+7%$(3x+48,000)$ = 37,110

We need to convert both interest rates to decimals, so we are going to divide both rates by 100%

6%$x$+7%$(3x+48,000)$ = 37,110

$\frac{6}{100} x+\frac{7}{100} (3x+48,000)=37,110$

$0.06x+0.07(3x+48,000)=37,110$

Now we can solve our equation, step-by-step, to find the amount of each investment.

Step 1. Distribute 0.07 to 3x and 48,000

$0.06x+0.07*3x+0.07*48,000=37,110$

$0.06x+0.21x+3,360=37,110$

Step 2. Combine like terms

$0.27x+3,360=37,110$

Step 3. Subtract 3,360 from both sides of the equation

$0.27x+3,360-3,360=37,110-3360$

$0.27x=35,750$

Step 4. Divide both sides of the equation by 0.27

$\frac{0.27x}{0.27x} =\frac{33,750}{0.27}$

$x=125,000$

We now know that he invested $125,000 at 6%. Since the amount invested at 7% is $3x+48,000$, we just need to replace x with 125,000 to find it:

Amount invested at 7% = $3(125,000)+48,000$

Amount invested at 7% = $375,000+48,000$

Amount invested at 7% = $423,000$

We can conclude that the actor invested $125,000 at 6% and $423,000 at 7%.