A total of $4100 was invested, part of it at 5% interest and the remainder at 10%. If the total yearly interest amounted to $370, how much was invested at each rate?

Question

A total of $4100 was invested, part of it at 5% interest and the remainder at 10%. If the total yearly interest amounted to $370, how much was invested at each rate?

Answer 1

If $x was invested at 5%, then the rest $(4100-x) was invested at 10%

So, just add up the interest, and you have

.05x + .10(4100-x) = 370

Now just find x ...

Answer 2

Thanks, Steve!

5%=800
10%=3,300

Answer 3

To solve this problem, we can set up a system of equations based on the given information.

Let's assume that the amount invested at 5% interest is "x" dollars. Therefore, the amount invested at 10% interest is "4100 - x" dollars.

The interest earned from the amount invested at 5% interest is given by: 0.05x (since 5% is equivalent to 0.05 in decimal form).

The interest earned from the amount invested at 10% interest is given by: 0.10(4100 - x) (since 10% is equivalent to 0.10 in decimal form).

According to the problem, the total yearly interest earned is $370, so we can set up the equation:

0.05x + 0.10(4100 - x) = 370

To solve this equation, we can simplify it:

0.05x + 410 - 0.10x = 370

-0.05x + 410 = 370

Now, let's isolate the variable "x" by subtracting 410 from both sides of the equation:

-0.05x = 370 - 410

-0.05x = -40

Finally, divide both sides of the equation by -0.05 to solve for "x":

x = -40 / -0.05

x = 800

Therefore, $800 was invested at 5% interest, and $4100 - $800 = $3300 was invested at 10% interest.