Question

Katy invests a total of $10,500 in two accounts paying 11% and 5% simple interest, respectively. How much was invested in each account if, after one year, the total interest was $945.00.

A) Enter an equation that uses the information as it is given that can be used to solve this problem. Use x as your variable to represent the amount of money invested in the account paying 11% simple interest.

Equation:___________

B) The answers are:
$____ was invested at 11% and
$____was invested at 5%

Answer 1

Equation: 0.11x + 0.05(10500 - x) = 945

$7000 was invested at 11% and

$3,500 was invested at 5%

Explanation:

There are two equations needed

Let x = amount of money invested at 11%
11% = 11/100 = 0.11 in decimal
Interest at 11% for 1 year = 0.11x

Amount of money invested at 5% interest. 5% = 0.05

= 10500 - amount invested at 11%= 10500 - x

Interest from this amount after 1 year = 0.05(10500 - x)

Total of these two = 945

So 0.11x + 0.05(10500 - x) = 945

The equation based on the information as it is given:

0.11x + 0.05(10500 - x) = 945

We know the total interest obtained was $945 so we have the equation
0.11x + 0.05(10500- 5x) = 945

Simplify

0.11x + (525 - 0.05x) = 945
==>0.11x + 525 - 0.05x = 945

==> 0.06x = 945 - 525

==> 0.06x = 420

==> x = 420/0/06 = $7000

Amount invested at 11% interest is $7,000

Amount invested at 5% interest = 10,500 - 7,000 = $3,500