Question

​$6300 is​ invested, part of it at 12​% and part of it at 9​%. For a certain​ year, the total yield is ​$nbsp 660.00. How much was invested at each​ rate?

Answer 1

$3,100 was invested at 12% and $3,200 at 9%

Explanation:

Let x = the amount invested at 12% and y = the amount invested at 9%

As you know 12% = 0.12 and 9% = 0.09

Given:

​$6300 is​ invested, part of it at 12​% and part of it at 9​%: x + y = 6300

The total yield is ​$660.00: 0.12x + 0.09y = 660

Now you have 2 equations:

x + y = 6300 -----> x = 6300 - y

0.12x + 0.09y = 660

Substitute x = 6300 - y into 0.12x + 0.09y = 660 to find y

0.12(6300 - y )+ 0.09y = 660

756 - 0.12y + 0.09y = 660

-0.03y = -96

y = 3,200

x = 6,300 - 3,200

x = 3,100

Answer:

$3,100 was invested at 12% and $3,200 at 9%

Hope that helps