$6300 is invested, part of it at 12%, and part at 6%. For a certain year, the total yield is $564.00. How much was invested at each rate?

x - first part of the investment (12% rate) y - second part of the investment (6% rate)
Total investment: x + y = 6300 Total yield: 564 = x ∗ 0 , 12 + y ∗ 0 , 06
Based on that we can calculate x and y: \left \{ {y=6300-x} \atop {0,12x+0,06(6300-x)=564}} \right.
\left \{ {y=6300-x} \atop {0,12x+378-0,06x=564}} \right.
\left \{ {y=6300-x} \atop {0,06x=186}} \right.
\left \{ {y=6300-x} \atop {x=3100}} \right.
\left \{ {y=3200} \atop {x=3100}} \right.
Answer: $3100 was invested at 12% and $3200 at 6%.
If you have any questions, please let me know!

Answered by voytek | 2024-06-10

The student invested $3100 at 12% and $3200 at 6%.
;

Answered by voytek | 2024-12-20