Mathematics
BulaRauda
3

$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?

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(1) Answers
hjd999

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

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