SOLUTION: If grade A coffee costs .95 cents a pound and grade b coffee costs .75 cents a pound, how many pounds of each grade is needed if the total cost is $50.50, and the amount of grade b
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Question 147450: If grade A coffee costs .95 cents a pound and grade b coffee costs .75 cents a pound, how many pounds of each grade is needed if the total cost is $50.50, and the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds? Found 2 solutions by ankor@dixie-net.com, checkley77:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! If grade A coffee costs .95 cents a pound and grade b coffee costs .75 cents a pound, how many pounds of each grade is needed if the total cost is $50.50, and the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds?
:
Let x = amt of grade a coffee required
:
It says,"the amount of grade b that is needed is twice the amount needed for grad A plus 5 pounds", therefore we can say:
2x + 5 = amt of grade b coffee required
:
total cost equation;
.95x + .75(2x+5) = 50.50
:
.95x + 1.5x + 3.75 = 50.50
:
2.45x = 50.50 - 3.75
:
2.45x = 46.75
x =
x = 19.08 lb of grade a
and
2(19.08) + 5 = 43.16.lb of grade b
:
:
Check solution on a calc:
.95(19.08) + .75(43.16) =
18.126 + 32.37 = 50.496 ~ 50.50
:
Did this make sense to you? Any questions?
You can put this solution on YOUR website! .95x+.75(2x+5)=50.50
.95x+1.5x+3.75=50.50
2.45x=50.50-3.75
2.45x=46.75
x=46.75/2.45
19.08 lbs. of $.95 coffee is used
2*19.08+5=43.16 lbs. of $.75 coffee is used
proof
.95*19.08+.75*43.16=50.50
18.13+32.37
50.50=50.50
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