SOLUTION: A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more
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-> SOLUTION: A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more
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Question 124553: A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more than the number of ponds of the $.50 coffee how many pounds of each kind did he use? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more than the number of ponds of the $.50 coffee how many pounds of each kind did he use?
:
Let x = no. of lbs of $.50 coffee
and
(x+10) = no. of lbs of $.80 coffee
Then
(2x+10) = total lbs of the final mixture
:
.50x + .80(x+10) = .70(2x+10)
:
.50x + .80x + 8 = 1.4x + 7
:
1.3x - 1.4x = 7 - 8
:
-.1x = -1
x =
x = +10 lbs of $.70 coffee
then
10 + 10 = 20 lbs of $.80 coffee
;
:
Check solution
.5(10) + .8(20) = .7(30)
5 + 16 = 21
