SOLUTION: thnax for helping me th e last time i inquire;
solution m is 10% sugar while solution a is 30% sugar. a technician wants to have 60 liters mixture of solution m and a with 18% su
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-> SOLUTION: thnax for helping me th e last time i inquire;
solution m is 10% sugar while solution a is 30% sugar. a technician wants to have 60 liters mixture of solution m and a with 18% su
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Question 149674: thnax for helping me th e last time i inquire;
solution m is 10% sugar while solution a is 30% sugar. a technician wants to have 60 liters mixture of solution m and a with 18% sugar. how many liters of each solution should be used.? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! solution m is 10% sugar while solution a is 30% sugar. a technician wants to have 60 liters mixture of solution m and a with 18% sugar. how many liters of each solution should be used.?
.
Let m = amt of solution m
and a = amt of solution a
.
since we have two unknowns we need two equations:
m + a = 60 (equation 1)
.10m + .30a = .18(60) (equation 2)
.
solving equation 1 for m:
m + a = 60
m = 60 - a
.
plug the above into equation 2 and solve for 'a':
.10m + .30a = .18(60)
.10(60 - a) + .30a = .18(60)
6 - .10a + .30a = 10.8
6 + .20a = 10.8
.20a = 4.8
a = 24 liters
.
to find m, plug the above into equation 1 and solve for m:
m + a = 60
m + 24 = 60
m = 60 - 24
m = 36 liters
