SOLUTION: A 40% dye solution is to be mixed with a dye solution to get 120 liters of a 50% solution. How many liters of the 40% and 70% solutions will be needed?

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Question 323455: A 40% dye solution is to be mixed with a dye solution to get 120 liters of a 50% solution. How many liters of the 40% and 70% solutions will be needed?
Found 2 solutions by josmiceli, modi.kk91:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In words:
(final liters of dye)/(final liters of solution) = 50%
Let a = liters of 40% solution needed
Let b = liters of 70% solution needed
given:
(1) a+%2B+b+=+120
liters of dye in a = .4a
liters of dye in b = .7b
liters of dye in final solution = .5%2A120+=+60
(2) .4a+%2B+.7b+=+60
Multiply both sides of (1) by 4
Multiply both sides of (2) by 10
Subtract (1) from (2)
(2) 4a+%2B+7b++=+600
(1) -4a+-+4b+=+-480
3b+=+120
b+=+40
and, since
a+%2B+b+=+120
a+=+120+-+40
a+=+80
80 liters of 40% solution and 40 liters of 70% solution are needed
check:
(2) .4a+%2B+.7b+=+60
(2) .4%2A80+%2B+.7%2A40+=+60
32+%2B+28+=+60
60+=+60
OK

Answer by modi.kk91(20) About Me  (Show Source):
You can put this solution on YOUR website!
very easy solution by allegation rule

40% 70%


50%

70-50=20 and 50-40=10


ratio=2:1=====2x+1x=3x
ans=2x=80
1x=40

3x=120
x=120/3=40