SOLUTION: Please help me on this problem. I will write it how it looks in my text book.The two jugs stood side by side on the shelf. One contained a solution that was 60% antiseptic, and the

Question 23368: Please help me on this problem. I will write it how it looks in my text book.The two jugs stood side by side on the shelf. One contained a solution that was 60% antiseptic, and the other contained a solution that was 90% antiseptic. How much of each should be used to get 50 milliliters of a solution that is 78% antiseptic?
Found 3 solutions by Paul, Earlsdon, stanbon:
Answer by Paul(988)   (Show Source): You can put this solution on YOUR website!
60x+90(-x+50)=78(50)
60x-90x+4500=3900
-30x=-600
x=20
Hence for 60%---> 20 solution and for 90%---> 30 solution.
Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
You can write this out in words then do the algebraic equation.
You want to add x ml of 60% solution to (50-x) ml of 90% solution to obtain 50 ml of 78% solution. So, after converting the percentages to their equivalent decimal values,in algebrese (I just invented this word) you can write:
Simplify and solve for x.
Collect like-terms.
Subtract 45 from both sides of the equation.
Divide both sides by -0.3

You would need to mix 20 ml of the 60% solution with (50-20 = 30) ml of the 98% solution to obtain 50 ml of 78% solution.
Check:
20(0.6) + 30(0.9) = 12 + 27 = 39
50(0.78) = 39.

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Let's follow the antiseptic in start, add, and finish.
Let amount of 60% anti. be "x"
Then amount of 90% anti. is "50-x"
Amount of anti in 60% is 0.60x
Amount of anti in 90% is 0.90(50-x)= 45-0.9x
EQUATION:
Amt. in 60% + Amt. in 90% = Amt in 78%
Amt. of anti in 60% + Amt. in 90% = Amt. in 78%
0.60x + 45-0.90x = 0.78(50)ml
60x + 4500-90x = 78(50)
-30x = -600
x = 20 ml
50-x = 30 ml
Cheers,
Stan H.

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