Question 134502
Let x=number of ml of 40% acid solution needed

Now we know that the amount of pure acid in the 40% solution (0.40x) plus the amount of pure acid in the 65% solution(0.65*150) has to equal the amount of pure acid in the final mixture (0.50(150+x)).  So our equation to solve is:

0.40x+0.65*150=0.50(150+x) simplify

0.40x+97.5=75-0.50x  subtract 75 and also 0.40x from each side

0.40x-0.40x+97.5-75=75-75+0.50x-0.40x collect like terms

22.5=0.10x  divide both sides by o.10
x=225 ml------------------amount of 40% solution needed

CK

0.40*225+0.65*150=0.50(225+150)
90+97.5=187.5

187.5=187.5


Hope this helps---ptaylor