SOLUTION: Solve A chemist needs 150 milliliters of a 43% solution but has only 13% and 58% solutions available. Find how many milliliters of each that should be mixed to get the desired

Algebra ->  Graphs -> SOLUTION: Solve A chemist needs 150 milliliters of a 43% solution but has only 13% and 58% solutions available. Find how many milliliters of each that should be mixed to get the desired      Log On


   



Question 760246: Solve
A chemist needs 150 milliliters of a 43% solution but has only 13% and
58% solutions available. Find how many milliliters of each that should be
mixed to get the desired solution.
Please show detailed steps. Thank you :)

Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Let the final mix have x ml of 13% solution and y ml of 58% solution.
x+%2B+y+=+150 (it is given that the final mix is 150 ml) +equation+%281%29
x ml of the mix is equivalent to a 0.13x solution (13% solution of say alcohol means that 100 ml of the solution will have 13 ml of alcohol and the rest as the solvent i.e. water).
Similarly, y ml of the mix is equivalent of a 0.58x solution and contains 0.58y ml of solute.
But the final mix (150 ml) is 43% strong. i.e it contains 150*0.43 ml of the solute.
Therefore, 0.13x+%2B+0.58y+=+150+%2A+0.43+=+64.5 +%28equation+2%29+

Solving for simultaneous equations for x and y from equations 1 and 2, we get that x = 50 and y = 100.
(If you don't know how to solve simultaneous equations, the steps are:
eq (1) rewritten as: +13x+%2B+13y+=+150%2A13+ or +13x+%2B+13y+=+1950+ eq (3)
eq (2) rewritten as: +13x+%2B+58y+=+6450+ eq (4)
Subtracting (3) from (4), +45y+=+4500+ or +y+=+100+
Since x + y = 150 and y = 100, +x+=+50+