Question 219408: how many liters of 20% alcohol solution must be mixed with 10L of a 50% solution to get a 30% solution?
can you please explain the process too?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! OK
Let x=amount of 20% solution needed
Now we know that the pure alcohol in the 20% solution that is needed (0.20x) plus the amount of pure alcohol in the 50% solution (0.50*10) has to equal the amount of pure alcohol in the final mixture(0.30(10+x)), so our equation to solve is:
0.20x + 0.50*10=0.30(10+x) get rid of parens and simplify
0.20x+5=3+0.30x subtract 0.30x and also 5 from each side
0.20x-0.30x+5-5=0.30x-0.30x-5+3 collect like terms
-0.10x=-2 divide each side by -0.10
x=20--------------------amount of 20% solution needed
CK
0.20*20+0.50*10=0.30*(10+20)
4+5=9
9=9
THE PROCESS IN SOLVING MIXTURE PROBLEMS IS TO LOOK FOR THINGS THAT REMAIN THE SAME (OR CAN BE MADE TO BE THE SAME) BEFORE AND AFTER THE MIXTURE TAKES PLACE. IN THIS PROBLEM, WE KNOW THAT THE PURE ALCOHOL DOES NOT CHANGE SO THAT GIVES US OUR EQUATION. IN THIS PARTICULAR MIXTURE PROBLEM, THE OTHER STUFF IN THE SOLUTION DOES NOT CHANGE EITHER AND WE CAN JUST AS EASILY SOLVE THE PROBLEM DEALING IN THE OTHER STUFF:
Let x=amount of 20% solution needed
Now we know that the OTHER STUFF in the 20% solution that is needed (0.80x) plus the amount of OTHER STUFF in the 50% solution (0.50*10) has to equal the amount of OTHER STUFF in the final mixture(0.70(10+x)), so our equation to solve is:
0.80x + 0.50*10=0.70(10+x) get rid of parens and simplify
0.80x+5=7+0.70x subtract 0.70x and also 5 from each side
0.80x-0.70x=7-5
0.10x=2
x=20-------------------same as before
Hope this helps---ptaylor
|
|
|
