SOLUTION: how many gallons of a 12% salt solution must be mixed with 6 gallons of a 20% salt soution to obtain a 15% salt solution?
100(.12)(x)+100(.20)(6)=.15%(100)
12x+20(6)=
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100(.12)(x)+100(.20)(6)=.15%(100)
12x+20(6)=
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Question 124343: how many gallons of a 12% salt solution must be mixed with 6 gallons of a 20% salt soution to obtain a 15% salt solution?
100(.12)(x)+100(.20)(6)=.15%(100)
12x+20(6)=15 12x+120=15
-120=-120
12x=-105 Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! I HAVE A SUGGESTION: WHEN WORKING WORD PROBLEMS, LAY OUT THE PROBLEM IN A VERY BASIC AND ORDERLY FASHION. I NEED TO KNOW WHAT x STANDS FOR, FOR EXAMPLE.
Let x=number of gal of 12% solution needed
Then (6+x) is the number of gal in the final 15% mixture
Now we know that the amount of pure salt in the 12% solution(0.12x)plus the amount of pure salt in the 6 gal of 20% solution (6*0.20) has to equal the amount of pure salt in the final mixture(6+x)*(0.15). So our equation to solve is:
0.12x+0.20(6)=0.15(6+x) get rid of parens
0.12x+1.2=0.9+0.15x subtract 0.9 and also 0.12x from both sides
0.12x-0.12x+1.2-0.9=0.9-0.9+0.15x-0.12x collect like terms
0.3=0.03x divide both sides by 0.03
x=10 gal--------------amount of 12% solution needed
CK
0.12*10+0.20*6=(6+10)*0.15
1.2+1.2=2.4
2.4=2.4
