SOLUTION: Ref. Question 1046126: .95(1) = .40(x + 1) .95 = .40x + 40 .95 - 40 = .40x + 40 - 40 .55 = .40x .55 / .40 = .40x / .40 1.375 = x Where am I correct ?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Ref. Question 1046126: .95(1) = .40(x + 1) .95 = .40x + 40 .95 - 40 = .40x + 40 - 40 .55 = .40x .55 / .40 = .40x / .40 1.375 = x Where am I correct ?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1046146: Ref. Question 1046126:
.95(1) = .40(x + 1)
.95 = .40x + 40
.95 - 40 = .40x + 40 - 40
.55 = .40x
.55 / .40 = .40x / .40
1.375 = x
Where am I correct ?

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A 1 qt. mixture of alcohol and water is 95% alcohol. The mixture is to be reduced to 40% alcohol. Determine amount of distilled water to be added.


A clear way to think of it is %28amountAlcohol%29%2F%28amountMixturePlusamountAddWater%29=ConcentrationAlcoholDesired;
and your variable will be amountAddWater or amount of water to add to the mixture.

%281%2A0.95%29%2F%281%2Bv%29=0.4, which is written using decimal fractions instead of percents. The amount of water to add is assigned as v.

SOLVE this equation for v.


-
There is a decimal mistake in the steps you asked about. Stepping through mine just two ,
0.95%2F%28v%2B1%29=0.4
0.95=0.4%28v%2B1%29
0.95=0.4v%2B0.4