Question 1135345
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Using the traditional algebraic method for solving the problem, write an equation that says x liters of pure (100%) antifreeze must be added to (4.6-x) liters of 45.4% antifreeze to get 4.6 liters of 55.1% antifreeze:<br>
{{{1(x)+.454(4.6-x) = .551(4.6)}}}   <--- corrected equation, thanks tutor @ikleyn; I previously showed .0454 and .0551....<br>
Ugly numbers; I'll let you finish the solution by that method.<br>
Here is a MUCH easier way to get the answer to a problem like this, based on the concept that the ratio in which the two ingredients need to be mixed is directly related to where the target percentage of 55.1% lies between the 45.4% and 100% percentages of the ingredients.<br>
Here without detailed explanation are the required calculations for your problem.<br>
(1) 55.1-45.4 = 9.7; 100-45.4 = 54.6
(2) 9.7/54.6 = 0.177655678... is the fraction of the way that 55.1% is from 45.4% to 100%, so it is the fraction of the mixture that needs to be the higher percentage ingredient
(3) 0.177688678*4.6 = 0.817216... is the number of liters of pure antifreeze that must be used in the mixture.<br>
ANSWER (to 3 significant digits): 0.817 liters