Question 1135345: An engine has a 4.6 litre capacity cooling system. If the most ideal solution of antifreeze for this engine is a 55.1 % solution (a mix of 55.1 % pure antifreeze and 44.9 % water). How much pure antifreeze (in litres) must be combined with a 45.4 % solution to achieve 4.6 litres of the ideal solution?
Express your answer to three significant digits.
Found 3 solutions by greenestamps, josgarithmetic, ikleyn: Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!
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:
<--- corrected equation, thanks tutor @ikleyn; I previously showed .0454 and .0551....
Ugly numbers; I'll let you finish the solution by that method.
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.
Here without detailed explanation are the required calculations for your problem.
(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.
ANSWER (to 3 significant digits): 0.817 liters
Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! M=4.6, volume capacity to meet, in litres
T=55.1, percent concentration for most ideal solution
H=100, percent pure antifreeze
L=45.4, percent of available lower conc. antifreeze
x, unknown volume of pure 100% antifreeze to use
M-x, unknown volume of L % antifreeze to use
Solve that equation for x.
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Answer by ikleyn(52787) (Show Source):
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