SOLUTION: A man has a 50 kilograms mixture that is one-third cement and two-thirds sand. How much of a mixture that half cement and half sand must be added to the 50 kilogram mixture to prod
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Question 552102: A man has a 50 kilograms mixture that is one-third cement and two-thirds sand. How much of a mixture that half cement and half sand must be added to the 50 kilogram mixture to produce a mixture that is one-fourth cement and three-fourths sand? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 50-50 mixture that needs to be added
Now we know that the amount of pure cement that exists before the mixture takes place has to equal the amount of pure cement that exists after the mixture takes place.
Pure cement before the mixture takes place: (1/3)*50+(1/2)*x
Pure cement after the mixture takes place:(1/4)(50+x)
Soooooo:
(1/3)*50+(1/2)x=(1/4)(50+x) multiply each term by 12
200+6x=150+3x collect like terms
6x-3x=150-200
3x=-50
x=-50/3------WAIT A MINUTE!!!! CAN'T HAVE NEGATIVE AMOUNTS
Sometimes it's better to analyze a problem from a "common sense" standpoint before jumping in with the math needed to solve it and this is a case in point.
Think about this problem:
We have a mixture of half orange juice and half water. How much pure orange juice do we add to the mixture in order to get a mixture that one-fourth orange juice and three-fourths water?? Do you see what I'm talking about?
