Question 1097844
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The weight of a plastic pail is 20% of the weight when it is filled with water.After a certain amount of water has been poured  out, 
the weight of the pail and the remaining water is 60% if the original total weight. What percent of water remained in the pail?
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Let P be the weight of the plastic pail alone and let W be the weight of water when it is fills the pail.


Then the first part of the condition gives you this equation:

P = 0.2*(P+W).        (1)


Let R be the weight of the remaining amount of water.
Then the second part of the condition gives you another equation 

P + R = 0.6*(P+W).    (2)


From (1), you have  

P = 0.2P + 0.2W  ====>  0.8P  = 0.2W  ====>  P = {{{(0.2/0.8)*W}}} = {{{(1/4)*W}}}.     (3)


From (2), you have 

P + R = 0.6*(P+W) = 3*0.2*(P+V) = 3P  ====>  R = 3P - P = 2P  ====>  substitute P = {{{(1/4)*W}}} from (3)  ====>  

R = {{{2*(1/4)*W}}} = {{{(1/2)*W}}} = 0.5W.


Hence, the weight of the remaining water R is 0.5W.


In other words, the percent of water remained in the pail is 50%.
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