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1. a solution of 61% fertilizer is to be mixed with a solution of 20% fertilizer to form 287 liters of a 57% solution. how many liters of the 61% solution must be used?

 

 

thank you so much if you asnwer

 Dec 21, 2016

Best Answer 

 #1
avatar+37153 
+5

287 l of 57%  = 163.59 l of pure fertilizer

 

x = amount of 20%       .20 x

287-x = amount of 61%  =  (287-x)(.61)

.20x + (287-x)(.61) = 287(.57)

.20x +175.07 - .61x = 163.59

-.41x = - 11.48

x = 28 liters of 20%      287-28 = 259 liters of 61%

 Dec 21, 2016
 #1
avatar+37153 
+5
Best Answer

287 l of 57%  = 163.59 l of pure fertilizer

 

x = amount of 20%       .20 x

287-x = amount of 61%  =  (287-x)(.61)

.20x + (287-x)(.61) = 287(.57)

.20x +175.07 - .61x = 163.59

-.41x = - 11.48

x = 28 liters of 20%      287-28 = 259 liters of 61%

ElectricPavlov Dec 21, 2016
 #2
avatar+129899 
+5

1. a solution of 61% fertilizer is to be mixed with a solution of 20% fertilizer to form 287 liters of a 57% solution. how many liters of the 61% solution must be used?

 

Let x be the  liters of 61%  solution.....then    287−  x  = the liters of 20% solution

 

And we have

 

.61x    +   .20 (287 − x)   =  .57 (287)     multiply through by 100 to clear the  decimals 

 

61x  + 20 (287 − x)   = 57 (287) simplify

 

61x + 5740 − 20x   =  16359

 

41x +  5740   = 16359     subtract 5740 from each side

 

41x   = 10619  divide both sides by 41

 

x = 259 liters  = 61%  solution   ...... and  287 −  259  = 28 liters of 20% solution

 

 

 

cool cool cool

 Dec 21, 2016

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