+127 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
+36919 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%
+36919 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%
+128707 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
