i've been working FOREVER on this qestion, and i keep getting really long decimal anwsers which i think are wrong. so i'm kinda giving up on this :(
can anyone please help me? thanks in advance! 10 points to best anwser
a lab technician has 20% of alchohol solution and a 40% alchohol solution. He wants to make 75 gallons of a 36% alchohol solution. how much of EACH solution should he use? use either eleiminition or susbistuation to solve this problem.
can anyone please help me? thanks in advance! 10 points to best anwser
a lab technician has 20% of alchohol solution and a 40% alchohol solution. He wants to make 75 gallons of a 36% alchohol solution. how much of EACH solution should he use? use either eleiminition or susbistuation to solve this problem.
-
Let x and (75-x) gallons of 20% and 40% alcohol solutions be mixed to produce 75 gallons of 36% alcohol.
Then the alcohol concentrations satisfy
(20% of x) +[40% of (75-x)]=(36% of 75)
i.e., (20/100)(x) + (40/100)(75-x) =(36/100)(75)
i.e., 0.2x +0.4(75-x) =0.36(75)
i.e., 0.2x+30-0.4x=27
i.e., x=15
Thus, 15 gallons of 20%alcohol and (75-15)=60 gallons of 40% solutions are required to prepare 75 gallons of 36% alcohol.
Then the alcohol concentrations satisfy
(20% of x) +[40% of (75-x)]=(36% of 75)
i.e., (20/100)(x) + (40/100)(75-x) =(36/100)(75)
i.e., 0.2x +0.4(75-x) =0.36(75)
i.e., 0.2x+30-0.4x=27
i.e., x=15
Thus, 15 gallons of 20%alcohol and (75-15)=60 gallons of 40% solutions are required to prepare 75 gallons of 36% alcohol.
-
29% and 57%