Carol D.

asked • 04/29/20

A chemist wants to create 20 liters of a solution that’s 14 Percent alcohol. If she has 8% solution and 24% solution how much of each would she need.

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Kathleen A.

You can create a system of equations. Define the variables as y represents the 24% solution x represents the 8% solution One equation is based on liters x+y=20 The second equation is based on the solution 0.08x+0.24y=0.14(20) 0.08x+0.24y=2.8 This is equated to .14(20) because we need 20 liters at 14% alcohol. Solving the first equation for a x and you get x=20-y then plug in to the second equation to get 0.08(20-y)+0.24y=2.8 Distribute 1.6-0.08y+0.24y=2.8 Solve 1.6+.016y=2.8 0.16y = 1.2 y =7.5 Recalling the second equation -->0.08x+0.24y=.14(20) y represents the 24% solution x represents the 8% solution Find x by plugging in y=7.5 in the solved equation --> x=20-y x=12.5 Thus, we need 7.5 liters of the 24% solution and 12.5 liters of the 8% solution.
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04/29/20

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