Grade 10 Mixture Problems. A simple presentation by Mr. Agostini. The Problem.
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Grade 10 Mixture Problems
A simple presentation by Mr. Agostini
( x -3)
( x 10)
-3x – 3y = -60
3x + 5y = 80
Back sub into x + y = 20
x + 10 = 20
x = 20 – 10
x = 10
Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution.
x + y = 20
0.3x + 0.5y = 8
+
2y = 20
y = 10
(1) x + y = 20
Back sub into x + y = 20
(2) 0.3x + 0.5y = 8
10 + y = 20
y = 20 – x sub into (2)
y = 20 - 10
0.3x + 0.5(20 – x) = 8
y = 10
0.3x + 10 – 0.5x = 8
Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution.
0.3x – 0.5x = 8 - 10
– 0.2x = - 2
x = 10