If x:y:z = 3:4:6, and (5x + 3z) = 132, then find the value of '3y'.
Let x = '3a' So, y = 3a X (4/3) = '4a' And z = 3a X (6/3) = '6a' ATQ; 3a X 5 + 6a X 3 = 132 Or, 15a + 18a = 132 So, 33a = 132 So, a = 4 So, required value = 4 X 4a = 4 X 3 X 4 = 48
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