1st drug

First drug pack has an active ingredient ratio of L1: L2 = 2: 1
Second drug pack have a ratio of active ingredients L1: L2 = 1: 3

In which ratio do we have to mix the two packages so that the ratio of substances L1: L2 = 1: 2?

Correct result:

r =  1:4

Solution:

 L1=x (2/(2+1))+y (1/(1+3)) L2=x (1/(2+1)+y 3/(1+3)  23x+14y=12 (13x+34y)  r=x/y 23r+14=12 (13r+34)   2/3 r+1/4=1/2 (r/3+3/4)  6r=1.5  r=14=0.25=14=1:4 \ \\ L_{1}=x \cdot \ (2/(2+1)) + y \cdot \ (1/(1+3)) \ \\ L_{2}=x \cdot \ (1/(2+1) + y \cdot \ 3/(1+3) \ \\ \ \\ \dfrac{ 2 }{ 3 } x + \dfrac{ 1 }{ 4 } y=\dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } x + \dfrac{ 3 }{ 4 } y) \ \\ \ \\ r=x/y \ \\ \dfrac{ 2 }{ 3 } r + \dfrac{ 1 }{ 4 }=\dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } r + \dfrac{ 3 }{ 4 } ) \ \\ \ \\ \ \\ 2/3 \cdot \ r + 1/4=1/2 \cdot \ (r/3 + 3/4) \ \\ \ \\ 6r=1.5 \ \\ \ \\ r=\dfrac{ 1 }{ 4 }=0.25=\dfrac{ 1 }{ 4 }=1:4



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