# 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?

Result

r =  1:4

#### Solution:

$\ \\ L_{ 1 } = x \cdot \ (2/(2+1)) + y \cdot \ (1/(1+3)) \ \\ L_{ 2 } = x \cdot \ (1/(2+1) + y \cdot \ 3/(1+3) \ \\ \ \\ \dfrac{ 2 }{ 3 } x + \dfrac{ 1 }{ 4 } y = \dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } x + \dfrac{ 3 }{ 4 } y) \ \\ \ \\ r = x/y \ \\ \dfrac{ 2 }{ 3 } r + \dfrac{ 1 }{ 4 } = \dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } r + \dfrac{ 3 }{ 4 } ) \ \\ \ \\ \ \\ 2/3 \cdot \ r + 1/4 = 1/2 \cdot \ (r/3 + 3/4) \ \\ \ \\ 6r = 1.5 \ \\ \ \\ r = \dfrac{ 1 }{ 4 } = 0.25 \ \\ = \dfrac{ 1 }{ 4 } = 0.25 = 1:4$

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