# 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:

$\ \\ L_{ 1 } = x \cdot \ (2/(2+1)) + y \cdot \ (1/(1+3)) \ \\ L_{ 2 } = x \cdot \ (1/(2+1) + y \cdot \ 3/(1+3) \ \\ \ \\ \dfrac{ 2 }{ 3 } x + \dfrac{ 1 }{ 4 } y = \dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } x + \dfrac{ 3 }{ 4 } y) \ \\ \ \\ r = x/y \ \\ \dfrac{ 2 }{ 3 } r + \dfrac{ 1 }{ 4 } = \dfrac{ 1 }{ 2 } \cdot \ (\dfrac{ 1 }{ 3 } r + \dfrac{ 3 }{ 4 } ) \ \\ \ \\ \ \\ 2/3 \cdot \ r + 1/4 = 1/2 \cdot \ (r/3 + 3/4) \ \\ \ \\ 6r = 1.5 \ \\ \ \\ r = \dfrac{ 1 }{ 4 } = 0.25 \ \\ = \dfrac{ 1 }{ 4 } = 0.25 = 1:4$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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