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not very big eqns are needed here, you can assume values starting from 5 ltrs as ratio is in multiple of 5 , do 2-3 values and you will fetch 8*3=24 -16=8 ,the required value it will save you from cumbersome eqns making
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Let orginal volume of mixture be x

Lime juice 0.2x

Soda 0.8x

Volume of pitcher is 3x

Volume left after mixture was poured

3x - x = 2x

Therefore 2x = 16

So x = 8

Original volume of lime juice=

0.2 x 8 = 1.6
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A full glass of lemonade is a mixture of 20% lime juice and 80% soda. The contents of the glass are poured into a pitcher that is 200% bigger than the glass. The remainder of the pitcher is filled with 16 liters of water. What was the original volume of lime juice in the mixture?

Let the volume of glass = v
then, the volume of lemonade class = 3v
Given that, the volume of lemonade class - the volume of glass = 16 litre
=> 2v = 16
=> v = 8

therefore volume of lime juice = (2/10)*8 = 1.6 Litre

Hence A
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Glass> L:S = 1:4 = 5 parts
Pitcher> L:S:W = 1:4:10 = 15 parts (Final mixture is 200% (or, 10parts) more than 5 parts and those 10 parts is filled with water)
10p water= 16L
1p water= 1.6L[Lime is 1 part in the final mixture]
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