A lime solution in jar A has 41% water while in jar B has 85% water. if the % of lime juice in mixture of Y litre of pure lime juice and X litre of lime solution from jar B is 83%. what would be the percentage of lime juice in solution of Y litre of solution from jar A as well as B is mixed with X …

Question

A lime solution in jar A has 41% water while in jar B has 85% water. if the % of lime juice in mixture of Y litre of pure lime juice and X litre of lime solution from jar B is 83%. what would be the percentage of lime juice in solution of Y litre of solution from jar A as well as B is mixed with X litre of pure lime juice.

Lelouch Vi Indica · Answer

@Monica G language is shit. You first need inverse ratio (100-83) and (83-15), as we are mixing the 2 in this ratio to get 83% final. that gives you 4:1. Next you mix 4lts of A, 4 lts of B and 1 lt of 100% lime juice. That's simple averages and you'll get 44%. The question is convoluted because of the language.

shivani s · Answer

Step 1 - (Y+0.15X)/(X+Y) = 0.83. Solve to get X=Y/4. Step 2 - (0.59Y + 0.15Y+Y/4)/(9Y/4) =&gt; 44%

Monica G · Answer

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Shubham Kapoor · Answer

44%

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