From a 54 litre vessel filled with acid, x litres was poured out and water was added instead, then the same quantity of the mixture was poured out again. after this mixture in the vessel contained 24 litres of pure acid. how much acid was poured out first time?
Answers
Remaining voulme of acid after first act is 54-x, which is distributed in 54 L of mixture, thus each L of mixture contains (54-x)/54 l of acid.
Now again x litres is poured out, in short, x*(54-x)/54 litres of acid is poured out.
Thus remaining quantitiy of acid is 54-x - (x*(54-x)/54)=24.
I hope you can solve this on your own, inform if you can't and I'd type out the whole solutions, and this is anyways a question of mathematics I guess.
Hope this helps, inform if wrong.
Best shortcut method
f/i=24/54=4/9 As this process is performed two times so Taking square root . If process is repeated three times we have taken out cube root and so on.
f/i= 2/3 this is remaining part
Removed portion =1/3
Hence required answer =54 of 1/3=18Lt.
OR
54*x*x=24
x^2=24/54
X=2/3
Taken out initially =54*1/3=18Lt
Detailed method
According to question if a container contains y liter of liquid & x liter of liquid is taken out. If this operation is repeated n times then the final quantity of the liquid in the container is {1-x/y}^n.
So 24=54{1-x/54}^2
{1-x/54}^2=24/54
{1-x/54}^2=4/9
{1-x/54}=2/3
X/54=1/3
X=18
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