Bunuel wrote:
A bartender removed from a bottle a certain quantity of a drink, which had 25% alcohol content and replaced the same quantity with another drink that had an alcohol content of 10%. The alcohol content in the drink after replacement is 20%. What was the quantity of the drink that he replaced, if there was 1 litre of the drink in the bottle?
A. 1/5 litres
B. 1/4 litres
C. 1/3 litres
D. 1/2 litres
E. 2/3 litres
Let 'x' litres of quantity be withdrawn / re.
Now we know -
Alcohol withdrawn = 0.25x (25% of withdrawn quantity)
Hence Alcohol remaining in bottle = 0.25 (1-x) (25% of remaining quantity. 1 Litre is the total quantity in bottle(given) ) -----
(1)We also know alcohol added is = 0.1x (10% of added quantity. Same quantity of total mixture is replaced) -----
(2)New total alcohol content = 0.2 (1L) (20% of Total quantity in bottle i.e. 1 Litre) ------
(3)Equation for alcohol content will be as follows -
Replaced alcohol + Remaining Alcohol = Total alcohol
i.e.
(2) + (1) = (3)=> 0.25(1-x) + 0.1x = 0.2(1L)
=> 0.25 - 0.25x + 0.1x = 0.2
=> 0.05 = 0.15x
Multiplying both sides by 100
=> 5 = 15x
therefore x = 1/3 Litres (which is the replaced quantity.)
Answer - C
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AD
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A Kudos is one more question and its answer understood by somebody !!!
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58% (01:42) correct 
