New Concept:
If the volume removed in every step is different, you can modify the formula.
Let the volume removed and replaced 3 times be v1 , v2, v3 like if in the above question 80L was removed first and replaced with water, followed by 90 l removed and replaced with water , then 100 l removed and replaced, the formula would become,
NewVolume = V*(1- v1/V)(1- v2/V) (1- v3/V)
A container contains 240 L of rum, 80L is taken out of the container every day and an equal quantity of water is put into it.Find the quantity of rum that remains in the container at the end of the fourth day
This is reply by AI
The container contains 240 L run
The easiest way to do such questions is to use the liquid replacement formula
If from V liquid , x amount is replaced everytime and replaced by other x then the concentration of the original liquid changes to
New Volume = V[1- (x/v)]^n
where x is the volume replaced, V is the total volume initial , n is the number of times it is done
NewVolume = 240*( 1- 80/240)^4
NewVolume = 240*2*2*2*2/3*3*3*3 = 80*16/27 = 47.40
New Concept:
If the volume removed in every step is different, you can modify the formula.
Let the volume removed and replaced 3 times be v1 , v2, v3 like if in the above question 80L was removed first and replaced with water, followed by 90 l removed and replaced with water , then 100 l removed and replaced, the formula would become,
NewVolume = V*(1- v1/V)(1- v2/V) (1- v3/V)
mention anyone using @username while adding comments