Question

A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit. the bottle then contained only 25% spirit. how much of the bottle did steal ?

Answer 1

Answer:

Let the champagne bottle has X liters of 50% spirit and the bartender steals Y liters from the bottle and replaces stolen quantity with the same Y liters of his 20% Spirit and the bottle is left with X liters of 25 % spirit

Now, Initial Quantity of spirit = Final Quantity (quantity of resultant mixture after replacement)

X liters of 50 % spirit = X liters of 25% spirit

Now, according to the condition given in the problem :

Quantity of spirit in the X liters of (50 % spirit) - Quantity of spirit in Y liters of (50 % spirit) + Quantity of spirit in Y liters of (20 % Spirit) = Quantity of spirit in X liters of (25 % Spirit)

$\implies\frac{50}{100}X-\frac{50}{100}Y+\frac{20}{100}Y=\frac{25}{100}X\\\\\implies \frac{X}{2}-\frac{Y}{2}+\frac{Y}{5}=\frac{X}{4}\\\\\implies -\frac{Y}{2}+\frac{Y}{5}=\frac{X}{4}-\frac{X}{2}\\\\\implies\frac{3Y}{10}=\frac{X}{4}\\\\\implies Y=\frac{5}{6}X$

where X = Quantity of 50 % spirit

Answer 2

by using mixtures and alligations the they could be in ratio of 5:25 that's is 1:5

so hence it is 1/1+5 = 1/6