Question 25
A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
Solution
Final quantity of alcohol in the mixture = $$\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$$ = 567 ml
Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml
Hence, we can say that the percentage of water in the mixture = $$\dfrac{308}{875}\times 100$$ = 35.2 %
Video Solution
CAT Quant Questions | CAT Quantitative Ability
CAT Percentages QuestionsCAT Averages, Ratio and Proportion QuestionsCAT Mensuration QuestionsCAT Quadratic Equations QuestionsCAT Remainders Questions
CAT DILR Questions | LRDI Questions For CAT
CAT Quant Based LR QuestionsCAT Special Charts QuestionsCAT Routes And Networks QuestionsCAT Coins and Weights QuestionsCAT 2D & 3D LR Questions
