Two vessels contain equal quantities of 40% alcohol Sachin changed the concentration of the first vessel to 50% by adding extra quantity of pure alcohol. Vivek changed the concentration of the second vessel to 50% replacing a certain quantity of the solution with pure alcohol. By, what percentage is the quantity of alcohol added by Sachin more/less than that replaced by Vivek?
Answers
Answer:
The answer is 20%.
Step-by-step explanation:
Let each vessel contain 100 liters of 40% alcohol.
Now according to question.
40 + x/ 100 + x= 50/100
80 + 2x = 100 + x
x = 20 liters
Suppose Vivel replaced y liters.
Alcohol in y liters.
40/100 x y = 2y/5
Now
40 - 2y/5 + y /100 = 50 / 100
80 - 4y/5 + 2y = 100
2y - 4y/5 = 20
6y/5 = 2=
6y = 100
Since y = 50/3
Required % = (20 - 50/3 / 50 /3 = 100)%
= 10/3 x 3/50 x 100% = 20%
Answer:
quantity of alcohol added by Sachin 20 % more than that replaced by Vivek
Step-by-step explanation:
Let say Both Solution = 100 L
Alocohol = 40L
Other Liquid = 60L
Sachin added S L of pure alochol
=> Alocohol= 40 + S L
Total = 100 + S L
40 + S = (50/100)(100 + S)
=> 80 + 2S = 100 + S
=> S = 20
Let Say Vivek Raplaced V Litre
V litre has 0.4V alocohol
So Alochol = 40 - 0.4V + V = 50
=> 0.6V = 10
=> V = 50/3
Sachin add more than Vivek = 20 - 50/3 = 10/3
% = 100 * (10/3)/(50/3)
= 20 %
quantity of alcohol added by Sachin 20 % more than that replaced by Vivek