There are two beakers containing equal quantities of
30% alcohol. Rahul changed the concentration of first
beaker to 40% by adding extra pure quality of alcohol.
Nikhil changed the concentration of second beaker
to 50% by replacing a certain quantity of the solution
with pure alcohol. By what percentage is the quantity
replaced by Nikhil more than that of alcohol added
by Rahul ?
a)500/7% b) 66 1/3% c) 50%
d) 72 2/3% e) Can’t be determined
Answers
Answer:
500/7 %
Step-by-step explanation:
Beaker Contains 30% alcohol
So other Liquid = 70 %
Let say Beaker has 30A alcohol
Other Liquid = 70A
& total solution = 100A
Rahul added X alcohol
Alcohol become = 30A + X
Solution become = 100A + X
((30A + X)/(100A + X)) * 100 = 40
=> ((30A + X)/(100A + X)) * 5 = 2
=> 150A + 5X = 200A + 2X
=> 3X = 50A
=> X = 50A/3
Let Say Rahul replaced Y
Total Solution = 100A ( as replacement)
Y what removed has 30% alcohol
=> (30/100)Y = 0.3Y alcohol reduced
Y alcohol added
Total net alcohol added = Y - 0.3Y = 0.7Y
Alcohol in solution = 30A + 0.7Y
(30A + 0.7Y)/(100A) * 100 = 50
=> 30A + 0.7Y = 50A
=> 20A = 0.7Y
=> Y = 200A/7
Y - X = 200A/7 - 50A/3 = (A/21)(600 - 350)
= (A/21)(250)
Y - X % of X = ((A/21)(250) / (50A/3) ) * 100
= 500/7 %
option A is correct