A barrel has 100 litres of pure alcohol. 20 litres of alcohol is remov...
The above solution of alcohol has been successively diluted three times. Hence we use the formula for successive dilution.
Where v is the initial volume of pure alcohol, c is the concentration of the alcohol after replacements and x is the volume of alcohol replaced.
51.2 litres
Hence, option 3.
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A barrel has 100 litres of pure alcohol. 20 litres of alcohol is remov...
To solve this problem, let's break it down into steps and analyze the changes happening to the alcohol content in the barrel after each step.
Step 1:
In the first step, 20 liters of alcohol is removed from the barrel and replaced with an equal amount of water. The total volume in the barrel remains the same at 100 liters. However, the alcohol content reduces by 20 liters.
Step 2:
In the second step, again 20 liters of the mixture is removed from the barrel and replaced with water. At this point, the total volume in the barrel is still 100 liters. However, the alcohol content reduces further by another 20 liters.
Step 3:
In the third step, the same process is repeated. 20 liters of the mixture is removed and replaced with water. The total volume in the barrel remains 100 liters, but the alcohol content reduces by another 20 liters.
Analyzing the changes:
After each step, the alcohol content reduces by 20 liters. So, after the first step, there are 100 - 20 = 80 liters of alcohol left in the barrel. After the second step, there are 80 - 20 = 60 liters of alcohol left. And after the third step, there are 60 - 20 = 40 liters of alcohol left in the barrel.
Therefore, the amount of alcohol in the resultant solution is 40 liters, which is option (c).
A barrel has 100 litres of pure alcohol. 20 litres of alcohol is remov...
Hey aarushi, there's a simple formula to this. Let a be the quantity of alcohol, b be the quantity removed and n be the no. of times operation is performed. Then,
fraction of original liquid(alcohol) left=(a-b/a) raised to the power n times.
Here, fraction of alcohol left= (100-20/100) to the power 3= 0.512
0.512 of 1= 51.2% of original amount.
Hence answer is c)
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