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kapil1995
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(1) If 50 liters alcohol is added, X will contain 60% alcohol.

After addition of 50 Liters alcohol , In 100 ml , 60 alcohol & 40 Water .

Means Orignally Its 50 Ml solution with 10 ml alcohol and 40 ml water

Means 20% Alcohol in solution X.

Sufficient

(2) If the volume of water, equivalent to that of the total solution, is added, X will contain 20% alcohol.

In final solution, 20 ml Alcohol and 80 ml Water .
Quantity of 20 ml Alcohol will remain same in orignal Solution .
Let W be orignal quantity of water.
So, Orignal solution = w +20

Now ,we are getting 80 ml water in final solution after adding equivalent to that of the total solution ,ie. w+20

80 = w + (w +20 )
w = 30

so orignal solution of 50 ml , water 30 ml and alcohol 20 ml

so, 40% Alcohol in solution X.

Sufficient

But As we are getting 2 different answer , from statement 1 and 2 , Final answer will be E

chetan2u @benuel bb What is wrong in my approach of 1st statement ???

I am sure experts will help you better understand the flaw in your reasoning but 1 eminent flaw which I found I am mentioning.

You have taken the constant value for the total quantity of the mixture instead if you would have taken it in variable form it would have solved the problem as in in place of 100ml if you would have taken 100x. |because we don't know what actual quantity was.

Now how that's a flaw.?

Let's take your reasoning only

After addition of 50 Liters alcohol , In 100 ml , 60 alcohol & 40 Water .

Means Orignally Its 50 Ml solution with 10 ml alcohol and 40 ml water

Means 20% Alcohol in solution X.
|Correct.

I am assuming after addition of 50 litres of alcohol the resultant solution becomes 150 (instead of 100), 90 alcohol & 60 Water.

Means Originally its 100ML solution with 40 alcohol and 60 Water.
Means 40% Alcohol in solution X. | Also correct.

How can there be two different values of alcohol concentration for the same solution X. | Ambiguity, which could have been solved by just taking a variable like 100x.
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kapil1995
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(1) If 50 liters alcohol is added, X will contain 60% alcohol.

After addition of 50 Liters alcohol , In 100 ml , 60 alcohol & 40 Water .

Means Orignally Its 50 Ml solution with 10 ml alcohol and 40 ml water

Means 20% Alcohol in solution X.

Sufficient

(2) If the volume of water, equivalent to that of the total solution, is added, X will contain 20% alcohol.

In final solution, 20 ml Alcohol and 80 ml Water .
Quantity of 20 ml Alcohol will remain same in orignal Solution .
Let W be orignal quantity of water.
So, Orignal solution = w +20

Now ,we are getting 80 ml water in final solution after adding equivalent to that of the total solution ,ie. w+20

80 = w + (w +20 )
w = 30

so orignal solution of 50 ml , water 30 ml and alcohol 20 ml

so, 40% Alcohol in solution X.

Sufficient

But As we are getting 2 different answer , from statement 1 and 2 , Final answer will be E

chetan2u @benuel bb What is wrong in my approach of 1st statement ???


60% of how much? We do not know, but you have taken it as 100L, a situation that is not correct.
Say initially it was t, so now total is t+50, as only alcohol is added.
The new % is 60%, so (a+50)/(t+50)=60/100
We cannot get a/t from this.
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