Question

A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol becomes 26%. The quantity of whisky replaced is: 

• A. $\dfrac{4}{3}$
• B. $\dfrac{3}{4}$
• C. $\dfrac{3}{2}$
• D. $\dfrac{2}{3}$

Hint:

For “replacement” problems, keep total volume constant and track the quantity of solute: new amount $=$ old amount $-$ removed $+$ added.

Answer 1

The Correct Option is D

Let total volume be $V$ and the replaced fraction be $x$.
Alcohol after replacement: \[ 0.40V-0.40xV+0.19xV = 0.40V-0.21xV. \] Given final concentration is $26\%$: \[ 0.40-0.21x=0.26 \;\Rightarrow\; 0.21x=0.14 \;\Rightarrow\; x=\frac{14}{21}=\boxed{\tfrac{2}{3}}. \] 

Answer 2

Instead of writing and solving a direct linear equation for the concentration, we can use the rule of alligation to find the replaced fraction, and check each option.

1. Option A (\(\tfrac43\)): By alligation, the ratio of the replaced part to the unreplaced part is \[ \frac{x}{1-x}=\frac{\text{initial}-\text{final}}{\text{final}-\text{replacement}}=\frac{40-26}{26-19}=\frac{14}{7}=2. \] Solving, \(x=2(1-x)\Rightarrow 3x=2\Rightarrow x=\tfrac23\), not \(\tfrac43\), so this option is incorrect (also, a fraction replaced cannot exceed \(1\)).
2. Option B (\(\tfrac34\)): This does not satisfy \(\dfrac{x}{1-x}=2\), so it is incorrect.
3. Option C (\(\tfrac32\)): A fraction replaced greater than \(1\) is not physically meaningful for "a part of the whisky," so this option is incorrect.
4. Option D (\(\tfrac23\)): Solving \(\dfrac{x}{1-x}=2\) gives exactly \(x=\dfrac23\), matching this option.

The alligation ratio \(14:7=2:1\) confirms that the replaced fraction is \(\dfrac23\).

Hence, the correct answer is option D: \(\dfrac23\).