Question

Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

• A. 94%
• B. 92%
• C. 90%
• D. 89%

Answer 1

The Correct Option is B

The correct answer is (B): \(92\%\)

(Let the quantity of solutions \(A\) and \(B\) mixed initially be \(p\) and \(3p\) respectively. 

After an additional \(4p\) of solution \(A\) is added \(60\%\) of \((1p+4p)+x\% \) of \(3p = 72\%\) of \((1p+4p+3p)⇒x=92\)

Answer 2

Let the quantity of alcohol in A and B initially in the mixture be x and 3x respectively. 
Let the concentration of alcohol in solution B be X%. 
Total quantity = \(x + 3x = 4x \)
Amount of solution A added to the mixture = \(4x \)
Total quantity now =\( 4x + 4x = 8x\) (total mixture doubled by adding solution A) 
Quantity of A and B in the new mixture =\( (x + 4x)\) and \(3x\) respectively 
Amount of alcohol in the mixture =\(\frac{(5x \times 60\% + 3x \times x)}{8x} = 72\%\)
 ⇒ \(300 + 3x = 576 \)
⇒ \(x= 276/3 = 92\) .. 
The concentration of alcohol in solution B is 92%.