The ratio of alcohol and water in solution A is 2:3; the ratio of the

Question

The ratio of alcohol and water in solution A is 2:3; the ratio of the same in solution B is 3:2. If A and B are mixed, which of the following may be the percentage of alcohol in the resultant solution?
 
I. 40%
II. 50%
III. 60%

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

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Mosaib · Answer

Solution A alcohol fraction = 2/(2+3)=2/5=40%
Solution B alcohol fraction = 3/(3+2)=3/5=60%

When you mix two solutions, the resulting percentage must lie between the two original percentages (weighted average rule).

So mixture alcohol % must be:
👉 greater than 40%
👉 less than 60%
(only equal if one solution amount is zero, which is not a mix)

Check options

I. 40% ❌
Cannot be exactly 40% unless only A is used.

II. 50% ✔
50% lies between 40% and 60% → possible.

III. 60% ❌
Cannot be exactly 60% unless only B is used.

✅ Correct Answer: B. II only

aditya1818 · Answer

Ratio of alcohol is 40% in the first and 60% in the second.

The mixture will have the alcohol anything between 40%-60%
So the answer should be (B)

But wait,
While I was analysing this question, I gave a thought to something else.
What if the solutions are in the ratio 1:10000000000000000000000000 or in the ratio 10000000000000000000000000:1

and so the alcohol combined percentage will be 40.0000000001% and might be 59.9999999999999% which is also 40% and 60% alcohol % combined.

I am confused as to what to consider in this case as "exact" 40% and 60% or the above ones which brings us down to option(E).

Technically speaking, option(E) is more correct as the ratio of (1:indefinite) or (indefinite:1) satisfies the statement "A and B are mixed" and somewhat satisfies alcohol % to 40 and 60 respectively.

I need others insights here.

My take is Option(E)

Dereno · Answer

Percentage of alcohol in Solution A = (2/5)*100 = 40%

Percentage of alcohol in Solution B = (3/5)*100 = 60%

So, the percentage of alcohol if both solutions A and B are mixed, we get

Since the question says, we mix A and B, then there is two solutions.

40% and 60% comes only when one of the solutions is NOT present.

So, with solution A and B , not equal to zero. We cannot reach a point where the percentages are 40% and 60%.

So only,  option B  satisfies.

Navdeep2000 · Answer

Let the solution A be totalled 5x. Then Alcohol will be 2x and Water will be 3x as per the ratio given to us.  
 Likewise, let the solution B be totalled 5y. Then Alcohol in it will be 3y and Water will be 2y as per the ration given to us.

Now let's plugin one by one each of the options to see if they work.
 I.  (2x+3y)/(5x+5y) = 40/100 or (2x+3y)/(5x+5y) = 2/5
Then, 10x+15y=10x+10y, which leads us to 15y=10y. However, this is not possible. Hence,  No .

II.  (2x+3y)/(5x+5y) = 50/100 or (2x+3y)/(5x+5y) = 1/2
Then, 4x+6y=5x+5y, which leads us to x=y. When we plug in this to our original equation, we would get (2x+3x)/(5x+5x) = 5x/10x or 1/2. Hence, it works.  Yes

III.  (2x+3y)/(5x+5y) = 60/100 or (2x+3y)/(5x+5y) = 3/5
Then, (10x+15y) = (15x+15y), which leads us to 10x=15x, which again is not possible. Hence,  No .

Thus, only option II works here.   Option B  .

Adit_ · Answer

I dont reallt get you. If you want 40% as the resultant mixture then you MUST have 0% of the 60% mixture but both of them are technically getting mixed so the 60% mixture cannot have 0. Same goes for 40%. Like you yourself mentioned if you mix both of them with a non-zero volume for both you will never get exactly 40% 40.0000001 or whatever is not 40%

Edskore · Answer

The concept being tested here is Mixture Problems — specifically, the weighted average boundary rule. The "which of the following MAY be true" format is deliberate: it punishes students who assume any value between 40% and 60% is automatically valid.

Step 1 — Find the alcohol percentage in each solution.
Solution A: ratio 2:3 → alcohol = 2/(2+3) = 2/5 = 40%
Solution B: ratio 3:2 → alcohol = 3/(3+2) = 3/5 = 60%

Step 2 — Apply the weighted average principle.
When two solutions are mixed, the resultant percentage is a weighted average of the two individual percentages, weighted by volume. If A has volume a and B has volume b (both > 0):

Resultant % = (0.4a + 0.6b) / (a + b)

This value is strictly between 40% and 60%. It approaches 40% only as b → 0, and approaches 60% only as a → 0. But the problem says A and B are mixed — meaning both are present in positive, non-zero quantities.

Step 3 — Evaluate each option.
I. 40% → Only if b = 0 (no B used). Not a genuine mix. ✗
II. 50% → Achieved when a = b (equal volumes). Valid. ✓
III. 60% → Only if a = 0 (no A used). Not a genuine mix. ✗

Answer: B (II only)

Common trap: Students see 40% and 60% as the endpoints of the possible range and include them. The word "mixed" is doing critical work here — it rules out using just one solution, which means the exact boundary values are unreachable. aditya1818 raised this edge case correctly; the issue is that "mixed" on the GMAT means a genuine combination of both, not a theoretical limit as one volume approaches zero.

Takeaway: On mixture "may be" questions, the resultant percentage lives in the open interval (lower%, upper%) — the endpoints are only reachable by using a single solution, which by definition isn't a mixture.

ExpertsGlobal5 · Answer

Video explanation:    https://www.youtube.com/watch?v=w3-uXE9Ia-o

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