12 Days of Christmas GMAT Competition - Day 10: Three ethanol-water

Question

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Three ethanol-water solutions are mixed in a 5:3:2 ratio by volume. Is the percentage of ethanol in the resulting mixture by volume greater than 75%?

(1) The percentage of ethanol by volume in the first solution is at most 80%.
(2) The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

Three ethanol-water solutions are mixed in a 5:3:2 ratio by volume. Is the percentage of ethanol in the resulting mixture by volume greater than 75%?

For simplicity, let’s assume the three ethanol-water solutions are 50, 30, and 20 liters, respectively. The question asks whether the resulting mixture of 100 liters contains more than 75 liters of ethanol.

(1) The percentage of ethanol by volume in the first solution is at most 80%.

This implies that the first solution contributes at most 50 * 80% = 40 liters of ethanol. If the first solution contains exactly 40 liters of ethanol and the combined second and third solutions contribute more than 35 liters of ethanol, the answer would be YES. However, if the first solution contains exactly 40 liters of ethanol and the combined second and third solutions contribute 35 liters or less, the answer would be NO. Not sufficient.

(2) The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

This implies that the second and third solutions together contribute at most (30 + 20) * 70% = 35 liters of ethanol. Similarly, depending on the contribution of the first solution, we can get both YES and NO answers to the question. Not sufficient.

(1) + (2) With both statements combined, the first solution contributes  at most  40 liters of ethanol, and the second and third solutions contribute  at most  35 liters of ethanol. This means the final mixture could contain  at most  40 + 35 = 75 liters of ethanol, which is not more than 75. Thus, the answer to the question is definitively NO. Sufficient.

Answer: C.

siddhantvarma · Answer

For a mixture ratio of 5:3:2, let v1, v2, v3 be ethanol % in each solution.
Final ethanol % = (5v1 + 3v2 + 2v3)/10

From (1): v1 ≤ 80%
From (2): v2, v3 ≤ 70%
Each statement alone is not sufficient since it only gives us partial information.

Taking both statements together, we can calculate the maximum possible ethanol % = (5×80 + 3×70 + 2×70)/10
= (400 + 210 + 140)/10 = 75%

Since question asks if % is greater than 75%, and maximum possible is 75%, answer must be "No"

Both statements together are sufficient =>  Answer: C

Kinshook · Answer

Three ethanol-water solutions are mixed in a 5:3:2 ratio by volume. Is the percentage of ethanol in the resulting mixture by volume greater than 75%?

For the percentage of ethanol in the resulting mixture by volume; the percentage of ethanol by volume in the first, second and third solutions will be required.

(1) The percentage of ethanol by volume in the first solution is at most 80%.

Since the percentage of ethanol by volume in each of the second and third solutions is not known;

so, the percentage of ethanol in the resulting mixture by volume can NOT be determined.

INSUFFICIENT

(2) The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

[color=#000000]Since the percentage of ethanol by volume in first solution is not known;[/color]

[color=#000000]so, the percentage of ethanol in the resulting mixture by volume can NOT be determined. [/color]

Also,

If the percentage of ethanol by volume in first solution is > 80 %; then, the percentage of ethanol in the resulting mixture by volume greater than 75%
If the percentage of ethanol by volume in first solution is <= 80 %; then, the percentage of ethanol in the resulting mixture by volume will NOT be greater than 75%

so, the percentage of ethanol in the resulting mixture by volume can NOT be determined.

[color=#000000]INSUFFICIENT[/color]

(1) & (2) combined

Given, the percentage of ethanol by volume in the first solution is at most 80%;
the percentage of ethanol by volume in each of the second and third solutions is at most 70 %

For the maximum percentage values possible for the three solutions ;

we get the percentage of ethanol in the resulting mixture by volume = 5(80) + 3 (70) + 2 (70) / (5 + 3 +2) = 750/10 = 75 %.........................(1)

For the lower percentage values possible for the three solutions;

we get the percentage of ethanol in the resulting mixture by volume will be less than 75 % ......................................(2)

Using (1) & (2),

so, the percentage of ethanol in the resulting mixture by volume can never be greater than 75%

NO is the answer

SUFFICIENT

(C) is the CORRECT answer

aviraj1703 · Answer

Let say 3 solutions are A, B and C

A : B : C = 5 : 3 : 2
A = 5x
B = 3x
C = 2x
Total = 10x

Ethanol in A = Ea
Ethanol in B = Eb
Ethanol in C = Ec
Ethanol in (A + B + C) = Ea + Eb + Ec = Et

(1)  The percentage of ethanol by volume in the first solution is at most 80%.
Ea = 5x*80/100 = 4x
=> Et% = (4x + Eb + Ec) * 100
This may change with Eb and Ec so  Statement (1) alone is insufficient.

(2)  The percentage of ethanol by volume in each of the second and third solutions is at most 70%.
Considering the max Ethanol % in B and C
Eb = 2.1x
Ec = 1.4x
Et% = (Ea + 3.5x) * 100
This may change with Ea  Statement (2) alone is insufficient.

Combining both the statements 
We get Ea = 4x
=> Et% = 75%

This is with the maximum concentration of Ethanol in B and C 
 Hence, Et% <= 75%

Answer: C.

missionmba2025 · Answer

Let's assume we use 5 gallons, 3 gallons and 2 gallons of liquids by volume.

(1) The percentage of ethanol by volume in the first solution is at most 80%.

Let's assume that the first solution is 80% pure

Ethanol in first solution = 0.8 * 5 = 4 gallons

a) Assume that the other two solutions are 99.99% pure.

Ethanol from both the solutions ~ 5 gallons

% of ethanol in resultant mixture ~90%

b) Assume that the other two solutions are 0.001% pure.

Ethanol from both the solutions ~ 0 gallons

% of ethanol in resultant mixture ~40%

Therefore, the percentage of ethanol in the resulting mixture can be between 0% to 90%. The statement alone is not sufficient.

(2) The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

Let's assume that the second and the third solutions are 70% pure

Ethanol from the second and third solution = 0.7 * 5 = 3.5 gallon

a) Assume that the first solution is 99.99% pure.

Ethanol from the first solution ~ 5 gallons

% of ethanol in resultant mixture ~85%

b) Assume that the first solution is 0.001% pure.

Ethanol from the first solution is ~ 0 gallons

% of ethanol in resultant mixture ~35%

Therefore, the percentage of ethanol in the resulting mixture can be between 0% to 85%. The statement alone is not sufficient.

(1) + (2)

Combining we know that the max ethanol from the second and third solution = 3.5 gallon, and the max ethanol from the first solution = 4 gallon.

Therefore, the percentage of ethanol in the resulting mixture can be between 0% to 75%. The statement is sufficient to answer that the resulting mixture by volume is  not  greater than 75%.

Option C

Krunaal · Answer

Let the percentage of ethanol in three ethanol solutions be E1, E2, E3, and Emix be the percentage of ethanol the mixture of the three.

Emix = (5*E1 + 3*E2 + 2*E3) / 10

We need to check whether, Emix > 75% or  5*E1 + 3*E2 + 2*E3 > 750?

Statement 1

E1 is less than or equal to 80, but without knowing E2 and E3, we can't get one answer.  INSUFFICIENT

Statement 2

E2 and E3 are less than or equal to 70, but without knowing E1, we can't get one answer.  INSUFFICIENT

Both Together

E1 is less than or equal to 80, and E2 and E3 are less than or equal to 70. Considering the maximum values, we get 5*E1 + 3*E2 + 2*E3 = 750 which is not more than 750.  SUFFICIENT

Answer C.

hr1212 · Answer

Let volumes for three solutions be 5V, 3V and 2V. Is ethanol percentage in final solution > 75% Let x, y and z be the ethanol percentages in each of this solution then, (5V/10)*(x/100) + (3V/10)*(y/100) + (2V/10)*(z/100) > 3V/4 5x + 3y +2z > 750, where 0<=x,y,z<=100 Statement 1 - x<=80 If x = 0, then 3y + 2z > 750 --- Not possible If x = 80, then 3y + 2z >350 --- Possible As we cannot confidently say Yes or No, this statement is insufficient. Statement 2 - y<=70 and z<=70 If y=0 and z=0 then 5x > 750 --- Not possible If y=70 and z=70 then 5x > 400 --- Possible As we cannot confidently say Yes or No, this statement is insufficient. Combining both these statements If x=y=z=0, then 0 > 750 --- Not possible If x=80 and y=z=70, then 750 > 750 --- Not possible We can now confidently say that ethanol cannot be more than 75%, so answer to the given question is No. Hence, both statements together are sufficient. Answer: C

AviNFC · Answer

let A=500 L, B=300L, C=200
so mix contains > 0.75*1000 ?
so ethanol >750 l?

1. Max in A =400
if A=200, B=100, C=100 .. NO
 A=350, B=250, C=175..YES.....NOT SUFFICIENT

2. B=210, C=140 Max

if A= 450, B=200, C=120..YES
if A=100, B=210, C=140...NO...NOT SUFFICIENT

together, 
A=400, B=210, C=140...MAX 750
So, not more than 750...SUFFICIENT

Answer C

twinkle2311 · Answer

We need to determine whether the ethanol percentage in the resulting mixture exceeds 75%.

Let the ethanol percentages in the three solutions be E1, E2, and E3, and the volumes mixed in a ratio of 5:3:2.

Ethanol % in the mixture = (5E1 + 3E2 + 2E3) / 10

We need to check if this is greater than 75%
--> (5E1 + 3E2 + 2E3) / 10 > 75
--> 5E1 + 3E2 + 2E3 > 750.

(A)   E1 is at most 80% 
If E1 = 80, and we maximize ethanol by setting E2 = E3 = 100, then

5(80) + 3(100) + 2(100) => 400 + 300 + 200 = 900, which is >750.

But if E1 < 75, it may not exceed 750.

Therefore, Statement (A) is not sufficient.

(B)   E2 and E3 are at most 70% 
If E1 = 100, and we maximize ethanol by setting E2 = E3 = 70, then

5(100) + 3(70) + 2(70) => 500 + 210 + 140 = 850, which is >750

But if E1 is much smaller, the ethanol percentage may fall below 75%.

Therefore, Statement (B) is not sufficient.

(A) + (B)  
We have E1 ≤ 80, and E2, E3 ≤ 70.

To maximize ethanol, set E1 = 80, E2 = 70, and E3 = 70
5(80) + 3(70) + 2(70) => 400 + 210 + 140 = 750, which is exactly 750.

Since it is not >750, the ethanol percentage does not exceed 75% under these conditions.

Answer: C

tgsankar10 · Answer

In the resulting mixture, first solution 50%, second solution is 30% & third solution is 20%

Ethanol percentage in resulting mixture x= 0.5*a+0.3*b+0.2*c

where  a, b, c  are ethanol percentage in the respective solutions in order

is  x>75   {%}

Statement 1:  The percentage of ethanol by volume in the first solution is at most 80%.

a\leq{80}

Not Sufficient 
 
  Statement 2:  The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

b\leq{70}  &  c\leq{70}

Not Sufficient

Combined:

x= 0.5*0.8+0.3*0.7+0.2*0.7=0.75

Even when  a, b, c  are at their maximum levels,  x=75   {%}  . So  x  cannot be greater than  75   {%}

Sufficient

Answer: C

OmerKor · Answer

Hi everyone

Last day! We are dealing here with a ratio problem. ---A----B----C----Total ---5----3----2-----10-- --50---30---20----100 - Easier work with (*10) Mix > 75% Yes = More than 75% = Sufficient No = Less than 75% = Sufficient Both = More&Less than 75% = Insufficient. (1) A is at most 80% ethanol, making 50 in total to 40 ethanol at most. A B C Total (Ethanol>75%?) Case 1: Total 50 30 20 100 Ethanol 40 30 20 90 = 90% YES Case 2: Total 50 30 20 100 Ethanol 1 1 1 3 = 3% NO Insufficient. (2) B&C is at most 70% ethanol, making 30&20 in total to 21&14 ethanol at most. A B C Total (Ethanol>75%?) Case 1: Total 50 30 20 100 Ethanol 50 21 14 85 = 85% YES Case 2: Total 50 30 20 100 Ethanol 1 1 1 3 = 3% NO Insufficient. (1)+(2) A B C Total (Ethanol>75%?) Case 1: Total 50 30 20 100 Ethanol (MAX) 40 21 14 75 = 75% NO Because I used the maximum volume of ethanol in each mix and still got less than 75%, There is no way to get a bigger number of ethanol so the answer is always NO (Less than 75%) Sufficient. Answer C

pintukr · Answer

Three ethanol-water solutions are mixed in a 5:3:2 ratio by volume. Is the percentage of ethanol in the resulting mixture by volume greater than 75%?

For the percentage of ethanol in the resulting mixture by volume; the percentage of ethanol by volume in the first, second and third solutions will be required.

(1) The percentage of ethanol by volume in the first solution is at most 80%.

Since the percentage of ethanol by volume in each of the second and third solutions is not known;

so, the percentage of ethanol in the resulting mixture by volume can NOT be determined.

INSUFFICIENT

(2) The percentage of ethanol by volume in each of the second and third solutions is at most 70%.

Since the percentage of ethanol by volume in first solution is not known;

so, the percentage of ethanol in the resulting mixture by volume can NOT be determined.

Also,
 
 If the percentage of ethanol by volume in first solution is > 80 %; then, the percentage of ethanol in the resulting mixture by volume greater than 75% 
  If the percentage of ethanol by volume in first solution is <= 80 %; then, the percentage of ethanol in the resulting mixture by volume will NOT be greater than 75%

so, the percentage of ethanol in the resulting mixture by volume can NOT be determined. 
 
  INSUFFICIENT

(1) & (2) combined

Given, the percentage of ethanol by volume in the first solution is at most 80%; 
  the percentage of ethanol by volume in each of the second and third solutions is at most 70 %

For the maximum percentage values possible for the three solutions ;

we get the percentage of ethanol in the resulting mixture by volume = 5(80) + 3 (70) + 2 (70) / (5 + 3 +2) = 750/10 = 75 %.........................(1)

For the lower percentage values possible for the three solutions;

we get the percentage of ethanol in the resulting mixture by volume will be less than 75 % ......................................(2)

Using (1) & (2),

so, the percentage of ethanol in the resulting mixture by volume can never be greater than 75%

NO is the answer

SUFFICIENT

(C) is the CORRECT answer

andreagonzalez2k · Answer

a,b and c: the three percentages

(5a+3b+2c)/10 > 75
5a+3b+2c > 750?

(1)
a<=80
if a=80 and b=c=0 the answer is no
if a=80 and b=c=100 the answer is yes

INSUFFICIENT

(2)
b<=70
c<=70
if b=c=70 and a=0 the answer is no
if b=c=70 and a=100 the answer is yes

INSUFFICIENT

(1)+(2)
a<=80
b<=70
c<=70

Taking their maximum values:
5a+3b+2c=5*80+3*70+2*70=750, that is not greater than 750

The answer is no.

SUFFICIENT

IMO C

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12 Days of Christmas GMAT Competition - Day 10: Three ethanol-water

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	A	B	C	Total (Ethanol>75%?)
Case 1: Total	50	30	20	100
Ethanol	40	30	20	90 = 90% YES
Case 2: Total	50	30	20	100
Ethanol	1	1	1	3 = 3% NO

	A	B	C	Total (Ethanol>75%?)
Case 1: Total	50	30	20	100
Ethanol (MAX)	40	21	14	75 = 75% NO

GMAT Data Sufficiency (DS) Questions

12 Days of Christmas GMAT Competition - Day 10: Three ethanol-water

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards