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Senior RC Moderator V
Joined: 02 Nov 2016
Posts: 4118
GPA: 3.39
Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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11 00:00

Difficulty:   45% (medium)

Question Stats: 66% (01:28) correct 34% (01:47) wrong based on 220 sessions

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Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4:1 respectively. X liters of the solution from bottle A are to be mixed with Y liters of the solution from bottle B. What will be the percentage of alcohol in the resultant solution?

(1) X = Y

(2) Total 10 liters of solutions from bottles A and B are mixed.

Source: Experts Global GMAT
Difficulty Level: 600

_________________

Originally posted by SajjadAhmad on 18 Oct 2018, 11:24.
Last edited by SajjadAhmad on 29 Dec 2018, 01:31, edited 7 times in total.
examPAL Representative P
Joined: 07 Dec 2017
Posts: 1153
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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2
Bottles A and 8 contain alcohol-water solutions in ratios of 1:3 and 4:1 respectively. X liters of the solution from bottle A are to be mixed with Y liters of the solution from bottle 8. What will be the percentage of alcohol in the resultant solution?

(1) X = Y

(2) Total 10 liters of solutions from bottles A and B are mixed.

EG

To find the percentage in the resulting solution, we need to know the ratio between X and Y (note - we do not need actual number of liters!)
We'll look for statements that give us this information, a Logical approach.

(1) This gives us the ratio (1:1), just what we need!
Sufficient.

(2) But what is the ratio between them?
Insufficient.

(A) is our answer.
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Manager  S
Joined: 05 Nov 2015
Posts: 62
Location: India
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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A
equal proportion from both mixture (any %) will have resultant ratio 5:4
Intern  B
Joined: 11 Sep 2018
Posts: 5
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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1
I dont understand how A is the correct answer. Can someone explain this approach step by step?
Director  V
Joined: 27 May 2012
Posts: 902
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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Bottles A and 8 contain alcohol-water solutions in ratios of 1:3 and 4:1 respectively. X liters of the solution from bottle A are to be mixed with Y liters of the solution from bottle 8. What will be the percentage of alcohol in the resultant solution?

(1) X = Y

(2) Total 10 liters of solutions from bottles A and B are mixed.

Source: Expert Global
Difficulty Level: 650

Dear Moderator,
Came across this question, instead of variable B number 8 has come up at 2 places, leading to some confusion.Hope you will correct the same. Thank you.
_________________
- Stne
Math Expert V
Joined: 02 Sep 2009
Posts: 58446
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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stne wrote:
Bottles A and 8 contain alcohol-water solutions in ratios of 1:3 and 4:1 respectively. X liters of the solution from bottle A are to be mixed with Y liters of the solution from bottle 8. What will be the percentage of alcohol in the resultant solution?

(1) X = Y

(2) Total 10 liters of solutions from bottles A and B are mixed.

Source: Expert Global
Difficulty Level: 650

Dear Moderator,
Came across this question, instead of variable B number 8 has come up at 2 places, leading to some confusion.Hope you will correct the same. Thank you.

____________________
Edited. Thank you.
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Experts' Global Representative S
Joined: 19 Feb 2010
Posts: 235
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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1
Neferteena wrote:
I dont understand how A is the correct answer. Can someone explain this approach step by step?

Hello Neferteena,

Statement 1 suggests that X = Y = x (say).
A contains 25% alcohol and B contains 80%.
Hence,
Alcohol in A = 0.25x
Alcohol in B = 0.8x
Total alcohol = 1.05x
Total volume = 2x (x each in A and B)
Percentage of alcohol can be found (1.05/2)
1 alone is sufficient.

2 alone isn't sufficient.

That's how the answer is A.

Please find a concise video explanation here-

Hope this helps.

All the best!
Experts' Global team
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GMAT Tutor B
Joined: 07 Nov 2016
Posts: 56
GMAT 1: 760 Q51 V42 Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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2
Neferteena wrote:
I dont understand how A is the correct answer. Can someone explain this approach step by step?

The ratio of alcohol to water in bottle A is 1: 3 , If the volume of solution in bottle A is X
The Volume of alcohol will be (1/4)* X = 0.25 X
Similarly the ratio of alcohol to water in bottle B is 4:1, If the volume of solution in bottle B is Y
The volume of alcohol will be (4/5) * Y = 0.8 Y
To find : Percentage of alcohol if the solutions of A and B are mixed

It will be $$\frac{( 0.25 * X + 0.8 * Y)}{(X + Y)}$$ * 100

St 1 : X = Y

Value will become $$\frac{(0.25 * X + 0.8 * X)}{(X + X)}$$ * 100

X will get cancelled in the numerator and denominator and a unique numerical value will result

Sufficient

St 2 : X + Y = 10

But if we substitute X = 3 and Y =7 in the general formula mentioned before we discussed St 1 ,we will get one value

If we substitute X = 5, and Y = 5 , we will get a different value

Not Sufficient

Choice A
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Intern  B
Joined: 19 May 2019
Posts: 10
Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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Maxximus wrote:
Neferteena wrote:
I dont understand how A is the correct answer. Can someone explain this approach step by step?

Hello Neferteena,

Statement 1 suggests that X = Y = x (say).
A contains 25% alcohol and B contains 80%.
Hence,
Alcohol in A = 0.25x
Alcohol in B = 0.8x
Total alcohol = 1.05x
Total volume = 2x (x each in A and B)
Percentage of alcohol can be found (1.05/2)
1 alone is sufficient.

2 alone isn't sufficient.

That's how the answer is A.

Please find a concise video explanation here-

Hope this helps.

All the best!
Experts' Global team

Is it possible to solve this question by setting up a weighted average equation? I was trying to set one up by starting off with o.25x + 0.8y but I didn't know what it would be equal to. Or would this just be the wrong way to approach this question? Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4   [#permalink] 07 Sep 2019, 17:03
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