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Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4

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Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post Updated on: 29 Dec 2018, 01:31
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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.

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Difficulty Level: 600

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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.
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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 18 Oct 2018, 11:30
1
SajjadAhmad 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.

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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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 18 Oct 2018, 11:33
A
equal proportion from both mixture (any %) will have resultant ratio 5:4
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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 11 Nov 2018, 19:39
1
I dont understand how A is the correct answer. Can someone explain this approach step by step?
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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 21 Dec 2018, 09:40
SajjadAhmad 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.
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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 21 Dec 2018, 09:48
stne wrote:
SajjadAhmad 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.


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Edited. Thank you.
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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 22 Dec 2018, 09:15
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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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4  [#permalink]

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New post 22 Dec 2018, 18:28
1
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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Re: Bottles A and B contain alcohol-water solutions in ratios of 1:3 and 4   [#permalink] 22 Dec 2018, 18:28
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