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A bartender mixes drink A with drink B. If drink A contains 24 percent

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A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 07 Jan 2015, 07:34
3
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A
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C
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Tough and Tricky questions: Mixture Problems.



A bartender mixes drink A with drink B. If drink A contains 24 percent alcohol, what percentage of the mixture of drink A and B is alcohol?

(1) The ratio of drink A and drink B is 1:4
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A.

Kudos for a correct solution.

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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 07 Jan 2015, 07:43
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Bunuel wrote:

Tough and Tricky questions: Mixture Problems.



A bartender mixes drink A with drink B. If drink A contains 24 percent alcohol, what percentage of the mixture of drink A and B is alcohol?

(1) The ratio of drink A and drink B is 1:4
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A.

Kudos for a correct solution.


Statement-1 : If Ratio is 1:4. Then we have total of 5 parts in the mixture. Then 1/5th of the mixture is A and 4/5 of mixture is B. No other information. SO Insufficient.

Statement-2 : B= 1.5 times of A in the concentration of alcohol. We can get the value by multiplying 1.5 with 0.24. But still this statement alone not sufficient to solve.

Combining.

We know the alcohol concentration of A from the question and alcohol concentration of B from Statement-2. And we know how many A and B are present in the mixture from Statement A. So we can find the alcohol percentage in the mixture.

So answer is C.

And I hope it is correct. Bunuel please post your version of explanation :).
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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 07 Jan 2015, 11:42
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Bunuel wrote:

Tough and Tricky questions: Mixture Problems.



A bartender mixes drink A with drink B. If drink A contains 24 percent alcohol, what percentage of the mixture of drink A and B is alcohol?

(1) The ratio of drink A and drink B is 1:4
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A.

Kudos for a correct solution.


(1) We know the ratio of the drink mixture but we dont know the alcohol content of Drink B - Insuff
(2) We can find the amount of alcohol in Drink B but we don't Know the mixture ratio = Insuff

Combining both we can find all the unknowns

Ans C
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A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 09 Jan 2015, 06:26
Bunuel wrote:

Tough and Tricky questions: Mixture Problems.



A bartender mixes drink A with drink B. If drink A contains 24 percent alcohol, what percentage of the mixture of drink A and B is alcohol?

(1) The ratio of drink A and drink B is 1:4
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A.

Kudos for a correct solution.


OFFICIAL SOLUTION:

Statement (1) gives us the quantity of each of the drinks in the new cocktail. Statement (2) gives us the connection between the alcohol concentration of Drink B and the mixed cocktail. To solve this question, try using the statements and see what happens.

Let us use the statement (1) by itself. If the ratio of A to B is 1:4, then there is one part A to 4 parts B. This means if there are five parts total, one part is A and 4 parts are B, or 1/5 is A and 4/5 is B. A = .2 and B = .8. This information doesn’t tell us enough to solve the question because we don’t know the alcohol concentration of drink B. Statement (1) is not sufficient (meaning choices A and D can be ruled out).

Let us use the statement (2) by itself. It tells us the concentration in drink B. It is 1.5 times greater than drink A. The concentration is drink A is 24%, therefore the concentration in drink B = .24 × 1.5 = .36 (36%) alcohol. This information doesn’t tell us enough to solve the question because we don’t know the quantity of each drink in the new cocktail. Statement (2) is not sufficient (meaning choice B can be ruled out).

If we use the both statements together, we can set up an equation for this, using the following amounts:
Suppose the new mixture is 1. Then Alcohol in Drink A = .24 × .2, Alcohol in Drink B = .36 × .8. We can get the percentage of alcohol in the new mixture: (.048 + .288) / 1 = .336 (33.6%).

Since we required both statements to answer this question, the correct answer is (C).
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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 09 Jan 2015, 20:17
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To answer, we must know the percentage alcohol in B as well as the ratio of A/B in the drink.

(1) Tells us the ratio, but not the % by vol. of alcohol. NOT sufficient.

Strike A/D.

(2) Gives us enough info to figure out the % by vol. in B, but we don't know the requisite ratio. NOT sufficient.

Strike B.

(1/2) We know both pieces of information necessary to solve the problem if need be.

The answer is C.
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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 12 Mar 2015, 23:58
from 2) find the alcohol percentage of second drink and from 1) we get the total mixture and after that we can find the total strength of alcohol of the mixture. hence C.
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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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New post 06 Jun 2015, 06:37
I got confused by marked line
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A

http://english.stackexchange.com/questi ... -than?rq=1
Both the usage conveys same meaning.
Jack has 3 times as many sweets as John.
Jack has 3 times more sweets than John.

If John has 5 sweets then john will have 15.
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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent  [#permalink]

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Re: A bartender mixes drink A with drink B. If drink A contains 24 percent   [#permalink] 22 Dec 2018, 16:31
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