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At a local coffee shop, pastries may have nuts, chocolate, both, or ne

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At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 15 Apr 2018, 09:42
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Question Stats:

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At a local coffee shop, pastries may have nuts, chocolate, both, or neither. If 400 pastries were sold Friday, and if of those, 60% contained chocolate how many of those sold contained only nuts?

(1) The number of pastries containing neither is one-fourth of the number containing chocolate.

(2) One third of pastries sold containing chocolate also contained nuts.

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At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post Updated on: 15 Apr 2018, 11:27
Bunuel wrote:
At a local coffee shop, pastries may have nuts, chocolate, both, or neither. If 400 pastries were sold Friday, and if of those, 60% contained chocolate how many of those sold contained only nuts?

(1) The number of pastries containing neither is one-fourth of the number containing chocolate.

(2) One third of pastries sold containing chocolate also contained nuts.


Items belonging to 2 overlapping sets can be in 4 distinct parts: the overlap between the sets, in only one set, in only the other set, or in neither set.
To calculate the part that is only nuts we need to know the other 3 (overlap, only chocolate, neither).
We'll look for a statement that gives us this information, a Logical approach.

Note that we're told the total amount of chocolate, that is the 'overlap' + 'only chocolate'. So all we need to know is the 'neither' part.

(1) this gives us the 'neither' part - enough!
Sufficient.

(2) this gives us the 'overlap' part - not enough.
Insufficient.

(A) is our answer.
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Originally posted by DavidTutorexamPAL on 15 Apr 2018, 10:20.
Last edited by DavidTutorexamPAL on 15 Apr 2018, 11:27, edited 1 time in total.
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 15 Apr 2018, 11:20
Bunuel wrote:
At a local coffee shop, pastries may have nuts, chocolate, both, or neither. If 400 pastries were sold Friday, and if of those, 60% contained chocolate how many of those sold contained only nuts?

(1) The number of pastries containing neither is one-fourth of the number containing chocolate.

(2) One third of pastries sold containing chocolate also contained nuts.


So basically there are four kinds of pastries: (1) having Only Chocolate, (2) having both Chocolate/Nuts, (3) having Only Nuts, (4) having Neither. We have to determine the number of type (3) pastries, i.e., having Only Nuts. And we know that (1) + (2) + (3) + (4) = 400

We are given that out of 400, 60% or 0.6*400 = 240 contain chocolates. So these 240 are sum of type (1) and (2). So we just need to know (4) to calculate the value of (3).

Statement 1 - gives us the number of type (4) pastries. these are 1/4 * 240 = 60. So we can find type (3). Sufficient.

Statement 2 - tells us that type (2) are 1/3 of type (1). But this is not sufficient to calculate type (3).

Hence A answer
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 16 Apr 2018, 01:33
Why the answer cannot be C?
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 16 Apr 2018, 05:09
Priyanka8793 wrote:
Why the answer cannot be C?


Hello

If you check the above two solutions, first statement is sufficient to answer the question asked. Second statement is not sufficient on its own to answer.

when one or more of the statements is sufficient to answer the question, then we do NOT combine the statements. Statements have to be combined only when none of the statements is sufficient on its own to answer the question.

Thus the answer is A.
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 16 Apr 2018, 06:36
Bunuel wrote:
At a local coffee shop, pastries may have nuts, chocolate, both, or neither. If 400 pastries were sold Friday, and if of those, 60% contained chocolate how many of those sold contained only nuts?

(1) The number of pastries containing neither is one-fourth of the number containing chocolate.

(2) One third of pastries sold containing chocolate also contained nuts.


Answer: A.

Please see the attachment!
Attachments

File comment: c-coffee
nc-not coffee
n-nuts
nn-not nuts
x - desired value.

Nuts&Choco.jpeg
Nuts&Choco.jpeg [ 589.41 KiB | Viewed 1168 times ]


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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 14 Dec 2018, 20:09
Hello dear,

A coffee based business is certainly one of the best trades that you can think of. We are living in a world dominated by coffee drinkers. In estimation, around one out of five people go into a coffee shop every day to drink their favorite coffee beverages. My favorite coffee shop in Davis is Trailhead Coffee Shop, which offers organic fair trade products & Internet Cafe: Coffee, Tea, & Chocolates.
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne  [#permalink]

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New post 02 Jan 2019, 12:39
This is Venn Diagram question.
Given, Universal set U = 400
Chocolate C = 0.6 * 400 = 240
N = nuts
To find : N-C

S1: not(C u N) = 0.25 * C = 60
now, C u N = 400-60 = 340
therefore, N-C = 340 - C = 100
Sufficient

S2: only information about the distribution in set C is given. i.e. C-N = 160, C intersection N = 80.
So, not sufficient.

Ans A
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne   [#permalink] 02 Jan 2019, 12:39
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