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At a local coffee shop, pastries may have nuts, chocolate, both, or ne
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15 Apr 2018, 09:42
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49% (01:21) correct 51% (01:32) wrong based on 47 sessions
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At a local coffee shop, pastries may have nuts, chocolate, both, or ne
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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 onefourth 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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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne
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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 onefourth 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
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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
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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
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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 onefourth of the number containing chocolate.
(2) One third of pastries sold containing chocolate also contained nuts. Answer: A. Please see the attachment!
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File comment: ccoffee ncnot coffee nnuts nnnot nuts x  desired value.
Nuts&Choco.jpeg [ 589.41 KiB  Viewed 332 times ]
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Re: At a local coffee shop, pastries may have nuts, chocolate, both, or ne &nbs
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