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.
OA A
Source: Veritas Prep
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At a local coffee shop, pastries may have nuts, chocolate,
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BTGmoderatorDC
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From Statement 1, since 60% contain chocolate, one quarter of that percentage, or 15%, contain neither chocolate nor nuts. That accounts for 75% of the pastries, and the rest, or 25%, must contain nuts but not chocolate. Since there are 400 pastries in total, 100 contain only nuts, and Statement 1 is sufficient.
Statement 2 tells us that 20% of the pastries contain both chocolate and nuts, which doesn't help us.
The answer is A.
Statement 2 tells us that 20% of the pastries contain both chocolate and nuts, which doesn't help us.
The answer is A.
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This is a Venn Diagram question.BTGmoderatorDC 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.
OA A
Source: Veritas Prep
Given, Universal set \(U = 400\)
Chocolate \(C = 0.6 * 400 = 240\)
\(N =\) nuts
To find: \(N-C\)
Statement 1: not\((C u N) = 0.25 \cdot C = 60\)
now, \(C\) u \(N = 400-60 = 340\)
therefore, \(N-C = 340 - C = 100\)
Sufficient. \(\color{green}\checkmark\)
Statement 2: only information about the distribution in set \(C\) is given. i.e. \(C-N = 160\), \(C\) intersection \(N = 80\).
So, not sufficient. \(\color{red}{\Large{\times}}\)
Hence, the correct answer is __A__
