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:
1. Pastries may have nuts, chocolate, both, or neither.
2. 400 pastries were sold Friday, and if of those, 60% contained chocolate.
=> Number of pastries with chocolate = 60% of 400 = 240
Question: How many of those sold contained only nuts?
Say,
N = # of pastries with nuts;
C = # of pastries with chocolate = 240 (given);
B = # of pastries with nuts & chocolate;
X = # of pastries with neither nuts nor chocolate
Thus,
400 = N + C - B + X
400 = N + 240 - B + X
160 = N - B + X
We have to determine the value of the number of pastries contained only nuts, it means we have to get the value of (N - B).
If we get the value of X, we get the value of (N - B).
Question rephrased: How many pastries have neither nuts nor chocolate or what is the value of X?
Let's take each statement one by one.
(1) The number of pastries containing neither is one-fourth of the number containing chocolate.
=> X = C/4 = 240/4 = 60. Sufficient.
(2) One-third of pastries sold containing chocolate also contained nuts.
=> B = C/3 = 240/3 = 80.
We can't get the value of X. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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