BTGmoderatorDC wrote: ↑Fri May 22, 2020 12:56 am
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
Say the # of pastries having only nuts = n; the # of pastries having only chocolate = c; the # of pastries having both nuts and chocolate = b; and the # of pastries having none = x
So, we have n + c + b + x = 400;
c + b = 60% pf 400
c + b = 240
Thus, from n + c + b + x = 400 and c + b = 240, we have n + x = 160.
We have to get the value of n.
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 + b)/4 = 240/4 = 60
Thus, from n + x = 160, we have n = 100. Sufficient.
(2) One-third of pastries sold containing chocolate also contained nuts.
=> b = (c + b)/3 = 240/3 = 80
However, with this information, we cannot get the value of n. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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