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At a local coffee shop, pastries may have nuts, chocolate,

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At a local coffee shop, pastries may have nuts, chocolate,

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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.

OA A

Source: Veritas Prep

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Let chocolate only pastry = c
Let both chocolate and nut pastry = x
Let Nut only pastry = n
Let neither nuts nor chocolate pastry = y
$$c+x+n+y=400$$
400 pastries for Friday sales
$$60\%of400=\frac{60}{100}\cdot400=240\ pastries\ contains\ chocolate$$
$$c+x=240$$

Question
How many of those sold contained only nuts? i.e find the value of n

Statement 1
The numbers of pastries containing neither is one-fourth of the number containing chocolate
$$g=\frac{1}{4}\left(c+x\right)$$
$$g=\frac{1}{4}\left(240\right)$$
$$=60\ pastes$$
Recall that
$$c+x+n+y\ =400\ $$ where c+x =240 and y = 60
$$240+n+60=400$$
$$300+n=400$$
$$n=400-300$$
$$n=100\ pastries$$
Statement 1 is SUFFICIENT.

Statement 2
One- third of pastes sold containing chocolates also contained nuts
$$x=\frac{1}{3}of\ \left(c+x\right)$$
$$where\ c+x=\ 240\ pastries\ sold\ contains\ chocolate$$
$$x=\frac{1}{3}\cdot240=80$$
This is not enough to calculate the value of n as the value of y is unknown hence statement 2 is INSUFFICIENT.
Statement 1 alone is SUFFICIENT.

$$Answer\ is\ Option\ A\ $$

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