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In a certain box of cookies, 3/4 of all the cookies have nuts and 1/3 [#permalink]
Bunuel wrote:
In a certain box of cookies, \(\frac{3}{4}\) of all the cookies have nuts and \(\frac{1}{3}\) of all the cookies have both nuts and fruit. What fraction of all the cookies in the box have nuts but no fruit?

(A) \(\frac{1}{4}\)

(B) \(\frac{5}{12}\)

(C) \(\frac{1}{2}\)

(D) \(\frac{7}{12}\)

(E) \(\frac{5}{6}\)



Let total cookies be 12x
Cookies with nuts = 9x
Cookies with nuts and fruits = 4x
Cookies with nuts but not fruits = 5x
Fraction of cookies with nuts but not fruits \(= \frac{5x}{12x} = \frac{5}{12}\)

IMO B

Originally posted by Kinshook on 17 Jul 2019, 00:53.
Last edited by Kinshook on 17 Jul 2019, 02:50, edited 1 time in total.
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Re: In a certain box of cookies, 3/4 of all the cookies have nuts and 1/3 [#permalink]
Kinshook, 1/3 of all the cookies has nuts and fruit, making it 4x. Let me know if I am incorrect.

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Re: In a certain box of cookies, 3/4 of all the cookies have nuts and 1/3 [#permalink]
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Re: In a certain box of cookies, 3/4 of all the cookies have nuts and 1/3 [#permalink]
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