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17 Jul 2019, 00:25
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
86% (01:37) correct
14%
(01:58)
wrong
based on 136
sessions
History
Date
Time
Result
Not Attempted Yet
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) \(\frac{5}{6}\)
(A) \(\frac{1}{4}\)
(B) \(\frac{5}{12}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{5}{6}\)
17 Jul 2019, 00:39
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Let number of cookies be 60
Cookies with nuts = 45
Cookies with nuts and fruits =20
Cookies with nuts but not fruits = 25
25/60 = 5/12
Hence B
Posted from my mobile device
Cookies with nuts = 45
Cookies with nuts and fruits =20
Cookies with nuts but not fruits = 25
25/60 = 5/12
Hence B
Posted from my mobile device
Let number of cookies be 60
Cookies with nuts = 45
Cookies with nuts and fruits =20
Cookies with nuts but not fruits = 25
25/60 = 5/12
Hence B
Cookies with nuts = 45
Cookies with nuts and fruits =20
Cookies with nuts but not fruits = 25
25/60 = 5/12
Hence B
