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In a certain box of cookies, of all the cookies have nuts an
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15 Jul 2019, 02:00
15 Jul 2019, 02:00
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Question Stats:
82% (01:46) correct
17% (01:17) wrong based on 104 sessions
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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) \(\frac{5}{6}\)
Re: In a certain box of cookies, of all the cookies have nuts an
[#permalink]
20 Jul 2019, 14:59
20 Jul 2019, 14:59
2
Let, total cookies 120. Cookies with Nut= 90, cookies with Nut and fruit= 40. So cookies with only Nut = 90-40=50.
So the fraction of the cookies in the box with Nut but no Fruit= 50/120=5/12
So the fraction of the cookies in the box with Nut but no Fruit= 50/120=5/12
Re: In a certain box of cookies, of all the cookies have nuts an
[#permalink]
21 Jul 2019, 19:55
21 Jul 2019, 19:55
1
Let's call the total number of cookies T. The number of cookies with nuts is 3/4 T, and the number of cookies with nuts+fruit is 1/3 T.
Cookies with nuts = (nuts with fruit) + (nuts without fruit). So:
3/4 T = 1/3 T + (fraction of nuts without fruit) T
3/4 T - 1/3 T = (fraction of nuts without fruit) T
5/12 T = (fraction of nuts without fruit) T
-Jordan Wolbrum
Cookies with nuts = (nuts with fruit) + (nuts without fruit). So:
3/4 T = 1/3 T + (fraction of nuts without fruit) T
3/4 T - 1/3 T = (fraction of nuts without fruit) T
5/12 T = (fraction of nuts without fruit) T
-Jordan Wolbrum
