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Of the students who eat in a certain cafeteria, each student [#permalink]
 04 Jun 2010, 00:29
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54% (02:44) wrong based on 0 sessions
Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans? (1) 120 students eat in the cafeteria (2) 40 of the students like lima beans
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dimitri92 wrote: What is the best approach to tackle questions like these ? The question is basically asking how many dislikes Lima beans but like Sprouts.. Given 2/3 of the entire student poplutation dont like LIMA.. of these 3/5 DONT like sprouts..so 2/5 like sprouts.. 1) Given total students = 120 so 2/3 * 120 = 80 who dislikes lima beans out of these 2/5* 50 are the ones who likes sprouts but dislikes beans ... Hence Sufficient 2) 40 Likes beans so in thats means 120 is the total number of students... same logic as 1 -- Hence Sufficient hope this helps..!
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dimitri92 wrote: What is the best approach to tackle questions like these ? Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans? I'd advise to make a table: Attachment: 
Lima-Sprouts.JPG [ 11.66 KiB | Viewed 1476 times ]
Note that: "2/3 dislike lima beans" means 2/3 of total dislike lima; " of those who dislike lima beans, 3/5 also dislike brussels sprouts" means of those who dislike lima 1-\frac{3}{5}=\frac{2}{5} like sprout, or \frac{2}{3}*\frac{2}{5}=\frac{4}{15} of total dislike lima but like sprouts. So to calculate # of students who dislike lima but like sprouts we should now total # of students (t). (1) 120 students eat in the cafeteria --> t=120 --> x=\frac{4}{15}t=32. Sufficient. (2) 40 of the students like lima beans --> total students who like lima + total students who dislike lima = total --> 40+\frac{2}{3}t=t --> t=120 --> x=\frac{4}{15}t=32. Sufficient. Answer: D.
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Cafeteria: overlapping sets (mba.com) [#permalink]
17 Jul 2010, 10:39
Of the students who it in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussel sprouts. Of these students 2/3 dislike beans; and of those who dislike lima beans, 3/5 also dislike sprouts. How many of the students like sprouts but dislike lima beans?
(1) 120 students eat in cafeteria
(2) 40 of the students like lima beans
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Of the students who eat in a cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes sprouts. Of these students, 2/3 dislike lima beans, 2/5 also dislike sprouts.How many of the studentslike sprouts but dislike lima beans?
1. 120 students eat in the cafeteria
2. 40 of the students like lima beans
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Bunuel wrote: dimitri92 wrote: What is the best approach to tackle questions like these ? Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans? I'd advise to make a table: Attachment: Lima-Sprouts.JPG Note that: "2/3 dislike lima beans" means 2/3 of total dislike lima; " of those who dislike lima beans, 3/5 also dislike brussels sprouts" means of those who dislike lima 1-\frac{3}{5}=\frac{2}{5} like sprout, or \frac{2}{3}*\frac{2}{5}=\frac{4}{15} of total dislike lima but like sprouts. So to calculate # of students who dislike lima but like sprouts we should now total # of students (t). Answer: D. I didn't understand this part... means of those who dislike lima 1-\frac{3}{5}=\frac{2}{5} like sprout, or
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onedayill wrote: Bunuel wrote: dimitri92 wrote: What is the best approach to tackle questions like these ? Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans? I'd advise to make a table: Attachment: Lima-Sprouts.JPG Note that: "2/3 dislike lima beans" means 2/3 of total dislike lima; " of those who dislike lima beans, 3/5 also dislike brussels sprouts" means of those who dislike lima 1-\frac{3}{5}=\frac{2}{5} like sprout, or \frac{2}{3}*\frac{2}{5}=\frac{4}{15} of total dislike lima but like sprouts. So to calculate # of students who dislike lima but like sprouts we should now total # of students (t). Answer: D. I didn't understand this part... means of those who dislike lima 1-\frac{3}{5}=\frac{2}{5} like sprout, or If " of those who dislike lima beans, 3/5 (40%) also dislike brussels sprouts", hence rest of of those who dislike lima beans or 2/5 (60%) must like sprouts. As "2/3 of total dislike lima beans" then 2/3*2/5=4/15 of total dislike lima but like sprouts. Hope it's clear.
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D both 1 and 2 independently tell us what the total number of students is which in turn lets us calculate what is needed in the double-set matrix
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