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Of the students who eat in a certain cafeteria, each student [#permalink]
03 Jun 2010, 23:29

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Difficulty:

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Question Stats:

64% (02:07) correct
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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

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

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 6311 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 lima1-\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.

Of the students who eat in a certain cafeteria, each student [#permalink]
17 Aug 2010, 13:19

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 lima1-\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 lima1-\frac{3}{5}=\frac{2}{5} like sprout, or _________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

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 lima1-\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 lima1-\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.

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 _________________

Re: Of the students who eat in a certain cafeteria, each student [#permalink]
29 Sep 2013, 08:51

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Of the students who eat in a certain cafeteria, each student [#permalink]
29 Sep 2014, 20:57

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:

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 lima1-\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.

Let Total Student are t. 2/3 t dislike lima bean so 1/3 likes lima bean Now 3/5 * 2/3 *t dislike sprout = 6/15*t dislike both LB and BS

Now we know that how many dislike and Like LB and that dislike both LB and BS But we do not know how many like BS. I struck here and selected E wrongly.

Can you please explain in easy language. I did not get the solution _________________

Re: Of the students who eat in a certain cafeteria, each student [#permalink]
30 Sep 2014, 00:17

Expert's post

him1985 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:

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 lima1-\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.

Let Total Student are t. 2/3 t dislike lima bean so 1/3 likes lima bean Now 3/5 * 2/3 *t dislike sprout = 6/15*t dislike both LB and BS

Now we know that how many dislike and Like LB and that dislike both LB and BS But we do not know how many like BS. I struck here and selected E wrongly.

Can you please explain in easy language. I did not get the solution

We need to find how many students like brussels sprouts but dislike lima beans (box in red in my solution). Each statement is sufficient to find this value as shown above. Can you please tell me what is unclear there? _________________

Re: Of the students who eat in a certain cafeteria, each student [#permalink]
22 Nov 2014, 08:45

I have a question , this is a subtle concept but i guess very important.

Like in this question , i was left little misled by the work either they like or dislike Limabean , and either they like or dislike Sproat. So i thought Neither will be 0

So when do we need to identify the Neither case . I thought here also there will be no neither case ie neither like limabean and sproat.

But i see all the 4 boxes in matrix are filled .

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 lima1-\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.

Re: Of the students who eat in a certain cafeteria, each student [#permalink]
22 Nov 2014, 09:00

Expert's post

hanschris5 wrote:

I have a question , this is a subtle concept but i guess very important.

Like in this question , i was left little misled by the work either they like or dislike Limabean , and either they like or dislike Sproat. So i thought Neither will be 0

So when do we need to identify the Neither case . I thought here also there will be no neither case ie neither like limabean and sproat.

But i see all the 4 boxes in matrix are filled .

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 lima1-\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.

Each student either likes or dislikes lima beans, means that there are students who does NOT like lima beans. Each student either likes or dislikes brussels sprouts, means that there are students who does NO like brussels sprouts.

Thus, there might be students who does NOT like either lima beans or brussels sprouts. _________________

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