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# Of the students who eat in a certain cafeteria, each student either

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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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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).

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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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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).

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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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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:

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.

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
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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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 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.

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?
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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 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.

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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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 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.

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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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Attached is a visual that should help. Remember that all sums of "X" and "not X" must equal 1, so when the question tells us that 2/3 of students dislike lima beans, it is of course also telling us that 1/3 of students like lima beans.

Likewise, when the question tells us "of those who dislike lima beans, 3/5 also dislike brussels sprouts," it is also telling us that of those who dislike lima beans, 2/5 also like brussels sprouts.

Other than that, getting this question right is mostly a matter of organizing your information well using a matrix. See below.
Attachments

Screen Shot 2017-09-14 at 11.59.34 AM.png [ 275.19 KiB | Viewed 121867 times ]

Originally posted by mcelroytutoring on 05 Apr 2016, 22:34.
Last edited by mcelroytutoring on 15 Sep 2017, 09:41, edited 8 times in total.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
@[quote="Bunuel"]
"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).

Hi Bunuel,

Can you please elaborate this part? I particularly did not understand the 2/3*2/5 part? Why are we doing this?
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
ashikaverma13 wrote:
@
Bunuel wrote:
"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).

Hi Bunuel,

Can you please elaborate this part? I particularly did not understand the 2/3*2/5 part? Why are we doing this?

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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
I have a question. Why does nobody take into account the ones who like lima beams, do they like the brussels sprouts or not? We do not know that and for me it is a msileading question because if some of the ones who like lia beams also like brussel sprouts then we do not know how many of them do like and how many do not. So the right anwser must be E
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
bernardoassensio wrote:
I have a question. Why does nobody take into account the ones who like lima beams, do they like the brussels sprouts or not? We do not know that and for me it is a msileading question because if some of the ones who like lia beams also like brussel sprouts then we do not know how many of them do like and how many do not. So the right anwser must be E

We have table which takes into account all possible cases:
1. Those who likes lima and likes sprouts;
2. Those who likes lima and do NOT like sprouts;
3. Those who do NOT like lima and likes sprouts;
4. Those who do NOT like lima and do NOT like sprouts.

You can check almost all the solutions from previous page for that. For example, this one: https://gmatclub.com/forum/of-the-stude ... ml#p733320 So, your doubt/question is not clear.

P.S. This is a GMAT Prep question, so it's as clear and unambiguous as it gets.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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dimitri92 wrote:
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

Attachment:
del5.jpg

"Of the students who eat in a certain cafeteria, ... "
Say T students eat in the cafeteria

"Of these students, 2/3 dislike lima beans"
(2/3)*T dislike Lima

"and of those who dislike lima beans, 3/5 also dislike brussels sprouts"
Of (2/3)T, (3/5) also dislike brussels so (3/5)*(2/3)T = (2/5)T dislike brussels
We don't know about the rest of the (T/3) that how many of them dislike brussels.

"How many of the students like brussels sprouts but dislike lima beans?"
(2/3)T dislike Lima and (2/5)th of these like Brussels (since (3/5)th of these do not like Brussels)
So (4/15)T dislike Lima but like Brussels.

(1) 120 students eat in the cafeteria
This gives us the value of T. We need to find (4/15)T which we can now. Sufficient

(2) 40 of the students like lima beans
(2/3)T dislike Lima so (1/3)T like Lima. If (1/3)T = 40, we get T = 120.
Again, we can now find (4/15)T. Sufficient.

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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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dimitri92 wrote:
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

This is overlapping sets, since all the students like and dislikes both foods. I drew a table with the givens and then filled in with information from 1) and 2)

Both 1) and 2) give us sufficient information to solve for x, so C
It's pretty quick to check the table to make sure everything matches up.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
We're told that 2/3 dislike lime beans. Therefore 1/3 like lima beans.

Out of the 2/3 who dislike lima beans, 3/5 also dislike brussels sprouts. This means of this subset of the 2/3 portion, 60% don't like brussels sprouts (we don't need to calculate the exact total)

How many of the students like brussels sprouts but dislike lima beans? To answer this question we need to know either the number of students that like lima beans (1/3 of the school population) or the total number of students (3/3 of the school population).

(1) Directly gives us our answer. SUFFICIENT.

(2) Directly gives us our answer. SUFFICIENT.

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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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For those who are looking for a simple logical solution,

condition 1): Total 120 students

120*2/3 = 80 dislike lima
80*3/5 = 48 dislike brussels

so people who dislike lima but like brussels = 80-48 = 32 like brussels.

condition 2): 40 students like lima

(1-2/3)*total = 40
total = 120 then it follows the same logic as condition 1

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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
Bunuel wrote:
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 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.

What should be kept in mind while choosing the heads of the table.

I made the table as follows after which I lost the question completely.

.............Lima Beans | Brussel Sprouts
Like
.............
Dislike
.............
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Of the students who eat in a certain cafeteria, each student either [#permalink]
The best representation for this that I have found is a layered representation. Think of each layer as a choice each student has to make (one for Lima Beans and the other for Brussels Sprouts)

................... Like .......... Dislike
Lima Beans...1/3 .............2/3
..................L....DL;;;.....L..........DL
Brussel.....na....na;;;;..4/15.....2/5 [(2/3) * (3/5)]

So if you know either the overall student population, or the population liking/disliking Lima Beans, you can get the answer, which is basically the ratio 4/15 (arrived at by subtracting 2/5 from 2/3)

Why the traditional matrix form is a little difficult to use for this is because, it does not neatly segregate into the 4 distinct segments LB (Like-Dislike) & BS (Like-Dislike). This question instead, makes you break down the decision choice into two 'levels'. That is why the above-mentioned sequential approach makes this question very easy to crack. Think about it and you will know the subtlety of the approach.
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