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

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Of the students who eat in a certain cafeteria, each student either [#permalink] New post   04 Jun 2010, 00:29
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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

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Of the students who eat in a certain cafeteria, each student either [#permalink] New post   04 Jun 2010, 07:42
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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?

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.

Answer: D.


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Of the students who eat in a certain cafeteria, each student either [#permalink] New post   Updated on: 16 Apr 2021, 06:24
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OluOdekunle 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


We can use the Double Matrix Method to solve this question. This technique can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of students, and the two characteristics are:
- like Brussels sprouts or dislike Brussels sprouts
- like lima beans or dislike lima beans

So, we can set up our diagram as follows:


Target question: How many of the students like Brussels sprouts but dislike lima beans?
Let's place a STAR in the box representing those students who like Brussels sprouts but dislike lima beans.


Since we don't know the TOTAL NUMBER of students, let's let x represent the total student population. So, we'll add that to our diagram as well.


Given: 2/3 dislike lima beans
So, (2/3)x = total number of students who dislike lima beans
This means the other 1/3 LIKE lima beans. In other words, (1/3)x = total number of students who LIKE lima beans.
We'll add that to the diagram:


Given: Of those who dislike lima beans, 3/5 also dislike Brussels sprouts
If (2/3)x = total number of students who dislike lima beans, then (3/5)(2/3)x = total number of students who dislike lima beans AND dislike Brussels sprouts.
(3/5)(2/3)x simplifies to (2/5)x, so we'll add that to our diagram:


Finally, since the two boxes in the right-hand column must add to (2/3)x, we know that the top-right box must = (4/15)x [since (2/3)x - (2/5)x = (4/15)x]
So, we can add that to the diagram:


Great! We're now ready to examine the statements.

Statement 1: 120 students eat in the cafeteria
In other words, x = 120
Plug x = 120 into the top-right box to get: (4/15)(120) = 32
So, there are 32 students who like Brussels sprouts but dislike lima beans.

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 40 of the students like lima beans.
The left-hand column represents students who like lima beans.

In total, (1/3)x = total number of students who LIKE lima beans.
So, statement 2 is telling us that (1/3)x = 40
We can solve the equation to conclude that x = 120
Once we know the value of x, we can determine the number of students who like Brussels sprouts but dislike lima beans (we already did so in statement 1)
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

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Originally posted by BrentGMATPrepNow on 25 Aug 2016, 06:51.
Last edited by BrentGMATPrepNow on 16 Apr 2021, 06:24, edited 2 times in total.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   17 Aug 2010, 14:19
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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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   18 Aug 2010, 04:35
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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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   29 Sep 2014, 21:57
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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.

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
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   30 Sep 2014, 01:17
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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.

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?
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   22 Nov 2014, 09:45
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 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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   22 Nov 2014, 10:00
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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.

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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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   Updated on: 15 Sep 2017, 09:41
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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
Screen Shot 2017-09-14 at 11.59.34 AM.png [ 275.19 KiB | Viewed 121368 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] New post   22 Apr 2017, 15:50
@[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] New post   23 Apr 2017, 03:32
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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] New post   30 Mar 2018, 02:08
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] New post   30 Mar 2018, 04:37
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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] New post   01 Dec 2018, 02:29
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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

Show Spoiler
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.

Answer (D)
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   19 Jan 2019, 09:13
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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] New post   15 Nov 2020, 15:33
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.

Answer is D.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   23 Nov 2020, 05:42
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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

Answer: Both alone are sufficent.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] New post   12 Apr 2021, 02:01
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.

Answer: D.





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] New post   21 Apr 2021, 08:57
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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Of the students who eat in a certain cafeteria, each student either [#permalink]
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