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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 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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04 Jun 2010, 06:42

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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:

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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04 Jun 2010, 02:58

3

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

Re: Of the students who eat in a certain cafeteria, each student either
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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 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
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18 Aug 2010, 03:35

3

1

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.

Re: Of the students who eat in a certain cafeteria, each student either
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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 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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30 Sep 2014, 00:17

1

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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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 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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22 Nov 2014, 09:00

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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25 Jul 2015, 02:06

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.

how can we solve this using formula of set theory.

total= not like lima + not like sprouts - not like both

Re: Of the students who eat in a certain cafeteria, each student either
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05 Dec 2015, 20:05

Bingo, I got it at my first attempt. its D. One rule: If Answer is D, both statements will never conflict. So, 2/3 dislikes lima beans; it means 1/3 likes lima beans. The number is then (1/3*120) 40. And statement 2 provides us same number. So answer is D even without much calculation. Kudos...........

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09 Dec 2015, 10:32

moinbdusa wrote:

Bingo, I got it at my first attempt. its D. One rule: If Answer is D, both statements will never conflict. So, 2/3 dislikes lima beans; it means 1/3 likes lima beans. The number is then (1/3*120) 40. And statement 2 provides us same number. So answer is D even without much calculation. Kudos...........

There are going to be cases where for value questions: Stmnt 1 gives x=10 Stmnt 2 gives x=20 But still the answer is D. @Bunuel...please correct me if i am wrong.
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Re: Of the students who eat in a certain cafeteria, each student either
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29 Dec 2015, 22:49

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:

The attachment Lima-Sprouts.JPG is no longer available

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.

I think drawing a correct table here is key. The table I drew was not wrong but it wasn't right for the question asked. So how to ensure that you make the right table?

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Updated on: 15 Sep 2017, 08:41

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1

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 25895 times ]

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Of the students who eat in a certain cafeteria, each student either
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Updated on: 16 Apr 2018, 11:57

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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

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22 Apr 2017, 14:50

@[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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23 Apr 2017, 02:32

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.

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18 Feb 2018, 02:01

The key thing to remember that each student either likes or dislikes lima beans and the same applies for brussels sprouts.

So if we have ⅔ of the total who dislike LB and ⅗ of whom also dislike BS. This means that another portion of people (⅖) who dislike LB like BS. That is what basically is required to know. If we can find the total amount of people then we can solve the problem.

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