It is currently 23 Feb 2018, 02:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 323
Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

03 Jun 2010, 23:29
10
This post received
KUDOS
57
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

64% (00:56) correct 36% (01:03) wrong based on 1305 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Attachments

del5.jpg [ 23.82 KiB | Viewed 51517 times ]

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Download the Ultimate SC Flashcards

Senior Manager
Joined: 25 Jun 2009
Posts: 294
Re: bowled [#permalink]

### Show Tags

04 Jun 2010, 02:58
2
This post received
KUDOS
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..!
Math Expert
Joined: 02 Sep 2009
Posts: 43891
Re: bowled [#permalink]

### Show Tags

04 Jun 2010, 06:42
19
This post received
KUDOS
Expert's post
43
This post was
BOOKMARKED
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 57921 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.
_________________
Senior Manager
Joined: 25 Feb 2010
Posts: 445
Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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
_________________

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

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

Math Expert
Joined: 02 Sep 2009
Posts: 43891
Re: bowled [#permalink]

### Show Tags

18 Aug 2010, 03:35
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________
Current Student
Joined: 20 Jan 2014
Posts: 175
Location: India
Concentration: Technology, Marketing
Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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
_________________

Consider +1 Kudos Please

Math Expert
Joined: 02 Sep 2009
Posts: 43891
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

30 Sep 2014, 00:17
Expert's post
1
This post was
BOOKMARKED
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?
_________________
Intern
Joined: 03 Jul 2014
Posts: 17
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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.

Answer: D.
Math Expert
Joined: 02 Sep 2009
Posts: 43891
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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.
_________________
Current Student
Joined: 14 May 2014
Posts: 45
Schools: Broad '18 (WA)
GMAT 1: 700 Q44 V41
GPA: 3.11
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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
Intern
Joined: 02 Dec 2015
Posts: 1
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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...........
Manager
Joined: 26 Jan 2015
Posts: 93
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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.
_________________

Kudos is the best way to say Thank you! Please give me a kudos if you like my post

Intern
Joined: 29 Nov 2015
Posts: 13
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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?
Attachments

lima beans.JPG [ 17.25 KiB | Viewed 38591 times ]

Director
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 515
Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

05 Apr 2016, 21:34
2
This post received
KUDOS
1
This post was
BOOKMARKED
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 9220 times ]

_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and online via Skype, since 2002.

You can download my official test-taker score report directly from the Pearson Vue website: https://tinyurl.com/y8zh6qby Date of Birth: 09 December 1979.

GMAT Action Plan - McElroy Tutoring

Last edited by mcelroytutoring on 15 Sep 2017, 08:41, edited 8 times in total.
Board of Directors
Joined: 17 Jul 2014
Posts: 2734
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

09 Apr 2016, 13:53
what is x?

1. x=120 sufficient.
2. x/3=40 sufficient.
Attachments

q1.jpg [ 19.54 KiB | Viewed 17593 times ]

Intern
Joined: 16 Jul 2016
Posts: 35
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

02 Aug 2016, 16:21
I took the table approach, but I used slightly different values as well as decimals.

If we can find x, we have the answer.
Attachments

GMAT - Sheet1.pdf [15.23 KiB]
Downloaded 82 times

 To download please login or register as a user

SVP
Joined: 11 Sep 2015
Posts: 2058
Location: Canada
Re: Of the students who eat in a certain cafeteria, each [#permalink]

### Show Tags

25 Aug 2016, 05:51
4
This post received
KUDOS
Expert's post
Top Contributor
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 =
[Reveal] Spoiler:
D

RELATED VIDEO

MORE PRACTICE

_________________

Brent Hanneson – Founder of gmatprepnow.com

Manager
Joined: 19 Aug 2016
Posts: 150
Location: India
GMAT 1: 640 Q47 V31
GPA: 3.82
Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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?
_________________

Consider giving me Kudos if you find my posts useful, challenging and helpful!

Math Expert
Joined: 02 Sep 2009
Posts: 43891
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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.

Hope it's clear.
_________________
Intern
Joined: 10 Jul 2017
Posts: 1
Re: Of the students who eat in a certain cafeteria, each student [#permalink]

### Show Tags

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.
Attachments

Image.png [ 33.37 KiB | Viewed 371 times ]

Re: Of the students who eat in a certain cafeteria, each student   [#permalink] 18 Feb 2018, 02:01
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.