[phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 168: array_key_exists() expects parameter 2 to be array, null given [phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 169: array_key_exists() expects parameter 2 to be array, null given [phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 170: array_key_exists() expects parameter 2 to be array, null given
Of the 60 animals on a certain farm, 2/3 are either cows or : GMAT Data Sufficiency (DS) - Page 3

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
11 Jul 2014, 17:20

Expert's post

rbaudoin10 wrote:

What about the 20 extra animals ? Why can't they be cows or pigs as well within that set ? Nothing is mentioned about these 20 animals. We just know that 40 animals are either pigs or cows. Does anyone get my point ?

Having 13 pigs, 27 cows, and 20 other animals is a possiblity.

Having 14 pigs, 30 cows and 16 other animals is another possibility.

Following that logic, E is the right answer.

To add to what Bunuel said:

It is similar to: Of the 100 people in a room, 40 are men. What does this mean to you? It means that 60 are not men, right?

Similarly, Of the 60 animals on a certain farm, 40 are either pigs or cows. This means that rest of the 20 are neither pigs nor cows!

When you are given concrete numbers, it implies that the group has been considered. If only some people were considered, you would have been given "Of the 100 people in a room, at least 40 are men." _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
30 Aug 2014, 10:31

Bunuel wrote:

shrivastavarohit wrote:

Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer.

Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS.

1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful.

Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient.

Posted from my mobile device

Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

(1) The farm has more than twice as many cows as it has pigs --> so we have is c>2p and not c=2p --> as c+p=40 (p=40-c and c=40-p) --> 40-p>2p, 13.3>p, p_{max}=13 and c_{min}=27. Many combinations are possible: (27,13), (28, 12), ... Not sufficient.

(2) p>12. Not sufficient

(1)+(2) p>12 but p_{max}=13, hence p=13 --> c=27. Sufficient.

Answer: C.

One more thing: if the ratio indeed were c=2p, then the question would be flawed as solving c=2p and c+p=40 gives p=13.3, but # of pigs can not be a fraction it MUST be an integer.

Hi Bunuel,

Two questions:

-How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct?

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
31 Aug 2014, 23:48

Expert's post

russ9 wrote:

Bunuel wrote:

shrivastavarohit wrote:

Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer.

Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS.

1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful.

Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient.

Posted from my mobile device

Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

(1) The farm has more than twice as many cows as it has pigs --> so we have is c>2p and not c=2p --> as c+p=40 (p=40-c and c=40-p) --> 40-p>2p, 13.3>p, p_{max}=13 and c_{min}=27. Many combinations are possible: (27,13), (28, 12), ... Not sufficient.

(2) p>12. Not sufficient

(1)+(2) p>12 but p_{max}=13, hence p=13 --> c=27. Sufficient.

Answer: C.

One more thing: if the ratio indeed were c=2p, then the question would be flawed as solving c=2p and c+p=40 gives p=13.3, but # of pigs can not be a fraction it MUST be an integer.

Hi Bunuel,

Two questions:

-How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct?

There are total of 40 pigs and cows. We know that there are at most 13 pigs (13 or less). Thus there are at least 27 cows.

Or: p + c = 40 and p <= 13 --> (40 - c) <= 13 --> c >= 27.

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
29 Sep 2014, 06:58

1

This post received KUDOS

Bunuel wrote:

ngoctraiden1905 wrote:

From your ans I think the stem means 2/3 of 60 animals are both pigs and cows, not either pigs or cows. I think there is some problem with stem here?

How it's possible for an animal to be BOTH a pig and a cow? 2/3 of 60 animals are either pigs or cows. So there are total of 40 cows and pigs. Yansta8's solution is correct.

C it is.

Maybe they are half cow, half bear, half pig.... Cowbearpig _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
25 Nov 2014, 03:17

I guess we cannot solve it using Matrix method as , Both cannot be Cow and Pig .

Is there any alternative method to solve this .

Thanks

Bunuel wrote:

Syed wrote:

yangsta8 wrote:

Question stem says that 2/3 of 60 are pigs or cows. That means 40 animals are pigs or cows. So all we need to have sufficiency is either number of pigs or number of cows.

1) Farm has more than 2 cows for 1 pig (at least 26.6 animals of the 40 are cows). This just tells us that the farm has at least 27 pigs and that the max number of cows is 13. For example the farm could have 30 pigs and 10 cows. Not a definitive number. Insufficient.

2) Farm has more than 12 pigs. Again not enough info. Insuff

1+2) Statements together tell us: 12 < number of pigs is <=13 Which means number pigs = 13 Number of cows = 27.

ANSWER = C.

I do agree with 'yangsta8' that the answer is 'C' but (if I am NOT wrong), Stmnt#1 says the number of Cows are more than twice than the pigs. Thus, the cows could be 27 and Pigs could be 13; and For example, the farm could have 30 cows and 10 pigs. ------------------------------- Hi Bunuel and Yangsta8 - Please correct me if I am wrong (which is very much possible)!

c+p=40

(1) c>2p --> min # of cows is 27 and max # pigs is 13, so there can be any combination not violating this and totaling 40. Not sufficient (2) p>12 Not sufficient

(1)+(2) p>12 but max of p is 13, hence p=13 --> c=27

You are right there can be 27 cows (min) and 13 pigs (max) or 30 cows and 10 pigs.

Think there was simple typo from yangsta8.

gmatclubot

Re: Of the 60 animals on a certain farm, 2/3 are either cows or
[#permalink]
25 Nov 2014, 03:17

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...

This is a long overdue post! A lot of Indian applicants, having scheduled interviews in March, reached out to me asking about my interview experience with Kellogg. I had a...