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Could we add across inequalities as I did below? If not, when is it allowed to add across inequalities? (1) c > 2p NS
(2) p > 12 NS
(1) + (2) c + p > 2p + 12 40 > 2p + 12 2p < 28 p < 14 Because (2) gives p > 12, with the above statement, p = 13 and we can solve for c. S
You can add inequalities as long as both the inequalities have the same sign. a > b c > d gives a+c > b + d Think logically: a is greater than b and c is greater than d so a+c will be greater than b+d because you are adding the larger numbers together. So what you have done above is correct.
If the inequality signs are different, you cannot add them.
Also, you can subtract inequalities when they have opposite signs but prefer not to do that to avoid confusion. Just flip the sign of one inequality by multiplying it by -1 and then add them up.
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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.
Hi bunnel,
I need clarification here
2/3 are either pigs or cows
i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P
i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P
Please clarify.
Thanks.
If I may add, 40 are either pigs or cows does not mean that either all 40 are pigs or all 40 are cows. It means of all the 40, some are pigs and the rest are cows.
Say, if you say that 90% of the students are either from Michigan or Ohio, it means that the rest of the 10% are from other states but 90% belong to these two states. How many of the 90% are from Michigan and how many are from Ohio, we don't know but we know that together they account for 90% of the class.
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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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11 Jul 2014, 10:25
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.
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.
Of the 60 animals on a certain farm, 40 are either pigs or cows means that from the remaining 20 animals neither is either pig or cow.
_________________
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."
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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30 Aug 2014, 11:31
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?
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]
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29 Sep 2014, 07:58
1
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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
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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09 Jul 2016, 18:54
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SC teaches us to avoid the ambiguous use of grammar idioms. Yet in this question GMAT makes a very unclear stem: 2/3 are either pigs or cows. Which can be interpreted as 2/3 pigs or 2/3 cows. In my opinion the question could have stated:
Of the animals on a certain farm, 2/3 are pigs AND cows to avoid such ambiguity
_________________
Never stop learning, because life never stops teaching.
Can I interpret Of the 60 animals on a certain farm, 2/3 are either pigs or cows. using combinations principle that EITHER / OR means + , AND means multiply. 2/3 of 60 are pigs or cows -> 14= p+c
Clearly an animal can not be pig and cow, and I am not concerned if there are other animals on field.
_________________
It's the journey that brings us happiness not the destination.
Can I interpret Of the 60 animals on a certain farm, 2/3 are either pigs or cows. using combinations principle that EITHER / OR means + , AND means multiply. 2/3 of 60 are pigs or cows -> 14= p+c
Clearly an animal can not be pig and cow, and I am not concerned if there are other animals on field.
Yes here it will be "+". but the highlighted value is incorrect
gmatclubot
Re: Of the 60 animals on a certain farm, 2/3 are either cows or
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03 Jan 2018, 03:29