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

Target question: How many of the animals are cows?

Given: Of the 60 animals in a certain farm, 2/3 are either pigs or cows
Let P = # of pigs
Let C = # of cows
2/3 of 60 = 40, so we can say that P + C = 40

Statement 1: The farm has MORE THAN twice as many cows as it has pigs.
In other words, 2P < C
If we know 2P < C and P + C = 40, do we have sufficient information to find the value of C?
No. Consider these 2 conflicting cases:
Case a: P = 1 and C = 39, in which case there are 39 cows
Case b: P = 2 and C = 38, in which case there are 38 cows
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The farm has more than 12 pigs.
There's no way we can use this information to determine the number of cows.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 says that P > 12. So, let's examine some possibilities.
If P = 13, then C > 26 (from statement 1). So, C must equal 27 (since P + C = 40)
If P = 14, then C > 28 (from statement 1). In this case, P+C will be GREATER THAN 40, but we need P+C to EQUAL 40 (from the given information). So, P cannot equal 14.
In fact, for the same reasons, P cannot equal 15, 16, 17, etc. . .

So, the only case that's possible is for there to be 13 pigs and 27 cows
Since we can now answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

Given: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?
This tells us that 40 of the animals are either pigs or cows, which means the remaining 20 animals are neither pigs nor cows.

How's that help?

HI all
I am facing little difficulty in interpretation of the question, My specifc doubts are:-
1. Here either or means C+p = 40 , since a pig cannot be a cow at the same time. What if the question had a scenario, where a case of both was possible.....should we consider c+p- both = 40 in that case. I am haveing a doubt with either or statement
2. In case of question stating "What was the number of cows", what should we infer that it requires number of only cow ie cow - both or only cow + both, i am facing difficulty......
3."I. The farm has more than twice as many cows as it has pigs "
Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1

Pls help me in clearing my doubts...

Regards
Archit

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

Please clarify.

Thanks.

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.

I believe answer should be E.

As question demands number of cows among animals, we can form many other combinations from given 1st and 2nd statement.
For ex. Pigs = 30, Cows = 10.

S1: More than twice as many cows as pigs. 30 > 10*2
S2: Number of pigs more than 12. 30>12

Similarly many other combinations are possible like, (28,12), (32,8) etc.

Kindly help.

I have prepared a youtube video to explain this problem,

Click here to see the detailed solution of this problem

Hope it is clear.

We are told
C + P = 40

We are asked to find how many animals are cows
(1) C>2p

From the stem: P = 40-c
substitute in
C>2(40-c)
C>80-2c
3C>80
C>80/3
C> 26 2/3
C can be 27, 28 or 29 as we are simply told that C is MORE THAN twice P

If C=27, P=13
If C=28, P=12
If C=29, P=11

(2) P>12
Insufficient - heaps of combinations are available.
12 Pigs + 28 Cows
20 Pigs + 10 Cows

(1)+(2) Based on what we know, if P>12 then C can only be equal to 27

Thus, sufficient

I completely agree, I also think this question is stated in a confusing way. I also thought 2/3 were either cows or pigs, which is why I picked the wrong answer.

Especially in quantitive, questions should be worded without ambiguity.

I agree. the question framing is a bit confusing. It appears as if 2/3 is either pings or cows, meaning either pigs are 40 in number or cows are 40 in number

How can the answer be C,isn't it wrong to assume there are no animals which are neither pigs nor cows?

Given: P + C = 40.

We are asked to determine the value of C.

(1) The farm has more than twice as many cows as pigs

C > 2P

Using the given condition and the condition from statement (1), we have:

C + P > 3P
40 > 3P
P is less than 13.3

That is, the maximum number of pigs is 13. We cannot determine the number of cows because the number of pigs is an unknown quantity. INSUFFICIENT

(2) The farm has more than 12 pigs

P > 12

We cannot determine the number of cows because the number of pigs is an unknown quantity. INSUFFICIENT

Statements (1) and (2):

The farm has more than twice as many cows as pigs AND the farm has more than 12 pigs

Combining the two results from above, we know that the number of pigs must be greater than 12 and less than or equal to 13. This can only mean that the number of pigs is 13. Therefore, the number of cows is 27. SUFFICIENT

ANSWER: (C)

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 3 variables and 1 Equation: Let the original condition in a DS question contain 3 variables and 1 Equation. Now, 3 variables and 1 Equation would generally require 2 more equations for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 2 more equations to match the numbers of variables and equations in the original condition, the logical answer is C.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find the number of cows.

=> Let us assign variables: Animals(a), pigs(p), cows(c).

=> a = 60 ; \(\frac{2}{3}\) are either pigs or cows: \(\frac{2}{3}\) * 60 = 40 => p + c = 40

Second and the third step of Variable Approach: From the original condition, we have 3 variables (a, p, and c) + 1 Equation (p + c = 40).To match the number of variables with the number of equations, we need 2 more equations. Since conditions (1) and (2) will provide 2 equations, C would most likely be the answer.

Let’s take look at both condition together.

Condition(1) tells us that the farm has more than twice as many cows as pigs.

Condition(2) tells us that the farm has more than 12 pigs .

=> If p = 13, then cows > 2 * 13 = 26 such that p + c = 40. Therefore, c = 27

Since the answer is unique , both conditions combined together are sufficient by CMT 2.

Both conditions combined together are sufficient.

So, C is the correct answer.

Answer: C


SAVE TIME: By Variable Approach, when you know that we need two equations, we will directly combine the conditions to solve. We will save time in checking the conditions individually.

Cow or Pigs=60*2/3=40
C+P=40

(1) C>2P; Pig is 5 then cow is more than 10, i.e, can be 30, Pig is 8, Can be more than 16 that is 34 so INSUFFICIENT.

(2) No information about Cow, cow can be any number, 27, 26,.25 with respect to the number of Pig>12

Using both information (1) and (2)
P>12 and C>2P total will must be 40.
if Pig 13; Cow 27 Possible [C.>2P]
If Pig 14; Cow 29 total 43 Not Possible
Sufficient.

From the question stem, we know that there are a total of 40 pigs or cows. Of these 40, we need to ascertain the number of cows.
Let the number of cows be represented by c and the number of pigs be represented by p. So, we know, p + c = 40

From statement I alone, the farm has more than twice as many cows as pigs. Therefore,
c > 2p.

Plugging values is the best way from here on.
If p = 10, c = 30. However, if p = 12, c = 28. Both these cases satisfy the constraint given in statement I alone and the question data.

We do not have a unique answer from the information given in statement I alone. Statement I alone is insufficient.
Answer options A and D can be eliminated. Possible answer options at this stage are B, C or E.


From statement II alone, the farm has more than 12 pigs. This means that the number of pigs can be any value from 13 to 40 and consequently the number of cows can be any value from 27 to 0.

Statement II alone is insufficient to give us a unique answer. Answer option B can be eliminated. Possible answer options at this stage are C or E.

Combining the information given in statements I and II, we have the following:
c > 2p and p > 12. The smallest value of p is 13 and for this value, c = 27 which is definitely more than twice of 13.

Can we take p as 14? We cannot because more than twice of 14 will give us a value of more than 28 and then the total of 40 will be breached.

Note that we cannot take fractional values here for p or c since they represent countable objects.

Combining statements I and II, we got a unique value for the number of cows. The combination of statements is sufficient to answer the question. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!
Aravind B T

The total number of pigs and cows is 40.
For1, C>2P
For 2, P>12
Combine 1 and 2, if P=13, C is 14;if P=14, C is 12, it is impossible.
Answer is C

Stem analysis:

We're provided in the stem that 2/3 are EITHER pigs or cows. Based on the stem that means that Cows + Pigs - Both = 2/3(60), or in a more shorthand way, C+P-B = 40. So if we can find the value of P and B we can find the answer to C.

Statement (1):
From this statement we can say that C>2P. But for arguments sake let's say that C=2P, but keep in mind that in 2P the 2 must be at least just a little larger.

So now we have that 2P+P-B = 40. Now remember we're assuming that in 2P we can have a 2, but in reality we could have 2.x, or 3, etc. But nonetheless we can see this isn't sufficient to determine P and B either individually or together.

Statement (1):
This alone doesn't give us much to work with.

Statement 1 and 2:
Ok so now we know that the value of P MUST be >12, let's put that back into our equation with an assumption that P = 12, and again keeping in mind that P actually must be >12, and in this case must be at least 13 since an animal can't be a fraction.

So 2P + P - B = 40. If P = 13 then let's see what we would have 3(13) - B = 40. 39 - B = 40. So this allows us to conclude that only ONE scenario is possible, # of pigs must be equal to 13, the value of both must be 0, and the value of cows must be 27. We know that B must be equal to 0 since 2P must be GREATER than 2P, if even by just a little bit.