Of the 60 animals on a certain farm

[minu]

Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of these animals are cows?
(1) The farm has more than twice as mnay cows as it has pigs.
(2) The farm has more than 12 pigs

Answer=C

Thanks!

[tmmyc]

Rephrase the question:
60 * 2/3 = 40 pigs or cows
p + c = 40

1)
Example 1: pigs = 2 cows = 38
Example 2: pigs = 4 cows = 36
Insufficient

2)
Example 1: pigs = 13 cows = 27
Example 2: pigs = 39 cows = 1
Insufficient

Together)
pigs = 13 cows = more than twice 13, or 26
13 + c = 40
c = 27
Valid, since 27 is greater than 26

pigs = 14 cows = more than twice 14, or 28
14 + c = 40
c = 26
Invalid, since 26 is not more than 28

Any larger number of pigs will also be invalid.

Sufficient. There are 13 pigs and 27 cows.

[StaceyKoprince]

nice!

[Jimmy]

But doesn't statement 1 mean: cows = 2(pigs). We have two equations...couldn't we solve for cows?

[Guest]

Doh...It's because it says MORE than twice right?

[StaceyKoprince]

Yep. We'd just know a range based on that and the question asks us for one specific number.

[andrern2000]

I only want to express my curiosity to know if the statement 2) helps? figured the amount of cows and pigs with step 1:

1) The farm has more than twice as many cows as it has pigs.
c > 2p.
Assume the equation for solution: c = 2p
The total animals that are either cows or pigs: 2/3 * 60 = 40

So c + p = 40
2p + p = 40
3p = 40
p = 13 1/3, substituting this into c = 2p yields p = 26 2/3

Since an animal exists as a whole we do rounding and get either 13-14 pigs and 26-27 cows.
To satisfy the 1) argument of c > 2p there's only one choice 13 pigs and 27 cows.

2) The farm has more than 12 pigs.
The range could be from (13 pigs - 27 cows) to (39 pigs - 1 cow), given this statement alone. Not sufficient.

So i picked option A) Statement 1 is sufficient and statement 2 is not sufficient?

[RonPurewal]

I only want to express my curiosity to know if the statement 2) helps? figured the amount of cows and pigs with step 1:

1) The farm has more than twice as many cows as it has pigs.
c > 2p.
Assume the equation for solution: c = 2p
The total animals that are either cows or pigs: 2/3 * 60 = 40

So c + p = 40
2p + p = 40
3p = 40
p = 13 1/3, substituting this into c = 2p yields p = 26 2/3

Since an animal exists as a whole we do rounding and get either 13-14 pigs and 26-27 cows.
To satisfy the 1) argument of c > 2p there's only one choice 13 pigs and 27 cows.

Whoa, no.

Your value of c = 26 2/3 is a threshold value, but there's no reason why you would have to pick one of the two neighboring integers.
When you get that value, you've discovered that any value of c > 26 2/3 will work.
You could have 27 cows and 13 pigs.
You could have 28 cows and 12 pigs.
You could have 29 cows and 11 pigs.
And so on.

This idea is already stated earlier in the thread, so it seems you haven't read the thread yet. Always read the threads before posting on them!