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carcass
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it is said that we do not have to solve the DS questions. As can be seen, however, this is not the case for most of the time, especially after going through Bunuel's solutions:) Thanks for the solution.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Fractions/Ratios/Decimals questions: search.php?search_id=tag&tag_id=36
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Fractions/Ratios/Decimals questions: search.php?search_id=tag&tag_id=36
All PS Fractions/Ratios/Decimals questions: search.php?search_id=tag&tag_id=57

Questtion :
Total no of animals in farm currently = 80
H=24, C=1, S=30, P=14

(1) The ratio of cows to pigs and the ratio of hens to sheep were the same at the end of 2004 and 2005.
C/P and H/S will be the same for 2004 and 2005 respectively.

Hence for 2004 :
C/P = 12/14 = 6/7 Hence, the multiplication factor for the ratio = x =2
H/S = 24/30 =4/5 Hence, the multiplication factor for the ratio = y = 6

Since the no of animals inc by 22 and none of the animals leave the farm, the new factors say x' and y' will be such that x'>=x or y'>=y

hence,
6x'+7x'+4y'+5y' = 80+22
13x'+9y'=102

trying with the first combination say, x'=3 and y'=7
39 + 63 = 102
102 = 102 (Bingo!!)

Thus, x' = 3 and y'=7
Therefore no of pigs = 7x' = 21
so, 7 new pigs were added
=>Sufficient

(2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005
We still know nothing about the other animals
=>Not sufficient

Ans: A
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At the end of 2004, a certain farm had 24 hens, 12 cows, 30 sheep, and 14 pigs. By the end of 2005, 22 new animals — each either a hen, cow, sheep or pig — were brought to the farm. No animals left the farm. How many pigs were there on the farm at the end of 2005?

(1) The ratio of cows to pigs and the ratio of hens to sheep were the same at the end of 2004 and 2005.
(2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005

Statement 1:
H:S = 24:30 = 4:5.
In this ratio, the sum of the parts = 4+5 = 9.
Implication:
For the ratio to stay the same, H+S must increase by a multiple of 9.

C:P = 12:14 = 6:7.
In this ratio, the sum of the parts = 6+7 = 13.
Implication:
For the ratio to stay the same, C+P must increase by a multiple of 13.

Since the total number of animals increases by 22, only one case is possible:
H+S increases by 9 and C+P increases by 13, for a total increase of 22.

Since C+P increases by 13, there are 6 more cows and 7 more pigs.
Thus:
New pigs = 14+7 = 21.
SUFFICIENT.

Statement 2:
Increase in the number of sheep \(= \frac{1}{6}* 30 = 5\).
Since the total number of animals increases by 22, the increase in H+C+P = 22-5 = 17.
No way to determine the increase in P alone.
INSUFFICIENT.

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I don't think anyone will have this much time to decipher that only 1 case exists for statement (1) but what I did was while solving (and i have seen this as a trend in DS questions) that often statement (2) tells about the particular scenario that will help us that only 1 case exists...

In this ques too, (2) gave us that only 1 case exists but then it is not required/sufficient.

Hence answer is (A)
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