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At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 54% (02:43) correct 46% (02:56) wrong based on 242 sessions

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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

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Originally posted by carcass on 07 Mar 2012, 14:23.
Last edited by Bunuel on 14 Jan 2015, 01:42, edited 3 times in total.
Edited the question and added the OA.
Math Expert V
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Posts: 59147
Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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11
5
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.

At the end of 2004 the ratio of cows to pigs was 12/14=6/7, in order the ratio to remain the same at the end of 2005, cows and pigs should be added in the same ratio (or not added at all). So, possible numbers of new cows and pigs are: (0, 0) or (6, 7). Notice that (12, 14) is not possible since 12+14=26, which is more than total number of new animals brought, 22.

The same for the ratio of hens to sheep: at the end of 2004 the ratio was 24/30=4/5. So, possible numbers of new hens and sheep are: (0, 0), (4, 5) or (8, 10).

Only one combination makes total of 22 animals: (6, 7) and (4, 5) --> 6+7+4+5=22. Hence, 7 new pigs were brought to the farm, so there were 14+7=21 pigs on the farm at the end of 2005. Sufficient.

(2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005. 5 sheep were brought to the farm, but we know nothing about the rest of 17 animals. Not sufficient.

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Board of Directors D
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Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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Cool

I changed the animals with variables to make it look more difficult, muddy the waters.

I think that when you realize that you have the relationship between ratios is already clear sufficient and move on is the right thing but your deep explanation is always useful to understand.

Next time I'll post the question correctly, eventhough it was the same thing.

Thanks Bunuel.

PS: OA is A
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Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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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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Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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1
Bunuel wrote:
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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Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30  [#permalink]

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carcass wrote:
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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Available for tutoring in NYC and long-distance. Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30   [#permalink] 13 Sep 2019, 11:02
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