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Source: — Data Sufficiency |
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by parallel_chase » Sun Sep 07, 2008 1:01 am
Well the answer is C. This is very tricky question.

If you look at the Statement I it says:

The cows are more than twice the number of pigs. But we dont know the value of more.
Pigs could be 10 then cows would be 10 more than twice the number of pigs.
Similarly pigs could be any value between 1-13. I say 13 because if there are more than 13 pigs then this statement will have no meaning.Therefore insufficient.

Statement II pigs are more than 12. It does not tell us anything about the cows. Therefore Insufficient.

If you combine both the statements, first statement says pigs 1-13, seond statement says more than 12. Hence C is the answer.

Let me know if you still have any doubts.
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by flystud777 » Sun Sep 07, 2008 11:54 am
i get it now. i was misreading the first statement. i thought it meant twice as many cows vs pigs instead of MORE than twice.

thanks for your input
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by muzali » Tue Nov 25, 2008 1:35 pm
Not clear to me....would like to get a dumbed down explanation.
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by cramya » Tue Nov 25, 2008 1:49 pm
Question stem:

Let x be the no of pigs
Let y be the number of cows

Given x+y=40

Stmt I

y>2x

x could be 10 y could be 30
x could be 12 y could be 28

INSUFF

Stmt II

x>12
x -13 y=27
x-14 y=26

INSUFF

Combining statements I and II

x+y=40 (1)

y>2x (2)

x>12 (3)

Adding (2) and (3) (inequalities facing same direction can be added)

x+y>2x+12
y>x+12

y>40-y+12 (since from (1) x= 40-y)

2y>52
y>26

The only number satisfying stmt I and II is x=13 y=27

Therefore there are 27 cows(more than twicw the number of pigs i.e more than 26 which is statement I) and 13 pigs (more than 12 pigs which is statement II)

One definite answer

C)

Hope I dint miss something here! Let me know if u still have questions.
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by muzali » Tue Nov 25, 2008 2:33 pm
My bad, I was reading "2/3 are either pigs or cows" as 40 animals are either cows or 40 animals are either pigs!!!

Thanks for the explanation though....
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