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Source: — Data Sufficiency |
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by kstv » Sat Jun 05, 2010 7:31 pm
Of 60 animals, 2/3 are either pigs or cows = 40
1) Twice as many cows as pigs, but 40 is not perfectly divisible by 3
so it can be 26 cows and 14 pigs , but 26 is not twice of 14
rather it must be 27 cows and 13 pigs . Suff.
2) More than 12 pigs , so less than 28 cows. Insuff.
Together also Insufficient.
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by Testluv » Sat Jun 05, 2010 10:55 pm
From the original, we know that 40 animals are either pigs or cows, or 40 = p + c

(2): clearly insufficient.

(1) "The farm has more than twice as many cows as it has pigs" means:

c>2p

(@kstv, I think you read (1) as "the farm has twice as many cows as it has pigs"!)

Insufficient.

Combined, we know that:

c and p are positive integers;
(I) c+p = 40;
(II) c>2p; and
(III) p>12

(we also know that c>p+12 but that doesn't matter).

(III) tells us p>12. Say p is 13. From (I), c would then be 27. Checking against (II), is 27>2*13? Yes. So, p = 13, and c = 27 is one solution set, but just barely. As you can see, if we were to increase p to 14, then from (I), c would have to be 26 but since 26 is not > 2*14, (II) would then be breached.

Thus, p is 13, and c is 27.

Choose C.
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by Patrick_GMATFix » Tue Jun 08, 2010 6:37 am
One mistake many test-takers make is to neglect to translate inequalities as they would translate equations (express relationships in algebraic terms)

60 animals. 2/3, or 40 are pigs or cows. So p+c=40. What is c?

(1) there coul be 39 cows and 1 pig, or 38 cows and 2 pigs. More than twice as many cows as pigs is not helpful to find the number of cows. NOT SUFFICIENT

(2) Essentially since the pigs & cows must add up to 40, because there are more than 12 pigs there must be fewer than 28 cows. The exact number of cows is unknown. NOT SUFFICIENT.

MERGE STATEMENTS.
There more than 12 pigs. (1) told us that cows is more than double of pigs. If there were 13 pigs, there would be 27 cows. There cannot be 14 pigs because the number of cows would have to be more than twice that, or more than 28. This would take the total over 40 and thus is not a possibility. There must be 13 pigs and 27 cows.

The answer is C.

This is GMATPrep question 1030. Have a look at the step-by-step video solution if this explanation doesn't make sense. You could also use the Drill Engine to generate timed drills with similar questions (set topic='inequalities' and difficulty='600-700')

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by Thouraya » Sat Jun 12, 2010 12:05 pm
I have two questions regarding this post:

1) isn't there only ONE combination of numbers for C & P that satisfy these two conditions at the same time: 40=c+p AND c>2P? (thus one is sufficient?)

2) The question does not say that we only have cows and animals in the farm, why are we assuming so?

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by Testluv » Sat Jun 12, 2010 12:35 pm
Hi Thouraya,
1) isn't there only ONE combination of numbers for C & P that satisfy these two conditions at the same time: 40=c+p AND c>2P? (thus one is sufficient?)
No, there are several! You can have 30 cows and 10 pigs. Is 30 + 10 = 40? Yes. Is 30>2*(10)? Yes. But we can also have 35 cows and 5 pigs. Is 35 + 5 = 40? Yes. Is 35>2*(5)? Yes. Remember, it's c>2p not c = 2p.

2) The question does not say that we only have cows and animals in the farm, why are we assuming so?
We're not! We know there are 60 animals. Of these, we know that 2/3 or 40 are either cows or pigs. So, 40 are either cows or pigs, so c + p = 40. We don't care about the other 20 animals, although we know that they are neither cows nor pigs.
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