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Grove and Pecan trees

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Grove and Pecan trees

by chidcguy » Thu Jun 19, 2008 4:11 pm
If a certain grove consists of 36 pecan trees, what was the yield per tree last year?

(1) The yield per tree for the 18 trees in the northern half of the grove was 60 kilos last year

(2) The yield per tree for the 18 trees in the eastern half of the grove was 55 kilos last year

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Re: Grove and Pecan trees

by Stuart@KaplanGMAT » Thu Jun 19, 2008 5:54 pm
chidcguy wrote:If a certain grove consists of 36 pecan trees, what was the yield per tree last year?

(1) The yield per tree for the 18 trees in the northern half of the grove was 60 kilos last year

(2) The yield per tree for the 18 trees in the eastern half of the grove was 55 kilos last year

Thanks
(1) only tells us about half the trees: insufficient.

(2) only tells us about half the trees: insufficient.

Together: at first glance it may seem that we have two halves; however, the northern half and the eastern half may overlap - we still have no clue about the yield in the SW quadrant (and we don't know how many trees are in the NE quadrant and have therefore been counted twice).

Choose (E).
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by chidcguy » Thu Jun 19, 2008 7:39 pm
I am trying to imagine N E W S Quadrants. I can imagine Quadrant 1-4 in CG. Can you explain how you arrived at the thought that there are trees that fall into both N and E Q's? A point will be only in one Q in CG correct? If we imagine trees as points, they can be either in Q 1 or Q 2. Are you saying that there will be points that will fall on the X axis or Y axis?

Your solution pretty much sounded like Northern Q is set A, Eastern Q is set B and we don't know A intersection B.
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by Stuart@KaplanGMAT » Fri Jun 20, 2008 9:56 am
chidcguy wrote:I am trying to imagine N E W S Quadrants. I can imagine Quadrant 1-4 in CG. Can you explain how you arrived at the thought that there are trees that fall into both N and E Q's? A point will be only in one Q in CG correct? If we imagine trees as points, they can be either in Q 1 or Q 2. Are you saying that there will be points that will fall on the X axis or Y axis?

Your solution pretty much sounded like Northern Q is set A, Eastern Q is set B and we don't know A intersection B.
The statements talk about the "Northern half" and the "Eastern half" - so statement (1) tells us the production of trees in the NW & NE, statement (2) tells us the production of trees in the NE & SE... we still have no clue about production in the SW or how many trees we've double counted by using NE twice.
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by ildude02 » Fri Jun 20, 2008 3:02 pm
I thought since total number of trees in a grove is given as 36, and since combining 1 and 2 gives a total of 36, C should be enough to solve the problem and we don't need to worry about how many trees are there in SW etc. I agree that if the trees mentioned in E and N intersect, then it might no tbe sufficient. But again, since it's given as trees in N half and E half, I never considered Nand NE here. Do we have to consider the intersection here?
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by san2009 » Fri Jul 09, 2010 2:53 am
Stuart Kovinsky wrote:
chidcguy wrote:I am trying to imagine N E W S Quadrants. I can imagine Quadrant 1-4 in CG. Can you explain how you arrived at the thought that there are trees that fall into both N and E Q's? A point will be only in one Q in CG correct? If we imagine trees as points, they can be either in Q 1 or Q 2. Are you saying that there will be points that will fall on the X axis or Y axis?

Your solution pretty much sounded like Northern Q is set A, Eastern Q is set B and we don't know A intersection B.
The statements talk about the "Northern half" and the "Eastern half" - so statement (1) tells us the production of trees in the NW & NE, statement (2) tells us the production of trees in the NE & SE... we still have no clue about production in the SW or how many trees we've double counted by using NE twice.


In the second statement if they had southern half...then 1 & 2 together wud've been sufficient, correct?
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by Stuart@KaplanGMAT » Fri Jul 09, 2010 4:35 pm
san2009 wrote:

In the second statement if they had southern half...then 1 & 2 together wud've been sufficient, correct?
Correct!
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