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In the rectangular coordinate system, is the point (m, n)

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Senior Manager
Joined: 09 Aug 2005
Posts: 286
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In the rectangular coordinate system, is the point (m, n) [#permalink]  05 Mar 2006, 14:14
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In the rectangular coordinate system, is the point (m, n) farther from the origin than point (p, q)?
(1) mn − pq = 12
(2) p + q = 21.

How do you attack this one

its between A, C and E all of us can conclude that.

do you go the algebraicaly creative 'walah!' approach or

pick numbers to disprove A, C and E conclusively ?
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
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My thoughts on this question. I could be wrong, geometry is not one of my strengths.

1) mn is always 12 units greater than pq. Assuming no matter how we flip it, (m,n) will always be farther than pq from the origian. If (m,n) is a point, it is straight forward. Even if it is (m,x) and (y,n), the intersection making (m,n) will still be father than pq.

Ans: A
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Joined: 24 Sep 2005
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Re: DS - approach 2mins [#permalink]  06 Mar 2006, 06:37
old_dream_1976 wrote:
In the rectangular coordinate system, is the point (m, n) farther from the origin than point (p, q)?
(1) mn − pq = 12
(2) p + q = 21.

How do you attack this one

its between A, C and E all of us can conclude that.

do you go the algebraicaly creative 'walah!' approach or

pick numbers to disprove A, C and E conclusively ?

IMO, whenever we encounter a problem which looks very time-consuming, it's highly possible to have E as OA ( this we can conclude if we skillfully substitute numbers)

1) stmt 1: mn-pq= 12 --> we're easily misled to think that m+n > p+q as well. What's about the case of p or q = 0??
Example: p=0 , q= 20 , m=4, n=3 ---> point (p,q) is farther from the origin than (m,n)
p= 1, q=1, m=13,n=1 ---> (m,n) is farther than (p,q)
--->insuff

2) nothing about m,n ---> can't conclude ---> insuff

1) and 2) :
in case: p=0, q=21 , m=4,n=3 ---> (p,q) is farther than (m,n)
in case: p= 15, q= 6, m= 17, n= 6 (satisfy stmt 1) ---> (m,n) is farther than (p,q)
---> insuff

---> E it is.
Manager
Joined: 20 Feb 2006
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Me too for E (considering negative numbers).
Substituting values is the best approach according to me. I first substituted the values and later tried using algebraic approach but, I could not solve.
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