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Manager
Joined: 16 Apr 2006
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for this please refer to OG 11
PS , Q 7
i know after seeing the solution u would say this question is a cake walk..but i have one confusion please help me wid it..
acc to GMAT , the figure is to scale, now can i assume this graph as a parabola ?? coz it looks to be a one...if it does then it can complicate the problem ....
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VP
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I do not have OG 11. can you please post the question?
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Director
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PS Instructions say clearly, that figures are drawan as acurately as possible, unless otherwise stated.
So the points 1,0 i would assume it is.
because thats where the parabola intersects.
if you read the previous page the isntructions are there..
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VP
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Can one of you post the question, please?
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Director
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question is...
on the graph above, when x = 1/2, y = 2; and x =1,y=1. The graph is symmetric with repsect to the vertical linke at x = 2. According to the graph, when x = 3, y = ?
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Manager
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ok we assume it to be a parabola then the eqn of graph can be taken as
Y= a x*x + b*x + c (i.e. the standard quadratic one..)
now i need to find a,b,c which is simple as
x=2 , y=0 as it appears from graph
x=1 , y=1 as given in question
x=.5 , y=2 as given in question
so from this i wud get a,b,c
now we need to find y when x=3 , put back in the equation we will get the value of y...
i know this is a longer method..no doubt..but why doesn't it give the same answer...can anyone please explain it...
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Manager
Joined: 20 Nov 2004
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beyondinfinity wrote: i know this is a longer method..no doubt..but why doesn't it give the same answer...can anyone please explain it...
You can't assume the graph to be a parabola because it isn't one. And
even if it were: the y-value for x= 1/2 is only given to confuse us, so
don't use a formula and take the y-value for x=1 .
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Manager
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but i think i found my mistake this curve may be a fourth degree , sixth degree or any equation with even powers of X , since that is not clear so we cannot take it to be a "quadratic" equation.
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close
