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algebra II qn8

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Manager
Joined: 24 Aug 2009
Posts: 149

Kudos [?]: 125 [0], given: 46

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23 Oct 2009, 18:11
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y

Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain

Kudos [?]: 125 [0], given: 46

Manager
Joined: 05 Jul 2008
Posts: 136

Kudos [?]: 124 [0], given: 40

GMAT 2: 740 Q51 V38

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23 Oct 2009, 19:17
ISBtarget wrote:
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y

Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain

Hey, never divide both sides for any thing that you are not sure whether or not it could be Zero. In both statements, y=1 or 0

Kudos [?]: 124 [0], given: 40

Manager
Joined: 24 Aug 2009
Posts: 149

Kudos [?]: 125 [0], given: 46

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24 Oct 2009, 12:25
DavidArchuleta wrote:
ISBtarget wrote:
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y

Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain

Hey, never divide both sides for any thing that you are not sure whether or not it could be Zero. In both statements, y=1 or 0

thanks for the info

Even otherwise, take statement 1

take y out of left and right handside y ( y^2+2) = y ( 2y+1)

cancel out y on both sides , we get y^2 -2y+1 which is (y-1)^2 = 0 , which is y =1.

Kudos [?]: 125 [0], given: 46

Intern
Joined: 16 Oct 2009
Posts: 11

Kudos [?]: 1 [0], given: 0

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24 Oct 2009, 13:44
Here it goes..
2) You Cannot divided both sides by y and say y>1 so,
y2 = y
y2-y = 0
y(y-1) = 0
Solution is either y= 0 or y=1

Similarly for condition 1)
y3 + 2y = y +2(y2)
y3 -2y2 + y = 0 ... taking y as common equation becomes...
y(y2-2y+1) = 0
y = 0 or
(Y-1)2 = 0

Solution is either y = 0 or y = 1

Both the condition together are not able to provide single value of Y....
Hope this clears your confusion now...

Kudos [?]: 1 [0], given: 0

Re: algebra II qn8   [#permalink] 24 Oct 2009, 13:44
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