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by naaga » Sat Feb 14, 2009 12:43 pm
If y≠0 and y≠-1, which is greater, x/y (or) x/(y+1)
(1) x≠0
(2) x > y


folks ...how to solve this kind of problems I need standard method


thanks in advance
Last edited by naaga on Sat Feb 14, 2009 1:11 pm, edited 1 time in total.
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Source: — Data Sufficiency |

by cramya » Sat Feb 14, 2009 1:05 pm
Hi Naga,

Please post the complete question. Thanks!


Regards,
CR
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by naaga » Sat Feb 14, 2009 1:10 pm
sorry cramya , I didn't copied the whole thing, here is the complete question


If y≠0 and y≠-1, which is greater, x/y (or) x/(y+1)
(1) x≠0
(2) x > y
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by cramya » Sat Feb 14, 2009 1:17 pm
I would go with E

Stmt I

x not equal to 0

x/y and x/y+1 values depend on the digns of x and y . Since we dont anything about sign of y anything can be greater than the other

INSUFF

Stmt II

x>y

x can be 0 which makes x/y and x/y+1 the same or one can be greater than the other because of the possible postive / negative combinations amongst x and y

INSUFF

Together

y <> 0 y <> -1 x<>0 x>y

x=-1 y=-1000 x/y+1 is greater

x=1000 y = 2 x/y is greater
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by shargaur » Wed Mar 04, 2009 11:33 am
PostPosted: Sat Feb 14, 2009 1:10 pm Post subject:
sorry cramya , I didn't copied the whole thing, here is the complete question


If y&#8800;0 and y&#8800;-1, which is greater, x/y (or) x/(y+1)
(1) x&#8800;0
(2) x > y

as both have same num so denominator will decide which is greater.

Case 1) if y < 0 then x/y is greater
case 2) if y > 0 then x/y+1 is greater

but as there is no information provided by any of Stmt 1 or 2 about y hence E
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