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# RENT ds

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Current Student
Joined: 11 May 2008
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30 Jul 2008, 10:33
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The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y

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SVP
Joined: 07 Nov 2007
Posts: 1791

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Location: New York

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30 Jul 2008, 12:41
arjtryarjtry wrote:
The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y

2 nd statement looks suspicious to me.. is there any type mistakes. . what is the source of this question.
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VP
Joined: 17 Jun 2008
Posts: 1374

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03 Aug 2008, 09:13
arjtryarjtry wrote:
The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y

=> given condns result in :
X9/X7 = [1+ (100(x-y)-xy)/10000] ----- condn I

even if x> y as stated in (1) THERE ARE CONDNS WHEN condn I is >1 or <1 say x=11,y=10 x9/x7 <1 and x=4,y=3 then x9/x7>1 => INSUFF

(2) says 100xy<x-y => xy<100(x-y) => x9/x7 >1 always
hence SUFF

IMO B
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Manager
Joined: 01 Aug 2007
Posts: 73

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03 Aug 2008, 09:28
spriya wrote:
arjtryarjtry wrote:
The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y

=> given condns result in :
X9/X7 = [1+ (100(x-y)-xy)/10000] ----- condn I

even if x> y as stated in (1) THERE ARE CONDNS WHEN condn I is >1 or <1 say x=11,y=10 x9/x7 <1 and x=4,y=3 then x9/x7>1 => INSUFF

(2) says 100xy<x-y => xy<100(x-y)
Quote:
=> x9/x7 >1 always ---can u explain this??
hence SUFF

IMO B

hi spriya, can u explain above 2nd point assumption.
thanks

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VP
Joined: 17 Jun 2008
Posts: 1374

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03 Aug 2008, 10:17
younggun044 wrote:
spriya wrote:
arjtryarjtry wrote:
The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) 100xy < x – y

=> given condns result in :
X9/X7 = [1+ (100(x-y)-xy)/10000] ----- condn I

even if x> y as stated in (1) THERE ARE CONDNS WHEN condn I is >1 or <1 say x=11,y=10 x9/x7 <1 and x=4,y=3 then x9/x7>1 => INSUFF

(2) says 100xy<x-y => xy<100(x-y)
Quote:
=> x9/x7 >1 always ---can u explain this??
hence SUFF

IMO B

hi spriya, can u explain above 2nd point assumption.
thanks

Oh ok
X9/X7 = [1+ (100(x-y)-xy)/10000] here if xy <100(x-y) then
x9/x7= [1+ some positive value] = x9/x7 > 1
hence x9 >x7

Hope i cleared the doubt .
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cheers
Its Now Or Never

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Re: RENT ds   [#permalink] 03 Aug 2008, 10:17
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