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The annual rent collected by a corporation from a certain

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The annual rent collected by a corporation from a certain  [#permalink]

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New post 01 Oct 2008, 11:03
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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) xy/100 < x – y

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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 11:32
1) Insufficient

Substitute 20 for x, 19 for y, and 100 for rent in 1997.
1997: 100
1998: 100+100*0.2 = 120
1999: 120-120*0.19 = 120-22.8 = 97.2 which gives an answer "NO' that 1999 rent > 1997 rent

Substitute 20 for x, 1 for y, and 100 for rent in 1997
1997: 100
1998: 100+100*0.2 = 120
1999: 120-120*0.01 = 120-1.2 = 118.8 which gives an answer "YES" that 1999 rent > 1997 rent

2) Insufficient

Substitute 20 for x, 19 for y, and 100 for rent in 1997
20*19/100 < 20-19
380/100 < 1
3.8 < 1
Not true for this condition

Substitiute 20 for x, 1 for y, and 100 for rent in 1997
20*1/100 < 20-1
1/5 < 19
True for this condition

When combining both conditions x>y and xy/100 < x-y, we plug in the number 20 for x, 1 for y, and 100 for rent because these numbers satisfies both conditions as shown above.
When x = 20, y = 1, x>y.
When x = 20, y = 1, xy/100 < x-y
20*1/100 < 20-1
20/100 < 19
1/5 < 19

Therefore, the answer is C.
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Re: DS: Annual Rent  [#permalink]

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New post Updated on: 01 Oct 2008, 12:32
vksunder 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) xy/100 < x – y


Answer is B.

Let rent in 1997 = 1
rent in 1998 = 1+(x/100)
rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1-(y/100)}].......coz [a + b% of a]
= [1+(x/100)] [1-(y/100)]
The question demands: Is [1+(x/100)] [1-(y/100)] > 1 ??

assume this to be true & solve it further:
=> (100+x)(100-y)> 10000
=> 10000 + 100x -100y -xy >10000
=> 100x-100y-xy>0
=> 100(x-y)>xy
=> x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.

Originally posted by jatinrai on 01 Oct 2008, 11:39.
Last edited by jatinrai on 01 Oct 2008, 12:32, edited 1 time in total.
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 12:15
jatinrai wrote:
vksunder 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) xy/100 < x – y


Answer is B.

Let rent in 1997 = 1
rent in 1998 = 1+(x/100)
rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1+(y/100)}].......coz [a + b% of a]
= [1+(x/100)] [1+(y/100)]

The question demands: Is [1+(x/100)] [1+(y/100)] > 1 ??

assume this to be true & solve it further:
=> (100+x)(100-y)> 10000
=> 10000 + 100x -100y -xy >10000
=> 100x-100y-xy>0
=> 100(x-y)>xy
=> x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.


I think you meant 1-(y/100) here.
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 12:32
scthakur wrote:
jatinrai wrote:
vksunder 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) xy/100 < x – y


Answer is B.

Let rent in 1997 = 1
rent in 1998 = 1+(x/100)
rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1+(y/100)}].......coz [a + b% of a]
= [1+(x/100)] [1+(y/100)]

The question demands: Is [1+(x/100)] [1+(y/100)] > 1 ??

assume this to be true & solve it further:
=> (100+x)(100-y)> 10000
=> 10000 + 100x -100y -xy >10000
=> 100x-100y-xy>0
=> 100(x-y)>xy
=> x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.


I think you meant 1-(y/100) here.

Yup. Solved it on paper then copied. Hence, error.
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 20:16
Can v hv sum short cut method for this question?
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 20:37
As far as I can understand [1+(x/100)] [1-(y/100)] is diference b/w rent in 1999 and 1998. Y did u consider it >1 ?
jatinrai wrote:
vksunder 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) xy/100 < x – y


Answer is B.

Let rent in 1997 = 1
rent in 1998 = 1+(x/100)
rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1-(y/100)}].......coz [a + b% of a]
= [1+(x/100)] [1-(y/100)]
The question demands: Is [1+(x/100)] [1-(y/100)] > 1 ??

assume this to be true & solve it further:
=> (100+x)(100-y)> 10000
=> 10000 + 100x -100y -xy >10000
=> 100x-100y-xy>0
=> 100(x-y)>xy
=> x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 21:28
x percent more, y percent less

So can we assume x, y >0??

If x,y > 0 then (2) <=> x - y>0 <=> (1)

annual rent in 1997 & 1999 are connected by rent in 1998, so (1) is sufficient
hence (2) is sufficient

My answer is D
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 21:50
vksunder 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) xy/100 < x – y


B

% increase from 1997 to 1999
P = (1+x/100)(1-y/100)
= 1+ x/100-y/100 -xy/(100*100)
= 1+ 1/100 ( x-y-xy/100)

from equ 2 xy/100 < x – y --> x-y-xy/100>0

so P>1
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 21:53
lylya4 wrote:
x percent more, y percent less

So can we assume x, y >0??

If x,y > 0 then (2) <=> x - y>0 <=> (1)

annual rent in 1997 & 1999 are connected by rent in 1998, so (1) is sufficient
hence (2) is sufficient

My answer is D


I don't think you can assume that y>0 since you are told that y is less then 0 (negative).

:(
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Re: DS: Annual Rent  [#permalink]

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New post 01 Oct 2008, 21:53
lylya4 wrote:
x percent more, y percent less

So can we assume x, y >0??

If x,y > 0 then (2) x - y>0 (1)

annual rent in 1997 & 1999 are connected by rent in 1998, so (1) is sufficient
hence (2) is sufficient

My answer is D



(1) is not suffcient

for e.g same intial ren 100. increased by 20.01% then rent would be ~120 decreased by 20%
120-24= 96
for e.g same intial ren 100. increased by 40% then rent would be ~140 decreased by 20%
140-28=116

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Smiling wins more friends than frowning

Re: DS: Annual Rent &nbs [#permalink] 01 Oct 2008, 21:53
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