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The annual rent collected by a corporation from a certain

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Senior Manager
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The annual rent collected by a corporation from a certain [#permalink] New post 11 Sep 2007, 12:33
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B
C
D
E

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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 - Percentages Set 4, Math 29 [#permalink] New post 11 Sep 2007, 16:59
ashkrs wrote:
I get D . Both are sufficient.


Could you please explain how?

(1) Lets say the rent for 1997 is 100
x=25%, y=20% gives Rent98 = 125, Rent99 = 100. So basically Rent97 = Rent99
x=25%, y=10% gives Rent98 = 125, Rent99 = 112.5. So basically Rent97 < Rent99
Hence, Insufficient

I am having trouble with (2). I am not sure where to even begin.
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Re: DS - Percentages Set 4, Math 29 [#permalink] New post 11 Sep 2007, 17:24
[quote="gluon"]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.

I say D too.

Look at the chart: Lets say rent started @ 100.
yr: 97 98 99
amnt 100 110 amnt>100.

X is 10% and Y has to be less than X. So even if Y is 9.9999999999999999999% In 99 its still greater by fractions of a penny.

or if Y is just 9% then the amount in 99 is 1 penny higher than 97.


S2: basically says the same thing as S1.

xy/100<x> xy<100x-100y. For this inequality to be true, x must be greater than Y. if it isn't then you wind up with a negative number on the right side.

Also the left side of the inequality has to be positive since we cannot have negative percentages in this problem (I think :P)
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 [#permalink] New post 11 Sep 2007, 17:27
Rent in 1998 = 1997 + (x/100)*1997 ---1

Rent in 1999= 1998-(y/100)*1998 -----2

Substituing eqn1 in eqn2, we have.. this is relation between Rent in 1997 and 1999

Rent in 1999 = 1997[1+(x/100)-(y/100)-(xy/10000)] = 1997[1+1/100 (x-y-xy/100)]

So, I get B
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 [#permalink] New post 11 Sep 2007, 17:32
GMATBLACKBELT, I still dont see how statement A can help solve the question.. can you explain furthur?
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 [#permalink] New post 11 Sep 2007, 17:34
OOOps, I say D as well... from my earlier equation, if X>Y, then rent in 1999 > rent in 1997


D!
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 [#permalink] New post 11 Sep 2007, 17:46
St1:
If:
Rent in 1997: 100
Rent in 1998 : 110 (x = 10%)
Rent in 1999: 100.1 (y = 9%)
Then rent collected in 1999 > rent collected in 1997

If:
Rent in 1997 = 100
Rent in 1998 = 110 (x = 10%)
Rent in 1999: 99.11 (y = 9.9%)
Then rent collected in 1999 < rent collectedin 1997

Insufficient.

St2:
Can use the same calculations in st1. Insufficient.

Both combined doesn't provide any extra information.

Ans E
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 [#permalink] New post 11 Sep 2007, 17:59
should be 'B'.
as per the stem:
let's assume that rent in 97 was R.
In 98 : (1 + x/100)R
In 99: (1 + x/100)(1 - y/100)R


STMT1: x > y
For x = 2, y=1: (1 + x/100)(1 - y/100)R > R

For x = 99, y=98: (1 + x/100)(1 - y/100)R < R
So INSUFF

Stmt2: xy/100 < x-y i.e. 100(x-y) -xy > 0

solving (1 + x/100)(1 - y/100)= 1+ [100(x-y)-xy]/10000, this is greater than 1
So (1 + x/100)(1 - y/100)R > R
Hence suff
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 [#permalink] New post 11 Sep 2007, 19:30
vshaunak@gmail.com wrote:
should be 'B'.
as per the stem:
let's assume that rent in 97 was R.
In 98 : (1 + x/100)R
In 99: (1 + x/100)(1 - y/100)R


STMT1: x > y
For x = 2, y=1: (1 + x/100)(1 - y/100)R > R

For x = 99, y=98: (1 + x/100)(1 - y/100)R < R
So INSUFF

Stmt2: xy/100 <x> 0

solving (1 + x/100)(1 - y/100)= 1+ [100(x-y)-xy]/10000, this is greater than 1
So (1 + x/100)(1 - y/100)R > R
Hence suff


Beautiful explanation for (2). There is no way I am able to solve such problems in 2 minutes currently. As explained earlier I cracked (1) but got stuck on (2) :(

OA is indeed B.
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 [#permalink] New post 12 Sep 2007, 11:46
i got this under 10 secs

remember formulas

if price went up X% and then went down y%

x-y-(xy/100).

if the above is positive then net still increased..if negative net decreased...

looking at (2)

x-y>xy/100 told me right away that net was going to be positive :)

WORD OF CAUTION...first understand the concepts then learn the formulas..to increase efficency...like i m doing these days..have already learnt the concepts..now just practicing using formulas where possible... to increase efficency..


gluon wrote:
vshaunak@gmail.com wrote:
should be 'B'.
as per the stem:
let's assume that rent in 97 was R.
In 98 : (1 + x/100)R
In 99: (1 + x/100)(1 - y/100)R


STMT1: x > y
For x = 2, y=1: (1 + x/100)(1 - y/100)R > R

For x = 99, y=98: (1 + x/100)(1 - y/100)R < R
So INSUFF

Stmt2: xy/100 <x> 0

solving (1 + x/100)(1 - y/100)= 1+ [100(x-y)-xy]/10000, this is greater than 1
So (1 + x/100)(1 - y/100)R > R
Hence suff


Beautiful explanation for (2). There is no way I am able to solve such problems in 2 minutes currently. As explained earlier I cracked (1) but got stuck on (2) :(

OA is indeed B.
  [#permalink] 12 Sep 2007, 11:46
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