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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] New post 28 Aug 2004, 10:10
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A
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


please explain, thank you.
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 [#permalink] New post 28 Aug 2004, 14:33
My guess is D..

if x > y then it should be yes

for 2nd statement.. its says the same thing x>y
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 [#permalink] New post 29 Aug 2004, 01:43
Hi, all, first of all, let me speak my thought.
In question, we can get each year's annual rent.

1997 1998 1999
M (M+x%M) (M+x%M-[M+x%M]y%)

compare 1997 with 1999:

( M+x%M-[M+x%M]y%) - (M) = M (x-y-xy)%

If x-y-xy >0, then 1999's rent > 1997's rent, or otherwise.

But in (1) and (2), it seems to be not of help to answer the question.
Hence, I choose (E).

Any other ideas?

thank you

Last edited by chunjuwu on 30 Aug 2004, 05:25, edited 1 time in total.
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 [#permalink] New post 29 Aug 2004, 21:57
Both statements by themselves are sufficient.


If the rent is n --- 1997
then the rent is (100+x)n --- 1998
and the rent is (100-y)(100+x)n = (100^2 + 100x - 100y -xy)n --- 1999

From (1), no matter how big or small the difference is between x and y, (100-y)(100+x) is always positive.
And so (100-y)(100+x)n > n

From (2), xy/100 < x - y
xy < 100x - 100y
also tells us (100^2 + 100x - 100y -xy)n is always positive.

So (100^2 + 100x - 100y -xy)n > n.

Therfore the answer should be (D)
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 [#permalink] New post 30 Aug 2004, 04:34
I would choose B

St.1
Consider when x = 20 and y = 10
1997 = 100
1999 = 108

Consider x = 50 and y = 49
1997 = 100
1999 = 76,5

St.2
xy/100 < x - y

x must be greater than y and x = 50 and y = 49 won't work.

My explanation might be not very well, but I think B is the answer.

Correct me if I am wrong.

Regards,

Alex
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Re: DS0829 [#permalink] New post 30 Aug 2004, 04:49
chunjuwu 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


I'll go for e.
1) let's say in 97' was 100, let x =10, and y=9.9
98' it was 110, and 99' it was 110(.901)= 99.11
another scenario, x=10, y=5, and 99' > 98
insufficient

2) tells you essentially the same thing, so insufficient
D does not elicit any new info, hence it's E

I think the key here is to test extreme values for % changes
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 [#permalink] New post 30 Aug 2004, 05:03
Answer is E.

If X is the amount perceived in 97, income in 99 is equal to X(1+x)(1-y)

id est is (1+x)(1-y) <> 1 <=> 1 +x-y -xy <> 1

1)x>y does not provide the answer

2)xy/100<x-y neither as it is not enough precise xy>xy/100 and x-y >xy/100

1+2) does not change anything
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 [#permalink] New post 30 Aug 2004, 05:59
Sorry, let me revise my talk

1997 1998 1999
M (M+x%M) (M+x%M-[M+x%M]y%)

Hence, in 1999, the rent would be
(M+x%M-[M+x%M]y%) = M(1+x%-y%-xy%%)

I think this is a little tricky, it should be xy%% instead of xy%

compare 1999' with 1997'
we get [M(1+x%-y%-xy%%)] - M = M(x%-y%-xy%%) = 1/10000(100x-100y-xy)

if 100x-100y-xy > 0 then 1999' >1997' , or otherwise

in (1) x>y it always positive. so, sufficient.

in (2) xy /100 < x-y => 100x-100y-xy >0 it always positive, so sufficient.

Therefore, I go with (D).
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 [#permalink] New post 30 Aug 2004, 06:04
Chunjuwu,

Is there an official answer for this question.
Thanks.

Regards,

Alex
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 [#permalink] New post 31 Aug 2004, 06:42
ywilfred wrote:
Both statements by themselves are sufficient.


If the rent is n --- 1997
then the rent is (100+x)n --- 1998
and the rent is (100-y)(100+x)n = (100^2 + 100x - 100y -xy)n --- 1999

From (1), no matter how big or small the difference is between x and y, (100-y)(100+x) is always positive.
And so (100-y)(100+x)n > n

From (2), xy/100 < x - y
xy < 100x - 100y
also tells us (100^2 + 100x - 100y -xy)n is always positive.

So (100^2 + 100x - 100y -xy)n > n.

Therfore the answer should be (D)


I don't agree with this, I think it should be:

If the rent is n --- 1997
then the rent is (100+x)/100 * n --- 1998
and the rent is [(100-y)/100] * [(100+x)/100] *n --- 1999

[(100-y) /100] * [(100+x) /100] can be >100 or <100. Therefore insufficient

2. is the same as 1, so also insufficient

My answer: E

It would have been D if the rent in 99 were y percent less than that of 97
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 [#permalink] New post 31 Aug 2004, 08:02
I will go with E on this one.

For 1., 2. and comnibed I used the following number.
X = 20
Y = 19
or
X = 20
Y = 10


Whats the OA?
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 [#permalink] New post 31 Aug 2004, 08:36
The OA is E,

but I don't think so

thank you.
  [#permalink] 31 Aug 2004, 08:36
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