|
Author |
Message |
|
TAGS:
|
|
|
VP
Joined: 26 Apr 2004
Posts: 1255
Location: Taiwan
Followers: 1
Kudos [?]:
7
[0], given: 0
|
The annual rent collected by a corporation from a certain [#permalink]
 28 Aug 2004, 11:10
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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.
|
|
|
|
|
|
|
Intern
Joined: 20 Jul 2004
Posts: 15
Followers: 0
Kudos [?]:
0
[0], given: 0
|
My guess is D..
if x > y then it should be yes
for 2nd statement.. its says the same thing x>y
|
|
|
|
|
|
VP
Joined: 26 Apr 2004
Posts: 1255
Location: Taiwan
Followers: 1
Kudos [?]:
7
[0], given: 0
|
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, 06:25, edited 1 time in total.
|
|
|
|
|
|
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5134
Location: Singapore
Followers: 9
Kudos [?]:
87
[0], given: 0
|
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)
|
|
|
|
|
|
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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
|
|
|
|
|
|
Senior Manager
Joined: 07 Oct 2003
Posts: 380
Location: Manhattan
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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
|
|
|
|
|
|
Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
WE: Accounting (Accounting)
Followers: 0
Kudos [?]:
32
[0], given: 54426
|
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
|
|
|
|
|
|
VP
Joined: 26 Apr 2004
Posts: 1255
Location: Taiwan
Followers: 1
Kudos [?]:
7
[0], given: 0
|
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).
|
|
|
|
|
|
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Chunjuwu,
Is there an official answer for this question.
Thanks.
Regards,
Alex
|
|
|
|
|
|
Manager
Joined: 15 Jul 2004
Posts: 75
Location: London
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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
|
|
|
|
|
|
SVP
Joined: 16 Oct 2003
Posts: 1957
Followers: 2
Kudos [?]:
14
[0], given: 0
|
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?
|
|
|
|
|
|
VP
Joined: 26 Apr 2004
Posts: 1255
Location: Taiwan
Followers: 1
Kudos [?]:
7
[0], given: 0
|
The OA is E,
but I don't think so
thank you.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
