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by Rahul@gurome » Fri Jul 09, 2010 4:32 am
Let the rent be R in 1997.
Then rent in 1998 = R + x% of R = R(1 + x/100)
Rent in 1999 = R - y% of R = R(1 - y/100)
New rent = R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10000)
The question becomes Is R(1 + x/100 - y/100 - xy/10000) > R ?

R(1 + x/100 - y/100 - xy/10000) > R
(1 + x/100 - y/100 - xy/10000) > 1
x - y - xy/100 > 0 implies (x - y) > xy/100

(1) From x > y we cannot find whether (x - y) > xy/100. So, (1) is NOT SUFFICIENT.
(2) In the above calculation we have found that (x - y) > xy/100 implies xy/100 < (x - y), which is statement (2).
So, (2) is SUFFICIENT.

[spoiler]The correct answer is (B).[/spoiler]
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by pradeepkaushal9518 » Fri Jul 09, 2010 5:13 am
thanks guru for nice explanations
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by Rahul@gurome » Fri Jul 09, 2010 5:16 am
pradeepkaushal9518 wrote:thanks guru for nice explanations
U r welcome!
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by gmatmachoman » Mon Jul 12, 2010 11:37 pm
Rahul@gurome wrote:Let the rent be R in 1997.
Then rent in 1998 = R + x% of R = R(1 + x/100)
Rent in 1999 = R - y% of R = R(1 - y/100)
New rent = R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10000)
The question becomes Is R(1 + x/100 - y/100 - xy/10000) > R ?

R(1 + x/100 - y/100 - xy/10000) > R
(1 + x/100 - y/100 - xy/10000) > 1
x - y - xy/100 > 0 implies (x - y) > xy/100

(1) From x > y we cannot find whether (x - y) > xy/100. So, (1) is NOT SUFFICIENT.
(2) In the above calculation we have found that (x - y) > xy/100 implies xy/100 < (x - y), which is statement (2).
So, (2) is SUFFICIENT.

[spoiler]The correct answer is (B).[/spoiler]
@rahul, wat wil be level of difficulty of this question??
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by sumanr84 » Tue Jul 13, 2010 4:03 am
Solution by Ian : https://www.beatthegmat.com/word-prob-t19707.html

With that correction: To increase something by x%, you can multiply by (1 + x/100). To decrease something by y%, you can multiply by (1 - y/100).

Here, the rent (let's call it R) increases by x%, then decreases by y%. So our new rent is just:

R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10,000)

We want to know if

R(1 + x/100 - y/100 - xy/10,000) > R

or equivalently, dividing by R (which is fine, because R must be positive) if

1 + x/100 - y/100 - xy/10,000 > 1

Subtracting 1 from both sides, and multiplying by 100 on both sides, the question becomes whether

x - y - xy/100 > 0

or if xy/100 < x - y. Since Statement 2 tells us exactly that, the answer to the question must be yes if we use Statement 2, and it's sufficient. Since Statement 1 is insufficient (try x = 60, y = 40 to see that), the answer is B.
I am on a break !!
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by Rahul@gurome » Tue Jul 13, 2010 6:46 am
gmatmachoman wrote: @rahul, wat wil be level of difficulty of this question??
Its around 500-600 level.
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