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# The annual rent collected by a corporation from a certain

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Intern
Joined: 22 Jul 2010
Posts: 32

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22 Jul 2010, 08:36
i think the answer is D because A can also give clue. Upon expansion we get:
r(1+(x-y)/100 - xy/10000)

given x>y, therefore (x-y)/100 is going to be positive and xy/10000< (x-y)/100, denominator comparison is enough. So (x-y)/100 - xy/10000 > 0. So rent collected will be greater. So D is the answer...
Pls correct me if I am wrong...
Manager
Joined: 17 Nov 2009
Posts: 239
Re: GMAT 12th ed. #120 DS [#permalink]

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08 Sep 2010, 13:11
Bunuel wrote:
thanks 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 is greater than Y
2) xy/100 is less than x-y

I'm hoping that someone can elaborate on the number property that gives answer b.

PS I'm assuming that there is no way of knowing what level the questions are from the Review books???

Given:

Rent in 1997 - $$r$$;
Rent in 1998 - $$r*(1+\frac{x}{100})$$;
Rent in 1999 - $$r*(1+\frac{x}{100})*(1-\frac{y}{100})$$.

Question is $$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$ true? --> $$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$ --> $$x-y>\frac{xy}{100}$$ true?

(1) $$x>y$$, based on this information we can not conclude whether $$x-y>\frac{xy}{100}$$ is true or not. Not sufficient.

(2) $$\frac{xy}{100} < x -y$$, directly states that the equation we were testing is true. Sufficient.

Well crafted solution a KUDOS!

BTW... Thank you for such wonderful explanations that make understanding tough problems like these easy.
Senior Manager
Joined: 12 May 2010
Posts: 288
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)

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14 Nov 2010, 07:55
so it seems you must simplify the 'rephrased' d.s. questions until you can't simplify any more... THEN consider the statements 1 and 2.,.. correct?
Manager
Joined: 01 Apr 2010
Posts: 164
Re: GMAT 12th ed. #120 DS [#permalink]

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16 Nov 2010, 04:41
Bunuel wrote:
thanks 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 is greater than Y
2) xy/100 is less than x-y

I'm hoping that someone can elaborate on the number property that gives answer b.

PS I'm assuming that there is no way of knowing what level the questions are from the Review books???

Given:

Rent in 1997 - $$r$$;
Rent in 1998 - $$r*(1+\frac{x}{100})$$;
Rent in 1999 - $$r*(1+\frac{x}{100})*(1-\frac{y}{100})$$.

Question is $$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$ true? --> $$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$ --> $$x-y>\frac{xy}{100}$$ true?

(1) $$x>y$$, based on this information we can not conclude whether $$x-y>\frac{xy}{100}$$ is true or not. Not sufficient.

(2) $$\frac{xy}{100} < x -y$$, directly states that the equation we were testing is true. Sufficient.

simple,clear explanation n solution... thanks brunel... +1 ..

i face problems in answering % problems... i solve them.. but i take lot of time... more than that i feel i am not good in % problems... if there are any good set of % problems for practice, please share them... any advice would be useful....
Manager
Joined: 01 Apr 2010
Posts: 164

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16 Nov 2010, 06:01
Senior Manager
Joined: 21 Mar 2010
Posts: 310
Re: Need help on DS: OG #120 [#permalink]

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19 Feb 2011, 20:57
VeritasPrepKarishma wrote:
mbafall2011 wrote:
I like VeritasPrepKarishma's answer- Karishma, you should start a thread with all these formulae in one place- especially the unconventional ones!

I discuss one topic every week in my blog (which I started recently) at:
[url]
http://www.veritasprep.com/blog/categor ... er-wisdom/[/url]
I plan to make it a collection of various conventional and unconventional formulas (with explanation of their derivation) and concepts that are relevant for GMAT. You might find something useful there... (In fact, the latest post discusses this very formula.)

Thanks! perfect solution. May be add this as a signature if its not already there! I was so confused in a percentage problem today in one of the sample tests. Ill see if there is already a thread for it else post one.

Thanks
Intern
Joined: 29 Jan 2011
Posts: 24

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24 Feb 2011, 22:41
1
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VeritasPrepKarishma wrote:
This is a question based on successive percentage changes i.e. a number changes by a certain factor, then it changes again by some other factor and so on. Take 100 for example. I first increase it by 10% so it becomes 110. I then decrease 110 by 20% so it becomes 88.
I personally favor multiplying the factor of change together e.g.
$$100 * (1 + \frac{10}{100}) * (1 - \frac{20}{100})$$ = 88

but a lot of my students like to use the simple formula of two successive percentage changes which is: $$a + b + \frac{ab}{100}$$
(You can derive it very easily. Let me know if you face a problem)

If a number is changed by a% and then by b%, its overall percentage change is as given by the formula.
Using it in the example above, a = 10, b = -20 (since it is a decrease)
$$a + b + \frac{ab}{100}$$ = $$10 - 20 - \frac{10*20}{100}$$ = -12%
So overall change will be of -12%. 100 becomes 88.

In our question above, since we have two successive percentage changes of x% and y% (which is a decrease) so the formula gives us $$x - y - \frac{xy}{100}$$ is the overall change. The question is, whether this change is positive i.e. whether $$x - y - \frac{xy}{100}$$ > 0 or
$$x - y > \frac{xy}{100}$$?
Since statement 2 tells us that $$x - y > \frac{xy}{100}$$, it is sufficient.

And yes, it helps to be clear about exactly what is asked before you move on to the statements.

Hi Karishma, can you please explain how you derived $$a + b + \frac{ab}{100}$$? I'm also not quite clear on why you used that second subtraction in $$10 - 20 - \frac{10*20}{100}$$. Thanks!
Senior Manager
Joined: 21 Mar 2010
Posts: 310

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24 Feb 2011, 22:44
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GSDster wrote:
VeritasPrepKarishma wrote:
This is a question based on successive percentage changes i.e. a number changes by a certain factor, then it changes again by some other factor and so on. Take 100 for example. I first increase it by 10% so it becomes 110. I then decrease 110 by 20% so it becomes 88.
I personally favor multiplying the factor of change together e.g.
$$100 * (1 + \frac{10}{100}) * (1 - \frac{20}{100})$$ = 88

but a lot of my students like to use the simple formula of two successive percentage changes which is: $$a + b + \frac{ab}{100}$$
(You can derive it very easily. Let me know if you face a problem)

If a number is changed by a% and then by b%, its overall percentage change is as given by the formula.
Using it in the example above, a = 10, b = -20 (since it is a decrease)
$$a + b + \frac{ab}{100}$$ = $$10 - 20 - \frac{10*20}{100}$$ = -12%
So overall change will be of -12%. 100 becomes 88.

In our question above, since we have two successive percentage changes of x% and y% (which is a decrease) so the formula gives us $$x - y - \frac{xy}{100}$$ is the overall change. The question is, whether this change is positive i.e. whether $$x - y - \frac{xy}{100}$$ > 0 or
$$x - y > \frac{xy}{100}$$?
Since statement 2 tells us that $$x - y > \frac{xy}{100}$$, it is sufficient.

And yes, it helps to be clear about exactly what is asked before you move on to the statements.

Hi Karishma, can you please explain how you derived $$a + b + \frac{ab}{100}$$? I'm also not quite clear on why you used that second subtraction in $$10 - 20 - \frac{10*20}{100}$$. Thanks!

My earlier post in this thread has a link to her blog which explains successive percents.
http://www.veritasprep.com/blog/2011/02 ... e-changes/
Intern
Joined: 29 Jan 2011
Posts: 24
Re: GMAT 12th ed. #120 DS [#permalink]

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24 Feb 2011, 22:47
mbafall2011, I'll check it out. Thanks!

Bunuel wrote:
thanks wrote:
Feeling a little slow right now, but I'm failing to get the algebra from when you restate the question. The yearly breakdown is clear, but would you mind explaining the math where you multiply it out?

$$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$, $$r$$ cancels out. Then $$(1+a)(1-b)=1-b+a-ab$$.

$$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$, 1 cancels out --> $$0<-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$, multiplu by 100 both sides --> $$0<-y+x-\frac{xy}{100}$$, rearrange -->$$x-y>\frac{xy}{100}$$.

Hope it's clear.

Banuel, how advice on "seeing" when a question requires you to do this much manipulation on the stem? I don't think I would have ever thought to take the approach that you did.
Senior Manager
Joined: 21 Mar 2010
Posts: 310
Re: GMAT 12th ed. #120 DS [#permalink]

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24 Feb 2011, 22:54
GSDster wrote:
mbafall2011, I'll check it out. Thanks!

Bunuel wrote:
thanks wrote:
Feeling a little slow right now, but I'm failing to get the algebra from when you restate the question. The yearly breakdown is clear, but would you mind explaining the math where you multiply it out?

$$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$, $$r$$ cancels out. Then $$(1+a)(1-b)=1-b+a-ab$$.

$$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$, 1 cancels out --> $$0<-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$, multiplu by 100 both sides --> $$0<-y+x-\frac{xy}{100}$$, rearrange -->$$x-y>\frac{xy}{100}$$.

Hope it's clear.

Banuel, how advice on "seeing" when a question requires you to do this much manipulation on the stem? I don't think I would have ever thought to take the approach that you did.

You are welcome. I tried this method suggested in the blog, i didnt bother bother manipulating it to match the answer and hence i wasted time before going back to Bunuels method!
Although the formula is very useful
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

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25 Feb 2011, 05:37
Expert's post
1
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BOOKMARKED
You can use the formula in the question in a very straightforward manner...
Rent charged in 1998 was x% more
Rent charged in 1999 was y % less
So overall % change in rent charged was x - y - xy/100 (since y is a decrease use (-y) in place of y)
Question: Was rent collected in 1999 > 1997 i.e. was overall percentage of change positive?
Was x - y - xy/100 > 0 ?

(1) x > y
Not enough info

(2) xy/100 *< x – y
which is basically just x - y - xy/100 > 0 when you re-arrange.
So this statement gives you 'Yes' immediately. Sufficient.

Though, make sure that you know how the formula was derived and the basic concept behind it... Its good to know short cut formulas but they are not applicable everywhere... in some questions you might need to use ingenuity...
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 05 Oct 2011 Posts: 171 Re: What is the answer? [#permalink] ### Show Tags 08 Nov 2011, 14:01 Great explanations Bunuel and Karishma. Thank you Director Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 520 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Re: GMAT 12th ed. #120 DS [#permalink] ### Show Tags 24 Oct 2012, 20:32 Bunuel wrote: thanks 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 is greater than Y 2) xy/100 is less than x-y I'm hoping that someone can elaborate on the number property that gives answer b. PS I'm assuming that there is no way of knowing what level the questions are from the Review books??? Given: Rent in 1997 - $$r$$; Rent in 1998 - $$r*(1+\frac{x}{100})$$; Rent in 1999 - $$r*(1+\frac{x}{100})*(1-\frac{y}{100})$$. Question is $$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$ true? --> $$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$ --> $$x-y>\frac{xy}{100}$$ true? (1) $$x>y$$, based on this information we can not conclude whether $$x-y>\frac{xy}{100}$$ is true or not. Not sufficient. (2) $$\frac{xy}{100} < x -y$$, directly states that the equation we were testing is true. Sufficient. Answer: B. Hi Bunuel, I didn't quite understand this : $$x>y$$, based on this information we can not conclude whether $$x-y>\frac{xy}{100}$$ is true or not. Not sufficient. If x>y, it indicates x-y >0 and xy/100 will anyway be less than x-y right? _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: GMAT 12th ed. #120 DS [#permalink] ### Show Tags 24 Oct 2012, 21:27 sachindia wrote: If x>y, it indicates x-y >0 and xy/100 will anyway be less than x-y right? Not necessary! Take for example, x = 40 and y = 30 x - y = 10 xy/100 = 12 xy/100 > x - y _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Manager
Joined: 16 Jan 2011
Posts: 103
Re: The annual rent collected by a corporation from a certain [#permalink]

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05 Jun 2013, 10:21
could you tell me please is it crucial to represent percents as x/100 or (1-x/100)? cannot we just say x or (1-x) etc.?

What im trying to say here is if take the rent in 1997 equal 100 then in 1999 it would be (1-y)(1+x)*100
is rent in 1999 > rent in 1997 e.g. (1-y)(1+x)*100>100 --> (1-y)(1+x)>1 --> (x-y)>xy

compared to the statement (2) i wouldnt recognize these two inequalities as similar ones. And would go with (E)
Could you please shed light on this? What would be my steps if i came up with such a situation in the exam?
Intern
Joined: 22 Jul 2010
Posts: 32
Re: GMAT 12th ed. #120 DS [#permalink]

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13 Aug 2013, 12:47
Hi Bunuel

I wen with option D although I boiled the question to the equation same as u had mentioned simply because I could not come up with any value for X and Y that violates the condition X>Y and X-Y>XY/100. Could you come up with an exception to this condition so that A could be struck off?

Bunuel wrote:
thanks 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 is greater than Y
2) xy/100 is less than x-y

I'm hoping that someone can elaborate on the number property that gives answer b.

PS I'm assuming that there is no way of knowing what level the questions are from the Review books???

Given:

Rent in 1997 - $$r$$;
Rent in 1998 - $$r*(1+\frac{x}{100})$$;
Rent in 1999 - $$r*(1+\frac{x}{100})*(1-\frac{y}{100})$$.

Question is $$r<r*(1+\frac{x}{100})*(1-\frac{y}{100})$$ true? --> $$1<1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000}$$ --> $$x-y>\frac{xy}{100}$$ true?

(1) $$x>y$$, based on this information we can not conclude whether $$x-y>\frac{xy}{100}$$ is true or not. Not sufficient.

(2) $$\frac{xy}{100} < x -y$$, directly states that the equation we were testing is true. Sufficient.

Intern
Joined: 22 Jul 2010
Posts: 32
Re: The annual rent collected by a corporation from a certain [#permalink]

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13 Aug 2013, 21:52
Thanks. I overlooked this condition.

Posted from my mobile device
Intern
Joined: 08 Oct 2011
Posts: 42

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27 Sep 2013, 11:15
Zorba wrote:
Can anybody please expalin how can we get the ans by substituting values ????
Like I assumed the following values
Let the annual rent collected in 1997 be 100
Assume x = 20 and y= 10
Rent in 1998 = 120 and 1999 = 108
Assume x = 10 and y = 20
Rent in 1998 = 110 and 1999 = 89

For 1st option whenever X>Y we get the rent in 1999>1997.
For 2nd option - Since these are %, they will always be +ve and to make xy/100<x-y, x will have to be greater than y and hence the rent in 1999>1997

I used the same approach to get to D

Let the annual rent collected in 1997 be 100

1) Assume x=10%, y=9%
Rent in 1998 = 110
Rent in 1999 = 100.1
For X>Y, rent in 1999 > rent in 1997

2) xy/100 < x-y

I used the same plug in values here as in the 1st condition.

To hold (x-y) greater than (xy/100), x has to be greater than y and rent collected in 1997 is therefore less than the rent collected in 1999.

Math Expert
Joined: 02 Sep 2009
Posts: 39609

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27 Sep 2013, 11:24
aakrity wrote:
Zorba wrote:
Can anybody please expalin how can we get the ans by substituting values ????
Like I assumed the following values
Let the annual rent collected in 1997 be 100
Assume x = 20 and y= 10
Rent in 1998 = 120 and 1999 = 108
Assume x = 10 and y = 20
Rent in 1998 = 110 and 1999 = 89

For 1st option whenever X>Y we get the rent in 1999>1997.
For 2nd option - Since these are %, they will always be +ve and to make xy/100<x-y, x will have to be greater than y and hence the rent in 1999>1997

I used the same approach to get to D

Let the annual rent collected in 1997 be 100

1) Assume x=10%, y=9%
Rent in 1998 = 110
Rent in 1999 = 100.1
For X>Y, rent in 1999 > rent in 1997

2) xy/100 < x-y

I used the same plug in values here as in the 1st condition.

To hold (x-y) greater than (xy/100), x has to be greater than y and rent collected in 1997 is therefore less than the rent collected in 1999.

You cannot get sufficiency based only on one set of numbers. Try some others.
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Intern
Joined: 08 Oct 2011
Posts: 42

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27 Sep 2013, 11:29
Bunuel wrote:
aakrity wrote:
Zorba wrote:
Can anybody please expalin how can we get the ans by substituting values ????
Like I assumed the following values
Let the annual rent collected in 1997 be 100
Assume x = 20 and y= 10
Rent in 1998 = 120 and 1999 = 108
Assume x = 10 and y = 20
Rent in 1998 = 110 and 1999 = 89

For 1st option whenever X>Y we get the rent in 1999>1997.
For 2nd option - Since these are %, they will always be +ve and to make xy/100<x-y, x will have to be greater than y and hence the rent in 1999>1997

I used the same approach to get to D

Let the annual rent collected in 1997 be 100

1) Assume x=10%, y=9%
Rent in 1998 = 110
Rent in 1999 = 100.1
For X>Y, rent in 1999 > rent in 1997

2) xy/100 < x-y

I used the same plug in values here as in the 1st condition.

To hold (x-y) greater than (xy/100), x has to be greater than y and rent collected in 1997 is therefore less than the rent collected in 1999.

You cannot get sufficiency based only on one set of numbers. Try some others.

So when I picked 10 & 9, I also picked the maximum difference 10 & 1 and got to the same answer

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