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The annual rent collected by a corporation from a certain building was

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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) \(\frac{xy}{100} < x-y\)

Originally posted by thanks on 02 Jan 2010, 15:10.
Last edited by Bunuel on 09 Sep 2019, 00:32, edited 3 times in total.
Edited the question and added the OA
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New post 02 Jan 2010, 15:16
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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.
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Re: The annual rent collected by a corporation from a certain building was  [#permalink]

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New post 14 Jan 2010, 15:09
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samrand 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
OFFICIAL Test:answer:B but why?


consider rent collected are variables as follows
1997 - a, 1998 - b, 1999 - c.

Given a[1 + (x/100)] = b, and b[1 - (y/100)] = c.
Therefore, a[1+(x/100)][1-(y/100)]=c.
For c to be more than a, a has to be multiplied with a positive term expressed as [1 + (m/100)]

Only stmt-2 helps us derive such an expression upon simplifying the above equation. Stmt-1 is definitely non sufficient.
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New post 02 Jan 2010, 15:51
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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.
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New post 08 Sep 2010, 13:11
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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.

Well crafted solution a KUDOS!

BTW... Thank you for such wonderful explanations that make understanding tough problems like these easy.
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New post 15 Nov 2010, 08:37
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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.
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New post 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.

Answer: B.



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....
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New post 16 Nov 2010, 04:46
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srivicool wrote:
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.



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....


Theory: math-number-theory-percents-91708.html

Practice DS: search.php?search_id=tag&tag_id=192
Practice PS: search.php?search_id=tag&tag_id=191

Hope it helps.
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New post 19 Feb 2011, 07:49
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I like VeritasPrepKarishma's answer- Karishma, you should start a thread with all these formulae in one place- especially the unconventional ones!
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New post 19 Feb 2011, 20:09
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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.)
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New post 25 Feb 2011, 05:37
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...
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New post 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?
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New post 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
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New post 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?
Thanks in Advance!!!



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.
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New post 13 Aug 2013, 21:03
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mitmat wrote:
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?
Thanks in Advance!!!



Given X > Y, X-Y>XY/100 may or may not hold

Say X = 25, Y = 20 (or X = 51, Y = 50 etc)
X > Y holds
X-Y>XY/100 does not hold

Say X = 20, Y = 10
X > Y holds
X-Y>XY/100 holds
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New post 02 May 2015, 11:28
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.


I'm abit uncertain about this approach. Forgive me, but all I'm getting out of this is you saying "Okay look at statement two: it is sufficient"

The whole 100 --> 110 --> 88 approach is not even applied in the problem? So we have a total negative percentage change in that issue, but how does that translate to statement (2)


If I pick other numbers, say: 1997 as 100. 1998 as 110 and 1999 as 104.5

We got a 10% increase followed by a 5% decrease.

\(x - y \frac{xy}{100}\) still holds? \(10-(-5) > \frac{10(-5)}{100}\)

Thanks in advance
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New post 06 May 2015, 01:06
Hi,
When I evaluated the question,I got reached the inequality in statement 2, hence clearly sufficient. However, in statement 1, I picked values for X and y. X=2 and y=1 and then I substituted them in the inequality xy/100<x−y when it held true, I assumed that the statement is correct. Why was I wrong here? Thanks to anyone who has an answer/insight to share.
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New post 06 May 2015, 02:12
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erikvm

So here is the breakdown of this approach:

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



Talks about the concept the question is testing - successive % changes and tells you what the concept is. Uses a random example.

VeritasPrepKarishma wrote:
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.


Talks about a formula related to this concept. Uses the same example as above.

VeritasPrepKarishma wrote:
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.


Here, we come to the given question. Shows you why statement 2 is sufficient. You just use the formula on x and y and you can see that statement 2 is sufficient.
Statement 1 is not sufficient because it tells you that x - y is positive. It doesn't tell you whether x - y - xy/100 is positive.

Pick some numbers to see how statement 1 is not sufficient:

If x = 10 and y = 5,
Rent in 1997 is 100, rent in 1998 is 110 and rent in 1999 is 104.5.
Rent in 1997 < Rent in 1999

But if

If x = 10 and y = 9.99999999 (almost 10 but still x > y)
Rent in 1997 is 100, then rent in 1998 is 110 and rent in 1999 is very slightly more than 99.
Rent in 1997 > Rent in 1999

So the rent in 1997 may be more or less than rent in 1999 if x > y. Hence statement 1 alone is not sufficient.
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New post 06 May 2015, 02:24
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naeln wrote:
Hi,
When I evaluated the question,I got reached the inequality in statement 2, hence clearly sufficient. However, in statement 1, I picked values for X and y. X=2 and y=1 and then I substituted them in the inequality xy/100<x−y when it held true, I assumed that the statement is correct. Why was I wrong here? Thanks to anyone who has an answer/insight to share.



Just because the inequality holds true for one set of values such that x > y, it doesn't mean it will hold for all such values. Look at this set of values:

x = 10, y = 9.5

x - y = 0.5

xy/100 = 0.95

Here, x - y is not greater than xy/100 though x > y.

In your example, you got that x - y is greater than xy/100. So stmnt 1 alone is not sufficient.

When you use number picking in DS, be very careful. It is very hard to prove that a statement holds based on some numbers. It is absolutely acceptable to prove that a statement IS NOT SUFFICIENT by assuming numbers because all you need is two set of numbers to do that. For example, to prove that stmnt 1 alone is not sufficient in this question, you just need two examples: one that I gave you and second that you used. We know that stmnt 1 alone is not sufficient. On the other hand, if you are trying to prove that stmnt 1 alone is sufficient, even if you check for 10 values, you cannot be sure that it will hold everytime x > y because there are infinite cases in which x is greater than y. It is possible that you miss one. Hence, to prove that a statement is enough, you will need to rely on logic/pattern recognition etc.
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New post 06 May 2015, 04:44
Thank you Karishma :) Got it. Do you have any recommendations about how to strategically pick values that could render different results when tested?
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Re: The annual rent collected by a corporation from a certain building was   [#permalink] 06 May 2015, 04:44

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The annual rent collected by a corporation from a certain building was

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