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The annual rent collected by a corporation from a certain

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Re: The annual rent collected by a corporation from a certain  [#permalink]

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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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06 May 2015, 02:12
1
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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06 May 2015, 02:24
1
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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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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06 May 2015, 20:13
naeln wrote:
Thank you Karishma Got it. Do you have any recommendations about how to strategically pick values that could render different results when tested?

Yes, focus on the transition points i.e. where two sides are equal.
Say you are given x > y. Find what happens at x = y.
For example, in the question above, I took x to be 10 and y to be infinitesimally close to 10 but still less than 10. This would give you the value at the extreme i.e. the lowest value of rent collected in 1999.

Here are a couple of posts on how to choose numbers:

http://www.veritasprep.com/blog/2013/05 ... on-points/
http://www.veritasprep.com/blog/2013/08 ... on-points/
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10 May 2015, 05:13
VeritasPrepKarishma wrote:
naeln wrote:
Thank you Karishma Got it. Do you have any recommendations about how to strategically pick values that could render different results when tested?

Yes, focus on the transition points i.e. where two sides are equal.
Say you are given x > y. Find what happens at x = y.
For example, in the question above, I took x to be 10 and y to be infinitesimally close to 10 but still less than 10. This would give you the value at the extreme i.e. the lowest value of rent collected in 1999.

Here are a couple of posts on how to choose numbers:

http://www.veritasprep.com/blog/2013/05 ... on-points/
http://www.veritasprep.com/blog/2013/08 ... on-points/

Thank you Karishma for sharing the invaluable posts. I truly appreciate it.
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09 Nov 2015, 14:47
Hi Karishma,

Did you manage to make a compilation of all conventional and unconventional formulas?

Thanks!
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09 Nov 2015, 20:23
1
1
MensaNumber wrote:
Hi Karishma,

Did you manage to make a compilation of all conventional and unconventional formulas?

Thanks!

I have discussed all relevant formulas on my blog: http://www.veritasprep.com/blog/categor ... er-wisdom/
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10 Jan 2016, 14:07
I tried the testing cases method. I ended up with D.

Could you give me a couple of cases where I can prove A to be insufficient? I am not able to think of any.
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10 Jan 2016, 22:15
1
Avinashs87 wrote:
I tried the testing cases method. I ended up with D.

Could you give me a couple of cases where I can prove A to be insufficient? I am not able to think of any.

If you are going to test cases to arrive at the answer in DS, you have to be very careful about checking all relevant ranges. In this post above (the-annual-rent-collected-by-a-corporation-from-a-certain-89184-40.html#p1523193), I have given an example of values which show that statement 1 alone is not sufficient. You must find the transition points and then ensure that you test cases around them. Of course, in most cases, it is far better to use logic.
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Re: The annual rent collected by a corporation from a certain  [#permalink]

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16 Jul 2016, 05:08
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.

Hi Bunuel,
Though I understand the above given solution, I am still confused why A is wrong.
If suppose anything is increased by X and that amount is then decreased by Y when X>Y where we will consider all positive values I can't seem to find a situation where 1997 amount would be less than 1999 amount.

e.g X=20% and Y=10% Let the 1997 amount be 100.
In 1997=100
In 1998= 120
In 1999= 108

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Re: The annual rent collected by a corporation from a certain  [#permalink]

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16 Jul 2016, 08:15
1
anurag16 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.

Hi Bunuel,
Though I understand the above given solution, I am still confused why A is wrong.
If suppose anything is increased by X and that amount is then decreased by Y when X>Y where we will consider all positive values I can't seem to find a situation where 1997 amount would be less than 1999 amount.

e.g X=20% and Y=10% Let the 1997 amount be 100.
In 1997=100
In 1998= 120
In 1999= 108

In your case consider y to be 19%.
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16 Jul 2016, 08:30
Quote:
In your case consider y to be 19%.

Ohhhh...Got it thanks!!
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07 Sep 2016, 09:57
1
Here is my two cents for those who are struggling odln how to build the comparison with the options.

I would NEVER pick numbers since it is extremely time consuming and mistake susceptible.

Posted from my mobile device
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17 Jan 2017, 19:14
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I do not recommend using algebra to evaluate condition #2--it's too complicated for most. Here is a visual that should help, showing you that you can evaluate condition #2 by plugging in to determine the "boundary line" (the value that separates the "Yes" answers from the "No" answers).
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Re: The annual rent collected by a corporation from a certain  [#permalink]

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21 Dec 2017, 06:44
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 > y
(2) xy/100 < x-y

Check out our detailed video solution of this problem here:
https://www.veritasprep.com/gmat-soluti ... ciency_382
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