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

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

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New post 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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New post 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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New post 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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New post 09 Nov 2015, 20:23
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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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New post 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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New post 10 Jan 2016, 22:15
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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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New post 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.

Answer: B.


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

Please explain!!
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New post 16 Jul 2016, 08:15
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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.

Answer: B.


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

Please explain!!


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

Ohhhh...Got it thanks!!
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New post 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.

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New post 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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New post 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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New post 08 Jan 2018, 12:06
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kirankp 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


We are given that the rent collected in a building was x percent more in 1998 than it was in 1997 and y percent less in 1999 than it was in was in 1998. Let’s start by defining some variables.

a = the annual rent collected in 1997

b = the annual rent collected in 1998

c = the annual rent collected in 1999

We can now create the following equations, using the "percent greater than" and "percent less than" formulas:

b = [(100+x)/100]a

c = [(100-y)/100]b

We need to determine whether the annual rent collected by the corporation was more in 1999 than in 1997. Thus, we need to determine: Is c > a?

Since b = [(100+x)/100]a and c = [(100-y)/100]b, that means

c = [(100-y)/100][(100+x)/100]a.

Now we can rephrase the question as:

Is [(100-y)/100][(100+x)/100]a > a?

Notice if we divide the entire inequality by a, we have:

Is [(100-y)/100][(100+x)/100] > 1?

Is (100-y)(100+x)/10,000 > 1?

Is (100+x)(100-y) > 10,000 ?

Is 10,000 – 100y + 100x – xy > 10,000 ?

Is -100y + 100x – xy > 0 ?

Is 100 x – 100y > xy ?

Is 100(x – y) > xy ?

Statement One Alone:

x > y

Knowing only that x is greater than y is not enough to determine whether 100(x – y) > xy. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:


(xy/100) < (x-y)

Multiplying both sides of the inequality by 100, we have:

xy <100(x – y)

xy < 100(x – y) is exactly the same as saying 100(x – y) > xy. Statement two alone is sufficient to answer the question.

Answer: B
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