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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.

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!

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.

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

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... _________________

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.

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?

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.

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 _________________

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

Please help !!

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.

Please explain why the answer is not D? Thank you.

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

Please help !!

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.

Please explain why the answer is not D? Thank you.

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

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

Please help !!

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.

Please explain why the answer is not D? Thank you.

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

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

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30 Sep 2014, 22:55

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Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

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30 Sep 2014, 23:05

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.

"R" cancels out because in this case we consider it +ve. As "Rents" are always +ve Correct?

Because we never cancel the variables in Inequalities unless we know the sign of that variable. Pls clarify my doubt. Thanks

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.

"R" cancels out because in this case we consider it +ve. As "Rents" are always +ve Correct?

Because we never cancel the variables in Inequalities unless we know the sign of that variable. Pls clarify my doubt. Thanks

We can safely assume here that the annual rent collected is some positive value. _________________

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

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29 Dec 2014, 15:18

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.

Can i state this:

1997 1998 1999

100 100+x 100+x-y

So: 100+x-y>100 and x>y.. What's wrong with this? E.G: If the price of a stock is x in 1997 and is 25 percent more in 1998 so the price in 1998 is 100+25=125 percent

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

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02 May 2015, 08:32

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, First, I would like to thank you for the huge support and the insightful GMAT techniques you have be providing; I have recently come across this forum, which, for me, is simply a gold mine. I hope you can help me learn where I went wrong with my analysis of this question. I simplified the inequality as you did in your answer. I could clearly see the statement 2 is sufficient.

However, my problem is with statement 1. x>y. I thought I could pick numbers so I picked x =2 and y =1 and then I substituted x and y in the inequality x−y>xy/100 and when it met, i thought the statement must be sufficient. Why was it wrong for me to do so?!

gmatclubot

Re: The annual rent collected by a corporation from a certain
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02 May 2015, 08:32

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