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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 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....

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.

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!

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

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.

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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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

gmatclubot

Re: The annual rent collected by a corporation from a certain
[#permalink]
30 Sep 2014, 23:05

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