Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 May 2015, 17:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The annual rent collected by a corporation from a certain

Author Message
TAGS:
Senior Manager
Joined: 21 Mar 2010
Posts: 316
Followers: 5

Kudos [?]: 22 [0], given: 33

1
This post was
BOOKMARKED
GSDster wrote:
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.

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!

My earlier post in this thread has a link to her blog which explains successive percents.
http://www.veritasprep.com/blog/2011/02 ... e-changes/
 Veritas Prep GMAT Discount Codes Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
Intern
Joined: 29 Jan 2011
Posts: 24
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: GMAT 12th ed. #120 DS [#permalink]  24 Feb 2011, 21:47
mbafall2011, I'll check it out. Thanks!

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.

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.
Senior Manager
Joined: 21 Mar 2010
Posts: 316
Followers: 5

Kudos [?]: 22 [0], given: 33

Re: GMAT 12th ed. #120 DS [#permalink]  24 Feb 2011, 21:54
GSDster wrote:
mbafall2011, I'll check it out. Thanks!

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.

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5449
Location: Pune, India
Followers: 1331

Kudos [?]: 6775 [1] , given: 177

1
KUDOS
Expert's post
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...
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Manager
Joined: 05 Oct 2011
Posts: 169
GMAT 1: 720 Q V
Followers: 9

Kudos [?]: 39 [0], given: 60

Great explanations Bunuel and Karishma. Thank you
Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 548
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 2

Kudos [?]: 37 [0], given: 562

Re: GMAT 12th ed. #120 DS [#permalink]  24 Oct 2012, 19:32
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.

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.

Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595

My GMAT Journey : end-of-my-gmat-journey-149328.html#p1197992

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5449
Location: Pune, India
Followers: 1331

Kudos [?]: 6775 [0], given: 177

Re: GMAT 12th ed. #120 DS [#permalink]  24 Oct 2012, 20:27
Expert's post
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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Manager
Joined: 15 Jan 2011
Posts: 107
Followers: 8

Kudos [?]: 73 [0], given: 13

Re: The annual rent collected by a corporation from a certain [#permalink]  05 Jun 2013, 09:21
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?
Intern
Joined: 22 Jul 2010
Posts: 30
Followers: 0

Kudos [?]: 6 [0], given: 11

Re: GMAT 12th ed. #120 DS [#permalink]  13 Aug 2013, 11:47
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?

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.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5449
Location: Pune, India
Followers: 1331

Kudos [?]: 6775 [2] , given: 177

Re: GMAT 12th ed. #120 DS [#permalink]  13 Aug 2013, 20:03
2
KUDOS
Expert's post
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?

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Intern
Joined: 22 Jul 2010
Posts: 30
Followers: 0

Kudos [?]: 6 [0], given: 11

Re: The annual rent collected by a corporation from a certain [#permalink]  13 Aug 2013, 20:52
Thanks. I overlooked this condition.

Posted from my mobile device
Intern
Joined: 08 Oct 2011
Posts: 42
Followers: 0

Kudos [?]: 9 [0], given: 38

Zorba wrote:
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

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.

Math Expert
Joined: 02 Sep 2009
Posts: 27170
Followers: 4226

Kudos [?]: 40945 [0], given: 5576

Expert's post
aakrity wrote:
Zorba wrote:
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

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.

You cannot get sufficiency based only on one set of numbers. Try some others.
_________________
Intern
Joined: 08 Oct 2011
Posts: 42
Followers: 0

Kudos [?]: 9 [0], given: 38

Bunuel wrote:
aakrity wrote:
Zorba wrote:
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

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.

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
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4764
Followers: 295

Kudos [?]: 52 [0], given: 0

Re: The annual rent collected by a corporation from a certain [#permalink]  30 Sep 2014, 21:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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.
_________________
Manager
Joined: 14 Jul 2014
Posts: 100
Followers: 0

Kudos [?]: 16 [0], given: 49

Re: The annual rent collected by a corporation from a certain [#permalink]  30 Sep 2014, 22:05
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
Math Expert
Joined: 02 Sep 2009
Posts: 27170
Followers: 4226

Kudos [?]: 40945 [0], given: 5576

Re: The annual rent collected by a corporation from a certain [#permalink]  30 Sep 2014, 23:27
Expert's post
Buddyisrael wrote:
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

We can safely assume here that the annual rent collected is some positive value.
_________________
Manager
Joined: 14 Jul 2014
Posts: 100
Followers: 0

Kudos [?]: 16 [0], given: 49

Re: The annual rent collected by a corporation from a certain [#permalink]  01 Oct 2014, 10:29
"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[/quote]

We can safely assume here that the annual rent collected is some positive value.[/quote]

Thanks Bunuel !! Your methods and strategies really help a lot !
Intern
Joined: 16 Nov 2013
Posts: 32
Followers: 0

Kudos [?]: 9 [0], given: 3

Re: The annual rent collected by a corporation from a certain [#permalink]  29 Dec 2014, 14:18
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.

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

Thank you
Intern
Joined: 05 Aug 2014
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 8

Re: The annual rent collected by a corporation from a certain [#permalink]  02 May 2015, 07:32
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.

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?!
Re: The annual rent collected by a corporation from a certain   [#permalink] 02 May 2015, 07:32

Go to page   Previous    1   2   3    Next  [ 41 posts ]

Similar topics Replies Last post
Similar
Topics:
2 The annual rent collected by a corporation from a certain 1 19 Sep 2008, 09:44
The annual rent collected by a corporation from a certain 8 22 Aug 2006, 09:58
The annual rent collected by a corporation from a certain 5 05 Aug 2006, 07:10
The annual rent collected by a corporation from a certain 3 12 May 2006, 10:58
The annual rent collected by a corporation from a certain 7 25 Apr 2006, 11:37
Display posts from previous: Sort by