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The annual rent collected by a corporation from a certain [#permalink]
02 Jan 2010, 14:10
1
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49
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00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
52% (03:11) correct
48% (02:23) wrong based on 966 sessions
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?
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.
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 -->\(x-y>\frac{xy}{100}\).
Re: What is the answer? [#permalink]
14 Jan 2010, 14:09
1
This post received KUDOS
samrand 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 OFFICIAL Test:answer:B but why?
consider rent collected are variables as follows 1997 - a, 1998 - b, 1999 - c.
Given a[1 + (x/100)] = b, and b[1 - (y/100)] = c. Therefore, a[1+(x/100)][1-(y/100)]=c. For c to be more than a, a has to be multiplied with a positive term expressed as [1 + (m/100)]
Only stmt-2 helps us derive such an expression upon simplifying the above equation. Stmt-1 is definitely non sufficient. _________________
Re: Percentiles confusion [#permalink]
25 May 2010, 13:59
So I solved it like this:
1998 = 1997 (1+X) 1998 (1-Y) = 1999
1999 = 1997 (1+X)(1-Y)
1999 = 1997 [1+ (x-y-xy)]
Meaning if (X-Y) - XY is positive 1999>1997 and if (X-Y) - XY is negative 1999<1997
The answer is B but If XY < 100(X - Y) and XY= XY/10000 and Y-X = Y-X/100 it seems like the 100 in part B is in the wrong place to solve the inequality:
XY/10,000 < 100(X-Y)/100
it just seems strange that the 100 is in there because what we are really trying to find is weather XY/10000< (Y-X)/100
Re: What is the answer? [#permalink]
27 May 2010, 04:42
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
Re: What is the answer? [#permalink]
29 May 2010, 05:29
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
Re: Successive percent problem from OG 12th edition DS 120 [#permalink]
22 Jul 2010, 00:16
This is a good example of a DS problem that can be manipulated a lot before getting to the two statements. It becomes a really simple problem once you have done all the manipulation, but getting there can take a while.
I worry on the actual test I will be wondering if I should simplify first, or be nervous with the real timer going and just jump right into the statements.
Re: What is the answer? [#permalink]
22 Jul 2010, 07:36
i think the answer is D because A can also give clue. Upon expansion we get: r(1+(x-y)/100 - xy/10000)
given x>y, therefore (x-y)/100 is going to be positive and xy/10000< (x-y)/100, denominator comparison is enough. So (x-y)/100 - xy/10000 > 0. So rent collected will be greater. So D is the answer... Pls correct me if I am wrong...
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.
Well crafted solution a KUDOS!
BTW... Thank you for such wonderful explanations that make understanding tough problems like these easy.
Re: What is the answer? [#permalink]
15 Nov 2010, 07:37
27
This post received KUDOS
Expert's post
16
This post was BOOKMARKED
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. _________________
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....
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....
Re: OG 12th DS #120 [#permalink]
23 Nov 2010, 17:14
1
This post received KUDOS
Expert's post
tonebeeze wrote:
I am having problems with the 700+ successive percents problems. Can you recommend any books or GMAT clubs links that can help remedy my issues? Thanks.
Below is a problem I had difficulty with today. The OG's explanation was a bit convoluted. I would appreciate other approaches or explanations to this problem. Thanks!
"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. Wass the annual rent collected by the corporation from the building more in 1999 than in 1997?"
1. \(x > y\)
2. \(xy/100 < x - y\)
Let us assume that the revenue collected in 1997 was 100. Then the revenue in 1998 was \(\frac{(100+x)*100}{100} = (100+x)\) and the revenue collected in 1999 was \(\frac{(100-y)*(100+x)}{100}\).
Thus the ratio of the revenues in 1999 to 1997\(= \frac{(100-y)(100+x)}{(100)(100)} = \frac{(10000 + 100 (x-y) - xy)}{10000}\)
Now if x > y it doesn't prove anything. So insufficient.
Looking at the second statement, it's given that \(\frac{xy}{100} < (x-y)\)
For conclusively evaluating if the numerator is greater than the denominator we need to find out whether \(100(x-y) > xy\)
We are given that \(\frac{xy}{100} < (x-y) i.e. (xy) < 100 (x-y)\) and hence we can conclusively say \(100 (x-y) - xy > 0\) and hence the numerator > 10000 and hence the ratio > 1.
Re: Need help on DS: OG #120 [#permalink]
19 Feb 2011, 19:09
2
This post received KUDOS
Expert's post
mbafall2011 wrote:
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.) _________________
Re: Need help on DS: OG #120 [#permalink]
19 Feb 2011, 19:57
VeritasPrepKarishma wrote:
mbafall2011 wrote:
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.
Re: What is the answer? [#permalink]
24 Feb 2011, 21:41
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!
gmatclubot
Re: What is the answer?
[#permalink]
24 Feb 2011, 21:41
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