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Easier approach

by tapanmittal » Mon Aug 17, 2015 10:26 am
127. 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

OA-B
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Source: — Data Sufficiency |
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by MartyMurray » Mon Aug 17, 2015 11:14 am
tapanmittal wrote:127. 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
Statement 1 is a trap here. You could almost convince yourself that it is sufficient, because given that x > y it seems that the rent collected went up more than it went down.

However, Statement 1 incorporates a trick commonly used in GMAT data sufficiency questions, describing a situation in percentage terms and then asking a question about actual, rather than percentage, change.

You could approach Statement 1 mathematically, but I prefer to just be aware that when something increases to a higher level there is more of a percentage change than there is when going the other direction.

For instance, going from 80 to 100 is a 25% increase, but going from 100 to 80 is a 20% decrease.

So Statement 1 could be true if the rent collected in 1999 were higher, the same, or lower than that collected in 1997. So Statement 1 is insufficient.

Statement 2 is interesting. It constrains the difference between x and y, and so it immediately starts to look sufficient, but is it?

Let's pick a number, 100, for the rent in 1997 and see what we can do.

If the rent collected in 1997 is 100, then the rent collected in 1998 is 100 + 100(x/100) = 100 + x. So the change is x.

If the rent collected in 1998 is 100 + x, then the rent collected in 1999 is
(100 + x) - (y/100)(100 + x)

So the change from 1998 to 1999 is y + xy/100.

So the question becomes is the change from 1997 to 1998, x, greater than the change from 1998 to 1999, y + xy/100.

We can manipulate Statement 2 to get x > y + xy/100.

So the change from 1997 to 1998 is greater, and Statement 2 is sufficient, and the correct answer is B.

Is that an "easier approach"? I guess that depends on what the alternative is.
Last edited by MartyMurray on Mon Aug 17, 2015 11:22 am, edited 1 time in total.
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by DavidG@VeritasPrep » Mon Aug 17, 2015 11:15 am
There's a nifty little formula for calculating the net percent change when you have two successive percent changes: x + y + xy/100.

(Try it with numbers. Say your starting value is 100, and that this quantity increases by 20% and then increases by 10%. After the 20% increase, you'll have 120. After this value increases by another 10%, you'll have 120 + 12 = 132. The net change is 32%. If we plug '20' in for x and '10' in for y, we get:
20 + 10 + 20*10/100 = 32.)

In this case, because we're told that rent increases by x percent in 1998 and then decreases by y percent in 1999, the net change will be x - y - xy/100. (We need to subtract y as this value represents a decrease.) If we want to know if rent has increased after the two successive changes, we really want to know:
Is x - y - xy/100>0? So that's our rephrased question.

1) Test simple cases. Say x = 50 and y = 10. If we start with 100 in 1997, after the 50% increase, we'll have 150 in 1998. After the 10% decrease, we'll have 150 - 15 = 135 in 1999. So now we have a YES. Rent collected is more in 1999 than in 1997.

Say x = 25 and y = 20. If we start with 100 in 1997, we'll have 125 in 1998 after the 25% increase. After the 20% decrease, we'll have 125 - 25 = 100 in 1999. So now we have a NO. Rent collected in 1999 was not more than rent collected in 1997.

2) Now we can just use our trusty formula. The rephrased question: Is x - y - xy/100>0?

We're told that xy/100< x - y. Subtract xy/100 from both sides, and we get 0< x - y - 100/xy. Because this tells us that our two successive changes will have a net positive change, we know that rent will be more in 1999 than in 1997. So this is sufficient on its own. Correct answer is B
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by GMATGuruNY » Mon Aug 17, 2015 11:44 am
I posted an approach similar to Marty's here:

https://www.beatthegmat.com/quick-way-to ... 82917.html

Another way to evaluate statement 2 is to plug in an easy value for x and solve for y.

Statement 2: (xy/100) < x-y
Let x=100%.
If we plug x=100 into (xy/100) < x-y, we get:
(100y)/100 < 100-y
2y < 100
y < 50%.

Let the 1997 rent = $100.
Since the 1998 rent increases by x=100%, the 1998 rent = $100 + 100% of 100 = $200.
Since the 1999 rent decreases by y<50%, the 1999 rent = $200 - (less than 50% of 200) = 200 - (less than 100) = MORE THAN $100.
Since the 1997 rent = $100, and the 1999 rent = more than $100, the 1999 rent is greater than the 1997 rent.
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by Max@Math Revolution » Thu Aug 20, 2015 9:45 am
In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and equations ensures a solution.



With R as the rent fee for 1997, and using the original condition and the question,
R==>R(1+x%)==>R(1+x%)(1-y%)
And once we substitute it into the question, we get R(1+x%)(1-y)%>R? ==> (1+x%)(1-y%)>1?
==> 1+x%-y%-(xy/10,000)>1?
==> x%-y%-(xy/10,000)>0?==> x-y-(xy/100)>0? ==> x-y>xy/100?
Thus the best answer is B.


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