## OG 13 127

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### OG 13 127

by oquiella » Tue Dec 22, 2015 3:39 pm
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 i 1997?

1. x>y
2. xy/100<x-y

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by [email protected] » Tue Dec 22, 2015 10:26 pm
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: x > y

This is a trap answer. One might think that because x > y the rent went up more than it went down, but since the rent was higher in 1998 than in 1997, a smaller percentage decrease from 1998 to 1999 can actually be a greater absolute decrease.

Use two extremes.

First extreme: x = 100 y = 0

When those numbers are used the 1999 rent will clearly be greater than the 1997 rent.

Second extreme: x = 101 y = 100

Using these numbers, the rent collected in 1999 goes to 0. So it can't be greater than that collected in 1997.

So we can get two different answers to the question.

Insufficient.

Statement 2: xy/100 < x - y

We know that the rent change from 1998 to 1999 was y percentage of a number affected first by x. So that xy/100 is interesting, and this might be sufficient.

Call the 1997 rent R. The 1998 rent is R + xR/100. The 1999 rent is R + xR/100 - (yR/100 + yxR/10,000).

So the question becomes the following. Is R < R + xR/100 - (yR/100 + yxR/10,000)?

Subtract R from both sides and the question becomes this. Is O < xR/100 - (yR/100 + yxR/10,000)?

This is starting to look like Statement 2.

Divide both sides by R. Is O < x/100 - (y/100 + yx/10,000)?

Is O < x/100 - y/100 - yx/10,000?

Is O < x - y - yx/100?

Is xy/100 < x - y?

We know from Statement 2 that the answer to the question is yes.

Sufficient.

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by GMATGuruNY » Wed Dec 23, 2015 4:40 am
I posted a solution 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.
SUFFICIENT.
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by oquiella » Thu Dec 24, 2015 4:46 am
Marty Murray 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
Statement 1: x > y

This is a trap answer. One might think that because x > y the rent went up more than it went down, but since the rent was higher in 1998 than in 1997, a smaller percentage decrease from 1998 to 1999 can actually be a greater absolute decrease.

Use two extremes.

First extreme: x = 100 y = 0

When those numbers are used the 1999 rent will clearly be greater than the 1997 rent.

Second extreme: x = 101 y = 100

Using these numbers, the rent collected in 1999 goes to 0. So it can't be greater than that collected in 1997.

So we can get two different answers to the question.

Insufficient.

Statement 2: xy/100 < x - y

We know that the rent change from 1998 to 1999 was y percentage of a number affected first by x. So that xy/100 is interesting, and this might be sufficient.

Call the 1997 rent R. The 1998 rent is R + xR/100. The 1999 rent is R + xR/100 - (yR/100 + yxR/10,000).

So the question becomes the following. Is R < R + xR/100 - (yR/100 + yxR/10,000)?

Subtract R from both sides and the question becomes this. Is O < xR/100 - (yR/100 + yxR/10,000)?

This is starting to look like Statement 2.

Divide both sides by R. Is O < x/100 - (y/100 + yx/10,000)?

Is O < x/100 - y/100 - yx/10,000?

Is O < x - y - yx/100?

Is xy/100 < x - y?

We know from Statement 2 that the answer to the question is yes.

Sufficient.

Can you explain in further detail this part

"Subtract R from both sides and the question becomes this. Is O < xR/100 - (yR/100 + yxR/10,000)?"

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by [email protected] » Thu Dec 24, 2015 5:01 am
oquiella wrote:Can you explain in further detail this part

"Subtract R from both sides and the question becomes this. Is O < xR/100 - (yR/100 + yxR/10,000)?
Equal amounts can be subtracted from both sides of an inequality.

Here's the inequality. Notice R on both sides.

R < R + xR/100 - (yR/100 + yxR/10,000)?

Now we subtract R from each side of the inequality to get the following.

O < xR/100 - (yR/100 + yxR/10,000)
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by [email protected] » Sun Oct 23, 2016 3:43 pm
oquiella 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 i 1997?

1. x>y
2. xy/100<x-y
We are given that the rent collected in a building was x percent more in 1998 than it was in 1997 and y percent less in 1999 than it was in was in 1998. Let's start by defining some variables.

a = the annual rent collected in 1997

b = the annual rent collected in 1998

c = the annual rent collected in 1999

We can now create the following equations, using the "percent greater than" and "percent less than" formulas:

b = [(100+x)/100]a

c = [(100-y)/100]b

We need to determine whether the annual rent collected by the corporation was more in 1999 than in 1997. Thus, we need to determine: Is c > a?

Since b = [(100+x)/100]a and c = [(100-y)/100]b, that means

c = [(100-y)/100][(100+x)/100]a.

Now we can rephrase the question as:

Is [(100-y)/100][(100+x)/100]a > a?

Notice if we divide the entire inequality by a, we have:

Is [(100-y)/100][(100+x)/100] > 1?

Is (100-y)(100+x)/10,000 > 1?

Is (100+x)(100-y) > 10,000 ?

Is 10,000 - 100y + 100x - xy > 10,000 ?

Is -100y + 100x - xy > 0 ?

Is 100 x - 100y > xy ?

Is 100(x - y) > xy ?

Statement One Alone:

x > y

Knowing only that x is greater than y is not enough to determine whether 100(x - y) > xy. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

(xy/100) < (x-y)

Multiplying both sides of the inequality by 100, we have:

xy <100(x - y)

xy < 100(x - y) is exactly the same as saying 100(x - y) > xy. Statement two alone is sufficient to answer the question.

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by [email protected] » Wed Jan 18, 2017 4:45 am
oquiella 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 i 1997?

1. x>y
2. xy/100<x-y

If the price of commodity increases by say 10% = 1/10, then its increased price should decrease by 1/(1+10) = 1/11 = 9.09% to match the initial price.

Or, we can say that if the price of commodity increases by say 1/a, then its price should decrease by 1/(1+a).

In the problem, say x% = 10%, thus to bring the increased price equal to the initial price, we must decrease the increased price by 9.09%.

You may try with these values; you would find that only statement 2 works.

Hope this helps!

-Jay
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