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Data Sufficiency questions

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Data Sufficiency questions

by steven7dong » Sat Apr 12, 2014 7:05 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

I saw that lots of post states that the 1999 rent collection is R(1+x/100)(1-y/100), however, I have problem understanding this, because I think the 1999 rent collection is R(1+x/100)(1+y/100).
According to the correct answer, looks like the equation they used is (1998 rent-1999 rent)/1998 rent. That's how they got R(1+x/100)(1-y/100)
The equation I used is (1999 rent-1998 rent)/1998 rent and that is how I got R(1+x/100)(1+y/100).
How do we determine which one minus which when doing this kind of questions.

Thanks for your time XP
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by ajaysingh24 » Sat Apr 12, 2014 9:45 pm
steven7dong 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

I saw that lots of post states that the 1999 rent collection is R(1+x/100)(1-y/100), however, I have problem understanding this, because I think the 1999 rent collection is R(1+x/100)(1+y/100).
According to the correct answer, looks like the equation they used is (1998 rent-1999 rent)/1998 rent. That's how they got R(1+x/100)(1-y/100)
The equation I used is (1999 rent-1998 rent)/1998 rent and that is how I got R(1+x/100)(1+y/100).
How do we determine which one minus which when doing this kind of questions.

Thanks for your time XP
Let the rent collection in 1997 as R
so rent collection in 1998 will be R(1 + x/100) ( as it is x percent more than in 1997)
rent collection in 1999 will be (rent collection in 1998)( 1 - y/100) ( as it is y percent less in comparison to 1998 )
so it effectively becomes R(1 + x/100)(1 - y/100) = R(1 -y/100 + x/100 - xy/10000)
Rent of 1999 - rent of 1997 = R[ (x-y)/100 - xy/10000 ]
so condition 2 gives the solutions
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by GMATGuruNY » Sun Apr 13, 2014 5:19 am
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

b
Let the 1997 rent = 100.
Increase in 1998 = (x/100) * 100 = x.
Thus, rent in 1998 = 100+x.
Decrease in 1999 = (y/100) * (100+x) = y + xy/100
The question stem asks whether the increase is greater than the decrease.

Question stem, rephrased: Is x > y + xy/100?

Statement 1: x > y
Case 1: x=20 and y=10
Here, y + xy/100 = 10 + (20*10)/100 = 12.
Is 20 > 12?
YES.

Case 2: x=200 and y=100
Here, y + xy/100 = 100 + (200*100)/100 = 300.
Is 200 > 300?
NO.

Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2: x-y > xy/100
Thus, x > y + xy/100.
SUFFICIENT.

The correct answer is B.
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