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Word prob

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by beater » Tue Sep 30, 2008 8:09 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


Could you please help me break down (2)
Last edited by beater on Wed Oct 01, 2008 9:26 am, edited 1 time in total.
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Source: — Data Sufficiency |
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Re: Word prob

by Morgoth » Tue Sep 30, 2008 10:01 am
beater 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) 100xy < x – y


Could you please help me break down (2)
let the rent = a

1997 = a

1998 = a + ax/100

1999 = a+ax/100 - y/100(a+ax/100)

question stem: 1999 > 1997 ?

a+ax/100 - y/100(a+ax/100) > a

Statement (1) x>y.

We can have both cases here 1999 > 1997, 1999<1997. Insufficient.

Statement (2)

1999>1997
a+ax/100 - y/100(a+ax/100) > a

(10000a + 100ax -100ay -ayx) / 10000 > a

10000a + 100ax - 100ay -ayx >10000a

100ax - 100ay - ayx > 0

100a (x-y) > ayx

100 (x-y) > yx

x-y > yx/100 [ this is what I am getting]
----------------------------------------------------

x-y>100xy [this is what the statement (2) is ]


are you sure that the statement is correctly posted!!
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Re: Word prob

by xunil56 » Tue Sep 30, 2008 2:08 pm
beater 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) 100xy < x – y


Could you please help me break down (2)
My answer is A, by plugging in numbers:

1) If x = 20% and y = 10% then
1997: $100 rent and 1998: $120 and 1999: $108

if you change this where x=y or x < y then rent collected in 1997 is more than in 1999.

2) If I plug in the values from above it is insuff.


What is the OA?
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by schumi_gmat » Tue Sep 30, 2008 2:21 pm
If u have 2 diff answers for same condition, then condition is insufficient

and hance A cannot be answer.

What is OA?
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Re: Word prob

by nitin86 » Tue Sep 30, 2008 7:47 pm
beater 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) 100xy < x – y


Could you please help me break down (2)
To calculate %change more than 1 time (say 2 times) over an amount, we use the following formula.

Say, x is the first %change on and y is the second %change on a number n,
then the final % change is given by

Final %change = X + Y + (X*Y) / 100

Now, in the above question, second %change was negative, so we can subsitute in the above formula, the negative value of y.

sO, its, Final %change = X + (-Y) + (X * (-Y))/100
which comes to X - Y - (X * Y)/100

Now use the above conditons to get the answer as B, for the 1999 rent to be greater than 1997, the % change over 1997 should be positive.

Hence, X - Y - (X * Y)/100 should be greater than zero.
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I messed up the question

by beater » Wed Oct 01, 2008 9:28 am
Guys, I messed up the question. There is a typo.

(2) is xy/100 < x-y NOT 100xy < x-y.

Sorry..Sorry for the mess!
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by ketkoag » Wed Apr 22, 2009 2:36 am
IMO B
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Re: I messed up the question

by Ian Stewart » Wed Apr 22, 2009 6:13 pm
beater wrote: Guys, I messed up the question. There is a typo.

(2) is xy/100 < x-y NOT 100xy < x-y.

Sorry..Sorry for the mess!
With that correction: To increase something by x%, you can multiply by (1 + x/100). To decrease something by y%, you can multiply by (1 - y/100).

Here, the rent (let's call it R) increases by x%, then decreases by y%. So our new rent is just:

R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10,000)

We want to know if

R(1 + x/100 - y/100 - xy/10,000) > R

or equivalently, dividing by R (which is fine, because R must be positive) if

1 + x/100 - y/100 - xy/10,000 > 1

Subtracting 1 from both sides, and multiplying by 100 on both sides, the question becomes whether

x - y - xy/100 > 0

or if xy/100 < x - y. Since Statement 2 tells us exactly that, the answer to the question must be yes if we use Statement 2, and it's sufficient. Since Statement 1 is insufficient (try x = 60, y = 40 to see that), the answer is B.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by wolfe05 » Mon Apr 27, 2009 9:48 am
That is a great solution, but how are we supposed to come to that conclusion in less than a minute?
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by Vemuri » Mon Apr 27, 2009 7:57 pm
Why is Statement 1 not sufficient? The question is asking if the rent collected in 1999 more than 1997. From the solutions discussed & from what Ian had to say, we can definitely answer the question.

If x=60 & y=40, then

Rent collected in 1998 = 1.6*Rent collected in 1997
Rent collected in 1999 = 1.6*0.6*Rent collected in 1997
Rent collected in 1999 = 0.96*Rent collected in 1997

Hence, rent collected in 1999 is less than rent collected in 1997. This is sufficient to answer the question (with a "No").
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by Ian Stewart » Tue Apr 28, 2009 2:06 am
Vemuri wrote:Why is Statement 1 not sufficient? The question is asking if the rent collected in 1999 more than 1997. From the solutions discussed & from what Ian had to say, we can definitely answer the question.

If x=60 & y=40, then

Rent collected in 1998 = 1.6*Rent collected in 1997
Rent collected in 1999 = 1.6*0.6*Rent collected in 1997
Rent collected in 1999 = 0.96*Rent collected in 1997

Hence, rent collected in 1999 is less than rent collected in 1997. This is sufficient to answer the question (with a "No").
Yes, if x = 60 and y = 40, the rent in 1999 is less than in 1997. But if x = 60, and y = 10, the rent is higher in 1999.
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by Vemuri » Tue Apr 28, 2009 2:46 am
Ian Stewart wrote: Yes, if x = 60 and y = 40, the rent in 1999 is less than in 1997. But if x = 60, and y = 10, the rent is higher in 1999.
Oh boy !!! Ian please lend me your brilliant brian on the day of my exam. :wink: You will do me a great favor.
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by aj5105 » Wed May 27, 2009 10:11 pm
How is this solution? Thoughts?


Let the rent in 1997 be R7 = 100

in 1998, R8 = 100 + ( x % of 100 ) = 100 + x

let rent in 1999 be R9

R9 = ( 100 + x ) - ( 100 + x )*y/100

= 100 + x - y - xy/100

so to compare R9 and R7 , we need to look for sign of ( x - y - xy/100 )

if it is +ve then R9 > R7 and vice versa

FIRST STATEMENT

x > y

say x = 3 , y = 1

then x - y - xy/100 = 2 - 0.03 = 1.97 (+VE )

say x = 20 , y = 19

then x - y - xy/100 = 1 - 3.8 = -2.8 (-VE )

SO NOT SURE

SECOND STATEMENT

xy/100 < x - y

This implies,

0 < x - y -xy/100

Hence,

x - y - xy/100 is +ve

AND we can say that R9 > R7

HENCE B alone is sufficient .
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by siddhans » Sun Jul 03, 2011 12:55 am
Ian Stewart wrote:
beater wrote: Guys, I messed up the question. There is a typo.

(2) is xy/100 < x-y NOT 100xy < x-y.

Sorry..Sorry for the mess!
With that correction: To increase something by x%, you can multiply by (1 + x/100). To decrease something by y%, you can multiply by (1 - y/100).

Here, the rent (let's call it R) increases by x%, then decreases by y%. So our new rent is just:

R(1 + x/100)(1 - y/100) = R(1 + x/100 - y/100 - xy/10,000)

We want to know if

R(1 + x/100 - y/100 - xy/10,000) > R

or equivalently, dividing by R (which is fine, because R must be positive) if

1 + x/100 - y/100 - xy/10,000 > 1

Subtracting 1 from both sides, and multiplying by 100 on both sides, the question becomes whether

x - y - xy/100 > 0

or if xy/100 < x - y. Since Statement 2 tells us exactly that, the answer to the question must be yes if we use Statement 2, and it's sufficient. Since Statement 1 is insufficient (try x = 60, y = 40 to see that), the answer is B.

If we substitute X=60 and Y=40 in x - y - xy/100 > 0


then we get

20- 24 which is not > 0 but <0

so doesnt it give us a definite answer than rent collected by corporation in 1999 is not more than rent collected in 1997?? So isnt A sufficient too?
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by Ian Stewart » Sun Jul 03, 2011 3:38 am
siddhans wrote:
If we substitute X=60 and Y=40 in x - y - xy/100 > 0


then we get

20- 24 which is not > 0 but <0

so doesnt it give us a definite answer than rent collected by corporation in 1999 is not more than rent collected in 1997?? So isnt A sufficient too?
You need to understand perfectly just what you're doing when you plug numbers into a DS question. If you plug in numbers, of course you are going to get *some* answer to the question. If, when evaluating Statement 1, you plug in x=60 and y=40 here, you find that the rent is lower in 1999. *All* you have proven is that the rent is *sometimes* less in 1999 than in 1997, nothing more. You have not proven that the rent is *always* less in 1999 than in 1997, which is what you would need to do to prove Statement 1 is sufficient.

If I were generating scenarios to evaluate Statement 1 here, I'd just imagine two extremes:

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

We know the rent first goes up by x%, then down by y%. If it increased by 1,000,000%, then dropped by 1%, clearly the rent is higher in 1999 than in 1997. If instead it increased by 200% then dropped by 100%, the rent is $0 in 1999, so must be lower than in 1997. So Statement 1 is not sufficient.Of course there are many other examples you could use; I just find these extreme examples easiest to work with.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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