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The annual rent collected by a corporation from a certain

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The annual rent collected by a corporation from a certain [#permalink]

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New post 29 Dec 2009, 04:55
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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
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 08:03
kirankp 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


will go with B

let annual rent collected be 20000 in 1997

stmnt1 - x>y. let x = 25% and y = 10% then amount collected in 1998 will be 25000 and amount collected in 1999 will be 22500. so annual rent collected in 1999 is more than 1997
let x=25% and y=20% then amount collected in 1998 will be 25000 and amount collected in 1999 will be 20000. so annual rent collected in 1999 is same as 1997
Hence Insuff

stmnt2 - here x cannot be less than y and therefore for any set of values satisfying this equation rent collected in 1999 will be more than 1997. hence suff
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 11:23
@ kp1811

please explain why you choose these values ?

I chose value = 100
x% = 10
y% = 5

my results were:

97 = 100
98 = 110
99 = 104.5

equation for 98 = 100 + x
equation for 99 = 100 + x - y(100+x)/100

elaborating eq 2; 100 +x-y-(xy)/100 (B) seems true but so does (A)

kindly elaborate.
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 12:01
kirankp 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


Hi,
My answer is C. Both statements together are sufficient.

Let us call rent in 1997 as p
Let us call rent in 1998 as q
Let us call rent in 1999 as r

q= p+(x/100)p --- (1)
r=q-(y/100)q --- (2)

We need to compare p and r?

replacing (1) in (2) we get:

r = p+(x/100)p - y/100(p+(x/100)p)
r = p+(x/100)p -(y/100)p - (xy/10000)p

x>y
implies (x/100)p - (y/100)p is positive, but we do not know if the result is greater than (xy/10000)p. Therefore insufficeint

xy/100 < x-y Insufficient since we do not know if x - y is positive or negative

Combining A and B
We get that x-y-xy/100 is a positive quantity
and hence r>p
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 12:14
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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.
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 18:56
hamza wrote:
@ kp1811

please explain why you choose these values ?

I chose value = 100
x% = 10
y% = 5

my results were:

97 = 100
98 = 110
99 = 104.5

equation for 98 = 100 + x
equation for 99 = 100 + x - y(100+x)/100

elaborating eq 2; 100 +x-y-(xy)/100 (B) seems true but so does (A)

kindly elaborate.


hamza....annual rent was taken randomly....100...20000...even A,R,B,C will suffice ...however values of x and y needs to be selected/checked in a way that we can prove if the statement is either suff or insuff

now in the example given by you "I chose value = 100
x% = 10
y% = 5

my results were:

97 = 100
98 = 110
99 = 104.5
"
this example of yours makes the statement sufficient but in the below examples we can prove that statement1 is insuff

if y% = 100/11 then for 99 we would get rent as 100 which is same as 97

if we take x = 25% and y=20% we get same value for 99 and 97
if we take x=50% and y = 100/3% again we get the same value for 99 and 97

in all these cases x>y. hope this helps
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Re: comparing rent [#permalink]

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New post 29 Dec 2009, 22:24
Thanks a lot for help.

The general point here is that 50% increase in A will always be less than 50% decrease in 1.5A.(Even 49% will be larger, or 45% too, if initial price is large)(now I see why kp1811 took 20,000 :) )

Value = 100
x = 50%
y = 40% (for ease of calculation, otherwise I would have liked to use 49.9% to prove it WRONG :) )

1997 = 100
1998 = 150 [1.5+50/100(100)]
1999 = 90 [150-40/100(150)]

value of 99 < value of 97

for B I see a sequence already, lets say value = 100

97 = 100
98 = 100+x
99 = (100+x) - (100+x)y/100

expanding 99 eq. => 100+x-y-xy/100

BUT, I would prefer keeping same approach to check both statements ( As i see a pattern in B, so plugging no.s later)

100+x-y-xy/100.
(x=50,y=40)

=100+50-40-(50*40)/100
=100+10-20
=90
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Re: comparing rent [#permalink]

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New post 01 Jan 2010, 16:50
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OA is B..

Explanation:

Let the rent paid in 1997 be R
Therefore rent paid in 1998 = \(R(1+\frac{X}{100})\)
and rent paid in 1999 = \(R(1+\frac{X}{100})(1-\frac{Y}{100})\)

Question ask us whether
\(R(1+\frac{X}{100})(1-\frac{Y}{100}) > R\)

\((1+\frac{X}{100})(1-\frac{Y}{100}) > 1\)

\(1+\frac{X}{100}-\frac{Y}{100} - \frac{XY}{10000} > 1\)

\(\frac{X}{100}-\frac{Y}{100} - \frac{XY}{10000} > 0\)

\(X-Y > \frac{XY}{100}\) ----- [1]

Statement 1:
x > y - Not sufficient as nothing can be said about equation [1]

Statement 2:
Sufficient as the same is given as true!

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Re: comparing rent [#permalink]

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New post 16 May 2011, 12:44
the solution by bunnuel is perfect it may take too much time. what to do?
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Re: comparing rent [#permalink]

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New post 16 May 2011, 19:19
B clearly gives the answer.
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The annual rent collected by a corporation from a certain building [#permalink]

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New post 31 Dec 2014, 02:27
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

Can i state this?

1997 1998 1999

100 100+x 100+x-y

So: 100+x-y>100 and x>y.. What's wrong with this? E.G: If the price of a stock is x in 1997 and is 25 percent more in 1998 so the price in 1998 is 100+25=125 percent

Thank you
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Re: The annual rent collected by a corporation from a certain building [#permalink]

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New post 31 Dec 2014, 02:49
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Answer B:

Statement 1: is obviously not enough. If rent goes up 51% this year and decreases 50% the next, we end up lower than where we started from, if it only decreases by 1%, we are still higher than before.

Statement 2: Sufficient.

\(R_{1998}=(1+\frac{x}{100})*R_{1997}\)

\(R_{1999}=(1-\frac{y}{100})*R_{1998}=(1-\frac{y}{100})*(1+\frac{x}{100})*R_{1997}\)

\(R_{1999}=(1-\frac{y}{100}+\frac{x}{100}-\frac{xy}{10000})*R_{1997}\)

\(R_{1999}=(1+\frac{x-y-0.01*xy}{100})*R_{1997}\)

With the statement we know that the fraction is positive. Therefore, the rent after the decrease of y% is still higher than it was before it increased by x%.

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Re: The annual rent collected by a corporation from a certain [#permalink]

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New post 31 Dec 2014, 03:11
gmatmania17 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

Can i state this?

1997 1998 1999

100 100+x 100+x-y

So: 100+x-y>100 and x>y.. What's wrong with this? E.G: If the price of a stock is x in 1997 and is 25 percent more in 1998 so the price in 1998 is 100+25=125 percent

Thank you


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Re: The annual rent collected by a corporation from a certain [#permalink]

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Re: The annual rent collected by a corporation from a certain [#permalink]

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New post 12 Jun 2016, 03:02
kirankp 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

Let the rent in 1997 be A.
(1)Let the amount be increased by x=25%
then if Y=22%(or 20<y<25%)
then we have a NO ans to our Ques.
But if Y=5%(or any smaller amount than 20)
then we have a YES ans to our Ques.
Insuff...
(2) rent in 1999=A(1+x/100)(1-y/100)
=A(1-y/100+x/100-xy/10000)
=A+A/100(x-y-xy/100)
given x-y-xy/100>0
so rent in 1999 is A+A/100(x-y-xy/100) must be greater than rent in 1997 which is A
suff..

Ans B
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Re: The annual rent collected by a corporation from a certain [#permalink]

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New post 08 Jan 2018, 11:06
kirankp 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


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.

Answer: B
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