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The annual rent collected by a corporation from a certain [#permalink]
01 Oct 2008, 10:03

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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?

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 10:32

1) Insufficient

Substitute 20 for x, 19 for y, and 100 for rent in 1997. 1997: 100 1998: 100+100*0.2 = 120 1999: 120-120*0.19 = 120-22.8 = 97.2 which gives an answer "NO' that 1999 rent > 1997 rent

Substitute 20 for x, 1 for y, and 100 for rent in 1997 1997: 100 1998: 100+100*0.2 = 120 1999: 120-120*0.01 = 120-1.2 = 118.8 which gives an answer "YES" that 1999 rent > 1997 rent

2) Insufficient

Substitute 20 for x, 19 for y, and 100 for rent in 1997 20*19/100 < 20-19 380/100 < 1 3.8 < 1 Not true for this condition

Substitiute 20 for x, 1 for y, and 100 for rent in 1997 20*1/100 < 20-1 1/5 < 19 True for this condition

When combining both conditions x>y and xy/100 < x-y, we plug in the number 20 for x, 1 for y, and 100 for rent because these numbers satisfies both conditions as shown above. When x = 20, y = 1, x>y. When x = 20, y = 1, xy/100 < x-y 20*1/100 < 20-1 20/100 < 19 1/5 < 19

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 10:39

vksunder 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

Answer is B.

Let rent in 1997 = 1 rent in 1998 = 1+(x/100) rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1-(y/100)}].......coz [a + b% of a] = [1+(x/100)] [1-(y/100)] The question demands: Is [1+(x/100)] [1-(y/100)] > 1 ??

assume this to be true & solve it further: => (100+x)(100-y)> 10000 => 10000 + 100x -100y -xy >10000 => 100x-100y-xy>0 => 100(x-y)>xy => x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.

Last edited by jatinrai on 01 Oct 2008, 11:32, edited 1 time in total.

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 11:15

jatinrai wrote:

vksunder 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

Answer is B.

Let rent in 1997 = 1 rent in 1998 = 1+(x/100) rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1+(y/100)}].......coz [a + b% of a] = [1+(x/100)] [1+(y/100)] The question demands: Is [1+(x/100)] [1+(y/100)] > 1 ??

assume this to be true & solve it further: => (100+x)(100-y)> 10000 => 10000 + 100x -100y -xy >10000 => 100x-100y-xy>0 => 100(x-y)>xy => x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 11:32

scthakur wrote:

jatinrai wrote:

vksunder 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

Answer is B.

Let rent in 1997 = 1 rent in 1998 = 1+(x/100) rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1+(y/100)}].......coz [a + b% of a] = [1+(x/100)] [1+(y/100)] The question demands: Is [1+(x/100)] [1+(y/100)] > 1 ??

assume this to be true & solve it further: => (100+x)(100-y)> 10000 => 10000 + 100x -100y -xy >10000 => 100x-100y-xy>0 => 100(x-y)>xy => x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.

I think you meant 1-(y/100) here.

Yup. Solved it on paper then copied. Hence, error.

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 19:37

As far as I can understand [1+(x/100)] [1-(y/100)] is diference b/w rent in 1999 and 1998. Y did u consider it >1 ?

jatinrai wrote:

vksunder 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

Answer is B.

Let rent in 1997 = 1 rent in 1998 = 1+(x/100) rent in 1999 = [1+(x/100)]+[{1+(x/100)}{1-(y/100)}].......coz [a + b% of a] = [1+(x/100)] [1-(y/100)] The question demands: Is [1+(x/100)] [1-(y/100)] > 1 ?? assume this to be true & solve it further: => (100+x)(100-y)> 10000 => 10000 + 100x -100y -xy >10000 => 100x-100y-xy>0 => 100(x-y)>xy => x-y > (xy/100)

This is exactly given by option 2. Therefore, option 2 is sufficient.

x>y i.e. option1 doesnt lead us anywhere. Therefore answer is B.

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 20:50

vksunder 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

B

% increase from 1997 to 1999 P = (1+x/100)(1-y/100) = 1+ x/100-y/100 -xy/(100*100) = 1+ 1/100 ( x-y-xy/100)

from equ 2 xy/100 < x – y --> x-y-xy/100>0

so P>1 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: DS: Annual Rent [#permalink]
01 Oct 2008, 20:53

lylya4 wrote:

x percent more, y percent less

So can we assume x, y >0??

If x,y > 0 then (2) <=> x - y>0 <=> (1)

annual rent in 1997 & 1999 are connected by rent in 1998, so (1) is sufficient hence (2) is sufficient

My answer is D

(1) is not suffcient

for e.g same intial ren 100. increased by 20.01% then rent would be ~120 decreased by 20% 120-24= 96 for e.g same intial ren 100. increased by 40% then rent would be ~140 decreased by 20% 140-28=116 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

gmatclubot

Re: DS: Annual Rent
[#permalink]
01 Oct 2008, 20:53