faster method plz2

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

faster method plz2

by raunekk » Thu Jul 03, 2008 11:22 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

any other approach ,other then pluggin numbers??

thank you

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Source: Beat The GMAT — Data Sufficiency | Thu Jul 03, 2008 11:22 am

gauravkhare: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Fri May 23, 2008 10:32 am; Thanked: 1 times

by gauravkhare » Thu Jul 03, 2008 11:36 am

i think 'D" is answer.

we can assume 1997 as 100, then 1998 becomes 100+x, and 1999 becomes 100 + (x-y) - xy/100. we need to find if term (x-y) - xy/100 is positive or not. both the statement are SUFF

gaurav

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Thu Jul 03, 2008 10:22 pm

sorry..its B

A is a trap!!

It took me 20 mins to solve this one.

Any other method
??
Stuart,Pls help!!

thanks

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ildude02: Master | Next Rank: 500 Posts; Posts: 320; Joined: Sun Jan 13, 2008 10:00 pm; Thanked: 10 times

by ildude02 » Fri Jul 04, 2008 7:51 am

D is the answer. Moreover, both A and B lead to the same conclusion that X> Y, so if B is the answer, A should be the answer as well. statement B doesn't hold true if X< y since it will give a positive value. Atleast I assumed that we can only consider positive values for both x and y since they are percentages. I hope it's an OK assumption.

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

Re: faster method plz2

by Ian Stewart » Fri Jul 04, 2008 9:06 am

raunekk 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

any other approach ,other then pluggin numbers??

thank you

This is quite an important one to understand for the GMAT. If anyone thinks Statement 1) is sufficient here, I'll make you an offer. Give me a thousand dollars. I'll increase your money by 60%, then decrease the result by 40%, then give you the final amount back. PM me if you're interested!

As for statement 2: If you start with, say, 100 dollars in 1997, you will have 100+x dollars in 1998 (starting value plus the increase). You will then have:

(100 + x) - (y/100)(100+x)

in 1999 (starting value minus the decrease). We want to know under what conditions this amount will be larger than the starting amount from 1997, which was 100:

(100 + x) - (y/100)(100+x) > 100
x - y - xy/100 > 0
xy/100< x – y

which is precisely what we're told in Statement 2. So 2) is sufficient, 1) is not. If you want, you can replace the starting value of '100' with some unknown 'C' here, though it won't make any difference to the conclusion.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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ildude02: Master | Next Rank: 500 Posts; Posts: 320; Joined: Sun Jan 13, 2008 10:00 pm; Thanked: 10 times

by ildude02 » Fri Jul 04, 2008 9:46 am

oops, my bad..I think I compared 1999 to 1998 instead of to 1997. In anycase, I want to understand if my approach is right or wrong.

Ian, can you please correct me if I'm wrong with the 1999> 1998 comparision. I chose some unknown value "a" for 1997(eventhough 100 would have made it easier

, so for 1998, we have "a + 0.0xa" and in 1999, we have (a+0.0xa)(1-0.0y). Comparing 1999 to 1998 =>

is (a+0.0xa)(1-0.0y) > (a+0.0xa), we can divide a + 0.0xa since both x and a are postive values and the should be a positive value without worrying about reversing the inequalities.,

=> is 1-0.0y > 1 ; since y is a positive value as well, would this equation be always true and so 1999 > 1998.

Now if I have compared 1999> 1997, this is how I would have went about solving it,

is (a+0.0xa)(1-0.0y) > a ; simplifying furthur,

is a(1+ 0.0x)(1-0.0y) > a => is (1+0.0x)(1-0.0y)> 1; which I think would be insufficient if x> y. But I did spend some time trying to figure out what values of x and y that can result in the value less then 1.