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Senior Manager
Joined: 18 Aug 2009
Posts: 313
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The price of a house decreased by x\% from 1998 to 1999 and increased by y\% from 1999 to 2000. If the house cost M dollars in 2000, how much did it cost in 1998?
* 100 * \frac{\frac{M}{1 + x}}{1 - y}
* \frac{\frac{M}{1 + \frac{x}{100}}}{1 - \frac{y}{100}}
* \frac{M}{1 + \frac{x}{100} - \frac{y}{100}}
* \frac{\frac{10,000M}{100 + y}}{100 - x}
* \frac{\frac{10,000M}{100 + x}}{100 - y}
Denote P as the price of the house in 1998. We know that M = P(1 - \frac{x}{100})(1 + \frac{y}{100}) . From this equation P = \frac{\frac{M}{1 - \frac{x}{100}}}{1 + \frac{y}{100}} or \frac{\frac{10,000M}{100 + y}}{100 - x} .
Is there something wrong in the solution or I'm misreading it?
Say X={1 - \frac{x}{100}} and Y={1 - \frac{y}{100}}
If I rewrite the middle step M=P(X)(Y) => P=\frac{M}{XY}
=> some representation of \frac{10,000M}{(100 + y)(100 - x)}
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Senior Manager
Joined: 18 Aug 2009
Posts: 313
Followers: 2
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oops... after posting, I realized both of the solution mentioned above are the same  Sorry abt that
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Senior Manager
Status: Happy to join ROSS!
Joined: 29 Sep 2010
Posts: 280
Concentration: General Management, Strategy
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gmattokyo, I had the same question as you did, but after your comment I realized the solution as well 
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Manager
Status: employeed
Joined: 01 Mar 2009
Posts: 56
Concentration: Economics, Technology
GMAT 1: 660 Q47 V34
GMAT 2: 680 Q46 V38
GPA: 3.2
WE: Consulting (Computer Software)
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Did anyone try to use the VIC (Variable In Choice) method? I got the same result for A as I did for D.
I Used the following values:
Original Price of House = 100
X = 10
Y = 10
Now, we can calculate M:
M= 100*(11/10)*(9/10)
M= 99
A) 100 * [(99 / 11) / 9] = 100 * [99/99] = 100
D) [ 10,000(99) / (100+10) ] / (100 - 10) = [ 10,000(99) / 90 * 110 ] = [ 10,000(99) / (9*10) * (11*10) ] = [ 10,000(99) / (9*11) * (10*10) ] = [ 10,000(99) / (99) * (100) ] = 100
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