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# m20 #13

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Senior Manager
Joined: 18 Aug 2009
Posts: 299

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19 Nov 2009, 06:46
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?

A. $$100 * \frac{\frac{M}{1 + x}}{1 - y}$$

B. $$\frac{\frac{M}{1 + \frac{x}{100}}}{1 - \frac{y}{100}}$$

C. $$\frac{M}{1 + \frac{x}{100} - \frac{y}{100}}$$

D. $$\frac{\frac{10,000M}{100 + y}}{100 - x}$$

E. $$\frac{\frac{10,000M}{100 + x}}{100 - y}$$

[Reveal] Spoiler:
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 $$\frac{10,000M}{(100 + y)(100 - x)}$$
Senior Manager
Joined: 18 Aug 2009
Posts: 299

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19 Nov 2009, 06:52
oops... after posting, I realized both of the solution mentioned above are the same
Sorry abt that
Senior Manager
Status: Happy to join ROSS!
Joined: 29 Sep 2010
Posts: 276
Concentration: General Management, Strategy
Schools: Ross '14 (M)

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14 Jan 2011, 13:28
gmattokyo, I had the same question as you did, but after your comment I realized the solution as well
Manager
Joined: 01 Mar 2009
Posts: 192
Location: United States
Concentration: Strategy, Technology
GMAT 1: 660 Q47 V34
GMAT 2: 680 Q46 V38
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WE: Consulting (Computer Software)

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03 Jul 2012, 06:34
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
Intern
Joined: 10 Jan 2014
Posts: 22

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22 Feb 2014, 02:46
Bump

I'm also having trouble trying to understand the solution.

I used M=1500; x=50; y=50. this gives a 1998 selling price of 2000 (2000*0.5*1.5=1500=M). If i use these values and plug them into both B and D i get the same result. wheres my mistake?

cheers
Math Expert
Joined: 02 Sep 2009
Posts: 39625

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22 Feb 2014, 03:45
damamikus wrote:
Bump

I'm also having trouble trying to understand the solution.

I used M=1500; x=50; y=50. this gives a 1998 selling price of 2000 (2000*0.5*1.5=1500=M). If i use these values and plug them into both B and D i get the same result. wheres my mistake?

cheers

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.
_________________
Re: m20 #13   [#permalink] 22 Feb 2014, 03:45
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# m20 #13

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