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Re M2013 [#permalink]
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16 Sep 2014, 00:08
Official Solution: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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\) 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 \(\frac{\frac{10,000M}{100 + y}}{100  x}\). Answer: D
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Re: M2013 [#permalink]
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18 Dec 2014, 06:51
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If we denote (1x/100) as A and (1+y/100) as B, then wouldn't M = P(A)(B) equate to P = M/(A*B) and not P=M/A/B?
What am I missing here?



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Re: M2013 [#permalink]
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26 Dec 2014, 05:02
I agree with codeblue, i think there's something wrong here



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Re: M2013 [#permalink]
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01 Mar 2015, 06:28
I think with approach mentioned in the previous posts answer should be as below.
M = P ( 1x/100)(1+y/100)
P = M /( 1x/100)(1+y/100)
P = 10000M /(100x)(100+y)
But this is not present in the answer choices.



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Re: M2013 [#permalink]
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11 Apr 2015, 09:18
abahl88 wrote: I think with approach mentioned in the previous posts answer should be as below.
M = P ( 1x/100)(1+y/100)
P = M /( 1x/100)(1+y/100)
P = 10000M /(100x)(100+y)
But this is not present in the answer choices. In this task present a little rewriting of answer and this is make task difficult. We have answer \(P = \frac{10000M}{(100x)(100+y)}\) and from this answer they make two step equation: \(\frac{10000M}{100+y}\) and result divide on 100x For example \(\frac{100}{10*5}=2\) it's the same as 100/10/5 = 2
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Re: M2013 [#permalink]
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20 Nov 2015, 10:10
i plugged in simple numbers of 100 for initial price in 1998, x=10% and y=20%. however when i was testing answer c already more than 3 min had elapsed
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Re: M2013 [#permalink]
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17 Feb 2016, 20:37
I worked this problem through logic and am wondering if it would be wrong in this case?
Since the final price is M dollars in 2000, we had to multiply the 1999 value by Y. so in order to back out of M, we would ultimately have to divide by Y first and then X. All of them except for C/D have you divide by X first so thats wrong, and then C looked clearly wrong so I was left with D.
Is there any problem to this approach that you may see?



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Re: M2013 [#permalink]
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04 Aug 2016, 17:50
I used the logic that we have to decrease the price by x% and later increase it by y%, so we should have X/100 or x subtracted in the equation.
All options have (1+x) or (100+x), only D has (100x).
Not sure if this logic is right.



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Re: M2013 [#permalink]
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10 Sep 2016, 14:24
PerseveranceWins wrote: I used the logic that we have to decrease the price by x% and later increase it by y%, so we should have X/100 or x subtracted in the equation.
All options have (1+x) or (100+x), only D has (100x).
Not sure if this logic is right. This logic is correct.



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Re: M2013 [#permalink]
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08 Oct 2016, 02:47
I think this is a poorquality question. Representation of the answer choices is complicated and not clear. $fact{M/(....}}}}]
Can be simply followed GMAT standards



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Re: M2013 [#permalink]
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08 Oct 2016, 17:17
I think this is a high quality question and I agree with the explanation



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Re: M2013 [#permalink]
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05 Mar 2017, 10:47
For problems like these it may be faster to pick "nice" numbers.
For example, let's say the house was $100 in '98, went down by 20%, and then up by 10%. $100 in '98 > $80 in '99 > $88
Plug these numbers into AE to see which gives you $100. C looked wrong right off the bat. A and B did not look like fun to calculate, but I would do so if D and E weren't correct.
I started with E, realized that it wasn't going to give me $100 after doing the first few calculations, and moved on to D. D gives (880,000)/110 = 8000. So far so good. 8000/80 =100.
E.



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Re: M2013 [#permalink]
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29 Sep 2017, 09:03
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Bunuel wrote: Official Solution:
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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\)
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 \(\frac{\frac{10,000M}{100 + y}}{100  x}\).
Answer: D Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100x) (100+y) became (100+y)/(100x)?



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Re: M2013 [#permalink]
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30 Sep 2017, 00:36
aashishagarwal2 wrote: Bunuel wrote: Official Solution:
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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\)
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 \(\frac{\frac{10,000M}{100 + y}}{100  x}\).
Answer: D Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100x) (100+y) became (100+y)/(100x)? \(M = P(1  \frac{x}{100})(1 + \frac{y}{100})\) \(M = P(\frac{100x}{100})(\frac{100+y}{100})\) \(10,000M = P(100x)(100+y)\) \(\frac{10,000M}{(100+y)} = P(100x)\) \(\frac{(\frac{10,000M}{(100+y)})}{(100x)} = P\)
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Re: M2013 [#permalink]
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30 Sep 2017, 02:43
Bunuel wrote: aashishagarwal2 wrote: Bunuel wrote: Official Solution:
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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\)
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 \(\frac{\frac{10,000M}{100 + y}}{100  x}\).
Answer: D Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100x) (100+y) became (100+y)/(100x)? \(M = P(1  \frac{x}{100})(1 + \frac{y}{100})\) \(M = P(\frac{100x}{100})(\frac{100+y}{100})\) \(10,000M = P(100x)(100+y)\) \(\frac{10,000M}{(100+y)} = P(100x)\) \(\frac{(\frac{10,000M}{(100+y)})}{(100x)} = P\) Sorry for troubling you. I always in other similar calculations take the entire (100x) (100+y) and divide it by 10,000M [this way: 10,000M / (100x) (100+y) = P]. Is this wrong? Have I been doing this wrong most of my life?



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Re: M2013 [#permalink]
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30 Sep 2017, 03:36
aashishagarwal2 wrote: Bunuel wrote: aashishagarwal2 wrote: Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100x) (100+y) became (100+y)/(100x)?
\(M = P(1  \frac{x}{100})(1 + \frac{y}{100})\) \(M = P(\frac{100x}{100})(\frac{100+y}{100})\) \(10,000M = P(100x)(100+y)\) \(\frac{10,000M}{(100+y)} = P(100x)\) \(\frac{(\frac{10,000M}{(100+y)})}{(100x)} = P\) Sorry for troubling you. I always in other similar calculations take the entire (100x) (100+y) and divide it by 10,000M [this way: 10,000M / (100x) (100+y) = P]. Is this wrong? Have I been doing this wrong most of my life? No, it's not wrong. \(\frac{(\frac{10,000M}{(100+y)})}{(100x)} = P\) is the same as \(\frac{10,000M}{(100+y)}*\frac{1}{(100x)}= \frac{10,000M}{(100+y)(100x)} =P\)
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Re: M2013 [#permalink]
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30 Sep 2017, 23:32
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Bunuel wrote: 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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\) Another appraoch Let original price =100 , x = 50%, y = 0%........ then final price M = 50 1) \(\frac{(100*50)}{150}\) (do not calculate) .................. Eliminate 2) \(\frac{50}{1.5}\) ............(do not calculate).................. Eliminate 3) \(\frac{50}{1.5}\) ............(do not calculate).................. Eliminate 4) \(\frac{(10,000*50)}{100}\) * \(\frac{1}{50}\) = 100................... Keep5) \(\frac{(10000*50)}{150}\) * \(\frac{1}{100}\).....(do not calculate)........ Eliminate Answer: D BunuelThe format of answer choices is not helpful and confusing a lot. Can'y you reformat to make some space between answer choices? Thanks



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Re: M2013 [#permalink]
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01 Oct 2017, 03:30
Mo2men wrote: Bunuel wrote: 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. \(\frac{\frac{10,000M}{100 + x}}{100  y}\) Another appraoch Let original price =100 , x = 50%, y = 0%........ then final price M = 50 1) \(\frac{(100*50)}{150}\) (do not calculate) .................. Eliminate 2) \(\frac{50}{1.5}\) ............(do not calculate).................. Eliminate 3) \(\frac{50}{1.5}\) ............(do not calculate).................. Eliminate 4) \(\frac{(10,000*50)}{100}\) * \(\frac{1}{50}\) = 100................... Keep5) \(\frac{(10000*50)}{150}\) * \(\frac{1}{100}\).....(do not calculate)........ Eliminate Answer: D BunuelThe format of answer choices is not helpful and confusing a lot. Can'y you reformat to make some space between answer choices? Thanks _______________ Edited. Thank you.
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