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# M20-13

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Math Expert
Joined: 02 Sep 2009
Posts: 49932

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16 Sep 2014, 01:08
2
10
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Difficulty:

75% (hard)

Question Stats:

64% (01:49) correct 36% (01:57) wrong based on 136 sessions

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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?

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

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Math Expert
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16 Sep 2014, 01:08
2
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 $$\frac{\frac{10,000M}{100 + y}}{100 - x}$$.

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Concentration: Finance
GMAT Date: 12-20-2014
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18 Dec 2014, 07:51
1
If we denote (1-x/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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Joined: 07 Dec 2014
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26 Dec 2014, 06:02
I agree with codeblue, i think there's something wrong here
Current Student
Joined: 28 Aug 2013
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Location: India
Concentration: Entrepreneurship, General Management
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01 Mar 2015, 07:28
I think with approach mentioned in the previous posts answer should be as below.

M = P ( 1-x/100)(1+y/100)

P = M /( 1-x/100)(1+y/100)

P = 10000M /(100-x)(100+y)

But this is not present in the answer choices.
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Posts: 1243
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Concentration: Entrepreneurship, Technology
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11 Apr 2015, 10:18
abahl88 wrote:
I think with approach mentioned in the previous posts answer should be as below.

M = P ( 1-x/100)(1+y/100)

P = M /( 1-x/100)(1+y/100)

P = 10000M /(100-x)(100+y)

But this is not present in the answer choices.

We have answer $$P = \frac{10000M}{(100-x)(100+y)}$$ and from this answer they make two step equation: $$\frac{10000M}{100+y}$$ and result divide on 100-x

For example $$\frac{100}{10*5}=2$$ it's the same as 100/10/5 = 2
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20 Nov 2015, 11:10
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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Intern
Joined: 19 Aug 2015
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17 Feb 2016, 21:37
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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04 Aug 2016, 18:50
1
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 (100-x).

Not sure if this logic is right.
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Joined: 18 Jun 2015
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10 Sep 2016, 15:24
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 (100-x).

Not sure if this logic is right.

This logic is correct.
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Joined: 28 Aug 2014
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08 Oct 2016, 03:47
I think this is a poor-quality question. Representation of the answer choices is complicated and not clear. $fact{M/(....}}}}] Can be simply followed GMAT standards Intern Joined: 17 Jan 2016 Posts: 19 Re: M20-13 [#permalink] ### Show Tags 08 Oct 2016, 18:17 I think this is a high quality question and I agree with the explanation Manager Joined: 23 Nov 2016 Posts: 76 Location: United States (MN) GMAT 1: 760 Q50 V42 GPA: 3.51 Re: M20-13 [#permalink] ### Show Tags 05 Mar 2017, 11:47 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 A-E 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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29 Sep 2017, 10:03
1
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 $$\frac{\frac{10,000M}{100 + y}}{100 - x}$$.

Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100-x) (100+y) became (100+y)/(100-x)?
Math Expert
Joined: 02 Sep 2009
Posts: 49932

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30 Sep 2017, 01:36
1
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 $$\frac{\frac{10,000M}{100 + y}}{100 - x}$$.

Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100-x) (100+y) became (100+y)/(100-x)?

$$M = P(1 - \frac{x}{100})(1 + \frac{y}{100})$$

$$M = P(\frac{100-x}{100})(\frac{100+y}{100})$$

$$10,000M = P(100-x)(100+y)$$

$$\frac{10,000M}{(100+y)} = P(100-x)$$

$$\frac{(\frac{10,000M}{(100+y)})}{(100-x)} = P$$
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30 Sep 2017, 03:43
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 $$\frac{\frac{10,000M}{100 + y}}{100 - x}$$.

Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100-x) (100+y) became (100+y)/(100-x)?

$$M = P(1 - \frac{x}{100})(1 + \frac{y}{100})$$

$$M = P(\frac{100-x}{100})(\frac{100+y}{100})$$

$$10,000M = P(100-x)(100+y)$$

$$\frac{10,000M}{(100+y)} = P(100-x)$$

$$\frac{(\frac{10,000M}{(100+y)})}{(100-x)} = P$$

Sorry for troubling you. I always in other similar calculations take the entire (100-x) (100+y) and divide it by 10,000M [this way: 10,000M / (100-x) (100+y) = P]. Is this wrong? Have I been doing this wrong most of my life?
Math Expert
Joined: 02 Sep 2009
Posts: 49932

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30 Sep 2017, 04:36
1
aashishagarwal2 wrote:
Bunuel wrote:
aashishagarwal2 wrote:

Hi Bunuel, Can you please explain the steps in more detail. I didn't understand how (100-x) (100+y) became (100+y)/(100-x)?

$$M = P(1 - \frac{x}{100})(1 + \frac{y}{100})$$

$$M = P(\frac{100-x}{100})(\frac{100+y}{100})$$

$$10,000M = P(100-x)(100+y)$$

$$\frac{10,000M}{(100+y)} = P(100-x)$$

$$\frac{(\frac{10,000M}{(100+y)})}{(100-x)} = P$$

Sorry for troubling you. I always in other similar calculations take the entire (100-x) (100+y) and divide it by 10,000M [this way: 10,000M / (100-x) (100+y) = P]. Is this wrong? Have I been doing this wrong most of my life?

No, it's not wrong.

$$\frac{(\frac{10,000M}{(100+y)})}{(100-x)} = P$$ is the same as $$\frac{10,000M}{(100+y)}*\frac{1}{(100-x)}= \frac{10,000M}{(100+y)(100-x)} =P$$
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01 Oct 2017, 00:32
1
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................... Keep

5) $$\frac{(10000*50)}{150}$$ * $$\frac{1}{100}$$.....(do not calculate)........Eliminate

Bunuel

The format of answer choices is not helpful and confusing a lot. Can'y you re-format to make some space between answer choices?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 49932

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01 Oct 2017, 04:30
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................... Keep

5) $$\frac{(10000*50)}{150}$$ * $$\frac{1}{100}$$.....(do not calculate)........Eliminate

Bunuel

The format of answer choices is not helpful and confusing a lot. Can'y you re-format to make some space between answer choices?

Thanks

_______________
Edited. Thank you.
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06 Jul 2018, 14:10
D can be picked by default as it's the only one which subtracts some multiple of x.

Not sure if this means the question is high quality or low quality, but it makes the answer very obvious either way!
Re: M20-13 &nbs [#permalink] 06 Jul 2018, 14:10

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# M20-13

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