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# M20-13

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Math Expert
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M20-13 [#permalink]

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16 Sep 2014, 00:08
Expert's post
12
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Difficulty:

85% (hard)

Question Stats:

61% (01:42) correct 39% (01:54) wrong based on 116 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. $$\frac{\frac{10,000M}{100 + x}}{100 - y}$$
[Reveal] Spoiler: OA

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Math Expert
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Re M20-13 [#permalink]

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16 Sep 2014, 00:08
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Expert's post
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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Intern
Joined: 20 Jun 2013
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Concentration: Finance
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Re: M20-13 [#permalink]

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18 Dec 2014, 06:51
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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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Intern
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Re: M20-13 [#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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Intern
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Re: M20-13 [#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 ( 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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Retired Moderator
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Re: M20-13 [#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 ( 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.

In this task present a little rewriting of answer and this is make task difficult.

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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WE: Management Consulting (Consulting)
Re: M20-13 [#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: M20-13 [#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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Manager
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Re: M20-13 [#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 (100-x).

Not sure if this logic is right.

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Re: M20-13 [#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 (100-x).

Not sure if this logic is right.

This logic is correct.

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Re: M20-13 [#permalink]

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08 Oct 2016, 02:47
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 Kudos [?]: [0], given: 0 Intern Joined: 17 Jan 2016 Posts: 22 Kudos [?]: [0], given: 12 Re: M20-13 [#permalink] ### Show Tags 08 Oct 2016, 17:17 I think this is a high quality question and I agree with the explanation Kudos [?]: [0], given: 12 Manager Joined: 23 Nov 2016 Posts: 61 Kudos [?]: 13 [0], given: 20 Location: United States (NY) GMAT 1: 740 Q50 V40 Re: M20-13 [#permalink] ### Show Tags 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 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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Re: M20-13 [#permalink]

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29 Sep 2017, 09:03
1
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BOOKMARKED
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 (100-x) (100+y) became (100+y)/(100-x)?

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Math Expert
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Re: M20-13 [#permalink]

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30 Sep 2017, 00:36
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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 (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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Re: M20-13 [#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 (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?

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Re: M20-13 [#permalink]

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30 Sep 2017, 03:36
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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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Re: M20-13 [#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................... Keep

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

Answer: D

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

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Re: M20-13 [#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................... Keep

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

Answer: D

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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Re: M20-13   [#permalink] 01 Oct 2017, 03:30
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# M20-13

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