Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 21 Jul 2019, 12:18 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M20-13

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56307

### Show Tags

2
10 00:00

Difficulty:   75% (hard)

Question Stats: 62% (02:43) correct 38% (02:49) wrong based on 93 sessions

### HideShow timer Statistics 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. $$\frac{\frac{10,000M}{100 + x}}{100 - y}$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

### Show Tags

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

_________________
Intern  B
Joined: 20 Jun 2013
Posts: 8
Concentration: Finance
GMAT Date: 12-20-2014
GPA: 3.71

### Show Tags

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?
Intern  Joined: 07 Dec 2014
Posts: 2

### Show Tags

I agree with codeblue, i think there's something wrong here
Current Student Joined: 28 Aug 2013
Posts: 8
Location: India
Concentration: Entrepreneurship, General Management
WE: Information Technology (Consulting)

### Show Tags

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.
Retired Moderator Joined: 06 Jul 2014
Posts: 1224
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 ### Show Tags

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
_________________
Senior Manager  Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37 GMAT 2: 650 Q43 V36 GMAT 3: 600 Q47 V27 GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

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
_________________
KUDO me plenty
Intern  Joined: 19 Aug 2015
Posts: 9

### Show Tags

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?
Manager  B
Joined: 23 Apr 2014
Posts: 57
Location: United States
GMAT 1: 680 Q50 V31 GPA: 2.75

### Show Tags

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.
Intern  B
Joined: 18 Jun 2015
Posts: 38

### Show Tags

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.
Intern  Joined: 28 Aug 2014
Posts: 2

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 I think this is a high quality question and I agree with the explanation Current Student B Joined: 23 Nov 2016 Posts: 72 Location: United States (MN) GMAT 1: 760 Q50 V42 GPA: 3.51 Re: M20-13 [#permalink] ### Show Tags 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.
Manager  B
Joined: 04 Jul 2017
Posts: 59
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 56307

### Show Tags

1
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)?

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

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

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

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

$$\frac{(\frac{10,000M}{(100+y)})}{(100-x)} = P$$
_________________
Manager  B
Joined: 04 Jul 2017
Posts: 59
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 56307

### Show Tags

1
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 $$\frac{10,000M}{(100+y)}*\frac{1}{(100-x)}= \frac{10,000M}{(100+y)(100-x)} =P$$
_________________
SVP  V
Joined: 26 Mar 2013
Posts: 2284

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 56307

### Show Tags

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.
_________________
Intern  B
Joined: 02 Jan 2018
Posts: 26
Location: United Kingdom
GMAT 1: 740 Q50 V41 GPA: 3.7

### Show Tags

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   [#permalink] 06 Jul 2018, 14:10

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

# M20-13

Moderators: chetan2u, Bunuel  