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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. \(\frac{M}{1 + \frac{x}{100} - \frac{y}{100}}\)

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

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
_________________

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
_________________

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?

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.

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)?

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)?

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?

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 \(\frac{10,000M}{(100+y)}*\frac{1}{(100-x)}= \frac{10,000M}{(100+y)(100-x)} =P\)
_________________

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

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