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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. \(\frac{M}{1 + \frac{x}{100} - \frac{y}{100}}\)
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 \(\frac{\frac{10,000M}{100 + y}}{100 - x}\) .
Is there something wrong in the solution or I'm misreading it? Say \(X={1 - \frac{x}{100}}\) and \(Y={1 - \frac{y}{100}}\) If I rewrite the middle step \(M=P(X)(Y)\) => \(P=\frac{M}{XY}\) => some representation of \(\frac{10,000M}{(100 + y)(100 - x)}\)
Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
I'm also having trouble trying to understand the solution.
I used M=1500; x=50; y=50. this gives a 1998 selling price of 2000 (2000*0.5*1.5=1500=M). If i use these values and plug them into both B and D i get the same result. wheres my mistake?
I'm also having trouble trying to understand the solution.
I used M=1500; x=50; y=50. this gives a 1998 selling price of 2000 (2000*0.5*1.5=1500=M). If i use these values and plug them into both B and D i get the same result. wheres my mistake?
cheers
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Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.
Hope it helps.
Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.