The price in June = \(p\).
The price in July = \(p(1+\frac{a}{100})(1-\frac{b}{100})\).

\(p = p(1+\frac{a}{100})(1-\frac{b}{100})\);

\(1 =\frac{100+a}{100}*\frac{100-b}{100}\);

\(b=100 - \frac{10,000}{100+a}=\frac{100a}{100+a}\).

Answer: B.
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let the value of land be : x
it is increased by a%, new value= (x)+(ax/100)
now, it is decreased by b%, such that it equals the initial value
hence final value of land= x(1+a/100) - x(1+a/100)(b/100)= x(1+a/100)(1-b/100)=x
solving for b you get b=100a/(100+a)

You can also use the formula of total % change in case of two successive % changes

Total % change = a + b + ab/100 where a and b are the successive % changes.

0 = a - b - ab/100 (because b is a decrease so we write it as -b)
Separating the variables, we get
b(1 + a/100) = a
b = 100a/(100 + a)

For more on this formula, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/
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Hi All,

This is a relatively rare concept on the GMAT, but there are certain percentages that will "offset" one another in the way that this question describes. The simplest pairing is a 25% increase will be offset by a 20% decrease.

Here's the proof:

Starting value = 100
25% increase = 100 + (.25)(100) = 125
20% decrease = 125 - (.2)(125) = 100
Ending value = 100

So we would be looking for an answer that equals 20 when A=25. While the answers might appear 'complex', you can actually avoid much of the 'math' if you just fill in the parts and take a good look at what you have...

Answer A: (200)(25)/(150) ... since 200>150, this answer will be greater than 25. ELIMINATE A.

Answer B: (100)(25)/125) = (100)(1/5) = 20 This is a MATCH.

Answer C: (25)(27)/(26)2 ... nothing 'cancels out', so this cannot = 20. ELIMINATE C.

Answer D: (25)(225)/10000... the 'units' digit of the numerator is a 5; when dividing by 10,000, this won't end in a 0. ELIMINATE D.

Answer E: (50/25) + 50 = 52. ELIMINATE E.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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B and C answers. Basically it's Who wants to be a millionaire.

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