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04 Apr 2012, 08:50
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E
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75% (02:02) correct
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If the population of city A is increased by A% in year 2006 from year 2005 and b% in 2007 from year 2006. If a+b=20, then what is the maximum percentage increase in population of city A from 2005 to year 2007?
A. 20.19%
B. 20.36%
C. 20.51%
D. 20.84%
E. 21%
A. 20.19%
B. 20.36%
C. 20.51%
D. 20.84%
E. 21%
04 Apr 2012, 09:17
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LM
Say the population of the city was 100 in 2005, then in 2007 it would be \(100(1+\frac{a}{100})(1+\frac{b}{100})=\frac{ab}{100}+a+b+100=\frac{ab}{100}+120\) (since \(a+b=20\)).
So, in order to maximize this expression we need to maximize the value of \(ab\), when given that \(a+b=20\). Useful property: for given sum of two numbers, their product is maximized when they are equal. Hence, \(ab\) will be maximized for \(a=b=10\) --> \(120+\frac{ab}{100}=120+1=121\).
Answer: E.
09 Aug 2017, 15:59
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Here is what I did on this one ->
I did not solve it to maximise but rather used hit and trial.
Let the population in 2005 be 100.
Since in most GMAT questions symmetry works perfectly what if a=b=10
Then 100 => 110 => 121
Hence 21% gain is possible.
Scanning the options => 21 is the highest possible value.
Hence E.
But I like what Bunuel did here.
To maximise ab given the sum a+b we must make a=b.
I did not solve it to maximise but rather used hit and trial.
Let the population in 2005 be 100.
Since in most GMAT questions symmetry works perfectly what if a=b=10
Then 100 => 110 => 121
Hence 21% gain is possible.
Scanning the options => 21 is the highest possible value.
Hence E.
But I like what Bunuel did here.
To maximise ab given the sum a+b we must make a=b.
