\(2005 ------2006-------------2007\)

\(x --------x(1+\frac{a}{100})----------x(1+\frac{a}{100})(1+\frac{b}{100})\)

\(% increase=\frac{x(1+\frac{a}{100})(1+\frac{b}{100})-x}{x}*100\)

\(=\frac{(100+a)(100+b)-100^2}{100^2}*100\)

\(=\frac{a+b+ab}{100}\)

as given a+b=20

\(=20+\frac{ab}{100}\)

To maximize \(\frac{ab}{100}\) we must keep a=b=10.

i.e. %increase = 20 + 1=21 (E)

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Piyush K

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