Stiv wrote:

A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12

B. \(33\frac{{1}}{{3}}\)

C. 36

D. 44

E. 50

\(? = p\)

\(97:\,\,n\,\,\,\,\,\, \to \,\,\,\,\,\,98:\,\,\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\,\,\,\, \to \,\,\,\,99:\,\,\,\frac{5}{6}\,\,\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\,\,\,\,\, \to \,\,00:\,\,\,{\left( {\frac{5}{6}} \right)^{\,2}}\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\)

\({\left( {\frac{5}{6}} \right)^{\,2}}\,\left( {1 + \frac{p}{{100}}} \right)\,\,n = n\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1 + \frac{p}{{100}} = {\left( {\frac{6}{5}} \right)^{\,2}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = p = 100\left( {\frac{{36}}{{25}} - \frac{{1 \cdot \boxed{25}}}{{\boxed{25}}}} \right) = 4 \cdot 11 = 44\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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