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.
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Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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