P(1)

A spinner has four sectors whose angles form a G.P . The smallest angle is $\displaystyle {{24}^{\circ }}$ and the sector are numbered as 1,2,3,4 in ascending order of their degree measures . If the arrow is spun 720 times what final score would you expect when all the individual scores are added together ?

Solution

Let the four angles be $ \displaystyle {{24}^{\circ }},{{24}^{\circ }}r,{{24}^{\circ }}{{r}^{2}},{{24}^{\circ }}{{r}^{3}}$ .

$ \displaystyle {{24}^{\circ }}+{{24}^{\circ }}r+{{24}^{\circ }}{{r}^{2}}+{{24}^{\circ }}{{r}^{3}}={{360}^{\circ }}$

$ \displaystyle {{1}^{\circ }}+r+{{r}^{2}}+{{r}^{3}}={{15}^{\circ }}$

$ \displaystyle {{r}^{3}}+{{r}^{2}}+r-{{14}^{\circ }}=0$

$ \displaystyle (r-{{2}^{\circ }})({{r}^{2}}+3r+{{7}^{\circ }})=0$

$ \displaystyle r-{{2}^{\circ }}=0\ \ \ \ (or)\ \ \ \ {{r}^{2}}+3r+{{7}^{\circ }}=0$

$ \displaystyle \ \ \ \ \ \ r={{2}^{\circ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (reject)\ (\because {{b}^{2}}-4ac<0)$

$ \displaystyle \text{The four angles are 2}{{\text{4}}^{\circ }},{{48}^{\circ }},{{96}^{\circ }}\text{  and  }{{192}^{\circ }}.$

$ \displaystyle \text{Final score = }(1\times \frac{1}{{15}}\times 720)+(2\times \frac{2}{{15}}\times 720)+$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3\times \frac{4}{{15}}\times 720)+(4\times \frac{8}{{15}}\times 720)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =720(\frac{1}{{15}}+\frac{4}{{15}}+\frac{{12}}{{15}}+\frac{{32}}{{15}})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =720\times \frac{{49}}{{15}}=2352$