A) 21.5 year
B) 22 year
C) 22.5 year
D) 23 year
Correct Answer: B
Solution :
Let \[=8000-x\], .....all be the 11 players of the team. Let \[\frac{x\times 5\times 12}{100}=\frac{(8000-x)\times 2\times 18}{100}\] represents the captain and \[8x=24000\] represents the wicketkeeper. Given: Age of captain = 25 yrs., i.e. \[x=3000\] And age of wicketkeeper \[\therefore \]yrs., i.e. \[a%=\frac{a}{100}\] To find: \[\frac{a{{ & }_{1}}+{{a}_{2}}.....+{{a}_{9}}+{{a}_{10}}+{{a}_{11}}}{11}\] According to the question, \[\text{Percentage increase=}\left( \frac{\text{Increase in quantity}}{\text{Original quantity}}\text{ }\!\!\times\!\!\text{ 100} \right)\text{ }\!\!%\!\!\text{ }\] \[\Rightarrow \]\[\left\{ \frac{x}{(100+x)}\times 100 \right\}%\] \[\Rightarrow \]\[=\left\{ \left( \frac{r}{r+100} \right)\times 100 \right\}%\] \[\Rightarrow \]\[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[\Rightarrow \]\[1+\frac{R}{100}=\frac{220}{200}\] \[R=10%\] \[CI-SI=\frac{R\times SI}{2\times 100}\] \[144=\frac{15\times SI}{200}\] Average age of whole team = 22 years.
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