COST \[C=av+\frac{b}{v}\] ACCORDING to given question, \[30a+\frac{b}{30}=75\] ?(i) \[40a+\frac{b}{40}=65\] ?(ii) On solving (i) and (ii), we get \[a=\frac{1}{2}\] and \[b=1800\] Now, \[C=av+\frac{b}{v}\Rightarrow \frac{DC}{dv}=a-\frac{b}{{{v}^{2}}}\] \[\frac{dC}{dv}=0\Rightarrow a-\frac{b}{{{v}^{2}}}=0\Rightarrow v=\sqrt{\frac{b}{a}}=\sqrt{3600}\] \[\Rightarrow v=60\,\,kmph\]

"> COST \[C=av+\frac{b}{v}\] ACCORDING to given question, \[30a+\frac{b}{30}=75\] ?(i) \[40a+\frac{b}{40}=65\] ?(ii) On solving (i) and (ii), we get \[a=\frac{1}{2}\] and \[b=1800\] Now, \[C=av+\frac{b}{v}\Rightarrow \frac{DC}{dv}=a-\frac{b}{{{v}^{2}}}\] \[\frac{dC}{dv}=0\Rightarrow a-\frac{b}{{{v}^{2}}}=0\Rightarrow v=\sqrt{\frac{b}{a}}=\sqrt{3600}\] \[\Rightarrow v=60\,\,kmph\]

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The cost of running a bus from A to B, is Rs. \[\left( av+\frac{b}{v} \right),\] where v km/h is the average speed of the bus. When the bus travels at 30 km/h, the cost comes out to be Rs. 75 while at 40 km/h, it is Rs. 65. Then the most economical speed (in km/ h) of the bus is:

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[c] Let COST \[C=av+\frac{b}{v}\] ACCORDING to given question, \[30a+\frac{b}{30}=75\] ?(i) \[40a+\frac{b}{40}=65\] ?(ii) On solving (i) and (ii), we get \[a=\frac{1}{2}\] and \[b=1800\] Now, \[C=av+\frac{b}{v}\Rightarrow \frac{DC}{dv}=a-\frac{b}{{{v}^{2}}}\] \[\frac{dC}{dv}=0\Rightarrow a-\frac{b}{{{v}^{2}}}=0\Rightarrow v=\sqrt{\frac{b}{a}}=\sqrt{3600}\] \[\Rightarrow v=60\,\,kmph\]

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Posted on 24 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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