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JEE Mains · Maths · STD 12 - 6. Application of derivatives
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
- A \(45\)
- B \(50\)
- C \(60\)
- D \(40\)
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
According to given question, \(30a + \frac{b}{{30}} = 75\,\,\,\,\,\,....\left( i \right)\) \(40a + \frac{b}{{40}} = 65\,\,\,\,\,\,....\left( {ii} \right)\) On solving \((i)\) and \((ii)\), we get \(a = \frac{1}{2}\) and \(b=1800\) Now, \(C = av + \frac{b}{v}\)…
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