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Assuming that petrol burnt per hour in driving a motor boat varies as the cube of its velocity. Show that the most economical speed when going against a current of k km/hour is \( \large\frac{3k}{2} \) km/hour.
cbse
class12
modelpaper
2012
sec-b
q13
math
Share
asked
Jan 31, 2013
by
thanvigandhi_1
edited
Jul 25, 2013
by
sreemathi.v
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Related questions
Assuming that petrol burnt per hour in driving a motorboat varies as the cube of its velocity. Show that the most economical speed when going against a current of p km/hour is \( \large\frac{3p}{2}\) km/hour.
cbse
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If a young man rides his motorcycle at 25 km/hour, he had to spend Rs.2 per km on petrol. If he rides at a faster speed of 40 km/hour, the petrol cost increases at Rs.5 per km. He has Rs.100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as L.P.P. and solve it graphically.
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(i) If a young lady drives her scooty at 30 km/hour, she had to spend Rs. 2.00 per km on petrol. If she drives at speed 50 km/h the petrol cost is Rs. 4.00 per km. She has Rs. 150 to spend on petrol and wants to describe the maximum distance she can travel within 2 hours. Express this as L.P.P and solve it graphically. (ii) Which mode of transport you suggest to a student and why?
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A man rides his motorcycle at the speed of 50km/hour.He has to speed Rs 2 per km on petrol.If he rides it at at a faster speed of 80km/hour,the petrol cost increases to Rs 3 per km.He has at most Rs120 to spend on petrol and one hour's time.He wishes to find the maximum distance that he can travel.Express this problem as a linear programming problem.
cbse
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A man rides his motorcycle at the speed of 50km/hour.He has to speed Rs 2 per km on petrol.If he rides it at at a faster speed of 80km/hour,the petrol cost increases to Rs 3 per km.He has at most Rs120 to spend on petrol and one hour's time.He wishes to find the maximum distance that he can travel.Determine the maximum distance that the man can travel.
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Prove that \( tan^{-1} \bigg(\frac{\large 1}{\large 4} \bigg) + tan^{-1} \bigg( \frac{\large 2}{\large 9} \bigg ) = \frac{\large 1}{\large 2} cos^{-1} \bigg( \frac{3}{5} \bigg). \)
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