OA Explanation:....while covering the same distance with each speed, means each portion is \(D\) Km long
\(6.5 = \frac{D}{30} + \frac{D}{45} + \frac{D}{60}\)
\(6.5 = \frac{D}{15}[\frac{1}{2} + \frac{1}{3} + \frac{1}{4}]\)
\((6.5)(15) = D(\frac{6 + 4 + 3)}{12})\)
\(D = 90\) Km
Total distance = \(3(90) = 270\) Km
Mileage for the first portion is 11 km/lt i.e. \(\frac{90}{11} = 8.18\) lt of fuel is being consumed
Mileage for the second portion is 14 km/lt i.e. \(\frac{90}{14} = 6.42\) lt of fuel is being consumed
Mileage for the third portion is 18 km/lt i.e. \(\frac{90}{18} = 5.00\) lt of fuel is being consumed
Total fuel consumption = \((8.18 + 6.42 + 5.0) = 19.6\) lt
Since the fuel consumption need to be minimised, Eden must drive his car at 60 Kmph through the entire distance of 270 Km
This means - Mileage for the entire distance will be 18 km/lt
i.e. \(\frac{270}{18} = 15.00\) lt of fuel will be consumed
Fuel saved = 19.6 - 15 = 4.6 lt
Hence, option B
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/