The fuel consumption of an engine is $\frac{1}{6, 000} \times ((S p e e d)^{3} \times T i m e)$ L. If a distance of 120 km has to be covered in exactly 4 h, what should be the minimum amount of fuel at the start of the journey so that the driver can make it? [Speed is in km/h and time is in h.]
18

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Solution

The correct option is A 18
Given that the distance of 120 km should be covered in 4 h.
$Speed = \frac{Distance}{Time taken} = \frac{120}{4} = 30 k m / h$
$∴ Fuel consumed = \frac{1}{6, 000} \times 30^{3} \times 4 = \frac{1}{6, 000} \times 27, 000 \times 4 = 18 L$

Hence, 18 L is the minimum amount of fuel at the start of journey, so that the driver can make it.

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The fuel consumption of an engine is 16,000×((Speed)3×Time) L. If a distance of 120 km has to be covered in exactly 4 h, what should be the minimum amount of fuel at the start of the journey so that the driver can make it? [Speed is in km/h and time is in h.]18

The correct option is A 18Given that the distance of 120 km should be covered in 4 h. Speed = DistanceTime taken =1204 =30 km/h ∴Fuel consumed = 16,000×303×4 =16,000×27,000×4 =18 L Hence, 18 L is the minimum amount of fuel at the start of journey, so that the driver can make it.

The fuel consumption of an engine is $\frac{1}{6, 000} \times ((S p e e d)^{3} \times T i m e)$ L. If a distance of 120 km has to be covered in exactly 4 h, what should be the minimum amount of fuel at the start of the journey so that the driver can make it? [Speed is in km/h and time is in h.]
18