The cost of the fuel for running the engine of an army tank is proport

Solution:

Let’s let c = the cost of the fuel per hour and s = the speed of the tank. We are told that cost is proportional to the square of the speed, so we have:

c = k x s^2,

where k = the constant of proportionality.

Substituting the given information, we have:

64 = k x 16^2

64/256 = k

1/4 = k

We must calculate the per-hour cost, the length of the trip (in hours), and the total cost for the trip for each of the answer choices.

Choice A: 10 kmph: c = ¼ x 10^2 = $25 per hour fuel cost + $400 = $425 per hour.
Length of trip: 400/10 = 40 hours, so the total trip cost is 425 x 40 = $17,000.

Choice B: 20 kmph: c = ¼ x 20^2 = $100 per hour fuel cost + $400 = $500 per hour.
Length of trip: 400/20 = 20 hours, so the total trip cost is 500 x 20 = $10,000.

Choice C: 35 kmph: c = ¼ x 35^2 = $306.25 per hour fuel cost + $400 = $706.25 per hour.
Length of trip: 400/35 = 11.43 hours, so the total trip cost is 706.25 x 11.43 ≈ $8,072.

Choice D: 40 kmph: c = ¼ x 40^2 = $400 per hour fuel cost + $400 = $800 per hour.
Length of trip: 400/40 = 10 hours, so the total trip cost is 800 x 10 = $8,000.

Choice E: 50 kmph: c = ¼ x 50^2 = $625 per hour fuel cost + $400 = $1025 per hour.
Length of trip: 400/50 = 8 hours, so the total trip cost is 1025 x 8 = $8,200.

We see that a speed of 40 kmph is the most cost-efficient speed of the given choices.

Answer: D