The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and cost ₹ 48 per hour at 16 miles per hour. What is the most economical speed if the fixed charges i.e. salaries etc. amount to ₹ 300 per hour.

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20 m.p.h

30 m.p.h

40 m.p.h

50 m.p.h

answer is C.

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Detailed Solution

Let us assume that the fuel charges be y
Let us assume that the speed generated in miles per hour be x
It is given in the question that the fuel charges for a running train are proportional to the square of the speed generated in miles per hour.
Thus, we can write Yαx2 By removing the proportionality sign, we can write
y = kx2 . . . (1)
Where, k is a proportionality constant.
Now, we have given that the train costs ₹ 48 per hour at 16 miles per hour.
Thus, when y = 48 , x = 16
48 = k162 ⇒ 48 = k × 256
k =

\frac{48}{256}

⇒ k =

\frac{3}{16}

By substituting this value in equation (1), we get
y =

\frac{3}{16}

x2 . . . (2)
Now, it is given that the fixed cost amounts to ₹300 per hour. Thus, the cost of running the train cannot exceed ₹300 per hour.
Mathematically, we can write it as
y ≤ 300
Therefore, equation (2) becomes y =

\frac{3}{16}

x2 ≤ 300
3x2 ≤ 300 × 16
x2 ≤ 100 × 16
⇒ x2 ≤ 102 × 42 ⇒ x2 ≤ 402 We know that, x2 ≤ a2 ⇒ x ∈ [−a, a]
By using the above property, we can write
x ∈ [−40, 40]
Therefore, the maximum possible speed at which the train could be run economically is 40 m.p.h.
So, the correct answer is “Option 3”.

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