Question

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.
(A) 20 m.p.h.
(B) 30 m.p.h.
(C) 40 m.p.h.
(D) 50 m.p.h.

Answer 1

Hint: Consider the cost of running the train to be $ y $ . And the speed of the train to be $ x $ . Then, find a relation between $ x $ and $ y $ , using the information given in the question. Once that relation is found. Use the condition that the cost of running the train cannot be more than ₹ 300

Complete step-by-step answer:
Let us assume that the fuel charges for a running train at any speed 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
 $ y \alpha {x^2} $
By removing the proportionality sign, we can write
 $ y = k{x^2} $ . . . (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 $
By substituting the value of in the above equation, we get
 $ 48 = k{16^2} $
 $ \Rightarrow 48 = k \times 256 $
By dividing both the sides by 256 and re-arranging it, we can write
 $ k = \dfrac{{48}}{{256}} $
 $ \Rightarrow k = \dfrac{3}{{16}} $
By substituting this value in equation (1), we get
 $ y = \dfrac{3}{{16}}{x^2} $ . . . (2)
Now, it is given that the fixed cost amount to ₹ 300 per hour. Thus, the cost of running the train cannot exceed ₹ 300 per hour.
Mathematically, we can write it as
 $ y \leqslant 300 $
Therefore, equation (2) becomes
 $ y = \dfrac{3}{{16}}{x^2} \leqslant 300 $
By cross multiplying, we get
 $ 3{x^2} \leqslant 300 \times 16 $
By dividing both the sides by 3 we get
 $ {x^2} \leqslant 100 \times 16 $
 $ \Rightarrow {x^2} \leqslant {10^2} \times {4^2} $
 $ \Rightarrow {x^2} \leqslant {40^2} $
We know that,
 $ {x^2} \leqslant {a^2} \Rightarrow x \in [ - a,a] $
By using the above property, we can write
 $ x \in [ - 40,40] $
Therefore, the maximum possible speed at which the train could be run economically is 40 m.p.h.
Therefore, from the above explanation, the correct answer is, option (C) 40 m.p.h.
So, the correct answer is “Option C”.

Note: Do not write $ y = {x^2} $ direction. Because, it is not given that the cost of running a train is equal to the square of the speed. It is given that they are proportional. Read questions carefully and do not forget to add the proportionality constant. The condition of the cost for one speed was given specifically because you were expected to add the proportionality constant and then find its value using the given condition.