Question

A train travels 360km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form a quadratic equation to find the speed of the train.

Answer 1

Hint: Assume that the fixed distance travelled by the train is ‘d’ km. Assume that the original speed of the train is ‘v’ km/hr and time taken to cover the distance at original speed is ‘t’ hours. Find the new speed and new time using this assumption. Now, form two equations in two variables ‘v’ and ‘t’ by applying the formula: distance = speed \[\times \] time. Solve the two equations to get the value of ‘v’ by obtaining a quadratic equation with the help of substitution of t.

Complete step by step answer:
Let us assume that the distance travelled by the train is ‘d’ km, the original speed of the train is ‘v’ km/hr and the time taken to cover the distance at original speed is ‘t’ hours.
It is given that, if the speed would have been 5 km/hr more, the time taken for the journey would have been 1 hour less. Therefore,
New speed = (v + 5) km/h
New time taken = (t - 1) hour
Now, applying the formula: - distance = speed \[\times \] time, we have,
\[\Rightarrow d=v\times t\]
Substituting d = 360, we get,
\[\Rightarrow 360=vt\]
\[\Rightarrow vt=360\] - (1)
Now, for new speed and new time, we have,
d = 360, ‘v’ is replaced by (v + 5) and t is replaced by (t - 1).
\[\Rightarrow d=\left( v+5 \right)\times \left( t-1 \right)\]
\[\Rightarrow 360=vt+5t-v-5\] - (2)
Substituting the value of vt from equation (1) in equation (2), we get,
\[\begin{align}
  & \Rightarrow 360=360+5t-v-5 \\
 & \Rightarrow v-5t+5=0 \\
\end{align}\]
Now, substituting the value of ‘t’ from equation (1) in above expression, we have,
\[\Rightarrow v-5\times \dfrac{360}{v}+5=0\]
\[\begin{align}
  & \Rightarrow {{v}^{2}}-1800+5v=0 \\
 & \Rightarrow {{v}^{2}}+5v-1800=0 \\
\end{align}\]
Therefore, the above expression is the quadratic equation to find the speed of the train.
Now, splitting the middle term, we get,
\[\begin{align}
  & \Rightarrow {{v}^{2}}+45v-40v-1800=0 \\
 & \Rightarrow \left( v+45 \right)\left( v-40 \right)=0 \\
\end{align}\]
\[\Rightarrow v=-45\] or v = 40
Since, speed cannot be negative, therefore v = -45 is rejected.

So, we have,
\[\Rightarrow v=40\] km/hr.

Note: One may note that it is necessary to convert the given equations into a quadratic equation because the two equations involve multiplication of variables which can only be solved by the substitution of one variable in terms of the other. One important thing to note is that while solving the quadratic equation we have used the middle term split method, you can use discriminant method for ease because the constant term is a larger number.