Question

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train forms the quadratic equation to find the average speed of the express train.

Answer 1

We use the speed- distance formula and formulate the relation among time taken for the journey ${{t}_{p}}$, the average speed ${{v}_{p}}$ and distance travelled ${{s}_{p}}$ by the passenger train as ${{t}_{p}}=\dfrac{{{s}_{p}}}{{{v}_{p}}}$. Similarly we formulate the relation among time taken for the journey ${{t}_{e}}$, the average speed ${{v}_{e}}$ and distance travelled ${{s}_{e}}$ by the express train. We are given that ${{t}_{p}}={{t}_{e}}+1$, ${{s}_{e}}={{s}_{p}}=132\text{km}$ and ${{v}_{e}}={{v}_{p}}+11$. We use the equation ${{t}_{p}}={{t}_{e}}+1$, express time in terms of distance and speed, put the given equation and obtain quadratic equation in ${{v}_{p}}$. We solve the quadratic equation by splitting the middle term method to get ${{v}_{p}}$ and then ${{v}_{e}}={{v}_{p}}+11$.$$$$

Complete step-by-step solution:
We know that the time taken $t$, the distance travelled $s$ and the average speed $v$ of any object are related by the equation

& v=\dfrac{s}{t} \\\ & \Rightarrow t=\dfrac{s}{v} \\\ \end{aligned}$$ Let us denote the average speed, time is taken for the journey and the distance traveled by the passenger train as ${{v}_{p}},{{t}_{p}},{{s}_{p}}$ respectively. Similarly, we denote the speed, time is taken for the journey and the distance traveled by the passenger train as ${{v}_{e}},{{t}_{e}},{{s}_{e}}$ respectively. We are given the question that the express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. Both the trains travel the same distance of 132 km. So we have $$\begin{aligned} & {{t}_{p}}={{t}_{e}}+1.....\left( 1 \right) \\\ & {{s}_{e}}={{s}_{p}}=132\text{km}.....\left( 2 \right) \\\ \end{aligned}$$ We are further given in the question that the average speed of the express train is 11 km/hr more than that of the passenger train. So we have $${{v}_{e}}={{v}_{p}}+11.....\left( 3 \right)$$ We express in terms of distance and speed from equation (1) and have, $$\begin{aligned} & {{t}_{p}}={{t}_{e}}+1 \\\ & \Rightarrow \dfrac{{{s}_{p}}}{{{v}_{p}}}=\dfrac{{{s}_{e}}}{{{v}_{e}}}+1 \\\ \end{aligned}$$ We put the values obtained in equation (2) and (3) in the above step and have, $$\begin{aligned} & \Rightarrow \dfrac{132}{{{v}_{p}}}=\dfrac{132}{{{v}_{p}}+11}+1 \\\ & \Rightarrow \dfrac{132}{{{v}_{p}}}-\dfrac{132}{{{v}_{p}}+11}=1 \\\ & \Rightarrow 132\left( \dfrac{{{v}_{p}}+11-{{v}_{p}}}{{{v}_{p}}\left( {{v}_{p}}+11 \right)} \right)=1 \\\ & \Rightarrow {{v}_{p}}^{2}+11{{v}_{p}}=132\times 11=1452 \\\ \end{aligned}$$ So the required quadratic equation as asked in the question in ${{v}_{p}}$ $$\Rightarrow {{v}_{p}}^{2}+11{{v}_{p}}-1452=0$$ We solve the above quadratic equation by splitting the middle term method and have, $$\begin{aligned} & \Rightarrow {{v}_{p}}^{2}+44{{v}_{p}}-33{{v}_{p}}-1452=0 \\\ & \Rightarrow {{v}_{p}}\left( {{v}_{p}}+44 \right)-33\left( {{v}_{p}}+44 \right)=0 \\\ & \Rightarrow \left( {{v}_{p}}+44 \right)\left( {{v}_{p}}-33 \right)=0 \\\ & \therefore {{v}_{p}}=-44,33 \\\ \end{aligned}$$ We reject the root of the quadratic equation ${{v}_{p}}=-44$ as average speed is always a positive quantity. So we have average speed of the passenger train as ${{v}_{p}}=33$ km/hr and the average speed of express train as $${{v}_{e}}={{v}_{p}}+11=33+11=44\text{ km/hr}$$ **Note:** We note that we have solved the problem without taking into consideration the time the trains stop in intermediate stations. The mistakes in this solution happens in ${{v}_{e}}={{v}_{p}}+11$ which students write as ${{v}_{p}}={{v}_{e}}+11$. We should remember that express trains are faster than passenger trains. The velocity unlike speed can take negative values. We can solve the quadratic equation using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ or the completing square method.