Express Train and Passenger Train Problem – Form the Quadratic Equation
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, form the quadratic equation to find the average speed.
Solution
Let the average speed of the passenger train be
\[ x \text{ km/hr} \]
Then the average speed of the express train is
\[ (x+11)\text{ km/hr} \]
Time taken by the passenger train:
\[ \frac{132}{x}\text{ hours} \]
Time taken by the express train:
\[ \frac{132}{x+11}\text{ hours} \]
Given that the express train takes 1 hour less than the passenger train,
\[ \frac{132}{x}-\frac{132}{x+11}=1 \]
Multiplying both sides by \(x(x+11)\),
\[ 132(x+11)-132x=x(x+11) \]
\[ 1452=x^2+11x \]
Bringing all terms to one side,
\[ x^2+11x-1452=0 \]
Required Quadratic Equation
\[ \boxed{x^2+11x-1452=0} \]
Answer
If \(x\) denotes the average speed of the passenger train, then the required quadratic equation is
\[ \boxed{x^2+11x-1452=0} \]
This equation can be solved to find the average speed of the passenger train and hence the express train.