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

An express train takes $1$ hour less than a passenger train to travel $132 \mathrm{~km}$ between Mysore and Bangalore. If the average speed of the express train is $11 \mathrm{~km} / \mathrm{hr}$ more than that of the passenger train, form the quadratic equation to find the average speed of express train.

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Write linear functions: word problems

Full solution

Q. An express train takes

1

hour less than a passenger train to travel

132 \mathrm{~km}

between Mysore and Bangalore. If the average speed of the express train is

11 \mathrm{~km} / \mathrm{hr}

more than that of the passenger train, form the quadratic equation to find the average speed of express train.

Denote average speed: Let's denote the average speed of the passenger train as ' $v$ ' km/hr. Therefore, the average speed of the express train would be ' $v + 11$ ' km/hr since it is given that the express train is $11$ km/hr faster than the passenger train.
Calculate time taken: The time taken by the passenger train to travel $132 \, \text{km}$ is $\frac{\text{distance}}{\text{speed}}$ , which is $\frac{132}{v} \, \text{hours}$ . Similarly, the time taken by the express train to travel the same distance is $\frac{132}{v + 11} \, \text{hours}$ .
Formulate equation for time difference: According to the problem, the express train takes $1$ hour less than the passenger train to travel the same distance. So, we can write the equation as: $\newline$ $\frac{132}{v} - \frac{132}{v + 11} = 1$
Clear fractions and simplify: To solve this equation, we need to find a common denominator for the fractions on the left-hand side. The common denominator is $v(v + 11)$ . We multiply both sides of the equation by this common denominator to clear the fractions: $\newline$ $v(v + 11)\left(\frac{132}{v}\right) - v(v + 11)\left(\frac{132}{v + 11}\right) = v(v + 11)(1)$
Simplify the equation further: Simplifying the equation, we get: $132(v + 11) - 132v = v^2 + 11v$
Cancel out like terms: Expanding and simplifying further, we get: $132v + 132\times 11 - 132v = v^2 + 11v$
Formulate quadratic equation: The ' $132v$ ' terms on both sides cancel out, leaving us with: $\newline$ $132\times 11 = v^2 + 11v$
Formulate quadratic equation: The $132v$ terms on both sides cancel out, leaving us with: $\newline$ $132\times 11 = v^2 + 11v$ Now we have a quadratic equation in terms of $v$ . To make it standard, we bring all terms to one side of the equation: $\newline$ $v^2 + 11v - 132\times 11 = 0$
Formulate quadratic equation: The $132v$ terms on both sides cancel out, leaving us with: $\newline$ $132\times 11 = v^2 + 11v$ Now we have a quadratic equation in terms of $v$ . To make it standard, we bring all terms to one side of the equation: $\newline$ $v^2 + 11v - 132\times 11 = 0$ Finally, we simplify the constant term to get the quadratic equation: $\newline$ $v^2 + 11v - 1452 = 0$

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