Question

A train travels at a certain average speed for a distance of 54km and then travels a distance of 63km at an average speed of 6kmph more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Answer 1

Hint: This is a time speed distance question. We will solve this question by using time = distance / speed formula. Assuming first speed as x we will obtain second speed and their respective time duration in terms of x. As the total time is given is the data, we will obtain a quadratic equation in terms of x. Doing middle term factorization or putting quadratic formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] we will get the answer.

Complete step-by-step answer:
A train travels at a certain average speed for a distance of 54km.
Let the average speed of the train x.
$\therefore $The time taken by train to cover the distance 54km is,
$T = \dfrac{D}{S}$
where T is time, D is distance, S is speed
$T = \dfrac{{54}}{x}$……….(1)
Again it travels a distance of 63km at an average speed of 6kmph more than the first speed.
The average speed of the train is x+6.
$\therefore $The time take by train to cover the distance 63km is,
$T = \dfrac{{63}}{{x + 6}}$…………….(2)
According to the question, train takes 3hr to complete the journey i.e. $T = 3$……..(3)
But the total time taken by the train is,
$T = \dfrac{{54}}{x} + \dfrac{{63}}{{x + 6}}$……….(4)
Comparing equation 3 and 4 we get that,
$\dfrac{{54}}{x} + \dfrac{{63}}{{x + 6}} = 3$
Simplifying the above equation we get,
$\dfrac{{54\left( {x + 6} \right) + 63x}}{{x\left( {x + 6} \right)}} = 3$
Doing cross multiplication in above equation we get,
$54\left( {x + 6} \right) + 63x = 3 \times x\left( {x + 6} \right)$
Expanding the above equation we get,
$54x + 324 + 63x = 3{x^2} + 18x$
Taking all the terms into one side and simplifying them we get,
$3{x^2} - 99x - 324 = 0$
Dividing 3 on both the side,
${x^2} - 33x - 108 = 0$
Doing middle term factorization we get,
${x^2} - 36x + 3x - 108 = 0$
Taking x common from first two terms and 3 from second two terms we get,
$x\left( {x - 36} \right) + 3\left( {x - 36} \right) = 0$
$\left( {x + 3} \right)\left( {x - 36} \right) = 0$
Taking $x + 3 = 0$
$x = -3$
Again taking $x - 36 = 0$
$x = 36$
X value can’t be -3, as speed can’t be negative.
Hence, $x = 36$
i.e. the first speed of the train is 36 kmph.

Note: The formula of time speed distance is given by,
$T = \dfrac{D}{S}$
where T is time, D is distance, S is speed
You should practice more questions of this type.
To solve the quadratic equation we use two methods. One is quadratic formula of obtaining roots i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and another is middle term factorization in which we split coefficient of middle term in such a way that its product is equal to the last term.
You should practice more and more quadratic equations using these two methods.