Question

A fats train takes 3 hours less than a slow train for a journey of ${\rm{600}}\;{\rm{km}}$ . If the speed of the slow train is $10\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$ less than that of the fast train, find the speeds of the two trains,

Answer 1

Hint: We have already given distance in the question and relation between the speed of fast and slow train is also given, we will assume the speed of slow and fast train. Then we will find the relation for the time required by each train. Since the difference between the time required is given in question, we will find the value of speed of train the relation of time.

Complete step-by-step answer:
We will assume the speed of the fats train is $x\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$ .
Then the speed of the slow train can be represented as $\left( {x - 10} \right)\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$
The distance is given as ${\rm{600}}\;{\rm{km}}$ .
Then we can say that time taken by the fast train to cover this distance is $\dfrac{{600}}{x}\;{\rm{hr}}$ and time taken by the slow train to cover this distance is $\dfrac{{600}}{{x - 10}}\;{\rm{hrs}}$ .
From the question we know that a fast train takes 3 hours less than the slow train, that means the difference between their time is 3 hours. This can be expressed as;
\[
\left( {\dfrac{{600}}{{x - 10}}} \right)\;{\rm{hrs - }}\left( {\dfrac{{600}}{x}} \right)\;{\rm{hrs}} = 3\;{\rm{hrs}}\\
\dfrac{{600x - 600\left( {x - 10} \right)}}{{x\left( {x - 10} \right)}} = 3\\
3\left( {{x^2} - 10x} \right) = 600x - 600x + 600\\
{x^2} - 10x = 2200
\]
On rearranging the above equation, we have
\[{x^2} - 10x - 2200 = 0\]
We will find the value of $x$ from the factorization method. For that we will have to split the coefficient of $x$ in such a way that the product of these two terms is equals to the product of the coefficient of ${x^2}$ and the constant term.
We know that 2200 has the factor 40 and 50 , and the difference between these two is equals to 10. Hence , we can rewrite our quadratic equation as:
 \[
\Rightarrow {x^2} + 40x - 50x - 2200 = 0\\
\Rightarrow x\left( {x + 40} \right) - 50\left( {x + 40} \right) = 0\\
\Rightarrow \left( {x - 50} \right)\left( {x + 40} \right) = 0
\]
We have value of $x$ as
$
\Rightarrow x - 50 = 0\\
\Rightarrow x = 50
$
And
$
\Rightarrow x + 40 = 0\\
\Rightarrow x = - 40
$
The speed of train can not be negative hence the value of$x$ should be $50\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$
 Hence we can say that the speed of slow train is $50\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$ and the speed the slow train is
$
 = x - 10\\
 = 50 - 10\\
 = 40
$
Hence the speed of slow train is $40\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$

Note: We are the formula of speed to express the relation for time, we know that speed can be defined as the ratio of distance and time. Hence, we will use this relation to express the relation for time. The quadratic equation formed during the calculation of speed can be either solved by factorization or by the quadratic formula.