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A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km per hour more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed ?

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Hint- Speed is equal to distance upon time and \[{t_1} + {t_2} = 3\] hours where ${t_1}\& {t_2}$ is time taken for complete journey
Let the initial speed be $x$ Km per hour.
As per the question,
Time taken at a speed of \[x\] km per hour \[\left( {{t_1}} \right){\text{ = }}\dfrac{{63}}{x}\] hr
Time taken at a speed of \[x + 6\] km per hour \[\left( {{t_2}} \right) = \dfrac{{72}}{{x + 6}}\] hr
Now, according to the question it is given that the total journey takes 3 hr.
\[{t_1} + {t_2} = 3\]
Put the values of \[{t_{_1}}\]and \[{t_2}\]
\[\dfrac{{63}}{x} + \dfrac{{72}}{{x + 6}} = 3\]
Take LCM
\[ \Rightarrow \dfrac{{63\left( {x + 6} \right) + 72x}}{{x\left( {x + 6} \right)}} = 3\]
With the help of cross multiplication, solve the above equation
\[
   \Rightarrow 63x + 378 + 72x = 3x\left( {x + 6} \right) \\
   \Rightarrow 378 + 135x = 3{x^2} + 18x \\
\]
Bring RHS term to LHS, we get
\[ \Rightarrow 135x + 378 - \left( {3{x^{^2}} + 18x} \right) = 0\]
Taking minus sign common, we get
\[
   \Rightarrow 3{x^{^2}} + 18x - 135x - 378 = 0 \\
   \Rightarrow 3{x^2} - 117x - 378 = 0 \\
\]
Taking $3$common, we get
\[
   \Rightarrow 3\left( {{x^2} - 39x - 126} \right) = 0 \\
   \Rightarrow {x^2} - 39x - 126 = 0 \\
\]
Now factorise the above equation
\[
   \Rightarrow {x^2} - 42x + 3x - 126 = 0 \\
   \Rightarrow x\left( {x - 42} \right) + 3\left( {x - 42} \right) = 0 \\
   \Rightarrow \left( {x - 42} \right)\left( {x + 3} \right) = 0 \\
\]
Now either $x = 42$ or $x = - 3$ and since speed cannot be negative.
So the speed of Train is $42$Km per hour.
Note - For such a type of question always begin with assuming the initial speed be $x$km per hour. And we know that speed is equal to distance upon time. So Calculate the time taken when speed is $x$ and $x + 6$ km per hour respectively.

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