Question

A journey 192 km from town A to town B takes 2 hours more by an ordinary passenger train than a dupe train. If the speed of the fright train is 16 km/hr more, find the speed of the passenger train.

Answer 1

Hint:
Here, we need to find the speed of the passenger train. Using the speed distance time formula, we can find the time taken by the passenger train to complete the journey, and the time taken by the faster train to complete the journey. Using this and the given information, we can form an equation. We will solve this equation to find the speed of the trains.

Formula Used:
We will use the formula for the time taken to cover a distance at a uniform speed which is given by \[{\rm{Time}} = \dfrac{{{\rm{Distance}}}}{{{\rm{Speed}}}}\].

Complete step by step solution:
Let the speed of the passenger train be \[x\] km/hr.
It is given that the speed of the faster train is 16 km/hr more than the passenger train.
Thus, we get the speed of the faster train as \[\left( {x + 16} \right)\] km/hr.
It is given that the length of the journey is 192 km.
Thus, both trains cover the distance of 192 km.
Now, we will use the speed distance time formula to find the time taken by the two trains to complete the journey.
Distance travelled by the passenger train \[ = \] 192 km
Speed of the passenger train \[ = x\] km/hr
Therefore using the formula \[{\rm{Time}} = \dfrac{{{\rm{Distance}}}}{{{\rm{Speed}}}}\], we get
Time taken by the passenger train to complete the journey \[ = \left( {\dfrac{{192}}{x}} \right)\] hours
Similarly, distance travelled by the faster train \[ = \] 192 km
Speed of the faster train \[ = \left( {x + 16} \right)\] km/hr
Therefore using the formula \[{\rm{Time}} = \dfrac{{{\rm{Distance}}}}{{{\rm{Speed}}}}\], we get
Time taken by the faster train to complete the journey \[ = \left( {\dfrac{{192}}{{x + 16}}} \right)\] hours
The time taken by the passenger train is 2 hours more than the time taken by the faster train.
Therefore, we get the equation
\[ \Rightarrow \dfrac{{192}}{x} = \dfrac{{192}}{{x + 16}} + 2\]
We need to solve this equation to get the value of \[x\] and hence, find the speed of the trains.
Simplifying the expression by cross-multiplying, we get
Rewriting the expression, we get
\[ \Rightarrow \dfrac{{192}}{x} - \dfrac{{192}}{{x + 16}} = 2\]
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{{192\left( {x + 16} \right) - 192x}}{{x\left( {x + 16} \right)}} = 2\]
Rewriting the expression by cross-multiplying, we get
\[ \Rightarrow 192\left( {x + 16} \right) - 192x = 2x\left( {x + 16} \right)\]
Multiplying the terms using the distributive law of multiplication, we get
\[ \Rightarrow 192x + 3072 - 192x = 2{x^2} + 32x\]
Subtracting the like terms, we get
\[ \Rightarrow 3072 = 2{x^2} + 32x\]
Subtracting 3072 from both sides, we get
\[ \Rightarrow 2{x^2} + 32x - 3072 = 0\]
Dividing both sides by 2, we get
\[ \Rightarrow {x^2} + 16x - 1536 = 0\]
This is a quadratic equation. We will factorise the equation to get the value of \[x\].
Factoring the quadratic equation, we get
\[\begin{array}{l} \Rightarrow {x^2} + 48x - 32x - 1536 = 0\\ \Rightarrow x\left( {x + 48} \right) - 32\left( {x + 48} \right) = 0\\ \Rightarrow \left( {x + 48} \right)\left( {x - 32} \right) = 0\end{array}\]
Therefore, either \[x + 48 = 0\] or \[x - 32 = 0\].
Simplifying the expressions, we get
\[x = 32, - 48\]
However, \[x\] cannot be \[ - 48\] since speed cannot be negative.
Therefore, we get the value of \[x\] as 32 km/hr.
Substituting 32 for \[x\] in the expression \[\left( {x + 16} \right)\], we get
Speed of faster train \[ = 32 + 16 = 48\] km/hr

Therefore, we get the speed of the passenger train as 32 km/hr, and the speed of the faster train as 48 km/hr.

Note:
We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. A quadratic equation has 2 solutions.
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
Here we have found out the speed of the faster train. We might get confused between velocity and speed. Velocity is a vector quantity while speed is a scalar quantity.