Question

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is
(a) 45 km/hr
(b) 50 km/hr
(c) 54 km/hr
(d) 55 km/hr

Answer 1

Hint: Here, we need to find the speed of the train. We will assume the speed of the train be \[x\] km/hr. We will use the relative speed, the formula for time and the given information to form an equation in terms of \[x\]. We will solve this equation to find the value of \[x\], and hence, the speed of the train.

Complete step by step solution:
Let the speed of the train be \[x\] km/hr.
We will use relative speed to find the speed of the train.
The speed of the train is \[x\] km/hr, and the speed of the man is 5 km/hr.
Since the man and the train are going in the same direction, we get
Relative speed \[ = \left( {x - 5} \right)\] km/hr
The distance travelled by the train is equal to its length.
Thus, we get
Distance travelled \[ = 125\] m
We will convert this to km using unitary method.
We know that 1000 m \[ = \] 1 km.
Dividing both sides by 1000, we get
1 m \[ = \dfrac{1}{{1000}}\] km
Multiplying both sides by 125, we get
125 m \[ = \dfrac{{125}}{{1000}}\] km
The time taken by the train to cover the 125 m is 10 seconds.
We will convert this to hours using unitary method.
We know that 3600 seconds \[ = \] 1 hour.
Dividing both sides by 3600, we get
1 second \[ = \dfrac{1}{{3600}}\] hr
Multiplying both sides by 10, we get
10 seconds \[ = \dfrac{{10}}{{3600}} = \dfrac{1}{{360}}\] hr
Now, the time taken is equal to the distance travelled, divided by the relative speed.
Therefore, we get the formula \[{\rm{Time}} = \dfrac{{{\rm{Distance}}}}{{{\rm{Relative speed}}}}\].
Substituting relative speed as \[\left( {x - 5} \right)\] km/hr, distance as \[\dfrac{{125}}{{1000}}\] km, and time as \[\dfrac{1}{{360}}\] hour, we get
\[ \Rightarrow \dfrac{1}{{360}} = \dfrac{{\dfrac{{125}}{{1000}}}}{{x - 5}}\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{1}{{360}} = \dfrac{{125}}{{1000\left( {x - 5} \right)}}\]
This is a linear equation in one variable in terms of \[x\]. We will solve this equation to find the value of \[x\].
Simplifying the expression by cross-multiplying, we get
\[ \Rightarrow 1000\left( {x - 5} \right) = 360 \times 125\]
Dividing both sides of the equation by 1000, we get
\[ \Rightarrow x - 5 = \dfrac{{360 \times 125}}{{1000}}\]
Simplifying the expression, we get
\[ \Rightarrow x - 5 = 9 \times 5\]
Multiplying the terms in the expression, we get
\[ \Rightarrow x - 5 = 45\]
Adding 5 on both sides of the equation, we get
\[ \Rightarrow x = 50\] km/hr
\[\therefore \] We get the speed of the train as 50 km/hr.
Thus, option (b) is the correct option.

Note: We have formed a linear equation in one variable in terms of \[x\] in the solution. A linear equation in one variable is an equation that can be written in the form \[ax + b = 0\], where \[a\] is not equal to 0, and \[a\] and \[b\] are real numbers. For example, \[x - 100 = 0\] and \[100P - 566 = 0\] are linear equations in one variable \[x\] and \[P\] respectively. A linear equation in one variable has only one solution.
A common mistake is to use the values as given in the question. Since the options are given in km/hr, we need to convert the distance to km, and the time to hours using a unitary method.
We used relative speed to find the solution. Relative speed is speed of an object, relative to the speed of another object. If the two objects are moving in the same direction, then the relative speed of an object is the difference in the speeds of the two objects. If the two objects are moving in the opposite direction, then the relative speed of an object is the sum in the speeds of the two objects.