Question

A train passes two bridges $104m$ and $194m$ long in $12$ seconds and $18$ seconds respectively. Find the speed of the train.
A. $16\text{ kmph}$
B. $72\text{ kmph}$
C. $\text{27 kmph}$
D. $54\text{ kmph}$

Answer 1

Hint: To find the speed of the train, let us assume the length of the train to be $l$ meters.
It is given that the length of the bridge $=104m$ . So, the length of the train can be written as $(104+l)m$ . Similarly, the length of the train passing the $104m$ long bridge is $(194+l)m$ . We will use the formula, $\text{Speed}=\text{ }\dfrac{\text{Distance}}{\text{Time}}$ . We will substitute the values of each bridge case in this equation. Since the speed of the train is constant, let us equate these two equations, that is, \[\dfrac{104+l}{12}=\text{ }\dfrac{194+l}{18}\] . Solving this gives the value of $l$ . Then, substitute this length in any of the obtained speed equations to find the speed of the train.

Complete step by step answer:
We need to find the speed of the train.
Let us assume the length of the train to be $l$ meters.
It is given that the length of the bridge $=104m$ .
Hence, the length of the train can be written as $(104+l)m$ .
We know that $\text{Speed}=\text{ }\dfrac{\text{Distance}}{\text{Time}}$
It is given that the train crosses $104m$ long bridge in $12$ seconds .
So, let us write the above equation as
$\text{Speed}=\text{ }\dfrac{104+l}{12}...(i)$
It is also given that the train crosses $194m$ long bridge in $18$ seconds.
From, this we get the length of the train as $(194+l)m$ and the speed can be written as
$\Rightarrow \text{Speed}=\text{ }\dfrac{194+l}{18}...(ii)$
The speed of the train is constant. Hence, we can equate the equations $(i)$ and $(ii)$ .
Hence, we get
\[\Rightarrow \dfrac{104+l}{12}=\text{ }\dfrac{194+l}{18}\]
Let us solve the equation to get the value of $l$ .
$\Rightarrow 18(104+l)=\text{ 12}(194+l)$
Let us do the multiplication operation. We will get
$\Rightarrow 1872+18l=2328+12l$
Let us collect constants to one side. We will get
$\Rightarrow 18l-12l=2328-1872$
Doing the subtraction operation, we will get
$\Rightarrow 6l=\text{ }456$
From this, we will get
$\Rightarrow l=\text{ }\dfrac{456}{6}=76$
Hence, the length of the train is $76$ meter.
Now, let us consider the equation $(i)$ .
$\Rightarrow \text{Speed}=\text{ }\dfrac{104+l}{12}$
Now substitute the value of $l$ in the above equation. We will get
$\Rightarrow \text{Speed}=\text{ }\dfrac{104+76}{12}=\dfrac{180}{12}=15\text{ m/s}$
Now, let us convert this to kmph.
We know that $1m=\dfrac{1}{1000}km$ and $1h=\dfrac{1}{3600}s$ .
Hence, $\Rightarrow 15\text{ m/s }=\dfrac{\dfrac{15}{1000}}{\dfrac{1}{3600}}$
$=\dfrac{15\times 3600}{1000}=54\text{ kmph}$
Hence, the speed of the train is $54\text{ kmph}$ .

So, the correct answer is “Option D”.

Note: Students can make mistakes in the equation of speed like $\text{Speed}=\text{ Distance}\times \text{Time}$ . Also, you can use the equation $\text{Speed}=\text{ }\dfrac{194+l}{18}$ to calculate the speed as shown below:
$\text{Speed}=\text{ }\dfrac{194+76}{18}=\dfrac{270}{18}=15\text{ m/s}$ .