# Easy Time, Speed & Distance Solved QuestionAptitude Discussion

 Q. A train 120 m in length passes a pole in 12sec and another train of length 100 m travelling in opposite direction in 10 sec. Find the speed of the second train in km per hour.
 ✔ A. 43.2 km/hr ✖ B. 43 km/hr ✖ C. 44 km/hr ✖ D. 43.5 km/hr

Solution:
Option(A) is correct

Let the speed of the train be $x$ km/hr

Then,

$120=x\times \dfrac{5}{18}\times 12$

$⇒ x = 36$ km/hr

Let speed of the other train be $y$ km/hr

Then, relative speed in opposite direction:

$=(y+36)\times \dfrac{5}{18}$

So total distance:

$(120+100)=(y+36)\times \dfrac{5}{18}\times 10$

$y = \textbf{43.2 km/hr}$

Edit: To understand how the figure $\dfrac{5}{18}$ is used, check explanation by Shireen.

## (6) Comment(s)

Abdul
()

I know the conversion but 18/5 should come but you multiply with 5/18. how?

Ritwik
()

If you are using 18/5 then you need to divide it from the speed of the train (km/hr). Shireen has already given the exact calculations so I am not repeating it here, kindly check her comment for the calculations.

Akshaya
()

Why do you multiply $\dfrac{5}{18}$ with 36kmph? it is already in kmph. isn't it? somebody clarify my doubt.

Deepti
()

It the other way around of your logic. You are correct that speed is already in kmph but other quantities are in mt, sec. units.

So in order to make the units compatible, speed is converted into mt/sec. units.

$=36 \times \dfrac{\text{ km}}{\text{ hour}}$

$=36 \times \dfrac{1000 \text{ mt}}{60 \times 60\text{ sec}}$

$=36 \times \dfrac{5 \text{ mt}}{18\text{ sec}}$

$=\left(36 \times \dfrac{5}{18} \right)\dfrac{\text{mt}}{\text{sec}}$

Dhruv Desai
()

I cannot understand the formula. What is $\dfrac{5}{18}$?

Shireen
()

It is conversion factor used for converting mt/sec to km/hr.

$\text{1 mt/sec} = \dfrac{18}{5}\text{ km/hr}$

and here is how it has come:

$1 \text{ mt} = \dfrac{1}{1000} \text{ km}$

$1 \text{ sec}= \dfrac{1}{60\times 60} \text{ hr}$ $= \dfrac{1}{3600} \text{ hr}$

$\Rightarrow \text{1 mt/sec}=\left(\dfrac{\frac{1}{1000}}{\frac{1}{3600}}\right) \text{ km/hr}$

$\Rightarrow \text{1 mt/sec}= \dfrac{3600}{1000} \text{ km/hr}$

$\Rightarrow \text{1 mt/sec}= \dfrac{18}{5} \text{ km/hr}$