Question

A journey of 300km from Pune to Bombay, takes two hours less by a fast train than by a slow train. If the average speed of the slow train is 25kmph less than that of fast train, find the average speed of each train?

Answer 1

Given:

Average speed of slow train is 25 kmph lower than fast train.

Time taken for Pune -> Bombay Journey by fast train is 2 hr less for fast train.

To find:

Average speed of each train

Calculation:

Let Speed of fast train be x , hence speed of slow train will be (x - 25) ; let time for slow train be t2 and time for fast train be t1 :

As per the question :

$t2 - t1 = 2$

$= > \dfrac{300}{(x - 25)} - \dfrac{300}{x} = 2$

$= > 300 \bigg \{ \dfrac{x - (x - 25)}{x(x - 25)} \bigg \} = 2$

$= > 300 \bigg \{ \dfrac{ 25}{x(x - 25)} \bigg \} = 2$

$= > {x}^{2} - 25x = (150 \times 25)$

$= > {x}^{2} - 25x - 3750 = 0$

$= > {x}^{2} - (75 - 50)x - 3750 = 0$

$= > {x}^{2} - 75x + 50x - 3750 = 0$

$= > x(x - 75) + 50(x -75) = 0$

$= > ( x + 50)(x - 75) = 0$

Either x = - 50 kmph or + 75 kmph

Since Speed is Always positive , we have to consider that :

$x = 75 \: kmph$

So speed of fast train = 75 kmph

Speed of slow train = 75 - 25 = 50 kmph

Answer 2

$\Large\underline{\underline{\sf{\color{lime}{GIVEN}}}}$

• A journey of 300 km from Pune to Bombay, takes two hours less by a fast train than by a slow train.  
• Average speed of the slow train is 25 kmph less than that of fast train.

$\Large\underline{\underline{\sf{\color{lime}{TO\ FIND}}}}$

→ The average speed of each train.

$\Large\underline{\underline{\sf{\color{lime}{SOLUTION}}}}$

◙ Let the average speed of the fast train be x kmph.

A/q, the average speed of the slow train is 25 kmph less than that of the fast train.

◙ Speed of the slow train = (x - 25) kmph

◙ Let the time taken by the fast train be t₁.

◙ Let the time taken by the slow train be t₂.

Again, according to the given conditions, t₁ > t₂.

→ t₂ - t₁ = 2 hours    . . . (i)

We know,

$\boxed{\sf{\dag\ Time\ taken=\frac{Distance\ covered}{Speed} }}$

We get the following :-

$\sf{t_1=\dfrac{300}{x} }$

and

$\sf{t_2=\dfrac{300}{x-25} }$

Putting these values in equation (i) :-

$\sf{\dashrightarrow \dfrac{300}{x-25}-\dfrac{300}{x}=2 }\\\\\\\sf{\dashrightarrow \dfrac{300x-300(x-25)}{x(x-25)}=2 }\\\\\\\sf{\dashrightarrow \dfrac{300x-300x+7500}{x^2-25x}=2 }\\\\\sf{\dashrightarrow \dfrac{7500}{x^2-25x}=2 }\\\\\sf{\dashrightarrow 2x^2-50x=7500}\\\\\sf{\dashrightarrow 2x^2-50x-7500=0}\\\\\sf{\dashrightarrow x^2-25x-3750=0}\\\\\sf{\dashrightarrow x^2-50x+75x-3750=0}\\\\\sf{\dashrightarrow x(x-50)+75(x-50)=0}\\\\\sf{\dashrightarrow (x-50)(x+75)=0}$

$\sf{\dashrightarrow (x-50)=0\:\:or\:\:(x+75)=0}\\\\\sf{\dashrightarrow x=50\ kmph\:\:or\:\:x=-75\ kmph}$

We know, speed can never be negative.

So, considering the values of x to be positive, we get :-

→ Speed of the fast train = 75 kmph

→ Speed of the slow train = (x - 25) = 50 kmph