Question

A passenger train takes two hour more for a journey of 300 km if its speed is decreased by 5km/hr from its usual speed. Find the usual speed of the train.

Answer 1

Let the usual speed of the train be 'x' km.
Then the time taken(t1) to cover 300 km will be 300/x hours.
And time taken(t2) to cover 300 km at speed decreased by 5 km will be 300/(x-5).

Now t2-t1 =2
So,300/(x-5)-300/x=2. ......substituting th values of t1 and t2
300[x-x+5]/([x(x-5)]=2
300 (5) = 2[x(x-5)]. ....multiplying both sides by 2[x(x-5)].
150(5=x(x-5). ....multipling both sides by 1/2
750 = x^2 - 5x

X^2 -5x -750 = 0. ......adding -750 to both sides
X^2 -30x+25x -750=0
X(x-30) +25(x-30) =0
(X+25)(x-30) =0
X=-25 and x=30

Speed cannot be negative.
Therefore the usual speed of the train is 30 km/h
And the decreased speed will be 25 km/h

Answer 2

Answer:

Let the usual speed of the train be y km / hr .

A.T.Q.

$\displaystyle{\frac{300}{y} -\frac{300}{y+5} =2 }$

300 ( y + 5 ) - 300 y = 2 y ( y + 5 )

y² + 5 y - 750 = 0

y² + 30 y - 25 y - 750 = 0

( y + 30 ( y -25 ) = 0

y = - 30 or y = 25

Since , speed of train can't be negative .

Therefore , the usual speed of the train is 25 km / hr .