Question

Q. A passenger train takes 2 hours less for a journey of 300 km. If the speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

Answer 1

Let usual speed = x km/hr
And time = y hr
Thus according to the question
x*y=300
and
(x+5)*(y-2)=300
(x+5)*(300/x-2)=300
300-2x+1500/x-10=300
-2x^2 +1500=10x
2x^2 + 10x-1500=0
x^2 +5x- 750=0
x^2 +30x-25x-750=0
x=-30,25
-30 is neglected
Therefore
x=25km/hr
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Answer 2

Answer:Given: Total distance of a journey = 150 km

Let the usual speed of the train be x km/h  and the increased  speed of the train is (x + 5) km)h.

Time taken by the train with usual speed to cover 150 km= 150/x hrs

Time taken by the train with increased speed to cover 150 km= 150/(x + 5) hrs  

[ Time = Distance/speed]

150/(x + 5) = 150/ x   - 1

150/ x -  150/(x + 5)  = 1

[150(x + 5) - 150x] /(x(x + 5) = 1

[By taking LCM]

150x + 750 - 150x /(x² + 5x) = 1

750 / x² + 5x = 1

(x² + 5x ) = 750

x² + 5x - 750 = 0

x² - 25x + 30x - 750 = 0

[By middle term splitting]

x(x - 25) + 30 ( x - 25) = 0

(x - 25) (x + 30) = 0

x = 25 or x = - 30  

Since, speed can't be negative, so x = - 30  

Therefore,  usual speed of the train be = x = 25 km/h