Question

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A train travel at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6km/hr more than the original speed . If it takes 3 hrs to complete total journey , what is its original average speed..??



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Answer 1

Hey !!!

Let the original speed is x
and time t

•°•speed = distance /time

so ..here distance = 63km
so x = 63/t

tx = 63

t = 63/x -------1)

and again ... here distance is 72km
and average speed increase 6km/h

so, x + 6 = 72/ t

=> 72/x+6 = t ------2)

adding both equation 1) and 2) we get

72/x + 6+ 63/x = 3

72(x) + 63(x+6)/x(x+6) = 3

72x + 63x + 378 /x² + 6x = 3

135x + 378 = 3x² +18x

0 =- 135x + 18x +3x² -378

0 = 3x² -117x - 378

0 = 3(x² - 39x - 126)

x² - 39x-126 =0

x² - 42x +3x -126 =0

x(x -42) +3(x-42) =0


(x - 42) (x +3) =0

x -42 =0

x = 42

and x+3 = 0

x = -3 (speed never in negative just neglect it )

so our original average speed of journey is 42km /h

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Hope it helps you !!!

@Rajukumar111

Answer 2

Solutions :-

Given

Distance travels on certain speed = 63 km
Let Original speed of train be x
Distance travels at speed of 6 km/hr = 72 km
Speed = (x + 6) km per hour
Time = 72/(x+6) hrs
Total journey completed in 3 hrs.

we know that

Distance = Speed × time
Speed = distance/time
Time = distance/speed


we have to Find the value of x :-

A/q

=> 63/x + 72/(x + 6) = 3
=> 63(x + 6) + 72x = 3x(x+ 6)
=> 21(x + 6) + 24x = x(x+6)
=> 45x + 21×6 = x2 + 6x
=> x^2 - 39x - 126 = 0
=> x^2 - 39x - 126 = 0
=> x^2-42x+3x -126=0
=> x(x-42)+3(x-42)=0
=> (x - 42)(x + 3) = 0
=> x = 42 or x = -3

Hence

Original average speed is 42 km / hr ( distance positive )