Question

a train travels at a certain average speed for a distance of 63km and then travels at a distance of 72km at an average speed of 6km/h more than ots orignal speed. if it takes 3h to complete total journey what is orignal average speed.

Answer 1

continuation.....
$3 {x}^{2} - 117x - 378 = 0$
${x}^{2} - 39x - 126 = 0$

${x}^{2} - 42x + 3x - 126 = 0$
$x(x - 42) + 3(x - 42)$
$(x + 3)(x - 42) = 0$
$x = - 3 \: \: \: \: \: \: or \: \: x = 42$
since speed cannot be negative,
$x = 42$

Answer 2

Solution :

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Given :

Condition 1 :
A train travels at a certain average speed for a distance of 63km and then travels at a distance of 72km at an average speed of 6km/h more than its original speed.

Let the speed of train be x kmph

Then, the time taken to travel 63 km is (time = $\frac{Distance}{Speed}$) = $\frac{63}{x}$ kmph)

The time taken to travel 72 kmph with raised speed of 6 kmph is $\frac{72}{x+6}$

Condition 2 ;

The TOTAL TIME TAKEN FOR THE JOURNEY TO COMPLETE : 3 hours,.

∴ $\frac{63}{x} + \frac{72}{x + 6} = 3$

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⇒ $\frac{63(x+6) + 72(x)}{x(x+6)} = 3$

⇒ $\frac{63x+ 378 + 72x}{x(x+6)} = 3$

⇒ $\frac{135x+ 378}{x^2+6x} = 3$

⇒ $135x+ 378 = 3(x^2+6x)$

⇒ $135x+ 378 = 3x^2+18x$

⇒ $135x+ 378 - 3x^2 - 18x = 0$

⇒ $-3x^2 + 117x + 378 = 0$ (Dividing this equation by - 3, we get)
 
⇒ $x^2 - 39x - 126 = 0$

⇒ $x^2 - 42x + 3x - 126 = 0$

⇒ $x(x - 42)+ 3 (x - 42) = 0$

⇒ $(x + 3) (x - 42) = 0$

For this equation, 0 is possible when,

x + 3 = 0 (or)  x - 42 = 0

⇒ x = -3 (or) x = 42

⇒ x = 42 (speed can't be negative.)

             ∴ The speed of train is 42 kmph

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                         Hope it Helps!!

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