Question

On a journey of 600 km, a train was delayed 1 hour and 30 min after having covered 1/4 of the way. To arrive on time at the destination, the engine driver had to increase the speed by 15 km/hour. How long did the train travel for?

Answer 1

Answer:

The train traveled for 32.66 hours.

Step-by-step explanation:

Let the speed of the train was initially for $\frac{1}{4}$ the distance was x km/hr.

So, 150 km the train moved with x km/hr speed and the remaining 450 km the train goes with the speed of (x + 15) km/hr.

Then, the time taken by the train to cover the first 150 km will be $\frac{150}{x}$ hrs.

Therefore, with the actual speed, the train would have covered the 150 km by $(\frac{150}{x} - 1\frac{1}{2}) = (\frac{150}{x} - \frac{3}{2}) = \frac{300 - 3x}{2x}$ hrs.

Now, from the given conditions, we can write the equation as

$4(\frac{300 - 3x}{2x}) = \frac{150}{x} + \frac{450}{x + 15}$

⇒ $2(\frac{300 - 3x}{x}) = 150 \times \frac{4x + 15}{x(x + 15)}$

⇒ (300 - 3x) (x + 15) = 75(4x + 15) {Since x ≠0}

⇒ 300x + 4500 - 3x² - 45x = 300x + 1125

⇒ 3x² + 45x - 3375 = 0

⇒ x² + 15x - 1125 = 0

⇒ $x = \frac{- 15 \pm \sqrt{15^{2} - 4(1)(-1125)}}{2}$ {Applying the quadratic formula}

⇒ x = 26.87 km/hr {Ignoring the negative root}

Therefore, the train traveled for  $4(\frac{300 - 3 \times 26.87}{26.87}) = 32.66$ hours. (Answer)