Question

a goods train leaves a station at 6 pm followed by an express train which leaves at 8 pm and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the trains remain constant between the two stations; calculate their speeds

Answer 1

Answers:

Goods train has a speed of 80 km per hour

Express train has a speed of 100 km per hour

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Step-by-step explanation:

Check out the diagram below to see the table. For now, refer to table 1.

Each row represents a different train.

We have x as the speed of the slower train (the goods train) and x+20 is the speed of the faster train (express train). These speeds are in km/hr or kph.

The y variable represents the number of hours it takes the goods train to arrive at its destination. The equation formed in the first row is 1040 = xy. This is because distance = rate*time.

Let's solve for y to get

1040 = xy

xy = 1040

y = 1040/x

We'll use this equation later.

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The second row of table 1 has x+20 as the speed and y-2-0.6 as the time.

The "-2" represents the fact that the express train is operating for 2 hours less compared to the goods train (because the goods train had a 2 hour head start). Then the "-0.6" indicates we're taking off another 36 minutes, which is equivalent to 36/60 = 0.6 hours.

So overall, the express train is traveling for 2+0.6 = 2.6 fewer hours compared to the goods train. The express train's time value is y-2.6 hours.

With this in mind, we can form this equation for the second row

distance = rate*time

1040 = (x+20)(y-2.6)

We'll now apply substitution to replace y with 1040/x. This is valid because earlier we found that y = 1040/x

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If we apply that substitution and solve for x, we get the following

`1040 = \left(x+20\right)\left(y-2.6\right)\\\\1040 = \left(x+20\right)\left((1040)/(x)-2.6\right)\\\\1040 = x\left((1040)/(x)-2.6\right) + 20\left((1040)/(x)-2.6\right)\\\\1040 = 1040 - 2.6x + (20800)/(x) - 52\\\\0 = -2.6x + (20800)/(x) - 52\\\\x*0 = x*\left(-2.6x + (20800)/(x) - 52\right)\\\\0 = -2.6x^2 + 20800 - 52x\\\\-2.6x^2 - 52x + 20800 = 0\\\\`

Let's now use the quadratic formula to finish things off.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-52)\pm√((-52)^2-4(-2.6)(20800)))/(2(-2.6))\\\\x = (52\pm√(219024))/(-5.2)\\\\x = (52\pm468)/(-5.2)\\\\x = (52+468)/(-5.2) \ \text{ or } \ x = (52-468)/(-5.2)\\\\x = (520)/(-5.2) \ \text{ or } \ x = (-416)/(-5.2)\\\\x = -100 \ \text{ or } \ x = 80\\\\`

Ignore the negative solution because we can't have a negative speed.

The only practical solution here is x = 80

This means the goods train has a speed of 80 km/hr.

The express train's speed must be x+20 = 80+20 = 100 km/hr.

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Now to verify these answers.

If the goods train travels 1040 km and does so at a speed of 80 km/hr, then it travels for 1040/80 = 13 hours. Add this onto 6:00 PM and we arrive at 7:00 AM. This is shown in the first row of table 2 (see below).

If the express train travels 1040 km and its speed is 100 kph, then it travels for 1040/100 = 10.4 hours. This is equivalent to 10 hours, 24 minutes because 0.4 hrs = 0.4*60 = 24 min

If we start at 8:00 PM and elapse 10 hours, then we'll arrive at 6:00 AM. Add on another 24 minutes, and we get to 6:24 AM. Notice how this is 36 minutes before 7:00 AM (because 24+36 = 60). So this confirms that the express train indeed arrives 36 minutes before the goods train does, despite the goods train having that 2 hour head start.