# A train leaves Canberra for Sydney at 12 noon and another train leaves Sydney forty minutes later. Both travel at the same constant speed. If it takes 3.5 hours for one trip, at what time will they...

A train leaves Canberra for Sydney at 12 noon and another train leaves Sydney forty minutes later. Both travel at the same constant speed. If it takes 3.5 hours for one trip, at what time will they pass each other.

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A train leaves Canberra for Sydney at 12 noon and another train leaves Sydney for Canberra 40 minutes later. Both travel at the *same constant* speed. The time taken by each to complete one trip is 3.5 hours.

Let the distance between the stations be D and the time after 12 that they meet is x hours. As they complete one trip in 3.5 hours their speed is `D/3.5`

Distance = speed*time; the time traveled by the train leaving Canberra is x and the time traveled by the other train is `x - 40/60` hours.

`(D/3.5)*x + (D/3.5)*(x - 40/60) = D`

Cancel D from both the sides

=> `x/3.5 + x/3.5 - 4/21 = 1`

=> `x*(2/3.5) = 25/21`

=> `x = 25/12` hours

=> x = 125 minutes

**The two trains meet after 125 minutes the first starts or at 2:05 pm**

**In my first answer i have made a mistake by taking 40minutes as 45 minutes. Appologize for that and this gives themodified answer.**

Let distance between canberra and sydney be X km.

The time for this trip = 3.5 hours

Then the travel speed of a train (V) = X/3.5 km/h

Say the trains meets at a distance Y km from canberra. Then from sydney it has (X-Y) km.

If the first train starts its journey at t=0; the second will start its journey at t=2/3h or t=40minutes.

Let the trains meet together at time t= t1

Then at time t=t1

first train has travelled Y km.

But distance = velocity*time

Therefore Y = v*t

** Y=(X/3.5)*t1**-------------(1)

When t=t1 the second train has only traveeled (t1-0.75) h since it was 45 minutes late.

The distance travelled by second train when trains meet together is (X-Y) km.

Then like first train;

**(X-Y) =(X/3.5)*(t1-2/3)**----------(2)

From (1) Y=(X/3.5)*t1

**Y/X = t1/3.5**-------------------(3)

From (2) (X-Y) =(X/3.5)*(t1-2/3)

(X-Y)/X = (t1-2/3)/3.5

(X/X-Y/X) = (t1-2/3)/3.5

1-Y/X = (t1-2/3)/3.5

**Y/X = 1-(t1-2/3)/3.5**----------------(4)

The left side of both (3) and (4) are equal.

Therefore **t1/3.5**= **1-(t1-2/3)/3.5**

t1/3.5= (3.5-(t1-2/3))/3.5

t1= 3.5-t1+2/3

t1 = (3.5+2/3)/2

t1= 2.083333 h

t1 = 2h and 5 min

**So the trains will meet 2 hours and 5 minutes after the first train starts its journey.**

**Assumption**

**Both trains travel at same speed right from the start.**

Ther has been a mistake in my previous answer where the time for secondtrain to leave is taken as 1 hour instead of 40 minutes. The corrected answer is as under:

As the speed of 2 trains are equal therefore the point of crossing will be midway between the point where the first train has reached after 40 minutesand Sydney and both the trains would have taken the same time to reach the midway.

The remaining time = 3.5 hours - 40 minutes = 170 minutes

The time taken be each train to pass = 170/2 = 85 minutes

the total time taken by first train to reach crossing = 40+85 = 125 minutes = 2 hours 5 minutes

The first train left Canberra at 12 noon and the time taken to reach crossing is 2 hour 5 minutes, therefore

**The Time when the two trains pass is 02:05 PM.**

Thanks to jeew-m for pointing out the error.