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Two trains leave towns 1136 kilometers apart at the same time and travel toward each other. One train travels 20km/h faster than the other. If they meet in hours, what is the rate of each train?

User LOTUSMS
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Answer: Let's assume the rate of the slower train is x km/h.

Then, the rate of the faster train would be x + 20 km/h, as it is traveling 20 km/h faster.

When the two trains meet, their combined distance traveled is equal to the total distance between the towns, which is 1136 kilometers.

Since distance = rate × time, the time taken for both trains to meet is t hours.

For the slower train, the distance traveled is x km/h × t hours = xt kilometers.

For the faster train, the distance traveled is (x + 20) km/h × t hours = (x + 20)t kilometers.

As they meet, their combined distance is 1136 kilometers, so we can write the equation:

xt + (x + 20)t = 1136

Now, let's solve for t:

xt + xt + 20t = 1136

2xt + 20t = 1136

t(2x + 20) = 1136

t = 1136 / (2x + 20)

Now that we have the value of t, we can find the rate of the slower train (x) by substituting the value of t back into either of the original equations. Let's use the equation for the slower train:

xt = 1136

Substitute the value of t:

x * (1136 / (2x + 20)) = 1136

Now, solve for x:

1136x = 1136(2x + 20)

1136x = 2272x + 22720

2272x - 1136x = 22720

1136x = 22720

x = 22720 / 1136

x = 20

So, the rate of the slower train is 20 km/h.

Now, to find the rate of the faster train, we can use the fact that it is 20 km/h faster than the slower train:

Rate of faster train = Rate of slower train + 20

Rate of faster train = 20 + 20

Rate of faster train = 40 km/h

Therefore, the rate of the slower train is 20 km/h, and the rate of the faster train is 40 km/h.

User Arijeet Saha
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