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A train travels a distance of 300 km at a constant speed. If the speed had been 10 km/hr faster, the train would have taken 1 hour less to cover the same distance. What is the original speed of the train? 

Created: 1 month ago | Updated: 1 month ago
Updated: 1 month ago

Let the original speed of the train be s'  km / hr and the original time taken be 't' hours. 

We know that distance = speed × time. 

Therefore: 300 = s×t ______ (i) 

If the speed had been 10 km/hr. faster, the new speed would be (s + 10) km / hr and the time taken would be (t - 1) hours. The distance remains the same: 

Therefore: 300 = (s + 10)×(t - 1) 

 

300 = 300 - s + 10t - 10 

 300 - 300 = - s + 10t - 10 

 s = 10t - 10 

t=s+1010

Now substitute this value of 't' back into equation (i) 

300 = s ×(s + 10)10

 3000 = s(s + 10) [Multiply both sides by 10] 

 3000 = s2 + 10s 

 s2 + 10s - 3000 = 0 

 s2 + 60s - 10s - 3000 = 0 

∴ (s + 60)(s - 50) = 0 

This gives us two possible solutions for 's': s = - 60 or s = 50 

Since speed cannot be negative, the original speed of the train is 50 km/hr. 

Answer: The original speed of the train was 50 km/hr.

1 month ago

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