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Q) A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed. If it takes 3 hours to complete total journey, what is its original average speed?

Ans:

Let’s consider the original average speed of the train: X km/hr.

Hence, to cover distance of 63 km, it will take \frac{63}{\times} hrs

Next, when its speed increased to (X + 6) km/hr,

the time taken to cover 72 km will be: \frac{72}{\times + 6} hrs

Now, total distance covered = 63 + 72 = 135 km

Total Time taken by train = Time taken to cover 63 kms + Time taken to travel 72 Kms

= \frac{63}{\times} hrs + \frac{72}{\times + 6} hrs

= \frac{63(\times + 6) + 72 \times}{\times (\times + 6)} hrs

= \frac{135 \times + 378}{\times (\times + 6)} hrs

Given that the total time taken is 3 hours

\therefore \frac{135 \times + 378}{\times (\times + 6)} = 3

\therefore (135 X + 378) = 3X (X + 6)

\therefore 135 X + 378 = 3X2 + 18 X

\therefore 3X2 – 117 X – 378 = 0

\therefore X2 – 39 X – 126 = 0

\therefore (X – 42) (X + 3) = 0

\therefore X = 42 and X = – 3

Since Speed can not be negative, therefore X = 42 will be taken

Hence, the original speed of the train is 42 km/hr.

Check: At 42 kmph, train will take 1.5 hrs to cover 63 km. At 6 kmph higher speed i.e. at 48 kmph, train will cover 72 km in 1.5 hrs. Total time will be 1.5 + 1.5 = 3 hrs

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