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

Q32 – Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans: 

Step 1: Let’s consider the original average speed of the car = X km/ hr

At this speed, time taken to cover 63 km = hrs

Next, Average speed of the train in 2nd part of journey = (X + 6) km/hr

∴ At this new speed, time taken to cover 72 km = hrs

Step 2: Given that total time taken to cover the full journey = 3 hrs

∴

∴

∴ 63 (X + 6) + 72 X = 3 X (X + 6)

∴ 21 (X + 6) + 24 X = X (X + 6)

∴ 21 X + 126 + 24 X = X ² + 6 X

∴ 45 X + 126 = X ² + 6 X

∴ X ² + 6 X – 45 X – 126 = 0

∴ X ² – 39 X – 126 = 0

Step 3: Let’s solve this equation to find the value of x:

∴ X ² – 42 X + 3 X – 126 = 0     (by mid-term splitting)

∴ X (X – 42) + 3 (X – 42) = 0

∴ (X – 42) (X + 3) = 0

∴ X = 42 or X = – 3

Here, we reject X = – 3 because X, being the speed, can not have negative value.

Therefore, the original average speed value is 42 km/hr.

Check:
Time taken in 1st part = = = = 1.5 hrs
Time taken in 2nd part = = =   = = 1.5 hrs
Total time taken in journey = 1.5 + 1.5 = 3 hrs
Since it matches with the given condition on the question, hence our solution is correct.

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