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 = 63 / X 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 = 72 / (X + 6) hrs
Step 2: Given that total time taken to cover the full journey = 3 hrs
∴ 63 / X + 72 / (X + 6) = 3
∴ [ 63 (X + 6) + 72 X] / X (X + 6) = 3
∴ 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 is the speed and speed can not have negative value
Therefore, the original average speed value is 42 km/hr.
Check:
Time taken in 1st part = 63 / X = 63 / 42 = 3 / 2 = 1.5 hrs
Time taken in 2nd part = 72 / (X + 6) = 72 / (42 + 6) = 72 / 48 = 3 / 2 = 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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