CBSE Class 10 Math Question Paper 2018 | Q23B

Question 23B: A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

Video Explanation

Explanatory Answer | CBSE Class 10 Math Board Paper 2018 Q23B

Distance for 1^st part = 63 km
Let the average speed for this part of the journey be s km/hr

Distance covered in 2^nd part = 72 km
Average speed for 2^nd part is 6 km/hr more than the first part.
i.e., Average speed for 2nd part = (s + 6) km/hr

Total time taken = 3 hr
Let the time taken for 1st part be t hr
∴ Time taken for second part = (3 - t) hr

Part 1 journey → s × t = 63 -----(1)
Part 2 journey → (s + 6) (3 - t) = 72 -----(2)

From (1) t = \\frac{63}{s})
Substitute t = \\frac{63}{s}) in (2)
(s + 6) (3 - \\frac{63}{s})) = 72
(s + 6) (\\frac{3s - 63}{s})) = 72
(s + 6) (3s - 63) = 72s
3s² - 63s + 18s - 378 = 72s
3s² - 117s - 378 = 0

Divide the equation by 3
s² - 39s – 126 = 0
s² - 42s + 3s – 126 = 0
(s - 42) (s + 3) = 0
s = 42 or s = -3

Speed has to be positive.
∴ Original average speed = 42 km/hr