Atrain p takes 1 hour less tha train q for journey 300 km. If speed of train is 15km/ hr more than the train q. Find the speed train p andq
Answers
Answer:-
Given:
A train "p" takes 1 hour less than the train "q" to travel a distance of 300 km
Let the time taken by train q be x hrs.
→ Time taken by train p = (x - 1) hrs.
And,
Speed of train p is 15 km/hr more than train q.
Let speed of train q = y km/h
Speed of train p = (y + 15) km/h
We know that,
speed = Distance/ time
Hence,
Speed of train p (y + 15) km/h = 300/(x - 1)
→ y + 15 = 300 / x - 1
On cross multiplication we get,
→ (x - 1)(y + 15) = 300 -- equation (1)
Speed of train q (y km/h) = 300/x
→ y = 300/x -- equation (2)
Substitute "y = 300/x" in equation (1)
→ (x - 1) (300/x + 15) = 300
→ x (300/x + 15) - 1( 300/x + 15) = 300
→ 300 + 15x - 300/x - 15 = 300
→ 15x - 300/x = 300 - 300 + 15
→ (15x² - 300)/x = 15
On cross multiplication we get,
→ 15x² - 300 = 15x
→ 15x² - 15x - 300 = 0
→ 15x (x - 1 - 20) = 0
→ x - 21 = 0
→ x = 21
→ Time taken by train q (x) = 21 hrs
Substitute the value of x in equation (2)
→ y = 300/x
→ y = 300/21
→ y = 14.28 km/hr (100/7 km/hr )
→ Speed of train q = 14.28 km/hr
→ Speed of train p = (14.28 + 15) = 29.28 km/hr