A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
Answers
Step-by-step explanation:
A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed? Hence, the speed of the train to cover 54 km or its first speed is 36 km/h.
Answer:
The first speed of the train is 36 km/hr.
Step-by-step explanation:
Let us assume that the first speed of the train is y km/hr
So, the speed for 63 km will be (y + 6) km/hr.
According to the question,
⋆ Distance covered first time d₁ = 54 km
⋆ Distance covered d₂ = 63 km
⋆ Total time taken for the journey t = 3 hrs
☀️ We know that, Speed = Distance/Time.
From the above formula, we get :
⇒ Distance/Speed = Time
☀️ Now, substituting the values :
⇒ d₁/First speed + d₂/Second speed = t
⇒ 54/y + 63/y + 6 = 3
☀️ Taking the LCM :
⇒ 54 (y + 6) + 63y/y (y + 6) = 3
⇒ 54y + 324 + 63y/y² + 6y = 3
⇒ 117y + 324/y² + 6y = 3
☀️ Cross multiplying them :
⇒ 3 (y² + 6y) = 117y + 324
⇒ 3y² + 18y = 117y + 324
⇒ 3y² + 18y – 117y – 324 = 0
⇒ 3y² – 99y – 324 = 0
☀️ Dividing the whole equation by 3 to get simplified form :
⇒ (3y² – 99y – 324)/3 = 0/3
⇒ y² – 33y – 108 = 0
☀️ Factorising by splitting the middle term :
⇒ [– 36y + 3y = – 33y]
⇒ [– 36y × 3y = – 108y²]
⇒ y² – 36y + 3y – 108 = 0
⇒ y (y – 36) + 3 (y – 36) = 0
⇒ (y – 36) (y + 3) = 0
⇒ y = 36 , y = – 3
________________________________
༜ The first speed of the train is 36 km/hr since, we cannot take y = – 3 [ speed can't be negative ].