mansi travel 300km to her native Party by train and Party by bus she taken four hour if she travel 60km by train and remaing by bus if she travel 100km By train and remaing by bus she taken ten minutes longer find the average speed of the train and the bus seprately
Answers
Solution :-
Let us assume that, speed of train is x km/h and speed of bus is y km/h.
Case 1 :- She takes 4 hours, if she travels 60 km by train and the remaining 240 km by bus.
→ Time = Distance / Speed .
So,
→ (60/x) + (240/y) = 4
Case 2 :- If she travels 100 kms by train and the remaining 200km by bus, she takes 10 minutes longer than the previous time.
So,
→ (100/x) + (200/y) = 4 + 10min. = 4 + (10/60) = 4 + (1/6) = (25/6)
Now, Lets Assume that, (1/x) = u and (1/y) = v .
Than,
→ (60/x) + (240/y) = 4
→ 60u + 240v = 4
→ 60(u + 4v) = 4
→ u + 4v = (4/60)
→ u + 4v = (1/15) --------- (1)
and,
→ (100/x) + (200/y) = (25/6)
→ 100u + 200v = (25/6)
→ 100(u + 2v) = (25/6)
→ u + 2v = (25/6) * (1/100)
→ u + 2v = (1/24) ---------- (2)
Now, Subtracting Equation (2) from Equation (1) , we get,
→ (u + 4v) - (u + 2v) = (1/15) - (1/24)
→ u - u + 4v - 2v = (8 - 5)/120
→ 2v = 3/120
→ 2v = 1/40
→ v = (1/80)
Putting this value in Equation (2) we get,
→ u + 2(1/80) = (1/24)
→ u + (1/40) = (1/24)
→ u = (1/24) - (1/40)
→ u = (5 - 3)/120
→ u = (2/120)
→ u = (1/60).
Therefore,
→ u = (1/x) = (1/60) => x = 60km/h .
→ v = (1/y) = (1/80) => y = 80km/h.
Hence, Speed of Train is 60km/h and speed of bus is 80km/h.
Let the speed of train and bus be u km/h and v km/h respectively.
According to the question,
....(i)
....(ii)
Let
The given equations reduce to:
60p + 240q = 4 ....(iii)
100p + 200q =
600p + 1200q = 25....(iv)
Multiplying equation (iii) by 10, we obtain:
600p + 2400q = 40....(v)
Subtracting equation (iv) from equation (v), we obtain:
1200q = 15
q =
Substituting the value of q in equation (iii), we obtain:
60p + 3 = 4
60p = 1
p =
:. p = , q =
u = 60 km/h , v = 80 km/h
Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.
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