A goods train starts from station A to the station B and at the same time a passenger train starts from station B to station A. After meeting they reached their destination in 9 hours and4 hours respectively. Find the ratio of their speed.
Answers
Answer:
680
Step-by-step explanation:
ANSWER
Solution:
Given Train starts at the same time.
Hence, Time remains constant.
Speed of Train 'T1' be 'S1' = 40 km\hr
Speed of Train 'T2' be 'S2' = 45 km\hr
Let X be the distance at which trains meet.
Train T2 travels X+40 km more than Train T1
According to the formula.
Speed=
Time
Distance
Time=
45
X+40
Time=
40
X
From above two equations.
40
X
=
45
X+40
45×X=40×X+40×40
5×X=1600
=320 km
Distance between two Stations = Distance at which train T1 meet + Distance at which Train T2 meet + Distance train T2 travels
= 320+320+40
= 680
- The ratio of speed of good trains to the passanger train is equal to 2 : 3 .
Given :- A goods train starts from station A to the station B and at the same time a passenger train starts from station B to station A. After meeting they reached their destination in 9 hours and 4 hours respectively.
To Find :- The ratio of their speeds ?
Formula used :-
- S(1) : S(2) = √T(2) : √T(1)
- S(1) = Speed of First Train.
- S(2) = Speed of Second Train.
- T(1) = After Crossing Each other Time taken by first Train to Reach opposite Point.
- T(2) = After Crossing Each other Time taken by second Train to Reach opposite Point.
Solution :-
Let us assume that, speed of goods train is equal to S(1) and speed of speed of passanger train is equal to S(2) .
So,
→ Ater meeting time taken by goods train to reach station B = 9 hours = T(1)
and,
→ Ater meeting time taken by passanger train to reach station A = 4 hours = T(2)
then, using above told formula we get,
→ S(1) : S(2) = √T(2) : √T(1)
→ S(1) : S(2) = √4 : √9
→ S(1) : S(2) = √(2)² : √(3)²
→ S(1) : S(2) = 2 : 3 (Ans.)
Note :- For proof of above told formula please refer to image . I have proved this formula .
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