A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.
Answers
SOLUTION :
Given: Total distance of a journey = 90 km
Let the original speed of the train be x km/h and the increased speed of the train is (x + 15) km/h
Time taken by the train with original speed to cover 90 km= 90/x hrs
[ Time = Distance/speed]
Time taken by the train with increased speed 90 km= 90/(x + 15) hrs
A.T.Q
90/ x - 90/(x + 15) = ½
[Time = 30/60 = 1/2h]
[90(x + 15) - 90x] /(x(x + 15) = 1/2
[By taking LCM]
90x + 1350 - 90x /(x² + 15x) = 1/2
1350 / x² + 15x = 1/2
x² + 15x = 2 × 1350
[By cross multiplication ]
x + 15x - 2700 = 0
x² - 45x + 60x - 2700 = 0
[By middle term splitting]
x(x - 45) + 60 ( x - 45) = 0
(x - 45) (x + 60) = 0
x = 45 or x = - 60
Since, speed can't be negative, so x ≠ - 60
Therefore, x = 45 km/h
Hence the original speed of the train 45 km/h .
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Question:-
→ A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.
Answer:
→ Original speed of the train = 45 km/hr .
Step-by-step explanation:
Given:-
→ Distance = 90 km .
Let the original speed of the train = x km/h .
→ Time taken to travel = 90/x hr .
∴ Then, new speed = ( x + 15 ) km/hr .
∵ Time taken to travel = 90/( x + 15 ) hr .
Now, A/Q,
∵ 90/x = 90/( x + 15 ) + 1/2 .
⇒ 90/x - 90/( x + 15 ) = 1/2 .
⇒ 90[ 1/x - 1/( x + 15 )] = 1/2 .
⇒ x + 15 - x/x( x + 15 ) = 1/2 × 1/90 .
⇒ x/15( x + 15 ) = 180 .
⇒ x( x + 15 ) = 15 × 180 = 2700 .
⇒ x² + 15x - 2700 = 0 .
⇒ x² + 60x - 45x + 2700 = 0 .
⇒ ( x + 60 )( x - 45 ) = 0 .
⇒ x + 60 = 0 or x - 45 = 0 .
⇒ x = 45 or - 60 .
[ ∵ Speed can't be negative. ]