A train covers a distance of 360 km at a uniform speed. If the train had the speed been 5km/hour more,
it would have taken 48 minutes less for the journey. Find the original speed of the train.
Answers
Let the original speed of train be x km/hr.
A train covers a distance of 360 km at a uniform speed.
Given distance = 360 km
time = distance/speed
→ 360/x
If the train had the speed been 5km/hour more, it would have taken 48 minutes less for the journey.
Now speed = (x + 5) km
Time = 48 min = 48/60 = 4/5
According to question,
→ 360/x - 360/(x + 50) = 4/5
→ 360[1/x - 1/(x + 50)] = 4/5
→ 360[5/(x² + 50x)] = 4/5
→ 90[5/(x² + 50x)] = 1/5
→ 5/(x² + 50x) = 1/450
Cross -multiply them
→ 2250 = x² + 50x
→ x² + 5x - 2250 = 0
Split the middle term
→ x² + 50x + 45 - 2250 = 0
→ x(x + 50) - 45(x + 50) = 0
→ (x - 45)(x + 50) = 0
→ x = 45, - 50
Speed of trcancan't be negative. So, negative one cancel out.
Original speed of train = 45 km/hr.
||✪✪ QUESTION ✪✪||
A train covers a distance of 360 km at a uniform speed. If the train had the speed been 5km/hour more,
it would have taken 48 minutes less for the journey. Find the original speed of the train. ?
|| ✰✰ ANSWER ✰✰ ||
Given ,
→ Total Distance = 360km
→ Let original Speed of Train = x km/h.
So,
→ Orginial Time = Distance/Speed = (360/x) hours.
Now, if Speed is 5km/h more ,
→ New speed = (x+5) km/h.
→ Less time = 48 minutes = 48/60 = 4/5 hours .
So , A/q,
→ (360/x) - 360/(x+5) = 4/5
Taking LCM ,
→ [360(x+5) - 360x ] / x(x+5) = 4/5
→ (360x + 360*5 - 360x) / (x² + 5x) = 4/5
Cross - Multiply Now,
→ 5*5*360 = 4(x² + 5x)
Dividing both sides by 4,
→ x² + 5x = 2250
→ x² + 5x -2250 = 0
Splitting The Middle Term Now,
→ x² + 50x - 45x - 2250 = 0
→ x(x+50) - 45(x+50) = 0
→ (x-45)(x+50) = 0
Putting both Equal to zero now,
→ x - 45 = 0. or,. => x +50 = 0
→ x = 45 or,. => x = (-50) .
Since, Speed cant be in Negative .