A train covered the distance of 400 km between A and B at a certain speed. On the way back it covered 2/5 of the distance at that same speed and then it decreased its speed by 20 km/hour. Find the speed of the train at the end of its journey from B back to A, if the entire trip took 11 hours.
400/x + 160/x + 240/(x-20) = 11
x = 80, 140/11
80 - 20 = 60
60 km/hr
Let's assume the initial speed of the train from A to B is "x" km/hour.
On the way back, the train covered 2/5 * 400 km = 160 km at the same speed of "x" km/hour.
Now, let's calculate the time taken for the train to travel 160 km at speed "x" km/hour.
Time = Distance / Speed
Time = 160 km / x km/hour
Next, the train decreased its speed by 20 km/hour, so the new speed is "x - 20" km/hour.
The remaining distance from B to A is 400 km - 160 km = 240 km.
Let's calculate the time taken for the train to travel 240 km at speed "x - 20" km/hour.
Time = Distance / Speed
Time = 240 km / (x - 20) km/hour
According to the problem, the entire trip took 11 hours. So, the total time for the journey is the sum of the time taken from A to B and the time taken from B to A.
Total Time = Time from A to B + Time from B to A
11 = 400/x + 160/x + 240/(x - 20)
To solve this equation, we can cross multiply and simplify:
11(x(x - 20)) = 400(x - 20) + 160x + 240x
11(x^2 - 20x) = 400x - 8000 + 400x + 240x
Expanding and simplifying the equation further:
11x^2 - 220x = 640x - 8000
Rearranging the terms:
11x^2 - 860x + 8000 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
x = (-(-860) ± √((-860)^2 - 4 * 11 * 8000)) / (2 * 11)
x = (860 ± √(739600 - 352000)) / 22
x = (860 ± √387600) / 22
Now, let's calculate the two possible values for x and determine the corresponding speed of the train at the end of its journey from B back to A.
1. x = (860 + √387600) / 22
x = (860 + 620) / 22
x = 1480 / 22
x ≈ 67.27 km/hour
2. x = (860 - √387600) / 22
x = (860 - 620) / 22
x = 240 / 22
x ≈ 10.91 km/hour
Since the speed of the train cannot be negative, the valid solution is x ≈ 67.27 km/hour. Therefore, the speed of the train at the end of its journey from B back to A is approximately 67.27 km/hour.
To solve this problem, we can use the formula Speed = Distance/Time.
Let's assume the speed of the train on its way from A to B is x km/hour.
Given that the distance between A and B is 400 km, we can calculate the time it took for the train to travel from A to B using the formula Time = Distance/Speed:
TimeAB = 400/x
Now, on the way back from B to A, the train covered 2/5 of the distance at the same speed of x km/hour. We can calculate the distance covered on the way back using the formula Distance = Fraction x Total Distance:
DistanceBA = (2/5) x 400 = 160 km
Since we know that the entire trip took 11 hours, we can calculate the time it took for the train to travel from B to A using the formula:
TimeBA = Total Time - TimeAB
TimeBA = 11 - (400/x)
It is given that the train decreased its speed by 20 km/hour on its way back. So, the speed of the train from B to A can be represented as (x - 20) km/hour.
Now we can calculate the time it took for the train to travel from B to A using the formula Time = Distance/Speed:
TimeBA = DistanceBA / (x - 20)
160 / (x - 20) = 11 - (400/x)
To solve this equation, we can cross-multiply and simplify:
160x = (x - 20) * (11x - 220)
160x = 11x^2 - 440x - 20x + 4000
0 = 11x^2 - 620x + 4000
We can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 11, b = -620, and c = 4000.
x = (-(-620) ± √((-620)^2 - 4 * 11 * 4000)) / (2 * 11)
x = (620 ± √(384400 + 176000)) / 22
x = (620 ± √560400) / 22
x = (620 ± 748.82) / 22
Now, we have two possible values for x:
x1 = (620 + 748.82) / 22 ≈ 64.9 km/hour
x2 = (620 - 748.82) / 22 ≈ -5 km/hour
Since speed cannot be negative, the speed of the train at the end of its journey from B back to A is approximately 64.9 km/hour.
if the initial speed is x, then since time=distance/speed,
400/s + 160/s + 240/(s-20) = 11