Let x be the speed from A to B
Since time = distance/speed,
120/x + 120/(x+10) = 240/x - 2/5
x = 50 km/hr
Since time = distance/speed,
120/x + 120/(x+10) = 240/x - 2/5
x = 50 km/hr
During the first trip from A to B, the automobile covered a distance of 240 km at a speed of "x" km/hour. The time taken can be calculated using the formula: Time = Distance/Speed.
So, Time taken from A to B = 240/x hours.
During the return trip, the automobile covered half the distance, which is 240/2 = 120 km, at the same speed "x" km/hour. The time taken for this half of the distance is 120/x hours.
For the remaining half distance, the automobile increased its speed by 10 km/hour. So, the speed for this part of the trip is (x+10) km/hour. The distance covered for this part is also 120 km.
The time taken for this part can be calculated as: Time = Distance/Speed = 120/(x+10) hours.
According to the given condition, the time taken for the return trip is 2/5 hours less than the time taken for the trip from A to B.
So we have the equation: 120/x + 120/(x+10) = 240/x - (2/5)
To solve this equation, we first find a common denominator:
(120(x+10) + 120x) / (x(x+10)) = (240 - (2/5)) / x
Expanding the numerator:
(120x + 1200 + 120x) / (x(x+10)) = (240 - (2/5)) / x
Combining like terms:
(240x + 1200) / (x(x+10)) = ((240 * 5) - 2) / (5x)
Cross multiplying:
(240x + 1200) * (5x) = (x(x+10)) * ((240 * 5) - 2)
Expanding both sides:
(1200x^2 + 6000x) = (x^2 + 10x) * (1200 - 2)
(1200x^2 + 6000x) = (x^2 + 10x) * 1198
Expanding further:
1200x^2 + 6000x = 1198x^2 + 11980x
Subtracting 1198x^2 and 11980x from both sides:
2x^2 - 5980x = 0
Factoring out 2x:
2x(x - 2990) = 0
Setting each factor equal to 0:
2x = 0 or x - 2990 = 0
Since the speed cannot be zero, we solve the second equation:
x - 2990 = 0
x = 2990
Therefore, the automobile's speed when driving from A to B is 2990 km/hour.
Let's assume the speed of the automobile while driving from A to B is 'x' km/hour.
On the way back from B to A, the automobile covered half of the distance, which is 240 km / 2 = 120 km at the same speed 'x' km/hour.
For the remaining distance, the automobile increased its speed by 10 km/hour. So, the speed from A to B on the return trip is 'x + 10' km/hour.
According to the given information, the drive back took 2/5 of an hour less than the drive from A to B. The time taken from A to B is denoted by 't' hours.
So, the time taken to drive back is (t - 2/5) hours.
Using the formula distance = speed × time, we can set up two equations:
Equation 1: 240 = x × t (distance = speed × time from A to B)
Equation 2: 120 = (x + 10) × (t - 2/5) (distance = speed × time from B to A)
Now, we can solve these equations to find the value of 'x'.
From Equation 1: x = 240 / t
Substituting this value of 'x' into Equation 2:
120 = (240 / t + 10) × (t - 2/5)
Now, we can simplify this equation and solve for 't'.
120 = (240 + 10t) × (t - 2/5)
120 = 240t - 4/5t^2 - 48
0 = 4/5t^2 - 240t + 168
To solve this quadratic equation, we can simplify it by multiplying through by 5:
0 = 4t^2 - 1200t + 840
Now, we can factorize this equation:
0 = 4(t^2 - 300t + 210)
Now, we have:
t^2 - 300t + 210 = 0
We can solve this quadratic equation using the quadratic formula or by factoring. The solutions for 't' will give us the time taken from A to B.
Once we find 't', we can substitute it back into Equation 1 to find the speed 'x' of the automobile when driving from A to B.