An automobile covered the distance of 240 km between A and B with a certain speed. On its way back, the automobile covered half of the distance at the same speed and for the rest of the trip he increased his speed by 10 km/hour. As a result, the drive back took him 2/ 5 of an hour less than the drive from A to B. What was the automobile’s speed when it was driving from A to B?
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
50km/hr
Let's assume the speed of the automobile when driving from A to B is "x" km/hour.
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.
To solve this problem, we can use the formula: distance = speed × time.
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.