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.
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I cant solve it though
400/s + 160/s + 240/(s-20) = 11
ok, hint:
multiply each term by s(s-20) to get a quadratic.
Hint: it comes out to be a nice whole number
400(s-20)+160(s-20)+240(s)=11(s)(s-20)
(400+160)(s-20)+240s=11(s^2-20s)
560(s-20)+240s=11s^2-220s
560s-11200+240s=11s^2-220
800s-11200=11s^2-220
−11s2+1020s−11200=0
(−11s+140)(s−80)=0
s=80? or 140/11?
It still does not work
To solve this problem, we need to set up a system of equations based on the given information.
Let's say the speed of the train on the first part of the journey from A to B is 's' km/hour. The time taken for this part of the journey can be calculated using the formula: time = distance / speed.
Given that the distance between A and B is 400 km, the time taken for the first part of the journey is 400 / s hours.
Now, for the second part of the journey from B back to A, the train covers 2/5 of the distance (2/5 * 400 = 160 km) at the same speed 's'. The time taken for this part of the journey is 160 / s hours.
Now, after covering this 160 km, the train decreases its speed by 20 km/hour. Let's say the new speed is 's - 20' km/hour. The remaining distance to be covered is (400 - 160) = 240 km. The time taken for this part of the journey can be calculated as 240 / (s - 20) hours.
The total time taken for the entire trip is given as 11 hours, so we can set up the following equation:
400 / s + 160 / s + 240 / (s - 20) = 11
Now, we need to solve this equation to find the value of 's'.
To solve this equation, we can multiply both sides by the LCD (least common denominator) to eliminate the fractions. In this case, the LCD is s(s - 20):
(400(s - 20) + 160s + 240s) / s(s - 20) = 11
Simplifying the equation further:
(400s - 8000 + 160s + 240s) / s(s - 20) = 11
(800s - 8000) / s(s - 20) = 11
800s - 8000 = 11s^2 - 220s
Rearranging the equation:
11s^2 - 1020s + 8000 = 0
Now, we can solve this quadratic equation for 's'. There are multiple methods to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. To find the speed of the train, we can either use factoring or the quadratic formula.
Once we find the value of 's', we can calculate the speed of the train at the end of its journey from B back to A by subtracting 20 km/hour from the value of 's'.