Math word problem!!?

Question

An engine pulls a train for 120 km. A second engine which travels 5 kph faster than the first then pulls the same train for 240 km. Find the speed of each train, if it takes 6 hrs. to cover the entire 360-km-stretch.

Answer 1

Speed of 1st train = x
Speed of 2nd train = x+5

Time = distance/sped
120/x + 240/(x+5) = 6
120(x+5) + 240x = 6x(x+5)
Divide by 6
20(x+5) + 40x = x(x+5)
20x + 100 + 40x = x^2 + 5x
x^2 - 55x - 100 = 0

Using the quadratic formula:
x = 56.76 (approx)
x+5 = 61.76 (approx)

Check:
120/56.76 + 240/61.76 = 6 (approx.)

The approximate speeds are 56.8km/hr and 61.8km/hr

Answer 2

Let the speed of the first engine be E. Let the speed of the second engine be E+5. the time taken by the first engine is 120/E and the time taken by the second engine is 240/E+5
so we have 120/E + 240/E+5 = 6 or 120E+600+240E= 6E(E+5) or 360E+600=6E^2 +30E or 60E+100=E^2+30E or
E^2 - 30E - 100=0 Now you have to solve for E.

Answer 3

distance = rate x time

time = distance/rate

Let x = rate of first engine
then x+5 = rate of 2nd engine

6 = 120/x + 240/(x+5)

6x(x+5) = 120(x+5) + 240x

This reduces to

6x² - 330x - 600 = 0

x² - 55x - 100 = 0

Solve with the quadratic formula

ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a

x = (55 ± √(3025 + 400)/2

x = (55 ± √(3425)/2

x = 56.76, x = -1.76

Disregard the negative solution

x = 56.76 km/hr (first engine)
x + 5 = 61.76 km/hr (second engine)
.

Answer 4

d1=120 km

Let the pulling speed of the first one be x kph

=>Second one (x+5) kph

d2=240 km

Total time taken=Time taken by first + Time taken by second

=>6=120/x + 240/(x+5)

=>6x(x+5) = [120(x+5) + 240x]

=>x(x+5) = 20(x+5) + 40x

=>x^2 + 5x = 20x + 100 + 40x

=>x^2 - 55x - 100=0

=>x=[55(+-)sqrt(3025 + 400)]/2

=>x=[55 (+-)58.52]/2

=>x=56.76 kph

Answer 5

Ta = 120/Va
Tb = 240/(Va+5)
Ta+Tb = 6

120/Va + 240/(Va+5) = 6
120(Va+5) + 240 Va = 6 * Va * (Va+5)
360 Va + 600 = 6 * Va^2 + 30 Va
Va^2 - 55 Va - 120 = 0
Va = 56.76174978 km/hr
Vb = 61.76174978 km/hr