Uniform Motion - Round Trip?

Question

My son had an Algebra question he was unable to answer. We were also unable to locate an example in his book or on the internet.

The question is:
If a plane flew 200 km/hr one way and 300 km/hr for the return trip and the total trip took 6 hrs, how far did the plane go? Any help is appreciated.

Answer 1

Let x = time to travel at 200 kph
6 - x = time for 300 kph
d = rt
200x = 300(6 - x)
200x = 1800 - 300x
500x = 1800
x = 3.6 hrs
and distance = 200(3.6) = 720 km

Answer 2

Since the distance for onward journey and return journey are equal ,
The ratio of time for these two journeys is in the inverse ratio to their speeds.
T1:T2 = 300: 200 = 3: 2.
T1: {T1 +T2} = 3: 5
Since the total time is 6 hours. T1 = [3/5] *6 =3.6h
Distance covered = 200 x 3.6 = 720 km
Total distance = 1440 km.

Answer 3

Let the distance covered in one trip be d. It covers the same distance in the return trip too.

Time = Distance/Speed

6 = d/200 + d/300
6 = 5d/600
5d = 600 * 6
5d = 3600
d = 720

It covers 720 km in one way, meaning a total distance of 1440 km for the original way and the return trip

Answer 4

r(1st trip)=200kph
r(2ND trip)=300kph
t(1st trip)=x
t(2ND trip)=6-x(that is because the total trip took 6hrs so the time in the 2ND trip should not exceed 6)
d=rt so
d1=200x
d2=1800-300x
because it is a round trip therefore the plane took the same distance in each trip so
d1=d2
substitute
200x=1800-300x
500x=1800
divide by 500
x=3.6
substitute the x in any equation on distance because its the same so
200(3.6)=720km
but since it is a round trip the distance is doubled making the total distance that the plane traveled is 1440 km

Answer 5

200X+300Y=6 ..............1st equation
200X=300y
so 200X-300Y=0...............2nd equation.