Question 558130: A passenger plane flew to moscow and back. The trip there took five hours and the trip back took nine hours. It averaged 144 km/h faster on the trip there than on the return trip. Find the passenger plane's average speed on the outbound trip.
O.k so I did have a go at it Trying to use a system of equations using the substitution method to eliminate one unknown but it didn't work. I will try and explain what I was playing around with:
D1 = Distance there D2 = Distance back ( but not sure if I needed this as it is same Distance )
Average Speed = S
D1/5 + D2/9 = S
D1/5 = D2/9+144 ( I tried substituting this in )
D2/9+144 + D2/9 = S ( then I combined the like terms )
D2/18+144 = s ( but this is where I realized I had failed as I still had 2 variables.
Please help me someone I am so bad at word problems and I would love to be confident at them.
P.S I wasn't sure about which was the outbound trip, I guessed it meant the trip back.
Found 5 solutions by Theo, KMST, ikleyn, MathTherapy, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the answer is:
plane speed equals 252 kmph
wind speed equals 72 kmph
distance equals 1620 km
the basic formula is rate * time = distance
let p = speed of the plane and let w = speed of the wind.
going there the formula becomes:
(p+w)*5 = d
coming back the formula becomes:
(p-w)*9 = d
since they're both equal to d (the distance is the same going and coming), you can set them equal to each other to get:
(p+w)*5 = (p-w)*9
simplify this equation to get:
5p + 5w = 9p - 9w
subtract 5p from both sides of this equation and add 9w to both sides of this equation to get:
14w = 4p
solve for p to get:
p = 14w/4
they are also telling you that the plane averaged 144 km/h faster on the trip there than on the return trip.
this means that:
they are talking about the speed of the plan going versus the speed of the plane coming back which is the rate part of the rate * time = distance equation.
this means that:
p + w = p - w + 144
the speed of the plane going is p + w
the speed of the plane coming back is p - w
subtract p from both sides of that equation and add w to both sides of that equation to get:
2w = 144
divide both sides of that equation by 2 to get:
w = 72
that's the speed of the wind.
now that you know the speed of the wind, you can go back to the equation of:
p = 14w/4 to get:
p = (14*72)/4 which becomes:
p = 252
the speed of the plane in still air is 252 km/h
the speed of the wind is 72 km/h
going, the formula of:
(p+w)*5 = d becomes:
(252+72)*5 = d which becomes:
d = 1620 km.
coming back, the formula of:
(p-w)*9 = d becomes:
(252-72)*9 = d which becomes:
d = 1620 km.
d is the same in both directions as it should be.
the answer is:
plane speed is 252 kmph
wind speed is 72 kmph
the passenger plane's average speed going is 252+72 = 324 kmph
the passenger plane's average coming back is 252-72 = 180 kmph
that's the net speed when you take the speed of the wind into account.
outbound depends on where you're looking from.
if you're looking from where you started from, then outbound would be going to moscow.
if you're looking from moscow, then outbound would be coming back.
i suspect they mean going to moscow but can't be sure the way it's worded.
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! You almost had it, and then lost your sense of direction.
You had  and you were saying it was the same distance both ways.
So you could have written 
Having fractions/denominators makes it look hard, but you could have solved it to find
D=1620 km
Now you have the distance. You already had the times.
The only question is the average speed in the outbound trip. I believe outbound is the trip going out there, to Moscow, so the time was 5 hours and the speed was 1620 km/5 h = 324km/h
Answer by ikleyn(52834)  (Show Source):
You can put this solution on YOUR website! .
A passenger plane flew to Moscow and back. The trip there took five hours and the trip back took nine hours.
It averaged 144 km/h faster on the trip there than on the return trip.
Find the passenger plane's average speed on the outbound trip.
~~~~~~~~~~~~~~~~~~~~~~~~~~
As I read the post by @Theo, I clearly see that his solution is overcomplicated
and his setup is over-sophisticated. Actually, this problem admits much more simple treatment.
It is not about a choice of the solution method - it is about how to teach
young students in a right way. It is impossible to teach them presenting Theo' approach.
Therefore, I developed and present here a simple solution, which is a standard
method (or one of standards methods) for this problem.
Outbound trip = the trip back (from Moscow).
Trip to Moscow is 5 hours.
The trip back is 9 hours.
Let x be the average speed on the outbound trip (from Moscow), in km/h.
It is the value, which is under the question of the problem.
Then the average speed on the trip to Moscow was (x-144) km/h, according to the problem.
The distance for the flight to Moscow is 5x kilometers (the product of the time and the speed).
The distance for the returning flight is 9(x-144) kilometers, similarly..
This distances are equal - so, we can write this equation
5x = 9(x-144). (1)
Simplify and find x
5x = 9x - 9*144
9*144 = 9x - 5x
9*144 = 4x
x = (9*144)/4 = 9*36 = 324 kilometers per hour.
ANSWER. The average speed on the outbound trip was 324 km/h.
Solved.
The solution, which I presented in this my post, follows a standard approach to the problem
without any sophistications. It provides the most simple and the most straightforward treatment of the problem.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
A passenger plane flew to moscow and back. The trip there took five hours and the trip back took nine hours. It averaged 144 km/h faster on the trip there than on the return trip. Find the passenger plane's average speed on the outbound trip.
O.k so I did have a go at it Trying to use a system of equations using the substitution method to eliminate one unknown but it didn't work. I will try and explain what I was playing around with:
D1 = Distance there D2 = Distance back ( but not sure if I needed this as it is same Distance )
Average Speed = S
D1/5 + D2/9 = S
D1/5 = D2/9+144 ( I tried substituting this in )
D2/9+144 + D2/9 = S ( then I combined the like terms )
D2/18+144 = s ( but this is where I realized I had failed as I still had 2 variables.
Please help me someone I am so bad at word problems and I would love to be confident at them.
P.S I wasn't sure about which was the outbound trip, I guessed it meant the trip back.
No!! The OUTBOUND trip is the one to Moscow, while the one back from Moscow is the INBOUND/RETURN trip.
Let OUTBOUND speed, be S
Then INBOUND/RETURN speed = S - 144
It took 5 hours to get there, so distance traveled to Moscow = 5S
It took 9 hours to get back, so distance on return trip = 9(S - 144)
Since distance there and back is the same, we get 5S = 9(S - 144)
5S = 9S - 9(144)
4S - 9S = - 9(144)
- 4S = - 9(144)
Average speed on the OUTBOUND, or 
Answer by greenestamps(13203)  (Show Source):
|
|
|
