The executives of All Australia Airlines are difficult masters! They proudly set the airline's timetable to show arrival time to the nearest minute and expect strict punctuality from their pilots. This is no easy task for the pilots, since, first, no allowance is made in the timetables for headwinds or tailwinds that might affect the duration of flights, and, second, pilots are instructed to fly at full throttle at all times. No one, obviously, pays much attention to the airline scheduled arrival times! At 7:30 one morning, an All Australia Airlines jet took off on time from Melbourne airport bound for Sydney. It arrived there at 8:30. At 9:15, again on time, took off for a return trip to Melbourne, arriving there at 10:45 after battling the same steady headwind it had taken advantage of on its way to Sydney. How late was the plane on arrival in Melbourne?
+129852 See if this makes sense, amnesia :
With the wind.....the plane covers (D/60) every minute and against the wind, the plane covers (D/90) every minute.....where D is the total disance beween the cities
So......the average rate per minute [ assuming no wind ] is given by :
[D / 60 + D / 90] / 2 =
[90D + 60D] / [ 60 * 90 *2] =
[150D] / [ 10800] =
[15D ] / 1080] =
D/72 ...... means that the plane would make the flight in a normal time of 72 minutes witn no wind
Thus....the plane is 90 − 72 = 18 min late
+259 7:30 start 1
8:30 arrival 1
9:15 start 2
10:45 arrival 2
I dont really get the question. Kinda sounds like a trick question, but if ur just asking for the time it was late in the 2nd arrival, wouldnt that just be
(2 time - arrival 2 time) - (start1 - arrival1) ?
1h30m - 1h = 30 minutes
Of cours by saying that the airplane was on time in the first flight, and not too early or also too late.
+259 Just want to point out that my answer is incorrect.
I really didnt read the topic correct, ones under me is correct thou.
+259 So I actually thought many times about the answer now and I got a question myself.
The answer under me seems mathmatic right, but it still wondered why the wind would hinder the plane in a ration of 12/18. ( As its 12minutes faster to the target ,and 18minutes late to get back to it)
If I wouldnt think about the speed of the plane in d/h, but in time I could create the following equations:
W = a negative number which shows how many minutes the winds let u reach ur destination faster
A = the time the Plane needs to reach its destination. always the same as stated in the topic
So with A being always the same + having the advantage of the wind it takes u 60mintues.
A+W = 60minutes
Now with A and the disadvantage its 90 minutes.
A-W = 90minutes
Means if i solve this equations:
A+w =60minutes
A-w =90minutes
2w = -30minutes II /2
w = -15minutes
putting this in the first equation
so A-15minutes = 60minutes
shows that A is 75minutes.
In the answer under me though its showed that A is 72minutes, when solved with involving the destination.
Im really confused here maybe someone can help
Let the airplane's speed = A
Let the wind's speed =W
Let the distance between Sydney and Melbourne =D
A + W =D/1 hour
A - W =D/1.5 hours, solve for A
A =5/6D hours - That is, the airplane absolute speed regardless of wind, is 5/6 the distance between Sydney and Melbourne. So the scheduled time for the trip must be:
D / (5/6D) = 6/5 hours
D =6/5 hours, or 72 minutes. But the returning plane took 90 minutes, therefore:
90 - 72 = 18 minutes too late.
amnesia:
Remember that the ABSOLUTE speed of the plane, going and coming, is 5/6 of the distance between the two cities, which worked out to be 6/5 hours, or 72 minutes. But the ougoing plane took 60 minutes, thereby shaving off 12 minutes, or 1 - (60/72) x 100 =16 2/3% off the 72 minutes(72 x 16 2/3%) =12 minutes.
Similarly, the incoming plane was supposed to take 72 minutes, but it took 90 minutes. Or 90/72 - 1 x 100 =25% longer. And 25% of 72 =18 minutes longer.
You are taking the AVERAGE time due to the wind, by basically adding the times up and dividing by 2, or 12+18 =30/2 =15 minutes. But you can't do that in figuring out the outgoing and incoming times, which worked out to be 12 and 18 minutes respectively.
+259 I appreciate ur try to explain me the math, but im stuck at the logic part :D
I just dont get the difference in the wind. I mean it has a different % of influence on the second flight then it has on the first,
even thought it doesnt changes its strength. Just -/+
First flight was 1/6 = 16.6% faster then normal. 12/72. And the second flight was 25% slower then average. 18/72.
But nevermind, I will just let that hole stay in my head.
amnesia:
No, it is the same basic percentage of 16 2/3%. The reason is very simple. On the outgoing flight, you have: 72 - 60 =12 minutes, or a reduction in time of 16 2/3%. On the returning flight, you have:90 - 72 =18 minutes delay, which is 50% greater(18/12 =1.5) than the reduction time of 16 2/3%. In other words, 16 2/3% x 1.5 =25% of 72 minutes =18 minutes.
+129852 See if this makes sense, amnesia :
With the wind.....the plane covers (D/60) every minute and against the wind, the plane covers (D/90) every minute.....where D is the total disance beween the cities
So......the average rate per minute [ assuming no wind ] is given by :
[D / 60 + D / 90] / 2 =
[90D + 60D] / [ 60 * 90 *2] =
[150D] / [ 10800] =
[15D ] / 1080] =
D/72 ...... means that the plane would make the flight in a normal time of 72 minutes witn no wind
Thus....the plane is 90 − 72 = 18 min late
