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Aug 7th, 2001, 02:31 AM
#1
Thread Starter
Conquistador
Not too hard: a maths problem
B1 and B2 went to visit each other. They left their houses at the same time. When they met on the way, they forgot they wanted to see each other and kept on walking. B1 reached B2's house 8 minutes after the meeting and B2 got to B1's house 18 minutes after the meeting. How long did it take B2 to reach B1's house from the time he left his own house?
...
Who else can work it out...
Kedaman? gooddreams?
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Aug 7th, 2001, 03:33 AM
#2
me being the simpleton that i am, i'm gonna say 26mins
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Aug 7th, 2001, 11:39 AM
#3
Frenzied Member
Is there enough data?
There does not seem to be enough data to solve this problem. I think for a given total time for the B1 trip, there is a fixed total time for the B2 trip.
When I assumed a B1 total time of 20 minutes, I got a B2 total time of 30 minutes not matter what B1 velocity I assumed. This agrees with the Kedaman answer.
When I assumed a B1 total time of 26 minutes, I got 26 miniutes for the total B2 time. this agrees with the Nullus answer.
When I assumed a B1 total time of 32 minutes, I got a B2 total time of 24 minutes.
Live long & prosper.
The Dinosaur from prehistoric era prior to computers.
Eschew obfuscation!
If a billion people believe a foolish idea, it is still a foolish idea!
VB.net 2010 Express
64Bit & 32Bit Windows 7 & Windows XP. I run 4 operating systems on a single PC.
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Aug 7th, 2001, 11:47 AM
#4
Hyperactive Member
30 mins by my calculations so keda is right. not a bad little puzzle though silv.
There are 10 types of people in the world - those that understand binary, and those that don't.
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Aug 7th, 2001, 12:54 PM
#5
Frenzied Member
Error.
Kedaman is right. I discovered an inconsistency in my calculations.
There seems to be enough data to solve this problem.
Live long & prosper.
The Dinosaur from prehistoric era prior to computers.
Eschew obfuscation!
If a billion people believe a foolish idea, it is still a foolish idea!
VB.net 2010 Express
64Bit & 32Bit Windows 7 & Windows XP. I run 4 operating systems on a single PC.
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Aug 7th, 2001, 03:57 PM
#6
transcendental analytic
Yep
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Aug 8th, 2001, 02:01 AM
#7
Thread Starter
Conquistador
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Aug 8th, 2001, 06:32 AM
#8
Frenzied Member
Originally posted by nullus
me being the simpleton that i am, i'm gonna say 26mins
That was my first thought too...but then i figured why would anyone post such a stupid question if the answer is that simple...how did u people figure that out?
You just proved that sig advertisements work.
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Aug 8th, 2001, 07:25 AM
#9
transcendental analytic
Code:
B1:
time: x 8
|--------|---------|
pos: S1 S2
B2:
time: x 18
|--------|-------------------|
pos: S2 S1
Speed:
B1=S2/8=S1/x
B2=S2/x=S1/18
S2/S1=8/x
S2/S1=x/18
x^2=8*18
x=12
total=12+18=30
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Aug 8th, 2001, 03:33 PM
#10
Fanatic Member
Kedaman's approach is the way to do it. However you can also do it this way too. I just assumed 10 km betweem B1 and B2.
VB Code:
Private Sub Command1_Click()
'Assumed 10 km between B1 and B2
Dim KmPerHourB1&, KmPerHourB2&, MinToMeet@
Dim KMLeftB1@, MinAfterMeetB1@, KMLeftB2@, MinAfterMeetB2@
Dim KMAfterMeetingB2@, TimeBeforeMeetingB2@
For KmPerHourB1 = 70 To 20 Step -1
For KmPerHourB2 = 70 To 20 Step -1
MinToMeet = 10 * 60 / (KmPerHourB1 + KmPerHourB2)
KMLeftB1 = 10 - MinToMeet / 60 * KmPerHourB1
MinAfterMeetB1 = KMLeftB1 / (KmPerHourB1 / 60)
KMLeftB2 = 10 - MinToMeet / 60 * KmPerHourB2
MinAfterMeetB2 = KMLeftB2 / (KmPerHourB2 / 60)
If MinAfterMeetB1 = 8 And MinAfterMeetB2 = 18 Then
Debug.Print "B1 kph: " & KmPerHourB1; " B2 kph: " & KmPerHourB2
KMAfterMeetingB2 = 18@ / 60 * KmPerHourB2
Debug.Print "B2 Travelled this many km after meeting B1: " & KMAfterMeetingB2
TimeBeforeMeetingB2 = (10 - KMAfterMeetingB2) / (KmPerHourB2 / 60)
Debug.Print "B2 travelled for this many minutes before meeting B1: " & TimeBeforeMeetingB2
Debug.Print "Total Time B2: " & 18@ + TimeBeforeMeetingB2
End If
Next KmPerHourB2
Next KmPerHourB1
End Sub
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