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Second hint. All you need are the equations given, and the two deducible facts:
1 None of the speeds can be less than zero
2 The third speed s3 is less than the first two speeds s1 & s2, so we can conclude that s3<(s1+s2)/2, because if it was not then it would be equal to or greater than one of them.
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The answer
Sorry, there is no easy way of hiding a multiple line post.
The first journey takes 1 hour and is driven at two speeds s1 and s2
We are given an equation for the average speed of this journey
A =(s1+ s2)/2
Since the time is 1 hour, if our speeds are in mph then
Distance in miles = D = (s1 + s2)/2
(We can also work out that the times driven at each speed must be equal, but that doesn’t help us)
For the second journey we are given
A2 = (s1 + s2 + s3)/3
Assuming the time for this journey is T we have another equation for distance travelled
D = T(s1 + s2 + s3)/3
So T(s1 + s2 + s3)/3 = (s1 + s2) /2
Rearranging
T = 3/2 * (s1 + s2)/(s1 + s2 + s3)
Assuming all speeds are positive, the minimum T happens with maximum s3, and the maximum T happens with minimum S3
For the minimum s3, all we can assume is s3>0, this gives
maxT < 3/2 * (s1 + s2)/(s1 + s2 + 0)
Max t < 3/2
For the minimum s3, since s3 is lower than s1 and s2, s3 MUST be less than the average of s1 and s2, whatever their values are.
s3 < (s1 + s2)/2
So
MinT > 3/2 * (s1 +s2)/(s1 + s2 + (s1 + s2)/2)
This rearranges to
MinT > 1
So my wife’s second journey must have taken longer than 1 hour, but less than 1 hour and 30 minutes.
Originally posted by @iamatigerInteresting stuff.The Algebra is a bit too complicated for me but I can just about see how it works.
The answer
Sorry, there is no easy way of hiding a multiple line post.
The first journey takes 1 hour and is driven at two speeds s1 and s2
We are given an equation for the average speed of this journey
A =(s1+ s2)/2
Since the time is 1 hour, if our speeds are in mph then
Distance in miles = D = (s1 + s2)/2
(We can also work out that the times ...[text shortened]... wife’s second journey must have taken longer than 1 hour, but less than 1 hour and 30 minutes.
I've got a book somewhere with some algebra problems that are way beyond my ability(I'm virtually self taught ,I couldn't be bothered at school and failed Maths).I only became interested in later life when I got this book.
Would you like me to find it and post one of the problems?
I assure you it'll be more challenging than the bloke filling a bath with a hole in it type of problem.The answers are in the back so I can provide an answer but I wouldn't be able to show you how to get to it!!
Originally posted by @iamatigerRight then.I remember spending hours on this one , but never worked out how to do it.Here goes:-
Sounds challenging!
An army 25 miles long starts on a journey of 50 miles just an an orderly at the rear starts to deliver a message to the General at the front.The orderly, travelling at uniform speed delivers his message and returns to the rear ,arriving there just as the army finishes the journey.
Q- How far does the orderly travel?
I remember thinking at the time that the question was a bit vague.Does it mean when the General at the front finishes , or when the orderly at the back finishes(assuming they are walking in one long column).?
Anyway,as i said, I know the answer but even with that knowledge I never worked out how it was arrived at.
Good luck.
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Looks like the back of the army has to travel 75 miles
Assume the army travels at speed a, and the messenger travels at ka
Army total time = 75/a
Messenger time to front = 25/(ka-a)
Messanger time to back = 25/(ka+a)
Messenger total time = (25/a)(1/(k-1)+1/(k+1))
Messenger time = army time so
75/a = (25/a)(1/(k-1)+1/(k+1))
Divide by 25/a
3 = 1/(k-1)+1/(k+1)
Multiply by (k-1)(k+1)= k^2 -1
3(k^2-1)= 2k
k^2 - 2k/3 = 1
k^2 - 2k/3 + 1/9 = 10/9
(k-1/3)^2 = 10/9
k=sqrt(10)/3 + 1/3 = (1+sqrt(10))/3
Orderly distance = orderly speed x time = ka*75/a=75a
Orderly distance = 25*(1+sqrt(10))=104.0569 miles
Originally posted by @iamatigerThe answer in the book isn't that far but as I said, the question is a bit vague and the book doesn't give any working out.I'll let you think about it and maybe make other assumptions and then give the quoted answer when you(or someone else) has had another go.
Looks like the back of the army has to travel 75 miles
Assume the army travels at speed a, and the messenger travels at ka
Army total time = 75/a
Messenger time to front = 25/(ka-a)
Messanger time to back = 25/(ka+a)
Messenger total time = (25/a)(1/(k-1)+1/(k+1))
Messenger time = army time so
75/a = (25/a)(1/(k-1)+1/(k+1))
Divide by 25/a
3 = 1/( ...[text shortened]... distance = orderly speed x time = ka*75/a=75a
Orderly distance = 25*(1+sqrt(10))=104.0569 miles
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Ok, let's assume that the army stops when the general has gone 50 miles.
This changes the calcs as follows:
Assume the army travels at speed a, and the messenger travels at ka
Army total time = 50/a
Messenger time to front = 25/(ka-a)
Messanger time to back = 25/(ka+a)
Messenger total time = (25/a)(1/(k-1)+1/(k+1))
Messenger time = army time so
50/a = (25/a)(1/(k-1)+1/(k+1))
Divide by 25/a
2 = 1/(k-1)+1/(k+1)
Multiply by (k-1)(k+1)= k^2 -1
2(k^2-1)= 2k
k^2 - k = 1
k^2 - k + 1/4 = 5/4
(k-1/2)^2 = 5/4
k=sqrt(5)/2 + 1/2 = (1+sqrt(5))/2
Orderly distance = orderly speed x time = ka*50/a=50k
Orderly distance = 25*(1+sqrt(5))= 80.9017 miles
is that right?
Originally posted by @iamatigerIt is!!.
Ok, let's assume that the army stops when the general has gone 50 miles.
This changes the calcs as follows:
Assume the army travels at speed a, and the messenger travels at ka
Army total time = 50/a
Messenger time to front = 25/(ka-a)
Messanger time to back = 25/(ka+a)
Messenger total time = (25/a)(1/(k-1)+1/(k+1))
Messenger time = army time ...[text shortened]... ly speed x time = ka*50/a=50k
Orderly distance = 25*(1+sqrt(5))= 80.9017 miles
is that right?
Well done!!
Your Algebraic talents are impressive.
I'll study your solution in detail when I have a bit of time and see what I can learn.
Perhaps when I'm trying to fill a bath with the plug out!!
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Originally posted by @vendaThanks 😀
It is!!.
Well done!!
Your Algebraic talents are impressive.
I'll study your solution in detail when I have a bit of time and see what I can learn.
Perhaps when I'm trying to fill a bath with the plug out!!
Got any other tricky ones?
I think the key was trying to solve for messenger_speed = k*army_speed.
At first I tried messenger_speed = k+army_speed
But there is no solution for k then, because the army speed does not cancel out.