Mixing
3 Towns and a Man problem
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
A, B, C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, c minutes respectively. If he starts from B he takes $a + c - b$ hours, if he starts from C, he takes $b + a - c$ hours, and if he starts from A he takes $c + b - a$ hours. Find the length of the circuit.
I did the following in an attempt to solve it:
Let $d = AB + BC + CA$, or the perimeter of the triangle.
Then, we have the rates in miles per minute:
$$fracd60(a + c - b)$$
$$fracd60(b + a - c)$$
$$fracd60(c + b - a)$$
At this point, I wasn't sure what else I could do, because in order to find the distance, I need to know the average rate from each starting point.
The only other thing I noticed is that the sum of the hours it takes from each starting point is $a + b + c$. However, I don't think this property seems relevant right now.
If anyone can help me deduce the rates or find another approach, then I can carry out the problem from there.
Thanks.
algebra-precalculus word-problem
asked Jul 21 at 21:41
A Silent Cat
1227
add a comment |Â
up vote
1
down vote
favorite
A, B, C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, c minutes respectively. If he starts from B he takes $a + c - b$ hours, if he starts from C, he takes $b + a - c$ hours, and if he starts from A he takes $c + b - a$ hours. Find the length of the circuit.
I did the following in an attempt to solve it:
Let $d = AB + BC + CA$, or the perimeter of the triangle.
Then, we have the rates in miles per minute:
$$fracd60(a + c - b)$$
$$fracd60(b + a - c)$$
$$fracd60(c + b - a)$$
At this point, I wasn't sure what else I could do, because in order to find the distance, I need to know the average rate from each starting point.
The only other thing I noticed is that the sum of the hours it takes from each starting point is $a + b + c$. However, I don't think this property seems relevant right now.
If anyone can help me deduce the rates or find another approach, then I can carry out the problem from there.
Thanks.
algebra-precalculus word-problem
asked Jul 21 at 21:41
A Silent Cat
1227
I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A, B, C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, c minutes respectively. If he starts from B he takes $a + c - b$ hours, if he starts from C, he takes $b + a - c$ hours, and if he starts from A he takes $c + b - a$ hours. Find the length of the circuit.
I did the following in an attempt to solve it:
Let $d = AB + BC + CA$, or the perimeter of the triangle.
Then, we have the rates in miles per minute:
$$fracd60(a + c - b)$$
$$fracd60(b + a - c)$$
$$fracd60(c + b - a)$$
At this point, I wasn't sure what else I could do, because in order to find the distance, I need to know the average rate from each starting point.
The only other thing I noticed is that the sum of the hours it takes from each starting point is $a + b + c$. However, I don't think this property seems relevant right now.
If anyone can help me deduce the rates or find another approach, then I can carry out the problem from there.
Thanks.
algebra-precalculus word-problem
asked Jul 21 at 21:41
A Silent Cat
1227
A, B, C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, c minutes respectively. If he starts from B he takes $a + c - b$ hours, if he starts from C, he takes $b + a - c$ hours, and if he starts from A he takes $c + b - a$ hours. Find the length of the circuit.
I did the following in an attempt to solve it:
Let $d = AB + BC + CA$, or the perimeter of the triangle.
Then, we have the rates in miles per minute:
$$fracd60(a + c - b)$$
$$fracd60(b + a - c)$$
$$fracd60(c + b - a)$$
At this point, I wasn't sure what else I could do, because in order to find the distance, I need to know the average rate from each starting point.
The only other thing I noticed is that the sum of the hours it takes from each starting point is $a + b + c$. However, I don't think this property seems relevant right now.
If anyone can help me deduce the rates or find another approach, then I can carry out the problem from there.
Thanks.
algebra-precalculus word-problem
asked Jul 21 at 21:41
A Silent Cat
1227
asked Jul 21 at 21:41
A Silent Cat
1227
asked Jul 21 at 21:41
A Silent Cat
1227
asked Jul 21 at 21:41
A Silent Cat
1227
1227
I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38
add a comment |Â
I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38
I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns. Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$ hours. Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\
xb+ yc+za=60(a+c-b)\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$. If we add these together we get
$$(x+y+z)left(a+b+cright)=60(a+b+c)\
x+y+z=60$$
answered Jul 21 at 22:32
Ross Millikan
276k21186352
add a comment |Â
up vote
1
down vote
You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
answered Jul 21 at 22:21
amd
25.9k2943
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
add a comment |Â
up vote
0
down vote
Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$frac3a + b + c = frac3d60(a + b + c) = fracd20(a + b + c)$$
Through basic algebra, we can then see that $d = 60$ miles.
answered Jul 21 at 22:40
A Silent Cat
1227
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
add a comment |Â
up vote
0
down vote
Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $frac Pc $ minutes
Equating the total time for the three trips ...
$$frac Pa +frac Pb +frac Pc =60(a+b+c)
\Pleft(fracbc+ac+ababcright) =60(a+b+c)
\P=frac60abc(a+b+c)bc+ac+ab$$
*EDIT (rates given in minutes per mile - thanks @amd )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\P=60$$
answered Jul 21 at 22:39
WW1
6,4571712
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns. Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$ hours. Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\
xb+ yc+za=60(a+c-b)\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$. If we add these together we get
$$(x+y+z)left(a+b+cright)=60(a+b+c)\
x+y+z=60$$
answered Jul 21 at 22:32
Ross Millikan
276k21186352
add a comment |Â
up vote
3
down vote
accepted
Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns. Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$ hours. Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\
xb+ yc+za=60(a+c-b)\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$. If we add these together we get
$$(x+y+z)left(a+b+cright)=60(a+b+c)\
x+y+z=60$$
answered Jul 21 at 22:32
Ross Millikan
276k21186352
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns. Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$ hours. Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\
xb+ yc+za=60(a+c-b)\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$. If we add these together we get
$$(x+y+z)left(a+b+cright)=60(a+b+c)\
x+y+z=60$$
answered Jul 21 at 22:32
Ross Millikan
276k21186352
Let $AB=x, BC=y, CA=z$ be the distances in miles between the towns. Then if he starts from $A$ he takes $xa+yb+ zc$ minutes, which we are told is $(c+b-a)$ hours. Using the other two as well we get $$xa+yb+ zc=60(c+b-a)\
xb+ yc+za=60(a+c-b)\
xc+ ya+ zb=60(b+a-c)$$
and we are asked to find $x+y+z$. If we add these together we get
$$(x+y+z)left(a+b+cright)=60(a+b+c)\
x+y+z=60$$
answered Jul 21 at 22:32
Ross Millikan
276k21186352
edited Jul 21 at 22:38
answered Jul 21 at 22:32
Ross Millikan
276k21186352
answered Jul 21 at 22:32
Ross Millikan
276k21186352
answered Jul 21 at 22:32
Ross Millikan
276k21186352
276k21186352
add a comment |Â
add a comment |Â
up vote
1
down vote
You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
answered Jul 21 at 22:21
amd
25.9k2943
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
add a comment |Â
up vote
1
down vote
You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
answered Jul 21 at 22:21
amd
25.9k2943
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
answered Jul 21 at 22:21
amd
25.9k2943
You’re on the right track. I would suggest keeping the three distances separate at first, so that your equations are $aBC+bCA+cAB = 60(a+c-b)$ and so on. If you add up the three equations, the left-hand side will be some multiple of $BC+CA+AB$ and the right-hand side looks like it will cancel nicely.
Looking at it from a different angle, the man effectively travels the circuit thrice: once walking, once riding, once driving. The total time for this is (a+b+c)d, and this is equal to the sum of the given times.
answered Jul 21 at 22:21
amd
25.9k2943
edited Jul 21 at 22:37
answered Jul 21 at 22:21
amd
25.9k2943
answered Jul 21 at 22:21
amd
25.9k2943
answered Jul 21 at 22:21
amd
25.9k2943
25.9k2943
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
add a comment |Â
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
@RossMillikan They’re trying to be tricky. The rates are given in minutes per mile.
– amd
Jul 21 at 22:34
add a comment |Â
up vote
0
down vote
Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$frac3a + b + c = frac3d60(a + b + c) = fracd20(a + b + c)$$
Through basic algebra, we can then see that $d = 60$ miles.
answered Jul 21 at 22:40
A Silent Cat
1227
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
add a comment |Â
up vote
0
down vote
Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$frac3a + b + c = frac3d60(a + b + c) = fracd20(a + b + c)$$
Through basic algebra, we can then see that $d = 60$ miles.
answered Jul 21 at 22:40
A Silent Cat
1227
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$frac3a + b + c = frac3d60(a + b + c) = fracd20(a + b + c)$$
Through basic algebra, we can then see that $d = 60$ miles.
answered Jul 21 at 22:40
A Silent Cat
1227
Alright, so I think I have an answer based upon what everyone has said until now:
In $a + b + c$ minutes, the man covers 3 miles.
Now, if he starts from each of those points and completes a circuit, of $d$ length, then he will cover $3d$ miles. His average time completing the circuit 3 times is equivalent to $60(a + c - b) + 60(b + a - c) + 60(c + b - a)$ minutes which is $60(a + b + c)$ minutes.
Therefore, the following ratios can be stated equal:
$$frac3a + b + c = frac3d60(a + b + c) = fracd20(a + b + c)$$
Through basic algebra, we can then see that $d = 60$ miles.
answered Jul 21 at 22:40
A Silent Cat
1227
answered Jul 21 at 22:40
A Silent Cat
1227
answered Jul 21 at 22:40
A Silent Cat
1227
answered Jul 21 at 22:40
A Silent Cat
1227
1227
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
add a comment |Â
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
Though I do like this solution, I think the one by Ross Millikan is more elegant.
– A Silent Cat
Jul 21 at 22:44
add a comment |Â
up vote
0
down vote
Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $frac Pc $ minutes
Equating the total time for the three trips ...
$$frac Pa +frac Pb +frac Pc =60(a+b+c)
\Pleft(fracbc+ac+ababcright) =60(a+b+c)
\P=frac60abc(a+b+c)bc+ac+ab$$
*EDIT (rates given in minutes per mile - thanks @amd )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\P=60$$
answered Jul 21 at 22:39
WW1
6,4571712
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
add a comment |Â
up vote
0
down vote
Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $frac Pc $ minutes
Equating the total time for the three trips ...
$$frac Pa +frac Pb +frac Pc =60(a+b+c)
\Pleft(fracbc+ac+ababcright) =60(a+b+c)
\P=frac60abc(a+b+c)bc+ac+ab$$
*EDIT (rates given in minutes per mile - thanks @amd )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\P=60$$
answered Jul 21 at 22:39
WW1
6,4571712
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $frac Pc $ minutes
Equating the total time for the three trips ...
$$frac Pa +frac Pb +frac Pc =60(a+b+c)
\Pleft(fracbc+ac+ababcright) =60(a+b+c)
\P=frac60abc(a+b+c)bc+ac+ab$$
*EDIT (rates given in minutes per mile - thanks @amd )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\P=60$$
answered Jul 21 at 22:39
WW1
6,4571712
Imagine taking three trips , one from each vertex, the total time for the three trips is $60(a+b+c)$ minutes
In the process of doing this you will end up travelling the entire perimeter using each mode of transportation
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $frac Pc $ minutes
Equating the total time for the three trips ...
$$frac Pa +frac Pb +frac Pc =60(a+b+c)
\Pleft(fracbc+ac+ababcright) =60(a+b+c)
\P=frac60abc(a+b+c)bc+ac+ab$$
*EDIT (rates given in minutes per mile - thanks @amd )
Let $P$ represent the perimeter of the triangle in miles, so the time spent driving is $Pc$ minutes
Equating the total time for the three trips ...
$$ P(a+b+c) =60(a+b+c)
\P=60$$
answered Jul 21 at 22:39
WW1
6,4571712
edited Jul 21 at 22:43
answered Jul 21 at 22:39
WW1
6,4571712
answered Jul 21 at 22:39
WW1
6,4571712
answered Jul 21 at 22:39
WW1
6,4571712
6,4571712
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
add a comment |Â
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
The rates are given in minutes per mile, not miles per minute.
– amd
Jul 21 at 22:40
add a comment |Â
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I am seeing equations of the form $frac BCa+frac CAb+frac ABc=a+c-b$ . (ignoring minutes to hours conversion). Adding these three equations gives a fairly simple expression.
– lulu
Jul 21 at 22:30
@RossMillikan I got exactly what you got (well, I persist in confusing $60$ and $frac 160$, but never mind). I don't think you can go further.
– lulu
Jul 21 at 22:34
@lulu: amd pointed out the rates are minutes per mile, not miles per minute. I'll fix my answer.
– Ross Millikan
Jul 21 at 22:35
@RossMillikan Ah, entirely missed that. Certainly simplifies things.
– lulu
Jul 21 at 22:38