Mentallic said:
Homework Statement
A particular type of tyre lasts 10,000km on a front wheel or 20,000km on a rear wheel. By interchanging the front and rear tyres, what is the greatest distance that can be traveled by a set of four of these tyres?
Well, the idea of the problem is that. When in the front, the wheel will be
damaged twice as fast as it's in the rear. We're considering a car (or some means of transportation) has 4 wheels. And it cannot move, when there are at least one wheel damaged.
So, here's my approach, it's pretty much like a guess, a mathematical guess.
Someone may come up with a better solution.
Let d_{i_1} , \ d_{i_2} correspondingly be the distance traveled by the wheel
i (in
km) when it's on the front, and on the rear, until one of the wheels is broken.
Because all 4 wheels must cover the same distance, so we'll have:
d_{1_1} + \ d_{1_2} = d_{2_1} + \ d_{2_2} = d_{3_1} + \ d_{3_2} = d_{4_1} + \ d_{4_2}
(1)
And, because the front wheels cover the same distance as the rear wheels, so we'll also have:
d_{1_1} + \ d_{2_1} + d_{3_1} + \ d_{4_1} = d_{1_2} + \ d_{2_2} + d_{3_2} + \ d_{4_2}
(2)
Now, assume that wheel 1 is the first wheel that is broken. So, that means:
2 d_{1_1} + d_{1_2} = 20000
(3) (you can get this equation by noticing that the wheel is damaged twice as fast when it's in the front)
So the distanced traveled until the wheel 1 is broken is: d_{1_1} + d_{1_2}
(*)
From
(1), we'll have:
\sum_{i = 1} ^ {4} \left( d_{i_1} + d_{i_2} \right) = 4 \times (d_{1_1} + \ d_{1_2})
Using
(2), the above equation boils down to:
2 \sum_{i = 1} ^ {4} d_{i_2} = 4 \times (d_{1_1} + \ d_{1_2})
\Rightarrow \sum_{i = 1} ^ {4} d_{i_2} = 2 \times (d_{1_1} + \ d_{1_2}) = ( 2 d_{1_1} + d_{1_2} ) + d_{1_2}
\Rightarrow \sum_{i = 1} ^ {4} d_{i_2} = 20000 + d_{1_2} (By using
(3))
\Rightarrow \sum_{i = 2} ^ {4} d_{i_2} = 20000
(4)
Since we have assumed that the wheel 1 is the first wheel to be broken. So, that means:
2 d_{i_1} + d{i_2} \leq 20000 , \forall i = \{ 2, 3, 4 \}
(5)
From
(5), we have:
\sum_{i = 2} ^ {4} ( 2 d_{i_1} + d_{i_2} ) \leq 60000
\Rightarrow 2 \sum_{i = 2} ^ {4} d_{i_1} + \sum_{i = 2} ^ {4} d_{i_2} \leq 60000
\Rightarrow 2 \sum_{i = 2} ^ {4} d_{i_1} \leq 40000 (using
(4))
\Rightarrow \sum_{i = 2} ^ {4} d_{i_1} \leq 20000
(6)
From
(4), and
(6), we can obtain:
3 ( d_{1_1} + d_{1_2} ) = \sum_{i = 2} ^ {4} \left( d_{i_1} + d_{i_2} \right) \leq 40000
\Rightarrow d_{1_1} + d_{1_2} \leq \frac{40000}{3}.
Ok, now, from the equation above what can possibly be the furthest distance that vehicle can travel? So, how can you switch the 4 wheels so that the furthest distance can be achieved? Can you go from here? :)
Oh well, I hope I'm being clear enough.. Not feeling very well today. :( So just shout if there's some step you don't understand. :)
Hopefully, someone'll come up with a better solution. Mine is pretty messy.. :(
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EDIT:
On reading the problem again, I'm pretty unsure on how many wheels this vehicle has. If it has 2 wheels, you can solve the problem, pretty much in the same manner. :)
I was also expecting an answer less than the average and VietDao's answer seems to support my gut feeling.
