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I just bought a car with $5$ identical tires on it, as you guess $4$ of them were already installed, just one is in the trunk.

The seller said that while the tires on the front, they can go $48k$ miles at most, whereas while on the rear, they can go $42k$ miles at most.

If you are allowed to change the tires as much as you want on the way,

1- What is the maximum distance you can go without damaging any tire?

If you are allowed to stop the car 3 times at most to change tires

2- What is the maximum distance you can go?

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  • $\begingroup$ Tires have equal amount of rubber and position determines wear? Or are position equal and tires have different amount of rubber? If so what is the tire in the rear. $\endgroup$ – paparazzo Apr 27 '18 at 11:08
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    $\begingroup$ @wearing is linear. tires are identical, car wears tire faster in the rear. $\endgroup$ – Oray Apr 27 '18 at 11:10
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Apr 29 '18 at 1:14
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For 1, if you keep rotating the tyres every mile or so, you get equal wear on each.

A kilomile eats 2/42 + 2/48 = 5/56 tyres, so 56 kilomiles would eat 5 tyres exactly

For 2, we have three changes, so one of the tyres won't get a turn in the trunk. Therefore, the answer cannot be more than

48 kilomiles.

So if we find a way to run each of the other tyres for that long, we are good.

Turns out it's quite possible:

Keeping one tyre in the front position, and rotating the other four every 12 kilomiles, each of the four tyres accumulates 24 kilomiles in the rear, 12 kilomiles in the front, and 12 kilomiles in the trunk. That makes for a total of 24/42 + 12/48 = 23/28 wear on each of the four tyres.

That solution has the added perk that you can continue on the remaining four tyres after the other one goes. If you don't care about that, here's another way:

After the first 6 kilomiles, swap the left rear tyre with the one in the trunk. After another 6, repeat with the right rear tyre. At that point, you have driven 12 kilomiles and have 36 kilomiles left in all 4 installed tyres.

The benefit of this other method is having balanced wear in the front all the time, and, except for the bit from 6 kilomiles to 12 kilomiles, in the rear as well. Also, after the first 12 kilomiles, the better pair will be the rear one, which is the recommended way of installing an unevenly worn set of old tyres. Since we are going to wear the tyres completely out anyway (not really recommended), this at least improves the survivability of the journey as much as possible.

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    $\begingroup$ Errr...what's a kilomile? Sorry, I'm from the US $\endgroup$ – NL628 Apr 21 '18 at 21:01
  • $\begingroup$ Sorry, I'm not from the US, so I assumed the "48k miles" would behave similarly to some other, saner, units of length. By a kilomile I of course mean "a 1000 miles"; for 1024 miles one should use "kibimile". ~~~ $\endgroup$ – Bass Apr 21 '18 at 21:33
  • $\begingroup$ ahhh nice :P lolol $\endgroup$ – NL628 Apr 21 '18 at 21:34
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For 1: assuming the wear is linear, we have 5 'full' tires and

every 1000 miles we deplete 1/42 + 1/42 + 1/48 + 1/48 = 30/336 of the tires.

That means

we can go 5 * 336 / 30 = 56k miles with them.

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  • $\begingroup$ The per mile is not correct when a spare is in the front $\endgroup$ – paparazzo Apr 27 '18 at 8:49
  • $\begingroup$ The spare isn't really different from the other wheels, though. $\endgroup$ – Glorfindel Apr 27 '18 at 10:24
  • $\begingroup$ Font and back are different so spare cannot be the same as other wheels. $\endgroup$ – paparazzo Apr 27 '18 at 10:57
  • $\begingroup$ I assume the wheel which isn't used does not wear. Remember that the problem states the wheels are identical; the wear is only dependent on the position. $\endgroup$ – Glorfindel Apr 27 '18 at 11:03
  • $\begingroup$ OK I can see that interpretation. I read it as the tire not position. $\endgroup$ – paparazzo Apr 27 '18 at 11:06
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Spare is same as front or rear? I am assuming same mileage for spare as rear (48) as trunk is in the rear.

As many changes as you want

You have 42 + 42 + 48 + 48 + 48 = 228
Divided by 4 is exactly 57 k miles
If the spare is 42 then 55.50

Only 3 changes

.
3 changes is 4 intervals
problem is one tire will stay on the car the whole time so 48 k miles
2 changes is same as 3
go 10 k spare in trunk
go 10 k spare front left
go 28 k spare front right

The more interesting question is 4 changes

In that case can get the optimal Each spare is in the trunk for 15 k miles
Each 48 k tire is in the turn 9 k miles 2 * 15 + 3 * 9 = 57

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    $\begingroup$ I... I don't think you should be doing those things to maths and logic. $\endgroup$ – Bass Apr 27 '18 at 9:42

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