# Furthest distance travelled?

1. Aug 5, 2009

### Mentallic

1. The problem statement, all variables and given/known data
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 travelled by a set of four of these tyres?

3. The attempt at a solution
I couldn't think of a way to solve this logically, so I attempted to use algebra to solve the problem and after lots of guessing at created equations that suits the problem, this is the best I was able to achieve...

let $$w_1=d-10000$$ and $$w_2=d-20000$$

w being the distance left in each tyre, and d being the distance already travelled.
Since the distance travelled must be equal, the variables d but must be equal.

Thus, $$w_2-w_1=10000$$

sweet... any help is appreciated

2. Aug 5, 2009

### VietDao29

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 travelled untill 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.. :(

-------------------

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. :)

Last edited: Aug 5, 2009
3. Aug 5, 2009

### Дьявол

I guess if there are 4 tires (2 front, 2 rear) and the car is balanced, the front and rear tires are less exposed so that the weight of the car is shared by all of them on equal parts.

Anyway, I do not understand the task completely. Is there any lack of information?

Are 10.000km (front) and 20.000km(rear) when the tires are already attached on the car, or separately?

Is the car back draft or front draft?

There are so many factors here.

Regards.

4. Aug 6, 2009

### Mentallic

Дьявол forget about any engineering aspects of the wear and tear a tyre experiences when fixed to a vehicle. There is enough information, all that needs to be known is that when the vehicle (lets consider it to be a bike just to simply have 1 front and back wheel) is in motion, its front tyre will be able to survive for 10000km before needing to be replaced while the back wheel would last for 20,000km. Now if you get a new set of tyres for both wheels and just start riding off, after 10,000km the front wheel will give way and the bike is useless, even though the back tyre has another 10,000km to give before deteriorating. Switching the front and back tyre around before this time will allow the initial back tyre to begin to deteriorate faster by being on the front wheel. Eventually it is possible to travel further than 10,000km but less than 20,000km.
Does this make more sense?

Sorry I can't follow what you've done vietdao, but I appreciate your efforts
Could this by any chance be the answer though?

$$\approx 13,300km$$ for the max distance seems pretty reasonable.

5. Aug 6, 2009

### symbolipoint

This feels like a series problem, but instead, if thought about in an overly simple way, imagine if the tires could be rotated between front and rear in infinitesimally small time increments. The fronts would wear the tires toward 10000 miles while the rears would wear the tires 20000 miles. The average is 15000 miles. Is this too simple to be sensible?

6. Aug 6, 2009

### VietDao29

There's no word in the problem that says that it's a bike. The "a set of four of these tyres" part makes me think that it's a car.

Yup, that's the answer. But btw, I wonder which part you don't understand.. :(

The main equations are (1), (2), and (3). Then, we just mess around with those, until we arrive at the final answer.

7. Aug 6, 2009

### Mentallic

I was also thinking it would eventually need to be converted into an infinite sum, but after a bit more pondering, I'm beginning to think otherwise. I believe there would only need to be one switch made in the entire journey to purposely let the front tyres wear down enough so that when they switch to the back, they end up degrading at the slower pace, allowing the now front tyres to catch up and finally break down with the back tyres.
This would mean both tyres would need to be on each wheel for an equal distance travelled.
Does this shed any light on the problem?

Oh and yes I think using logic and thus taking the average is too simplistic and very unmathematical :tongue: I was also expecting an answer less than the average and VietDao's answer seems to support my gut feeling.

Its pretty obvious though that it doesn't make a difference if it is a set of four wheels or two wheels. The only significance is the front and back wheel and each tyre that fits in with those. Thus, two tyres are sufficient.

Oh well I don't understand your solution because I haven't used sums all that much. I cannot seem to figure out what the notation represents and such and it is also pretty difficult to follow what each term means. e.g. $$d_i_1,d_i_2,d_1_1,d_1_2$$
it's probably really obvious to some, but I just don't see it.

Also, this problem was from a practice test, so I'd like a quicker or less complicated solution?

8. Aug 6, 2009

### VietDao29

Well, using the Capital Sigma notation makes the text/formulae clearer, and much shorter. Instead of writing a line full with numbers, one can shorten it with only a few symbols. So, if you're interested, you can have a quick glance http://en.wikipedia.org/wiki/Summation#Capital-sigma_notation". It's pretty exciting to learn something new, each day, huh.

Well yeah, actually, I'm not very satisfied with my own solution. It seems kind of messy. I'll have another look at it this evening. And hopefully some other HHs will pop-in, and give us some new approach.

Last edited by a moderator: Apr 24, 2017
9. Aug 6, 2009

### Дьявол

I think I understand now.

I got one pattern.

min(20000-x) and max(10000-x)

start: 20000km 10000km

----6000km has passed----

14000km 4000km

-----switch-----

8000km 7000km

----6500km has passed----

1500km 500km

----switch-----

1000km 750 km

----750 km has passed----

250km 0 km

As you can see I came up with 250km left which is not the optimal solution.

And the optimal solution is when you can came up with 0km 0km so both tires are completely used.

But as you can see to find the optimal solution we need equal damage on both tires, so that when tire1 km and tire2 km is tire1=tire2, we can subtract the rest damage of both of the tires and come up with the final solution.

As I said we need

$$\frac{20000-x}{2}=2*(10000-x)$$

By finding x you find the tires switched after x km and with equal damage (when switched).

Let me check it.

x=6666 km

So

20000-6666=13334 km and 10000-6666=3334 km

Now lets switch the tires.

6666 km and 6666 km.

As we can see, now they have equal damage, so that,

6666 has passed, and after the switch there are 6666 to go, and that is 2*6666=13332km (the final solution)

Regards.

10. Aug 6, 2009

### Dick

It's sort of right, but you are averaging the wrong thing. Optimal use of the tires means only that each one spends an equal distance on the front and back. It doesn't really matter how you accomplish the rotation. But what you want to average is the wear RATES, not the mileage. I.e. average 1/10000 and 1/20000 and get 3/40000 (rate units tire/mi). So the maximum distance satisfies 3*x/40000=1 or x=40000/3=13333.3..., as has already been worked out for a specific rotation scheme, where you switch the tires once after 6666.6... miles.

Last edited: Aug 6, 2009