Math Quandary

[Dr. Bill Kobb]

The following was posed to me by a friend. Anyone know the solution?

On a 20,000 mile trip when five (5) tires are used equally, how many miles does each tire sustain.

Math is NOT my specialty. 😛

 

[Budweiserbob!]

Is this some sort of riddle? 😛

Does the car have four or five wheels?

Are they on the road at the same time or is the driver prone to doing wheelies?

Are these typical tyres or special everlasting ones? XD

 

[melanie2]

*head spinning out of control...*

Milagros would know this..

 

[jts963]

16,000 miles on each tire, if you rotate every 4,000 miles.

 

[Budweiserbob!]

See, that's assuming he swaps the tyres at constant times. >.>
The question was too flawed. 😛

EDIT: After re-reading the question, I see that's what he meant. In which case 16,000 is right.

 

[Frowg]

yes

16,000 miles on each tire, if you rotate every 4,000 miles.

this is correct

 

[ttheocjones]

I'd hate to disagree but the answer is 4,000 miles. The problems says all 5 tires are used equally. 20,000 divided by 5.

 

[Cokecan]

I'd hate to disagree but the answer is 4,000 miles. The problems says all 5 tires are used equally. 20,000 divided by 5.

Except 4 of the 5 tires have to be in use at all times...

 

[FeatherDaemon1]

Well you would just need to keep a running tab on the mileage for each tire and rotate accordingly. The answer would be 4000 miles per tire.

 

[jts963]

Look at it this way. You have a 4-wheeled car, 5 tires and a total distance traveled of 20,000 miles. Each tire spends 4,000 miles riding in the trunk as the spare, which means each tire spends the remaining 16,000 miles on the road.

 

[BigNorm]

4,000 miles each tire. simple as pie.

I'd hate to disagree but the answer is 4,000 miles. The problems says all 5 tires are used equally. 20,000 divided by 5.

your right. the question doesn't state it's a normal care, a 3 wheeler, whatever. it says 5 tires are being used in the trip.

 

[jts963]

4,000 miles each tire. simple as pie.

I'd hate to disagree but the answer is 4,000 miles. The problems says all 5 tires are used equally. 20,000 divided by 5.

your right. the question doesn't state it's a normal care, a 3 wheeler, whatever. it says 5 tires are being used in the trip.

If you have an x-wheeled vehicle and y spare tires with the x+y tires to be used equally over distance z, the distance sustained by each tire is

z*x/(x+y)

What we know for sure: z = 20,000 miles and x+y = 5.

So

for a 4-wheeler, the distance per tire is 20,000*4/(4+1) = 16,000 miles;

for a 3-wheeler, the distance per tire is 20,000*3/(3+2) = 12,000 miles.

 

[Budweiserbob!]

jts is right and guessing from the logic and notation he's used he's studied maths at a higher level before.

The 20,000 miles divided by 5 tyres calculation assumes that "Only one tyre is touching the road at any point in time". If the car only had one wheel, this would be true. But assuming that this is a typical, realistic car and that all four tyres are touching the road at any one point in time, we have to consider the collective distances travelled. This is where the rotation logic that jts suggested comes into play.

Let's assume:

Front Left Tyre = A
Front Right Tyre = B
Back Left Tyre = C
Back Right Tyre = D
Spare Tyre = E

Also assuming: Each tyre is rotated over an equal period of time.

Trip 1 - 4000 miles - ABCD (E)
Trip 2 - 4000 miles - BCDE (A)
Trip 3 - 4000 miles - CDEA (B)
Trip 4 - 4000 miles - DEAB (C)
Trip 5 - 4000 miles - EABC (D)

Tyre A travelled on Trips 1, 2, 3 and 4. Tyre B travelled on trips 1, 2, 4 and 5 ... and so on.

So on average, if all the assumptions are true, over a 20,000 mile period each tyre would travel 16,000 miles.

I'm 99% with jts on this one.

The only thing I disagree with is the last part of his latest post...

Although the question states there's 5 tyres, a three wheeler car like a Reliant Robin wouldn't have two spare tyres. There'd be no room. 😛

 

[Cbarton]

Hey, JTS..we finally agree on something. I concur with the answer of 16,000 miles

 

[Dr. Bill Kobb]

Wow! I get home tonight and see that you guys have really put your heads together on this thing. You all frigging RAWK! 😛aw:

 

[Budweiserbob!]

Ahem. Anno. I think he was talking about five tyres you have with a car, four in use and one spare. 😛

When he said "equally", he probably meant with equal time on the road. So Time A = Time B = Time C = Time D = Time E where each of the letters corresponds to a certain tyre. XD

No, because all five tyres can't be on the road at the same time, division does come into play. ^_^

The question was phrased awkwardly, but I believe 16,000 miles is the correct answer. ^^

 

[AnnoXanti]

LOL, I'm aware of the "spare", like I said I was just trying to be stubborn. :ggrin: My last answer is still 20,000 miles. You see, my analogy here is compared to boiling an egg. Whether you boil 1 or 100 eggs, they will still get done in 3 minutes. I'll stand corrected though if I'm really wrong.

Ahem. Anno. I think he was talking about five tyres you have with a car, four in use and one spare. 😛

When he said "equally", he probably meant with equal time on the road. So Time A = Time B = Time C = Time D = Time E where each of the letters corresponds to a certain tyre. XD

No, because all five tyres can't be on the road at the same time, division does come into play. ^_^

The question was phrased awkwardly, but I believe 16,000 miles is the correct answer. ^^

 

[Budweiserbob!]

Ok, ok. Fine, I see your logic. x]

 

[jts963]

^ Thanks. If the real answer to this creepy question will really come, lol, I'll really stand corrected.

This is not about math but analysis. It is all about how many miles the tires sustained at a distance of 20,000 miles, whether used or not.

Actually I find it really silly.

You're confused on the meaning of "sustained." Obviously every part of the vehicle and its contents traveled the full 20,000 miles. But not every part of the vehicle sustained (that is, helped achieve) the distance of 20,000 miles. A spare riding in the trunk doesn't sustain anything, it only sustains when it stops being a spare and becomes one of the tires holding up the vehicle and leaving tracks on the road.

 

[AnnoXanti]

You're confused on the meaning of "sustained." Obviously every part of the vehicle and its contents traveled the full 20,000 miles. But not every part of the vehicle sustained (that is, helped achieve) the distance of 20,000 miles. A spare riding in the trunk doesn't sustain anything, it only sustains when it stops being a spare and becomes one of the tires holding up the vehicle and leaving tracks on the road.

^ Go back to the question and you'll see that it's not about me getting confused by the meaning of "sustained". :ggrin:

I'll wait then for the right answer.

 

[Budweiserbob!]

I don't think there is actually an answer. >.>
I think Rick's friend genuinely wanted to know. XD

 

[Dr. Bill Kobb]

No no, he likes to pose these odd math/logic questions to me now and again. I just e-mailed him on another matter, and mentioned the 16,000 answer(which sounds the most logical to me). He's usually pretty good about getting back with things promptly. I'll report on his answer as soon as I hear it.

 

[Duke Diablo]

I believe this is basic, really.

Let's assume that the word "car" is, in interpretation, able to be realized as a four-wheeled transportation device. "Sustained" in this probably means the amount of mileage that a tire went through.

Now, since it says 'each tire used equally', it means that the time the tire and the road touch is the same with each one. So t1=t2=t3=t4=t5 (where ti is the time each tire spends on the road). Now, we need to have 4 tires at any given time, which means 4 of these increase at all times, simultaneously. So, we have 4 tires, but 5 time intervals, which need to be separated equally. Let's call the amount of time needed to pass the complete distance of 20 000 miles t+.

An equation clearly comes out:

t1+t2+t3+t4+t5=4t+

since all the ti are equal, we can substitute them with t:

5*t=4t+

Thus t=4/5t+

Knowing that this is the percentage of time a tire travels, and making an assumption that the car traveled with equal speed at all times, we can say that vi (speed at which the wheel i traveled) is constant, and equals that of v+ (average speed on the trip). Thus, knowing that in inert systems (those that stand still or do not have acceleration, positive or negative), the path s equals time * speed, we get the next equations:

(s+) = 20 000 miles (that is the complete distance)

(s+) =(v+) * (t+)

si = vi * ti (that is the distance one wheel passed)

Now, being that ti=(4/5)*t+ and vi=v+, we write:

si = (v+)*(4/5)*(t+)

Now, being that we know that (v+)*(t+)=(s+)=20 000 miles, we can substitute:

si = 20 000 *(4/5) = 16 000

Thus, 16 000 is the answer I'm giving. Sorry if anyone didn't understand this, but I just wanted to give it such an answer that can leave no doubts to it being a correct one.

 

[Budweiserbob!]

Same conclusion, but involving mechanics. XD
Nicely done Diablo. ^^

 

[N/A]

Straightforwardly-worded question:
If, on average, a hen-and-a-half lays an egg-and-a-half in a day-and-a-half,
then, on average, how long does it take one hen to lay one egg?