Circumference- Wheels

[Guest]

The front wheel of a carriage makes 6 revolutions more than the rear wheel in traveling 360 feet. If the circumference of each wheel was 3 feet larger, the front wheel would make 4 revolutions more than the rear wheel in traveling the same distance. Determine circumference of each wheel.

Let x= circumference feet of front wheel.
Let y= circumference feet of rear wheel

The front wheel makes 360/x revolutions.
The rear wheel makes 360/y revolutions.

First situation:

360/x = 360/y + 6


If 3 feet were added to circumference of each wheel:

360/x + 3 = front wheel revolutions.
360/y + 3 = rear wheel revolutions.

Second situation:

360/x + 3 = 360/y + 3 + 4

How do I proceed ?

[Guest]

Do you want it all to work out evenly...integer values and whole revolutions.
Then if that is the case all wheels need to be factors of 360
So the rear original size must be a factor...... x feet giving 360/x revolutions
And the front original must be a factor ....allowing 6 more revolutions.....(360/x) + 6 revolutions
the new rear 3 feet longer must be a factor .....(x + 3) feet giving 360/(x + 3) revolutions
and new front 3 feet longer must be a factor.....allowing 4 more revolutions...(360/(x + 3)) + 4 revolutions

Taking factors of 360....that work out evenly for all cases in question.....
If rear orig = 15 feet then does 24 revs
If front orig = 18 feet then does 30 revs.....this is 6 more as required...OK
Add 3 feet to each wheel.......
New rear now = 18 feet and does 20 revs
New front now = 15 feet and does 24 revs .....this is 4 more as required

[Guest]

Error in last post...sorry.......corrected below....front orig - 12 feet....

Do you want it all to work out evenly...integer values and whole revolutions.
Then if that is the case all wheels need to be factors of 360
So the rear original size must be a factor...... x feet giving 360/x revolutions
And the front original must be a factor ....allowing 6 more revolutions.....(360/x) + 6 revolutions
the new rear 3 feet longer must be a factor .....(x + 3) feet giving 360/(x + 3) revolutions
and new front 3 feet longer must be a factor.....allowing 4 more revolutions...(360/(x + 3)) + 4 revolutions

Taking factors of 360....that work out evenly for all cases in question.....
If rear orig = 15 feet then does 24 revs
If front orig = 12 feet then does 30 revs.....this is 6 more as required...OK
Add 3 feet to each wheel.......
New rear now = 18 feet and does 20 revs
New front now = 15 feet and does 24 revs .....this is 4 more as required

[Guest]

The previous post is only a trial and error solution based on the fact that there are not many factors in 360 so a trial solution is easily found.....an algebraic solution is given below............

The circumference of all wheels need to be factors of 360 to give a number of complete revolutions and a simple answer to the problem.

So the rear original size must be a factor...... x feet giving 360/x revolutions
And the front original must be a factor ....allowing 6 more revolutions.....(360/x) + 6 revolutions
the new rear 3 feet longer must be a factor .....(x + 3) feet giving 360/(x + 3) revolutions
and new front 3 feet longer must be a factor.....allowing 4 more revolutions...(360/(x + 3)) + 4 revolutions

From the above expressions we can form equations relating the size of the wheels to the number of revolutions.
There is only 1 unknown.....X = the size of the rear wheel. All the other wheel sizes can be derived from that.

Distance travelled (360 feet) divided by the number of revolutions equals the size of the rim of the wheels.

We will set up an equation that contains the relationships between "the rear size and front size" and the change in size from "Original front size to the New +3 front size".

(Dist. Travelled)/(Orig. front wheel revolutions) + 3 = (Dist. Travelled)/(New front wheel revolutions)

(((360)/((360/X) + 6)) + 3) = ((360)/((360/(X+3)) + 4))

Divide across by 360 and further simplify....gives

((X)/(360 + 6X)) + (1/120) = ((X + 3)/(372 + 4X))

Simplifying further gives.....

(X + 3)(40X + 2400) = (4X + 372)(7X + 20)

Gives.... 12X^2 - 164X - 240 = 0

OR..... 3X^2 - 41X - 60 = 0

Factorise to solve quadratic......

(X - 15)(3X + 4) = 0

So..... X = 15 OR X = - 4/3

Take the value X = 15

This is the circumference of the original sized rear wheel.... 15 feet.

This gives the revolutions of rear orig. as 360/15 = 24 revolutions

And front orig. has 6 revs more so front orig has... 30 revolutions.

And 30 revolutions for front original means it has a circumference od 360/30 = 12 feet.

The new rear size circumference is 15 + 3 = 18 feet

And that means it has 360 /18 = 20 revolutions

And the new front wheel then must have 20 + 4 = 24 revolutions

So that means the circumference of the new front wheel is 360/24 = 15 feet.

[Guest]

"(((360)/((360/X) + 6)) + 3) = ((360)/((360/(X+3)) + 4))

Divide across by 360 and further simplify....gives

((X)/(360 + 6X)) + (1/120) = ((X + 3)/(372 + 4X))

Simplifying further gives.....

(X + 3)(40X + 2400) = (4X + 372)(7X + 20)"

I don't understand your solution.

[Guest]

working out some of the stages.......

Divide by 360.............

Gives..... (((1)/((360/X) + 6)) + 3/360) = ((1)/((360/(X+3)) + 4))

You need to carefully read the brackets to see what is combined with what....

and if we take out the first bit as below and combine it to one fraction...

((360/X) + 6) equals.... ((360 + 6X)/X) and in the line above it is actually reciprocal of it.....1/(((360 + 6X)/X))

and do same action for the RHS of the above line...gives...1/((372 + 4X)/(X+3))

So re-writing the line above...gives.....

((X)/(360 + 6X)) + (1/120) = ((X + 3)/(372 + 4X))

Now you have to do same with the LHS again of above to combine the ((X)/(360 + 6X)) and the 1/120 into one fraction.

After that you will have a single fraction on each side and you can cross multiply to get the line given below

(X + 3)(40X + 2400) = (4X + 372)(7X + 20)

Now all you have to do is multiply out the brackets and get the quadratic and then solve for X.

You can see it is a very simple quadratic the factorises and solves easily.

You need to do it yourself on a clean page well spaced out to let you see the fractions over fractions...

Its difficult to type and see on this post from a line of brackets.