Question 1209274: A tractor has a rear wheel with a circumference of x m. The circumference of the front wheel is 5 m less. If the front wheel makes 15 more revolutions than the rear wheel every 450 m, what is the circumference, in m, of the front wheel?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
A tractor has a rear wheel with a circumference of x m. The circumference of the front wheel is 5 m less.
If the front wheel makes 15 more revolutions than the rear wheel every 450 m,
what is the circumference, in m, of the front wheel?
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The rear wheel has the circumference of x meters.
Hence, the front wheel has the circumference of (x-5) meters, according to the problem.
The rear wheel makes  rotations every 450 m.
The front wheel makes  rotations every 450 m.
The problem says
- = 15.
It is your setup equation.
To solve it, first multiply both sides by x*(x-5); then simplify
450x - 450(x-5) = 15x(x-5)
450x - 450x + 2250 = 15x(x-5)
2250 = 15x(x-5)
150 = x*(x-5).
At this point, you may guess that x = 15.
Indeed, 15*(15-5) = 15*10 = 150.
In addition, the function x*(x-5) is monotonic at x > 5,
so, this guessed solution is UNIQUE in the domain x > 5.
Alternatively, you may solve equation x^2 - 5x - 150 = 0 factoring or by using
the quadratic formula.
So, the circumstance of the rear wheel is 15 meters.
Hence, the circumstance if the front wheel is 15-5 = 10 meters. ANSWER
Solved.
You may check the answer on your own, by substituting the numbers into the problem.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 10 meters
Explanation
x = circumference of rear wheel
x-5 = circumference of front wheel
Each measurement is in meters.
Each value is a positive real number.
x-5 > 0 leads to x > 5
So if any x value 5 or smaller shows up, we simply ignore it.
The goal is to compute x-5.
One revolution of the rear wheel will mean that it travels x meters because it's the perimeter of the wheel.
This assumes the wheel does not slip.
Two revolutions, the wheel travels 2x meters.
Three revolutions, the wheel travels 3x meters.
And so on.
After traveling 450 meters, the rear wheel makes 450/x revolutions.
Meanwhile the front wheel makes 450/(x-5) revolutions which is 15 more compared to the previous amount.
(450/x) + 15 = 450/(x-5)
x(x-5)*( (450/x) + 15 ) = x*(x-5)*( 450/(x-5) )
450(x-5) + 15x(x-5) = 450x
450x-2250 + 15x^2-75x = 450x
450x-2250 + 15x^2-75x - 450x = 0
15x^2 - 75x - 2250 = 0
15(x^2 - 5x - 150) = 0
x^2 - 5x - 150 = 0
(x+10)(x-15) = 0
x+10 = 0 or x-15 = 0
x = -10 or x = 15
Ignore the negative value since we stated earlier that x > 5.
If the real wheel has circumference x = 15 meters then it makes 450/x = 450/15 = 30 revolutions in the 450 meter distance.
Meanwhile the front wheel makes 450/(x-5) = 450/(15-5) = 45 revolutions which is 15 more compared to the previous value.
This helps verify we have the correct x value.
Then the last thing to do is compute x-5 to get x-5 = 15-5 = 10 meters as the circumference of the front wheel.
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