Question

The fore wheel of a tractor makes 120 revolutions more than the rear wheel going 720 meters. If the diameter of the fore wheel is doubled and the diameter of the rear wheel is increased by \[1\dfrac{1}{2}\] times the present diameter, then the fore wheel makes 20 revolutions more than the rear wheel in going the same distance. Find the circumference of each wheel.

Answer 1

Hint: In this question, we need to determine the circumference of each wheel of the tractor such that the fore wheel of a tractor makes 120 revolutions more than the rear wheel in going 720 meters. For this, we will establish the relation in accordance with the conditions given in the question and solve them simultaneously.

Complete step-by-step answer:
Let the diameter of the fore wheel of the tractor be \[d\]
The diameter of the rear wheel of the tractor be \[D\]
So the circumference of the fore wheel be \[ = \pi d\]
The circumference of the rear wheel be \[ = \pi D\]
Now since the tractor is traveling a distance of 720 m, so the number of revolution made by fore wheel will be \[ = \dfrac{{720}}{{\pi d}}\]
And the number of revolution made by rear wheel will be \[ = \dfrac{{720}}{{\pi D}}\]
It is also said that fore wheel of a tractor makes 120 revolutions more than the rear wheel in while going 720 m, hence we can write this as
 \[\dfrac{{720}}{{\pi d}} = \dfrac{{720}}{{\pi D}} + 120\]
This can also be written as
 \[
  \dfrac{{720}}{{\pi d}} - \dfrac{{720}}{{\pi D}} = 120 \\
   \Rightarrow \dfrac{1}{d} - \dfrac{1}{D} = \dfrac{{120\pi }}{{720}} \\
   \Rightarrow \dfrac{1}{d} - \dfrac{1}{D} = \dfrac{\pi }{6} - - (i) \\
 \]
Now it is said that the diameter of the fore wheel is doubled and the diameter of the rear wheel is increased by 1.5 times
So to cover 720m, the number of revolution made by new fore wheel will be \[ = \dfrac{{720}}{{\pi \left( {2d} \right)}}\]
And the number of revolution made by new rear wheel will be \[ = \dfrac{{720}}{{\pi \left( {\dfrac{3}{2}} \right)D}} = \dfrac{{480}}{{\pi D}}\]
Now it is said when the diameter of the fore wheel is doubled and the diameter of the rear wheel is increased by \[1\dfrac{1}{2}\] times the fore wheel makes 20 revolutions more than the rear wheel, hence we can write this as
 \[\dfrac{{720}}{{\pi 2d}} = \dfrac{{720}}{{\pi \left( {\dfrac{3}{2}} \right)D}} + 20\]
By solving this we get,
 \[
  \dfrac{{720}}{{\pi 2d}} - \dfrac{{720}}{{\pi \left( {\dfrac{3}{2}} \right)D}} = 20 \\
   \Rightarrow \dfrac{1}{{2d}} - \dfrac{2}{{3D}} = \dfrac{{20\pi }}{{720}} \\
   \Rightarrow \dfrac{1}{{2d}} - \dfrac{2}{{3D}} = \dfrac{\pi }{{36}} - - (ii) \\
 \]
Now multiply the denominator of the equation (i) by 2 and subtract equation (ii) from it, we get
 \[\dfrac{1}{{2d}} - \dfrac{1}{{2d}} - \dfrac{1}{{2D}} + \dfrac{2}{{3D}} = \dfrac{\pi }{{12}} - \dfrac{\pi }{{36}}\]
By further solving this we get
 \[
   - \dfrac{1}{{2D}} + \dfrac{2}{{3D}} = \dfrac{\pi }{{12}} - \dfrac{\pi }{{36}} \\
   \Rightarrow \dfrac{{ - 3 + 4}}{{6D}} = \dfrac{{3\pi - \pi }}{{36}} \\
   \Rightarrow \dfrac{1}{{6D}} = \dfrac{{2\pi }}{{36}} \\
   \Rightarrow D = \dfrac{{36}}{{12\pi }} \\
   \Rightarrow D = \dfrac{3}{\pi } \;
 \]
Now substitute the value of D in equation (i), we get
 \[
  \dfrac{1}{d} - \dfrac{\pi }{3} = \dfrac{\pi }{6} \\
   \Rightarrow \dfrac{1}{d} = \dfrac{\pi }{6} + \dfrac{\pi }{3} \\
   \Rightarrow \dfrac{1}{d} = \dfrac{{\pi + 2\pi }}{6} \\
   \Rightarrow \dfrac{1}{d} = \dfrac{\pi }{2} \\
   \Rightarrow d = \dfrac{2}{\pi } \;
 \]
Hence, the circumference of the fore wheel is given as:
  \[
  {C_f} = \pi d \\
   = \pi \times \dfrac{2}{\pi } \\
   = 2 \;
 \]
So, the correct answer is “2”.

The circumference of the rear wheel is given as:
  \[
  {C_r} = \pi D \\
   = \pi \times \dfrac{3}{\pi } \\
   = 3 \;
 \]
So, the correct answer is “3”.

Note: It is interesting to note that the diameter of the wheels of the vehicle plays a vital role in covering a specified distance within the time frame. Moreover, the circumference of the wheel depends directly on the diameter of the wheel as pi times. Here, if the diameter is increased then, obviously the number revolutions to go to the fixed distance will decrease.