While Lilo was playing with a toy tractor, she rolled it over an ink drop. The image on the right shows the ink drop and the marks on the floor left by tractor's wheels. The diameter of the back wheel of the tractor is 1.5 times that of its front wheel. What is the number of revolutions made by the front wheel between the ink drop and the last mark?
Show Hint
For problems involving wheels and revolutions, remember that the number of revolutions is inversely proportional to the wheel's diameter (or radius/circumference). A smaller wheel must turn more times to cover the same distance as a larger wheel.
Step 1: Understanding the Concept:
This problem relates the number of revolutions of a wheel to the distance it travels. The distance covered in one revolution is equal to the wheel's circumference. Both the front and back wheels of the tractor cover the same linear distance between two points.
Step 2: Key Formula or Approach:
The key formulas are:
Circumference of a wheel, \( C = \pi \times D \), where \( D \) is the diameter.
Distance traveled, \( L = N \times C \), where \( N \) is the number of revolutions.
Since the distance \( L \) is the same for both wheels:
\[ L = N_{front} \times C_{front} = N_{back} \times C_{back} \]
\[ N_{front} \times (\pi \times D_{front}) = N_{back} \times (\pi \times D_{back}) \]
\[ N_{front} \times D_{front} = N_{back} \times D_{back} \]
Step 3: Detailed Explanation: 1. Determine the number of revolutions of the back wheel:
The image shows the initial ink drop and the marks left by a wheel. Each mark corresponds to one full revolution. There are 4 marks after the initial ink drop. This means the wheel that made these marks completed 4 full revolutions. Since the back wheel is larger, it will make fewer revolutions for a given distance, resulting in more spaced-out marks, which matches the diagram. So, we can conclude that the back wheel made 4 revolutions.
\[ N_{back} = 4 \]
2. Use the given diameter relationship:
We are given that the diameter of the back wheel is 1.5 times that of the front wheel.
\[ D_{back} = 1.5 \times D_{front} \]
3. Calculate the number of revolutions of the front wheel:
Using the relationship from Step 2:
\[ N_{front} \times D_{front} = N_{back} \times D_{back} \]
Substitute the known values:
\[ N_{front} \times D_{front} = 4 \times (1.5 \times D_{front}) \]
The term \( D_{front} \) cancels out from both sides:
\[ N_{front} = 4 \times 1.5 \]
\[ N_{front} = 6 \]
Step 4: Final Answer:
The front wheel made 6 revolutions between the ink drop and the last mark.
