Percy sells paint in standard tins and large tins. The standard tin covers 40m^2 and the large tin cover 60m^2. The standard and large tins are mathematically similar. The volume of the large tin is 50% more than the volume of the standard tin. Both tines are cylinders. The radius of the standard tin is 10 cm. Calculate the radius of the large tin

The problem states that the large and standard tins are mathematically similar, meaning their dimensions scale proportionally. Additionally, the volume of the large tin is 50% more than that of the standard tin. Both tins are cylindrical, and the radius of the standard tin is given as 10 cm.

Let's break this down step by step:

### 1. Relationship Between Volumes
The formula for the volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height,
- \( \pi \) is the constant \( \pi \approx 3.14159 \).

Since the tins are mathematically similar, the ratio of the volumes is proportional to the cube of the scale factor \( k \) (since volume scales with the cube of linear dimensions):
\[
\frac{V_{\text{large}}}{V_{\text{standard}}} = k^3
\]
We are told that the volume of the large tin is 50% more than the volume of the standard tin, so:
\[
V_{\text{large}} = 1.5 \times V_{\text{standard}}
\]
Thus, the ratio of the volumes is:
\[
\frac{V_{\text{large}}}{V_{\text{standard}}} = 1.5
\]
This gives:
\[
k^3 = 1.5
\]
Taking the cube root of both sides:
\[
k = \sqrt[3]{1.5} \approx 1.1447
\]

### 2. Radius Scaling
The radius of the large tin \( r_{\text{large}} \) is proportional to the radius of the standard tin \( r_{\text{standard}} \), with the scaling factor \( k \):
\[
r_{\text{large}} = k \times r_{\text{standard}}
\]
Given that \( r_{\text{standard}} = 10 \, \text{cm} \):
\[
r_{\text{large}} = 1.1447 \times 10 \approx 11.447 \, \text{cm}
\]

### Final Answer
The radius of the large tin is approximately **11.45 cm**.

Would you like more details or have any other questions?

Here are some related questions:
1. How does the height of the large tin compare to the standard tin?
2. How do surface areas of the two tins compare mathematically?
3. How can you compute the height of each tin if the volume is known?
4. What is the total paint coverage difference between the two tins?
5. What would happen to the volume ratio if the large tin covered 80% more area?

**Tip:** When dealing with similar shapes, remember that surface area scales with the square of the scaling factor, while volume scales with the cube.

Percy sells paint in standard tins and large tins. The standard tin covers 40m^2 and the large tin covers 60m^2. The standard and large tins are mathematically similar. The volume of the large tin is 50% more than the volume of the standard tin. Both tins are cylinders. The radius of the standard tin is 10 cm. Calculate the radius of the large tin.

Discover how to calculate the radius of a larger tin when given the volume proportions and similarity of cylindrical shapes. This problem explores the relationship between volume, radius, and scaling in mathematically similar cylinders. Ideal for students in Grades 10-12.

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