In a street of houses, all the front doors are
either green, blue, brown or black. More
houses have green doors than brown doors.
Half as many have green doors as have blue
doors. There are 12 houses with black doors
which is the most common colour on the
street. How many houses have a brown door?

Let's break down the problem step by step.

Let:
- \( G \) be the number of houses with green doors.
- \( B \) be the number of houses with blue doors.
- \( Br \) be the number of houses with brown doors.
- \( Bl \) be the number of houses with black doors.

From the problem, we have the following information:

1. **More houses have green doors than brown doors**:
   \[
   G > Br
   \]

2. **Half as many have green doors as have blue doors**:
   \[
   G = \frac{1}{2}B
   \]

3. **There are 12 houses with black doors, which is the most common colour**:
   \[
   Bl = 12
   \]

Since black is the most common color and there are 12 houses with black doors, no other door color can have more than 12 houses.

So, considering the blue doors:
\[
B \leq 12
\]

From the equation \( G = \frac{1}{2}B \), and knowing that \( G > Br \), let's analyze possible values of \( B \) and \( G \).

Given \( B \) can be a maximum of 12:
- If \( B = 12 \), then \( G = \frac{1}{2} \times 12 = 6 \).
- If \( B = 10 \), then \( G = \frac{1}{2} \times 10 = 5 \).
- If \( B = 8 \), then \( G = \frac{1}{2} \times 8 = 4 \).

Since \( G > Br \), the only feasible value that satisfies all conditions is \( G = 6 \). Therefore:
- \( B = 12 \) and \( Br < 6 \).

Given \( Br \) is less than 6, it must be less than 6 but not more than 5 (since \( G > Br \)).

So, \( Br = 5 \) is the only solution that fits all the criteria.

**Therefore, the number of houses with brown doors is 5.**

Would you like more details on any part of this solution or have any questions?

Here are some related questions:

1. What is the total number of houses on the street?
2. If there were 3 more houses with blue doors, how would that affect the other numbers?
3. How would the problem change if the number of green doors was equal to the number of brown doors?
4. What if half as many houses had brown doors as had black doors, how many brown doors would there be?
5. How would you solve this problem if the number of black doors were unknown?

**Tip:** When solving word problems, translate each piece of information into an equation or inequality to systematically work towards the solution.

Explore how to solve a problem involving the distribution of house door colors, specifically focusing on determining the number of houses with brown doors based on given conditions. Discover step-by-step calculations and logical deductions. Suitable for Grades 7-9.

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