In a car park at east coast park, there are 125 cars, 3p motorcycles, 2q lorries and 20 buses. One of the behicles leaves the  car park at random. Given the probability that the vehicle is a motorcylce is 3/40, form an equation in p and q. Given the probability that the vehicle is a bus 1/10, form an equuation in p and q. Hnece find the value of p and of q.

Let's break down the problem step by step and formulate the required equations.

### Problem Breakdown

1. **Total number of vehicles**:  
   - Cars: 125  
   - Motorcycles: \(3p\)  
   - Lorries: \(2q\)  
   - Buses: 20  
   
   Therefore, the total number of vehicles in the car park is:
   \[
   \text{Total Vehicles} = 125 + 3p + 2q + 20 = 145 + 3p + 2q
   \]

2. **Probability that the vehicle is a motorcycle**:
   - Given probability: \(\frac{3}{40}\)
   
   The probability of selecting a motorcycle is given by:
   \[
   \frac{\text{Number of motorcycles}}{\text{Total number of vehicles}} = \frac{3p}{145 + 3p + 2q}
   \]
   Equating this to the given probability:
   \[
   \frac{3p}{145 + 3p + 2q} = \frac{3}{40}
   \]

3. **Probability that the vehicle is a bus**:
   - Given probability: \(\frac{1}{10}\)
   
   The probability of selecting a bus is given by:
   \[
   \frac{\text{Number of buses}}{\text{Total number of vehicles}} = \frac{20}{145 + 3p + 2q}
   \]
   Equating this to the given probability:
   \[
   \frac{20}{145 + 3p + 2q} = \frac{1}{10}
   \]

### Solving the Equations

Now let's solve these two equations to find the values of \(p\) and \(q\).

#### **First Equation:**

\[
\frac{3p}{145 + 3p + 2q} = \frac{3}{40}
\]

Cross-multiply to solve for \(3p\):
\[
3p \times 40 = 3 \times (145 + 3p + 2q)
\]
\[
120p = 435 + 9p + 6q
\]
Bring all terms involving \(p\) and \(q\) to one side:
\[
120p - 9p - 6q = 435
\]
\[
111p - 6q = 435
\]
Divide the entire equation by 3:
\[
37p - 2q = 145 \quad \text{(Equation 1)}
\]

#### **Second Equation:**

\[
\frac{20}{145 + 3p + 2q} = \frac{1}{10}
\]

Cross-multiply to solve for 20:
\[
20 \times 10 = 145 + 3p + 2q
\]
\[
200 = 145 + 3p + 2q
\]
Bring all terms involving \(p\) and \(q\) to one side:
\[
3p + 2q = 200 - 145
\]
\[
3p + 2q = 55 \quad \text{(Equation 2)}
\]

### Solving the System of Equations

Now we have two equations:
1. \(37p - 2q = 145\)
2. \(3p + 2q = 55\)

Add these two equations to eliminate \(q\):
\[
(37p - 2q) + (3p + 2q) = 145 + 55
\]
\[
40p = 200
\]
\[
p = \frac{200}{40} = 5
\]

Substitute \(p = 5\) into Equation 2:
\[
3(5) + 2q = 55
\]
\[
15 + 2q = 55
\]
\[
2q = 40
\]
\[
q = \frac{40}{2} = 20
\]

### Final Answer

- \(p = 5\)
- \(q = 20\)

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

Here are some additional questions related to this problem:

1. How would the equation change if the number of cars increased?
2. What if the probability of selecting a lorry was given instead?
3. How can we verify the solution using the values of \(p\) and \(q\)?
4. What is the probability of selecting a lorry given these values of \(p\) and \(q\)?
5. How would the problem change if another type of vehicle was added?

**Tip:** When forming equations from word problems, ensure each term in the equation accurately reflects the scenario described, especially in terms of probabilities and total quantities.

Learn how to solve a probability problem involving motorcycles and buses in a car park. Formulate equations based on given probabilities and total vehicle counts. Discover the values of p and q to determine the number of motorcycles and lorries. Suitable for Grades 9-12.

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