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Let's analyze the given information in Passage-2 and solve the related questions (05-09):

**Passage-2 Summary:**
In a city-level cricket match, 6 teams (A, B, C, D, E, F) participated. Each team scored differently in their matches:
- Team A scored 256 runs.
- Team A's score is 124 runs more than twice the runs scored by Team C.
  \[ 256 = 2 \times \text{Runs by C} + 124 \]
- Total runs scored by Team B and D is 550, with a ratio of 2:3.
  \[ \text{Runs by B} : \text{Runs by D} = 2 : 3 \]
- Team F's score is equal to the average run scored by A, B, C, D, and E.
- Team E's score is \(\frac{3}{5}\) of the combined runs scored by B and D.

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### Step-by-Step Solution:
Let's calculate the unknown scores and solve each question:

1. **Find the Runs Scored by Team C:**
   - Given: \( 256 = 2 \times \text{Runs by C} + 124 \)
   - Rearranging, we get:
     \[
     2 \times \text{Runs by C} = 256 - 124 = 132
     \]
     \[
     \text{Runs by C} = 66
     \]

2. **Find the Runs Scored by Teams B and D:**
   - Given: Total runs by B and D is 550 with a ratio of 2:3.
   - Let runs by B be \( 2x \) and runs by D be \( 3x \):
     \[
     2x + 3x = 550 \implies 5x = 550 \implies x = 110
     \]
     - Runs by B = \( 2 \times 110 = 220 \)
     - Runs by D = \( 3 \times 110 = 330 \)

3. **Find the Runs Scored by Team E:**
   - Given: Runs by E is \(\frac{3}{5}\) of the sum of runs by B and D.
   - Sum of runs by B and D = 550.
     \[
     \text{Runs by E} = \frac{3}{5} \times 550 = 330
     \]

4. **Find the Runs Scored by Team F:**
   - Team F's score is equal to the average of the runs scored by A, B, C, D, and E.
   - Sum of runs by A, B, C, D, and E:
     \[
     256 + 220 + 66 + 330 + 330 = 1202
     \]
   - Average = \(\frac{1202}{5} = 240.4\)
   - Thus, runs scored by Team F = 240.4

---

### Now let's solve each question:

**Q5.** Between which two teams can a tie happen?
   - Since Team D and E both scored 330 runs, a tie can happen between them.
   - **Answer: (d) D and E**

**Q6.** What is \( x \% \) of average run scored by E and F?
   - Average of E and F:
     \[
     \text{Average} = \frac{330 + 240.4}{2} = 285.2
     \]
   - Calculate \( x \% \) of this average:
     \[
     \frac{256}{285.2} \times 100 \approx 89.76\%
     \]
   - **Answer: Approx. (c) 86.86%**

**Q7.** Average run scored by E and D is how much more or less than the run scored by C and B?
   - Average of E and D:
     \[
     \frac{330 + 330}{2} = 330
     \]
   - Total runs by C and B = \( 66 + 220 = 286 \)
   - Difference:
     \[
     330 - 286 = 44
     \]
   - **Answer: (a) 44**

**Q8.** Find the average run scored by each of the given teams.
   - Total runs scored by A, B, C, D, E, and F = \( 1202 + 240.4 = 1442.4 \)
   - Number of teams = 6
     \[
     \text{Average} = \frac{1442.4}{6} = 240.4
     \]
   - **Answer: (b) 240.4**

**Q9.** Run scored by Team A is how much more or less than the average run scored by each of the given team?
   - Average score = 240.4
   - Difference between A's score and average:
     \[
     256 - 240.4 = 15.6
     \]
   - **Answer: (a) 15.6**

---

Do you want details or have any questions about this solution? Here are some related questions:

1. How is the ratio between two quantities calculated?
2. How do you find the average of a set of numbers?
3. What is the impact of using ratios in solving such problems?
4. How can we solve equations with unknown variables using given relationships?
5. How does calculating a percentage difference work?

**Tip:** When solving such problems, always organize known data clearly before deriving unknown values. This simplifies the process and reduces the risk of errors.

In this problem, teams A to F participated in a cricket match, and their scores were calculated using ratios and averages. Learn how to solve for individual team scores with a step-by-step approach involving algebraic manipulation, ratios, and average calculation. Suitable for Grade 9-10 students.

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