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What is the difference between average male participants in marathon and average female participants in marathon ?

660

550

450

350

None of the above

Correct Answer

📐 MA • Math Direct FormulaMathematicsData Interpretation

Option B

550

Quick Summary:

Calculate the difference for each year first: 300, 500, 700, 800, 700. Sum them up ( $2750$ ) and divide by $5$ ( $2750/5 = 550$ ).

📐MAMath SolutionDirect Formula

📋 Given

Data represents marathon participants: 2021(M:1500, F:1200), 2022(M:2000, F:1500), 2023(M:2500, F:1800), 2024(M:3000, F:2200), 2025(M:3500, F:2800).

🔢 Formula Used

Average=∑Valuesn,Difference=∣AvgM−AvgF∣\text{Average} = \frac{\sum \text{Values}}{n}, \quad \text{Difference} = |\text{Avg}_M - \text{Avg}_F|Average=n∑Values​,Difference=∣AvgM​−AvgF​∣

⚡ Exam Hall Shortcut / Speed Trick

Calculate the difference for each year first: 300, 500, 700, 800, 700. Sum them up ( $2750$ ) and divide by $5$ ( $2750/5 = 550$ ).

⚠️ Common Student Trap / Pitfall

Many students calculate the sum of total participants instead of separating male and female, or perform incorrect year-by-year subtraction.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Calculate average of male participants

Sum of male participants from 2021 to 2025 is $1500 + 2000 + 2500 + 3000 + 3500 = 12500$ . Divide by $5$ years.

AvgM=125005=2500\text{Avg}_M = \frac{12500}{5} = 2500AvgM​=512500​=2500

Calculate average of female participants

Sum of female participants from 2021 to 2025 is $1200 + 1500 + 1800 + 2200 + 2800 = 9500$ . Divide by $5$ years.

AvgF=95005=1900\text{Avg}_F = \frac{9500}{5} = 1900AvgF​=59500​=1900

Calculate the difference

Subtract the average female participants from the average male participants.

2500−1900=5502500 - 1900 = 5502500−1900=550

✅

B is correct because the average male participant count is $2500$ and female is $1900$ , resulting in a difference of $550$ .

Core Concepts Used

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Arithmetic Mean Data Interpretation Bar Graph Analysis

💡 EXAM TIP

For large datasets, calculating the difference per period before averaging can save significant time and reduce potential arithmetic errors.

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