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A class consists of 2020 boys and 3030 girls. In the mid-semester examination, the average score of the girls was 55 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 33 while the average score of the entire class increased by 22. The increase in the average score of the boys is

Solution

✅ Correct Option: 1

We can see this solution has the right approach but needs significant improvement to meet AfterBoards standards. Let us rewrite it to be crystal clear for students who might be new to weighted average problems.


Let the average score of boys in the mid-semester exam = bb

Given information:

  • Girls' average was 5 higher than boys
  • Girls' average in mid-semester = b+5b + 5
  • Class has 20 boys and 30 girls (total 50 students)

When we have different groups with different averages, we use a weighted average formula:

Class Average = (Boys × Boys’ Average) + (Girls × Girls’ Average)Total Students\tfrac{\text{(Boys × Boys' Average) + (Girls × Girls' Average)}}{\text{Total Students}}

Original class average = 20×b+30×(b+5)50\dfrac{20 \times b + 30 \times (b + 5)}{50}

= 20b+30b+15050\dfrac{20b + 30b + 150}{50}

= 50b+15050\dfrac{50b + 150}{50}

= b+3b + 3

Key insight: The original class average was b+3b + 3


In the final exam:

  • Girls' average dropped by 3: (b+5)−3=b+2(b + 5) - 3 = b + 2
  • Entire class average increased by 2: (b+3)+2=b+5(b + 3) + 2 = b + 5

Here's the crucial step: We know the new class average and the girls' new average, so we can find the boys' new average.

Using the weighted average formula again:

New class average = 20×(Boys’ new average)+30×(Girls’ new average)50\dfrac{20 \times \text{(Boys' new average)} + 30 \times \text{(Girls' new average)}}{50}

b+5=20×(Boys’ new average)+30×(b+2)50b + 5 = \dfrac{20 \times \text{(Boys' new average)} + 30 \times (b + 2)}{50}


50(b+5)=20×(Boys’ new average)+30(b+2)50(b + 5) = 20 \times \text{(Boys' new average)} + 30(b + 2)

50b+250=20×(Boys’ new average)+30b+6050b + 250 = 20 \times \text{(Boys' new average)} + 30b + 60

50b+250=20×(Boys’ new average)+30b+6050b + 250 = 20 \times \text{(Boys' new average)} + 30b + 60

20b+190=20×(Boys’ new average)20b + 190 = 20 \times \text{(Boys' new average)}

Boys’ new average=20b+19020=b+9.5\text{Boys' new average} = \dfrac{20b + 190}{20} = b + 9.5


Boys' original average = bb

Boys' new average = b+9.5b + 9.5

Increase in boys' average = (b+9.5)−b=9.5(b + 9.5) - b = 9.5


Answer: The increase in the average score of the boys is 9.5

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