Problem Analysis
The question asks for the number of boys in Section B, given that the ratio of boys to girls is the same in both Section A and Section B. We are provided with the number of boys and girls in Section A and the number of girls in Section B.
Given Information
- Section A: 36 boys, 24 girls.
- Section B: Unknown number of boys, 28 girls.
- Condition: Ratio of boys to girls is the same for both sections.
Step-by-Step Solution
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Calculate the ratio of boys to girls in Section A:
Ratio (A) = \(\frac{\text{Number of boys in A}}{\text{Number of girls in A}}\) = \(\frac{36}{24}\)
Simplify the ratio: \(\frac{36}{24} = \frac{3 \times 12}{2 \times 12} = \frac{3}{2}\)
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Set up the ratio for Section B:
Let \(B_{boys}\) be the number of boys in Section B.
Ratio (B) = \(\frac{\text{Number of boys in B}}{\text{Number of girls in B}}\) = \(\frac{B_{boys}}{28}\)
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Equate the ratios from both sections:
Since the ratio is the same: Ratio (A) = Ratio (B)
\(\frac{3}{2} = \frac{B_{boys}}{28}\)
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Solve for the number of boys in Section B (\(B_{boys}\)):
Multiply both sides by 28: \(B_{boys} = \frac{3}{2} \times 28\)
\(B_{boys} = 3 \times \frac{28}{2}\)
\(B_{boys} = 3 \times 14\)
\(B_{boys} = 42\)
Conclusion
There are 42 boys in Section B.
Important Questions from Ratio and proportion
A’s marks in Mathematics are directly proportional to practice time. In 6 hours of practice, A gets 70 marks. What should be the practice time (approximately) to get 90 marks?
The average of the areas of 2 similar triangles is 706.5 m2 whose perimeters are in the ratio of 6 : 11. What is 20% of the difference (in m2) in areas of both triangles?
In a triangle ABC, D and E are two points on sides AB and AC, respectively, such that DE is parallel to BC and AD : DB = 3 : 5. If AC = 5.6 cm, then find the value (in cm) of AE.
In a triangle ABC, P and Q are two points on AB and AC, respectively, such that PQ is parallel to BC. If AC = 5QC, then the ratio PQ : BC is equal to:
Find the mean proportional between 25 and 81.
