A school has 391 boys and 323 girls. They are to be divided into the largest possible number of identical classes so that each class has the same number of boys and the same number of girls. How many classes are formed, and how many boys and girls are in each class?
Aptitude
Problems on H.C.F and L.C.M
Difficulty: Medium
Choose an option
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A17 classes; 23 girls and 19 boys in each
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B17 classes; 23 boys and 19 girls in each
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C23 classes; 17 boys and 19 girls in each
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D19 classes; 23 boys and 17 girls in each
Answer
Correct Answer: 17 classes; 23 boys and 19 girls in each
Explanation
Introduction / Context:This item focuses on using the highest common factor (HCF) to form the maximum number of identical classes from two groups. The number of classes must divide each of the two populations so that every class has the same count of boys and the same count of girls.
Given Data / Assumptions:
- Boys = 391.
- Girls = 323.
- All classes must be identical with equal composition.
Concept / Approach:If the number of classes is k, then k must divide both 391 and 323. The largest possible number of classes is thus k = HCF(391, 323). Then the number of boys per class is 391 / k and the number of girls per class is 323 / k.
Step-by-Step Solution:
Compute HCF(391, 323): 391 − 323 = 68.323 mod 68 = 323 − 4*68 = 323 − 272 = 51.68 mod 51 = 17; 51 mod 17 = 0 ⇒ HCF = 17.Number of classes = 17.Boys per class = 391 / 17 = 23; Girls per class = 323 / 17 = 19.Verification / Alternative check:
Total check: 17 classes * (23 boys) = 391 boys; 17 * (19 girls) = 323 girls; composition identical for each class.Why Other Options Are Wrong:
- Only option B correctly states both the number of classes (17) and the per-class composition (23 boys, 19 girls). The others swap labels or give incorrect class counts.
Common Pitfalls:
- Mistaking “23 classes” for “23 per class”, or mixing the counts of boys and girls per class.
Final Answer:
17 classes; 23 boys and 19 girls in each