Problem 20 · 2008 AMC 8
Medium
Algebra & Patterns
common-numeratorsmallest-integer-total
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 3/4 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
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Answer: B — 17.
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Hint 1 of 2
The phrase "an equal number of boys and girls passed" is the hinge — name that shared count and build everything from it.
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Hint 2 of 2
Both the boy-count and girl-count must be whole numbers, so the shared passing count has to be divisible by the right denominators; take the smallest one that works.
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Approach: name the shared passing count, then make the head-counts whole
- Let p = the number who passed in each group. Since 2/3 of boys passed, boys = p ÷ (2/3) = (3/2)p; since 3/4 of girls passed, girls = p ÷ (3/4) = (4/3)p.
- For both head-counts to be whole numbers, p must be divisible by 2 (for the boys) and by 3 (for the girls) — smallest such p is 6. Then boys = 9, girls = 8.
- Total = 9 + 8 = 17.
- Why this transfers: when fractions of two groups must come out whole, the unknown has to clear every denominator at once — the smallest case is their LCM.
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