Problem 16 · 2015 AMC 8
Medium
Algebra & Patterns
fraction-word-problemsubstitution
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 13 of all the ninth graders are paired with 25 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Show answer
Answer: B — 4/11.
Show hints
Hint 1 of 2
Buddies come in pairs, so the count of paired ninth-graders equals the count of paired sixth-graders — one body on each side of every handshake. That hidden equality links the two unknown group sizes.
Still stuck? Show hint 2 →
Hint 2 of 2
There are no actual totals given, so invent friendly ones: with no specified size, pick numbers that make both fractions land on whole people, then just count heads.
Show solution
Approach: pick concrete sizes so the fractions are whole numbers
- Pairs are one-to-one, so paired ninth-graders = paired sixth-graders. Pick small numbers that make both fractions whole: ninth = 6, sixth = 5.
- Then paired ninth = (1/3)(6) = 2 and paired sixth = (2/5)(5) = 2. ✓
- Total students = 6 + 5 = 11. Buddied students = 2 + 2 = 4. Fraction = 4/11.
- Why this transfers: when a problem gives only ratios and no totals, the answer can't depend on the actual size — so plug in the smallest sizes that make every fraction whole and read off the result.
Another way — algebra:
- Let n = ninth-graders, s = sixth-graders. (1/3)n = (2/5)s ⇒ 5n = 6s.
- Buddied / total = ((1/3)n + (2/5)s) / (n + s) = (2 · (1/3)n) / (n + s) (since the two numerators are equal).
- Using 5n = 6s, ratio simplifies to 4/11.
Mark:
· log in to save