AMC 10 · 2007 · #16
Grade 6 arithmeticPick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Percentages with no actual head count feel abstract, so replace the class with a convenient concrete size (Tool #9): pick 10 students, which turns 10% and 90% into a clean 1 junior and 9 seniors. An average is just a total divided by a count, so turn each average into a total number of points (Tool #7). The class total minus the seniors' total is exactly the one junior's score, recovered by working backwards (Tool #11).
Pick a convenient class size
Nothing fixes the class size, so take 10 students: 10% is 1 junior and 90% is 9 seniors.
Swapping vague percentages for a real head count of 10 makes every quantity something you can just add up.
Swapping vague shares for a real head count makes every quantity something you can just add up.
▸ Why?
A share is a count out of a hundred, so a class sized to match turns each share into whole people.
▸ Why?
Only the relative sizes matter, so any convenient class size gives the same final answer.
Turn averages into total points
An average times its count is a total: the class holds 84 × 10 = 840 points, the 9 seniors hold 83 × 9 = 747.
Turning each average back into a raw total lets you add and subtract groups of scores like ordinary numbers.
6.SP.B.5Identify SubproblemsPeel off the seniors to find the junior
Take the seniors' 747 points out of the class's 840 and the lone junior's score is left: 840 - 747 = 93, choice (C).
The class total is the seniors' points plus the junior's, so subtracting the seniors leaves the junior's score alone.
4.NBT.B.4Work BackwardsTurn percentages into a real head count, change each average into a total number of points, then subtract the group you know to uncover the one you want.
- Pick a convenient class size
- Turn averages into total points
- Peel off the seniors to find the junior