AMC 10 · 2009 · #2
Grade 6 arithmeticPick an answer.
A stacked fraction like this cannot be attacked from the top, because the top 1/… needs its denominator finished first. Tool #7 (Identify Subproblems) says: break the tower into layers and solve the smallest, innermost one first. Here the innermost subproblem is just 1 + 1. Once that is a number, the next fraction becomes computable, and so on outward. That inside-out order is also Tool #11 (Work Backwards) — the deepest layer is really the starting point. Tool #3 (Eliminate Possibilities) gives a quick sanity net: the outer "1 + (something between 0 and 1)" forces the answer to sit strictly between 1 and 2.
Peel to the innermost layer
The innermost layer resolves to 2.
Anything sitting under a fraction bar is really inside parentheses, so you finish it before dividing.
5.OA.A.1Identify SubproblemsClimb up one floor
One floor up gives 3/2.
To add a whole number to a fraction, rewrite the whole number with the same denominator so the pieces are the same size.
5.NF.A.1Identify SubproblemsFlip the middle fraction
Dividing by it is a flip.
Dividing by a fraction is the same as multiplying by its flip, so 1 over 3/2 just turns into 2/3.
Dividing by a fraction is the same as multiplying by its flip.
▸ Why?
A fraction and its flip multiply back to one, so the flip is exactly what undoes it.
▸ Why?
Multiplying top and bottom by the same amount renames a fraction without changing its size.
Add the outermost 1
Adding the outer whole gives 5/3, choice (C).
One last same-denominator add finishes the climb: 3/3 + 2/3 = 5/3.
5.NF.A.1Identify SubproblemsA stacked fraction is a tower — start at the bottom, finish each denominator before you divide, and climb out one floor at a time to reach 5/3.
- Peel to the innermost layer
- Climb up one floor
- Flip the middle fraction
- Add the outermost 1