AMC 10 · 2019 · #18
Grade 8 rate-ratioPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Diagram): a number line from 0 to 2 with A and B marked makes the geometry obvious. Tool #13 (Algebra): the limit condition gives two clean linear equations in A and B. Tool #11 (Work Backwards): instead of simulating forward from the first walk, use the fixed-point relations A and B must satisfy at the limit and solve directly.
On a line from 0 (home) to 2 (gym), from B walking toward home leaves of the way, landing at B — that point is A.
Walking of the way toward 0 leaves only of the original distance left — so you land at of the starting position.
5.NF.B.4Draw A DiagramFrom A, walking toward the gym at 2 lands at A + (2 - A) = A + , which is B.
Same remainder trick from the home side, but the gym is at 2 instead of 0.
6.EE.A.2Convert To AlgebraSubstitute A = B into B = A + , giving B = B + , so B = .
Substitute one equation into the other — a single equation in B.
8.EE.C.7Convert To AlgebraDivide: B = · = , then A = B = .
One-step division gives B; then A = .
8.EE.C.7Convert To AlgebraThe gap: |A - B| = | - | = = 1 — choice (C).
Subtract two fractions with the same denominator — easy.
5.NF.A.1Work BackwardsThis AMC 10 problem only needs Grade 8 linear-equation skills you already know — at the limit, Henry's two turning points A and B satisfy A = and B = + , giving B = and A = , so the gap is = 1 . The answer is (C).