Competition · AMC preparation · step 4 of 4
AMC 10 · 2011B · #10
Grade 6 rate-ratioPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Eleven powers of ten look scary, so Tool #9 (Solve an Easier Related Problem) shrinks the set to two or three powers first, where the ratio is easy to compute by hand. Tool #5 (Look for a Pattern) then watches how that ratio behaves as the set grows, revealing that it always sits just above 9. Tool #3 (Eliminate Possibilities) finishes: once we know the value is a hair over 9, choices 1, 10, 11, and 101 are all clearly wrong.
Shrink the set and divide
Try the tiny twin {1,10,100}: the largest, 100, over 1+10=11 gives 100/11≈9.09.
A hard problem often has an easy twin with tiny numbers; solve the twin to see how the machine works.
6.RP.A.3Solve An Easier Related ProblemGrow it one step and compare
Grow it to {1,10,100,1000}: 1000/(1+10+100) = 1000/111≈9.009, still just above 9.
Watching how the answer changes as the problem grows tells you where it is heading.
4.OA.C.5Look For A PatternWrite the real denominator
Back to the full set: the ten smaller powers put a 1 in each place, summing to 1,111,111,111, while 10¹⁰ is 10,000,000,000.
A sum of different powers of ten just writes a 1 in each place, so the total is a neat string of ones.
A sum of different powers of ten just writes a one in each place, so the total is a string of ones.
▸ Why?
A number is its digits weighted by their places, so each power fills its own place exactly once.
▸ Why?
Multiplying by the same factor each step is what builds that run of powers in the first place.
Nail down the exact ratio
9 × 1,111,111,111 = 9,999,999,999 is exactly 1 less than 10¹⁰, so the ratio is 9.000000009 — a hair over 9.
If nine copies of the bottom miss the top by only 1, the ratio must be a whisker above 9.
5.NBT.B.5Look For A PatternRound and pick the answer
9.000000009 rounds to 9; 1 is far too small and 10, 11, 101 far too big, so it is (B) 9.
A number that sits a hair above 9 is closest to 9, so every other choice drops away.
6.RP.A.3Eliminate PossibilitiesWhen one number is a bit more than nine copies of another, their ratio is just over 9 — so it rounds to 9.
- Shrink the set and divide
- Grow it one step and compare
- Write the real denominator
- Nail down the exact ratio
- Round and pick the answer
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