Competition · AMC preparation · step 4 of 4
AMC 8 · 2006 · #16
Grade 6 rate-ratioPick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The three readers all share one unknown — the time T each spends reading. Tool #4 (Introduce a Variable) names that unknown so we can describe each person's contribution in pages. Tool #3 (Set Up an Equation) ties everything together: the three page counts must add to 760. Once T appears in a single linear equation, solving for it is straightforward.
Name the shared time
Name the shared time: let T be the seconds each person reads.
When a problem says "the same amount of time" without giving the number, that quantity gets a letter.
6.EE.B.6Introduce A VariableWrite each person's pages
Invert "seconds per page" to pages: in T seconds each person reads T divided by their per-page time.
Unit rates work both ways. If one page takes 20 seconds, then T seconds covers T/20 pages.
6.RP.A.3Introduce A VariableSet up the equation
Parts make the whole: the three page counts add up to 760.
"Parts add to the whole" is the standard work-problem equation.
The pages the three readers cover in the shared time T — T/20 for Alice, T/45 for Bob, T/30 for Chandra — must add up to the whole 760-page book.
▸ Why?
The book is split into three sections that together cover every page exactly once, with no gaps and no overlap, so the three readers' page counts must total the whole 760 pages.
▸ Why?
A reader who spends a fixed number of seconds on each page finishes, in T seconds, exactly T divided by that per-page time — for Alice at 20 seconds per page that is T/20 pages.
▸ Why?
At a steady 20 seconds per page, the pages read stack up as equal 20-second groups, so the total reading time equals the number of pages times 20 seconds.
▸ Why?
Since the total time equals the number of pages times the seconds spent per page, dividing the total time by the seconds per page recovers the number of pages.
Clear the denominators
Clear denominators: the LCM of 20, 45, 30 is 180, so multiplying through gives 9T+4T+6T = 19T.
Multiplying by the LCM turns the fraction equation into a tidy whole-number one.
6.NS.B.4Eliminate PossibilitiesSolve for the time
Solve: since 760 = 19 · 40, the 19 cancels, giving T = 180 · 40 = 7200.
Spotting that 760 is a multiple of 19 avoids the big multiplication.
6.EE.B.7Eliminate PossibilitiesWhen the same time hides behind different speeds, name it once and let one equation do the work.
- Name the shared time
- Write each person's pages
- Set up the equation
- Clear the denominators
- Solve for the time
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