AMC 10 · 2008 · #1

Grade 4 rate-ratio
rateratio-proportionunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: rate
📏 Short solution 💡 1 insight
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Problem
A doughnut machine starts at 8:30 AM. By 11:10 AM it has finished one third of the day's job. Find the clock time when it finishes the whole job, assuming it keeps working at the same steady rate.

Pick an answer.

(A)
$\ \text{1:50}\ {\small\text{PM}}$
(B)
$\ \text{3:00}\ {\small\text{PM}}$
(C)
$\ \text{3:30}\ {\small\text{PM}}$
(D)
$\ \text{4:30}\ {\small\text{PM}}$
(E)
$\ \text{5:50}\ {\small\text{PM}}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Analyze the Units

The job runs at a steady rate, so a chunk of time is a unit of work. First measure how long one third took (a time interval), then scale that interval up to the whole job, then add it back onto the start time. Splitting it into these three time-pieces keeps the clock arithmetic clean.

1STEP 1

Measure the one-third interval

From 8:30 AM to 11:10 AM the machine ran 2 hours 40 minutes, and that stretch finished one third of the job.

11{:}10 - 8{:}30 = 2 h 40 min
2STEP 2

Scale up to the whole job

Each third takes the same time, so the whole job is 3 times 160 minutes = 480 minutes, exactly 8 hours.

3 × (2 h 40 min) = 3 × 160 min = 480 min = 8 h
3STEP 3

Add the total time to the start

Add 8 hours to the 8:30 AM start: 4 hours reaches 12:30 PM, and 4 more lands on 4:30 PM — choice (D).

8{:}30 AM + 8 h = 4{:}30 PM
Answer
4:30 PM
One third took 2 hours 40 minutes, so two thirds more is another 5 hours 20 minutes past 11:10 AM. 11:10 AM plus 5 hours 20 minutes lands on 4:30 PM, matching the answer from adding 8 hours to the start. Both routes agree on (D), and 4:30 PM is a sensible full workday of 8 hours.
💡Key takeaway

If one third of a steady job took some time, the whole job takes three times that, so measure the piece and multiply.

  • Measure the one-third interval
  • Scale up to the whole job
  • Add the total time to the start