Reasoning · Grade 6-2 Optimal Design
Problem
Multiples of one glass never reach 15.
Listing the times one timer can produce is just skip-counting by 11 and by 7, and seeing that 15 is missing from both lists tells you at once that the answer must come from combining them.
3.MD.A.1Guess And CheckStart both and draw the two time lines.
Elapsed time on a number line is a length, so the amount of the 11-minute glass still left when the 7-minute glass finishes is just the difference of the two lengths, 11 minus 7.
3.MD.A.1Draw A DiagramBetween minutes 7 and 11 sits a 4-minute piece.
This is the same trick as measuring 4 cups with a 7-cup and an 11-cup jug: what you can measure is not only the two sizes but the gap between them.
3.MD.A.1Draw A DiagramStarting both glasses together creates a 4-minute piece neither glass could measure alone.
Both glasses start at the same moment, so the gap between when they run out is exactly the difference of their two times.
The target is then built from those pieces laid end to end, so any total made of them can be measured exactly.
15 breaks into 4 plus 11.
Splitting a total into two friendly parts is ordinary addition thinking -- the only new idea is that the parts here are stretches of time rather than numbers of objects.
4.OA.A.3Identify SubproblemsStart at minute 7 and turn the big glass at minute 11.
Every instruction in the plan is tied to a moment Ethan can actually see -- a glass running out -- so there is never a point where he has to judge how much sand is left.
3.MD.A.1Draw A DiagramUp to minute 22 is exactly 15 minutes.
Putting real clock times on the plan is the quickest way to be sure no minute has been double-counted or dropped, because the subtraction at the end has to land on 15 exactly.
3.MD.A.1Guess And CheckTwo timers give you more than two lengths -- the gap between them, 11 - 7 = 4 minutes, is a third length you can build with!