1.1 Small Group · Group 2
Second Challenge Form · Non-Routine Extension · Same Standard, New Problems
doer of math · math story · Procedural Fluency · Real-World Applications
- You already can do this. So we go up a level, not up a number: find what is always true, show it a second way, and be ready to defend it with a reason instead of an answer.
- The Ferris wheel has 20 cars, and about 4 people fit in each car. About how many people can ride at one time?
- I count the cars: 20. Then I estimate how many riders fit in one car: about 4.
- I do not count every person — an estimate is a reasoned answer that is close enough to be useful.
- 20 × 4 = 80, so I would say “about 80 people.”
- 5 = 70 + 8.5.
- 5 = 78 + 0.5.
- 5 = 80 − 1.5.
- Every one of those is the same number wearing different clothes. Find two more.
- Now prove it: say why that move had to work at all — not just that it did.
- Sentence frame — convince a skeptic: "My method works because ___ . It would stop working if ___ ."
- doer of math (persona que hace matemáticas) — Anyone who uses mathematical thinking — which is everyone, every day.
- math story (historia matemática) — Your own history with mathematics — what you have done, felt, and learned so far.
- strength (fortaleza) — Something you already do well and can share with others.
- decompose (descomponer) — To break a number into parts that add, subtract, multiply, or divide back to it.
- estimate (estimar) — A reasoned answer that is close enough to be useful, without counting every one.
- Believing that 'doing math' only counts when it happens on paper in a classroom.
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1 MULTIPLE CHOICE
Which set shows FIVE different correct decompositions of 78.5?
- A70 + 8.5, 78 + 0.5, 80 − 1.5, 39.25 × 2, 157 ÷ 2
- B70 + 8.5, 78 + 5, 80 − 1.5, 39.25 × 2, 157 ÷ 2
- C70 + 8.5, 78 + 0.5, 80 + 1.5, 39.25 × 2, 157 ÷ 2
- D70 + 8.5, 78 + 0.5, 80 − 1.5, 39.25 × 3, 157 ÷ 2
✏️ Workspace & Solution Steps -
2 MULTIPLE CHOICE
A wheel has 23 cars holding about 4 riders each. Dev rounds 23 to 20 and says “about 80.” Priya rounds to 25 and says “about 100.” The operator needs to know whether one full turn can clear a line of 95 people. Whose estimate is more useful here, and why?
- APriya's — a bigger estimate is always the safer one to plan with
- BNeither — only the exact answer 92 can be used for a decision
- CDev's — it is below the exact 92, so it will not promise the operator more room than the wheel has
- DBoth equally — they are both estimates of the same quantity
✏️ Workspace & Solution Steps -
3 MULTIPLE CHOICE
Claim: “If you round BOTH numbers up before multiplying, your estimate can never come out below the exact answer.” For counts like cars and riders, is this claim always, sometimes, or never true?
- ASometimes — it depends which number you round first
- BSometimes — it fails when the two numbers are close together
- CNever — rounding up changes the answer unpredictably
- DAlways — each factor got bigger, so the product cannot get smaller
✏️ Workspace & Solution Steps -
4 MULTIPLE CHOICE
Which decomposition of 78.5 uses multiplication correctly?
- A39 × 2
- B39.25 × 2
- C39.25 × 3
- D78.5 × 2
✏️ Workspace & Solution Steps
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5 ERROR ANALYSIS
Fix our table's thinking
- 1The problem:A wheel has 23 cars holding about 4 riders each. Dev rounds 23 to 20 and says “about 80.” Priya rounds to 25 and says “about 100.” The operator needs to know whether one full turn can clear a line of 95 people. Whose estimate is more useful here, and why?
- 2A classmate at our table answered:Neither — only the exact answer 92 can be used for a decision
Which step contains the error? Explain the mathematical misconception and write the correct calculation below.
✏️ Corrected Mathematical Work & Explanation -
6 OPEN RESPONSE
The wheel has 23 cars holding about 4 riders each. Write an estimate that is deliberately a little LOW, and say how you made it low on purpose. Then say when a low estimate would be the responsible one to give.
✏️ Mathematical Justification & Response
Writing Task: Justify why your mathematical solution is accurate and complete.