Prime Time
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Same Yard, New Fence
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Data Tracker
100

Tell whether 29 is prime or composite. Explain how you know.

  • Answer: 29 is prime.
    Why: A prime number has exactly two factors (1 and itself). Try dividing 29 by 2, 3, 4, 5 — none divide evenly. So its only factors are 1 and 29.
100

Compute 1,248 ÷ 6. Show the quotient and remainder if any.

  • Work: 6 goes into 12 two times → 2; bring down 4 → 6 goes into 4 zero times → 0; bring down 8 → 6 goes into 48 eight times → 8. Quotient = 208. Remainder = 0.
  • Answer: 208
100

Draw two different rectangles that both have area 12 square units. Write their side lengths and perimeters.

  • Example 1: 3 by 4 rectangle: area = 3 × 4 = 12. Perimeter = 2(3 + 4) = 14.
  • Example 2: 2 by 6 rectangle: area = 2 × 6 = 12. Perimeter = 2(2 + 6) = 16.
  • Note: Same area (12) but perimeters 14 and 16 (different).
100

Draw two different rectangles that both have perimeter 14 units. Write their side lengths and areas.

  • Perimeter 14 → length + width = 7. Possible integer pairs: (1,6) and (2,5) (also (3,4)).
  • Example 1: 1 by 6 → area = 6.
  • Example 2: 2 by 5 → area = 10.
  • Both have perimeter 14 but areas 6 and 10 (different).
100

For the list {3, 5, 5, 7, 9}, identify the mode, median, and range.

  • Mode: 5 (it appears most often).
  • Median: Put numbers in order (already ordered). Middle number is 5.
  • Range: Largest − smallest = 9 − 3 = 6.
200

List all factor pairs of 36. Based on these pairs, explain whether 36 is prime or composite.

  • Factor pairs: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6).
  • Answer: 36 is composite because it has more than two factors (examples: 2 × 18, 3 × 12).
200

Solve 3,564 ÷ 7 using a strategy based on place value. Show your steps and explain why each step is valid.

  • Steps: 7 into 35 goes 5 (5×7=35); subtract → 0; bring down 6 → 7 into 6 goes 0; bring down 4 → 7 into 64 goes 9 (9×7=63); subtract → remainder 1.
  • Answer: Quotient 509, remainder 1. (So 3,564 ÷ 7 = 509 R1.)
200

Show two different rectangles with area 24 square units that have different perimeters. Explain how the area can stay the same while the perimeter changes.

  • Example 1: 3 by 8 → area 24, perimeter 2(3 + 8) = 22.
  • Example 2: 4 by 6 → area 24, perimeter 2(4 + 6) = 20.
  • Explanation: Multiplying side lengths gives the same area, but adding side lengths (for perimeter) gives different totals, so perimeters can differ.
200

Give two rectangles with perimeter 20 units that have different areas. Show calculations of area for each and explain why areas differ.

  • Perimeter 20 → length + width = 10. Choose (4,6) and (1,9).
  • (4,6): area = 24.
  • (1,9): area = 9.
  • Same perimeter, different areas because the product length × width differs.
200

The test scores are {78, 82, 85, 85, 90, 92}. Find the mode, median, and range. Explain how you found each.

  • Mode: 85 (appears twice; others appear once).
  • Median: There are 6 numbers (even). Median is the average of the middle two. Middle two are 85 and 85 → median = (85 + 85) ÷ 2 = 85.
  • Range: 92 − 78 = 14.
300

A student claims 51 is prime because it is not even and not divisible by 3. Evaluate the student's reasoning, determine whether 51 is prime or composite, and explain your conclusion using divisibility tests and factor pairs.

  • Check divisibility: 51 is not even, but test 3: Add the digits: 5 + 1 = 6. Because 6 is divisible by 3, 51 is divisible by 3. 51 ÷ 3 = 17.
  • Answer: The student is wrong. 51 is composite because 3 × 17 = 51.
300

A student divides 4,321 ÷ 5 and gets 864 with a remainder of 1. Check the calculation, correct any errors, and explain how place value informs your correction. Show work.

  • Check: 864 × 5 = 4,320. Then add remainder 1 → 4,321. That equals the dividend.
  • Answer: The student is correct. 4,321 ÷ 5 = 864 R1.
300

Given area 36 square units, find all pairs of whole-number side lengths for rectangles with that area. For two of these rectangles, compute and compare perimeters, and explain why the perimeters differ even though area is the same.

  • Factor pairs for 36: (1,36), (2,18), (3,12), (4,9), (6,6).
  • Pick (4,9): perimeter = 2(4+9) = 26.
  • Pick (6,6): perimeter = 2(6+6) = 24.

Explanation: Both have area 36, but because their side lengths are different, the sums of the side lengths (used for perimeter) are different, so perimeters differ.

300

For perimeter 24 units, find at least three different rectangles (with whole-number side lengths) and compute their areas. Explain the pattern you observe and why some rectangles have larger areas than others.

  • Perimeter 24 → length + width = 12. Possible pairs: (1,11) area 11; (2,10) area 20; (3,9) area 27; (4,8) area 32; (5,7) area 35; (6,6) area 36.
  • Pattern: As the length and width get closer to each other, the area gets larger. The square (6,6) has the largest area among these.
300

Given the data set {12, 15, 15, 18, 21, 21, 24}, a classmate says the median is 18 and the mode is 21. Evaluate their claim, compute the correct median and mode, and explain any mistakes.

  • Sorted list: already sorted. There are 7 numbers (odd), so median is the middle one: the 4th number = 18. That part is correct.
  • Mode: Which number(s) appear most often? 15 appears twice, 21 appears twice. This data set is bimodal (two modes): 15 and 21.
  • Answer: Median = 18 (correct). Mode = 15 and 21 (not only 21).
400

Given the numbers 2–50, describe a strategy to find all prime numbers in that range. Apply your strategy to identify the primes between 30 and 50 and justify why your method works.

  • Quick strategy (sieve idea): Start with numbers 2–50. Cross out multiples of 2 (except 2), then cross out multiples of 3 (except 3), then multiples of 5, then multiples of 7. Remaining numbers are prime.
  • Primes between 30 and 50: 31, 37, 41, 43, 47.
  • Why: Each of these is not divisible evenly by any smaller prime (2, 3, 5, or 7), so they have only 1 and themselves as factors.
400

Create a word problem that results in the division 7,238 ÷ 4. Include a context where the remainder must be interpreted (e.g., items leftover, people sharing), solve it, and explain how you decided to interpret the remainder.

  • Example problem: "A library has 7,238 new bookmarks to pack into boxes that hold 4 bookmarks each. How many full boxes can they make? How many bookmarks will be left over?"
  • Division: 7,238 ÷ 4 = 1,809 remainder 2 (because 4 × 1,809 = 7,236; leftover 2).
  • Interpretation: They can fill 1,809 full boxes and will have 2 bookmarks left over (not enough to fill another full box).
400

Explain why fixing area does not determine perimeter. Provide a general argument using algebra for rectangles with area A where side lengths are whole numbers, and give a specific example with A = 48 to illustrate.

  • Idea for 4th grade: Area is length × width. Many different length × width pairs can multiply to the same area. Perimeter is 2(length + width). Since length + width can change even when the product stays the same, perimeter can change.
  • Example: Area 48 pairs: (6,8) → perimeter 2(6+8)=28. (4,12) → perimeter 2(4+12)=32. Both area 48, perimeters different (28 and 32).
400

Prove that among all rectangles with the same perimeter, the square has the largest area. Provide an explanation suitable for 4th-grade reasoning (use numbers and informal algebra or logical argument).

  • Simple explanation: For a fixed perimeter, you split that perimeter into two equal side lengths to make a square. Example with perimeter 24: split into two equal sides → each side pair sum is 12, so for the square both sides are 6 → area 36. Try any pair that adds to 12 but is not equal (like 5 and 7) → area 35, which is smaller. So making sides equal gives the largest area.
400

Create a data set of seven whole numbers between 10 and 30 that has a median of 18, a mode of 15, and a range of 12. Show your set and explain why it meets all three conditions.

  • Requirements: median 18 → middle (4th) number must be 18 when ordered. Mode 15 → 15 must appear more than any other number. Range 12 → max − min = 12.
  • One possible set (ordered): {12, 15, 15, 18, 20, 23, 24}
    Check: median = 18 (4th number), mode = 15 (appears twice, others not more), range = 24 − 12 = 12. All conditions met.
500

You are given a mystery number between 40 and 80. It is divisible by 2, 3, and 5. Could this number be prime? Explain using reasoning about prime and composite definitions and list all possible numbers that match those criteria.

  • If a number is divisible by 2, 3, and 5 then it is divisible by their least common multiple, which is 30. Numbers between 40 and 80 that are multiples of 30 are 60 (30×2). Also check multiples of 2, 3, and 5 simultaneously: any such number must be composite because it has at least 1, itself, and 30 as factor components (for 60, for example: 2 × 30 or 3 × 20 or 5 × 12).
  • Answer: It cannot be prime. The possible number in that range is 60.
500

Given 9,999 ÷ 9, predict the quotient without performing full standard algorithm steps, explain your reasoning using properties of numbers and place value, then perform the division to confirm.

  • Reasoning: 9,999 is 10,000 − 1. 10,000 ÷ 9 = 1,111 remainder 1 (because 9 × 1,111 = 9,999). Another way: dividing each 9s-place pattern gives 1,111.
  • Compute: 9,999 ÷ 9 = 1,111 exactly (since 9 × 1,111 = 9,999). Remainder 0.

Answer: 1,111

500

Two rectangles both have area 60 square units. One rectangle has integer side lengths and the other has non-integer side lengths. Construct one example of each, compute their perimeters, and analyze how side length choices affect perimeter while keeping area constant.

  • Integer example: 5 by 12 → area 60. Perimeter = 2(5 + 12) = 34.
  • Non-integer example: 7.5 by 8 → area 7.5 × 8 = 60. Perimeter = 2(7.5 + 8) = 31.
  • Analysis: Both areas are 60, but sums of side lengths differ, so perimeters differ. Choosing different side lengths (including decimals) changes perimeter while keeping area constant.
500

A farmer has 40 meters of fencing and wants to make a rectangular pen. Explore how changing one side length changes the area. Find the dimensions (with whole numbers) that give the greatest area and explain why this choice maximizes area compared to other whole-number rectangles with the same perimeter.

  • Perimeter 40 → length + width = 20. Try whole-number pairs that add to 20: (10,10) area 100; (9,11) area 99; (8,12) area 96; (7,13) area 91.
  • Best choice: 10 by 10 (a square) gives the largest area 100. Explanation: When length and width are as close as possible (equal), area is largest.
500

Original data set: {72, 75, 75, 78, 80, 82, 85, 88, 90}.

Change three students' scores (not five) so that the class median increases by 2 but the mode stays the same. Show the new list, the new median, and confirm the mode is still 75.

  • Original median = 80 (5th number). We want new median = 82.
  • Keep both 75s so the mode stays 75.
  • Change these three scores:
    1. Change 72 → 81
    2. Change 78 → 82
    3. Change 80 → 83
  • New list (sorted): {75, 75, 81, 82, 83, 82, 85, 88, 90} → properly sorted: {75, 75, 81, 82, 82, 83, 85, 88, 90}
  • New median (5th number) = 82 (increase of 2).
  • New mode = 75 (still appears twice; no other number appears more than twice).
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