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Mixed fruit

These are level 3 statistics, geometry, algebra, and number problems from the Figure It Out series. This focuses on finding outcomes using a diagram, interpreting three dimensional drawings, using algebraic thinking to solve problems, and exploring average. A PDF of the student activity is included.

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Tags

  • AudienceKaiako
  • Learning AreaMathematics and Statistics
  • Resource LanguageEnglish
  • Resource typeActivity
  • SeriesFigure It Out

About this resource

​​​​​Figure It Out is a series of 80 books published between 1999 and 2009 to support teaching and learning in New Zealand classrooms.

This resource provides the teachers' notes and answers for one activity from the Figure It Out series. A printable PDF of the student activity can be downloaded from the materials that come with this resource.

Specific learning outcomes:

  • Find outcomes using a diagram (Problem 1).
  • Interpret three dimensional drawings (Problem 2).
  • Use algebraic thinking to solve problem (Problem 3).
  • Explore averages (Problem 4).
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    Mixed fruit

    Achievement objectives

    GM3-4: Represent objects with drawings and models.

    NA3-1: Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.

    NA3-6: Record and interpret additive and simple multiplicative strategies, using words, diagrams, and symbols, with an understanding of equality.

    S3-3: Investigate simple situations that involve elements of chance by comparing experimental results with expectations from models of all the outcomes, acknowledging that samples vary.

    Required materials

    • Figure It Out, Level 3, Problem Solving, "Mixed fruit", page 20

    See Materials that come with this resource to download:

    • Mixed fruit activity (.pdf)

    Activity

     | 

    There are two ways to establish the most likely outcomes. One way is to roll two dice many times and record the differences. This could be done using a tally chart like this:

    Tally chart with dice throws recorded.

    Provided that enough dice throws are recorded, this will give a strong indication of the frequency of these differences.

    Another strategy is to use a systematic approach to find all the possible outcomes. A table is possibly the best way to do this.

    Table using a systematic approach to find all the possible outcomes.

    This table shows how each difference can be generated.

    From the 36 possible outcomes when two dice are rolled, 10 outcomes give a difference of one, and eight outcomes give a difference of two.

    Students may choose to model this problem with multilink cubes.

    Encourage students to solve this problem by using each view to draw a bird’s-eye plan, numbering the squares so that each number refers to the number of cubes in the column.

    For example:

    Multilink cubes. 1, 2, 3, 1, 2.

    In Problem 2, the bird’s-eye view can be developed like this:

    Birds eye viewc, front view and left view of multilink cubes.
    • The right view confirms the left view
    Birds eye viewc, front view and left view of multilink cubes.
    • So, filling in the squares in a way that satisfies the maximum height with the smallest number of cubes gives this bird’s-eye view:
    Birds eye view of multilink cubes.
    • If students assume that the model must be a connected whole, this diagram gives a minimum solution with eight cubes:
    Multilink cubes.

    Students could use trial and improvement to solve this problem. This method will be more efficient if students notice that an orange must cost 20 cents less than an apple. Trading an orange for an apple reduces the cost by 20 cents ($2.35 – $2.15).

    Students could use a table to organise the possibilities:

    Cost of apple

    Cost of orange

    3 apples + 2 oranges

    2 apples + 3 oranges

    50c

    30c

    $2.10

    $1.90

    60c

    40c

    $2.60

    $2.40

    55c

    35c

    $2.35 ✓

    $2.15 ✓

    Alternatively, students could use a pattern:

    Pattern used to compare the cost of apples and oranges.
    • If five oranges cost $1.75, each orange must cost 35 cents because $1.75 ÷ 5 = $0.35.

    A common property of the sets of five numbers is that they will have an average (mean) of 15 because 75 ÷ 5 = 15. Working with 15 as the central number gives many solutions.

    For example:

    Diagram exploring the average means.

    Other solutions that include 15 are:

    Chart of numbers as an explanation on solutions exploring the averages.

    Note that the balancing pairs of numbers add to 30. Another way is to consider two balancing pairs of numbers where the pairs add to 60 in a different way.

    Example of where the pairs add to 60 through different equations.

    If 15 is not chosen as the central number, the others must be weighted to give an average of 15.

    For example:

    Number chart used as an example.

    1.

    a. 0, 1, 2, 3, 4, or 5

    b. The most likely answer is 1, and the next most likely is 2.

    2.

    6 (8 if all the blocks are connected)

    3.

    $0.35

    4.

    a. Answers will vary, but most solutions will include 15 as the central number, for example,

    • 7, 9, 15, 21, 23, or
    • 7, 11, 15, 19, 23.

    Solutions that do not include 15 must give an average of 15, for example,

    • 9, 11, 13, 19, 23, or
    • 7, 9, 17, 19, 23.

    b. Some other solutions include:

    • 9 + 11 + 15 + 17 + 23 = 75
    • 9 + 13 + 15 + 17 + 21 = 75
    • 11 + 13 + 15 + 17 + 19 = 75

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