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.
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).
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:
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.
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:
In Problem 2, the bird’s-eye view can be developed like this:
- The right view confirms the left view
- 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:
- If students assume that the model must be a connected whole, this diagram gives a minimum solution with eight 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:
- 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:
Other solutions that include 15 are:
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.
If 15 is not chosen as the central number, the others must be weighted to give an average of 15.
For 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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