Mixing Fruit

Problem One

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.

Problem 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 Two, 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:

Problem Three

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:

Alternatively, students could use a pattern:

If five oranges cost $1.75, each orange must cost 35 cents because $1.75 ÷ 5 = $0.35.

Problem Four

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:

Answers to Problems

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