The sample is a collection of related NAPLAN 2008–2010 Year 3 items that required students to construct and solve simple number sentences.
The following sample illustrates the kinds of responses that students typically produce when they have achieved the element of the standard addressed by this task.
Structure – Level 3
The student correctly completes all of the following questions, without the assistance of technology:
- NAPLAN 2009 Year 3 Numeracy Question 9: the fourth card [6 + 6 + 6]
- NAPLAN 2010 Year 3 Numeracy Question 5: the fourth card [500 + 60 + 4]
- NAPLAN 2009 Year 3 Numeracy Question 7: 13 stickers
- NAPLAN 2008 Year 3 Numeracy Question 11: [4] x 3 = [12]
- NAPLAN 2009 Year 3 Numeracy Question 31: 80 cents for each girl
- NAPLAN 2009 Year 3 Numeracy Question 25: 21
- NAPLAN 2009 Year 3 Numeracy Question 18: 10
- NAPLAN 2010 Year 3 Numeracy Question 32: $10
- NAPLAN 2010 Year 3 Numeracy Question 31: 6 students in each team
- NAPLAN 2003 Year 3 Numeracy Question 19: 1010
- NAPLAN 2010 Year 3 Numeracy Question 26: 31
Question 9 (2009) and Question 5 (2010) require students to recognise equality of different representations both visual and numerical. No computation is required to recognise the equality.
Question 7 (2009) requires students to check the truth or otherwise of several worded number statements for different numbers. This could be done by informal use of a list or table corresponding to:
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Kate
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John
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Lisa
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More than Kate?
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Less than Lisa?
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Satisfies both?
|
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11
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8
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16
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False
|
|
No
|
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11
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10
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16
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False
|
|
No
|
|
11
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13
|
16
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True
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True
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Yes
|
|
11
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17
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16
|
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False
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No
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Question 11 (2008) requires students to complete a number sentence involving equality so that it correctly represents a practical multiplicative situation: 4 tri-cycles (the 3 for the tri-cycle is given) have a total of 12 wheels.
Question 31 (2009) requires students to complete a number sentence involving equality so that it correctly represents a practical division (sharing) situation. This could be done by an informal division based on something like '$4 is the same as 40 lots of 10 cents, so each girl gets 40 ÷ 5 = 8 lots of 10 cents = 80 cents'. Alternatively students may approach the problem by sharing, such as that indicated in the following table:
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Girl
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A
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B
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C
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D
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E
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Amount in cents
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100
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100
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100
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100
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0
|
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90
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90
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90
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90
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40
|
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80
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80
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80
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80
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40 + 40 = 80
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Question 25 (2009) and Question 18 (2009) both involve recognition of equivalence. Students do not need to carry out calculations (although the questions can also be answered by this means) but to identify the structure relation involved. For Question 25 (2009), an increase of 1 in the first term of the second expression relative to the first term of the first expression requires a corresponding decrease of 1 in the second term of the second expression relative to the second term of the first expression to maintain the same value. Alternatively, by calculation a student would evaluate 19 + 22 = 41, then solve the equation 41 = 20 + ? to identify ? = 21. Question 18 (2008) can involve application of the associative (grouping) property for multiplication, 6 × 5 = (3 × 2) × 5 = 3 × (2 × 5) = 3 × 10. Students could also do this by visual modelling (I have six groups of five pens, so I need two of each of these to make a group of ten pens, and I can form three such groups). Alternatively, by calculation, a student would evaluate 6 × 5 = 30, then solve the equation 30 = 3 × ? to identify ? = 10.
Question 32 (2010) requires students to formulate a number sentence (equation) and solve it. This could be done with informal algebraic reasoning such as 'cap + bag = $ 25, bag is cap + $5 so cap + cap + $5 = $ 25 so two caps = $ 20 so cap = $10', or similar. Alternatively, students could reason that as the bag is more expensive the cap must be less than half of the total price and use systematic trial and error. For example:
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Cap $
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Bag $
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Total $
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Equal to $ 25?
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|
8
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8 + 5 = 13
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8 + 13 = 21
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No
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9
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9 + 5 = 14
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9 + 14 = 23
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No
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10
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10 + 5 = 15
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10 + 15 = 25
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Yes
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Question 31 (2010) requires students to formulate and solve a division (sharing) problem in a practical context: 30 students ÷ 5 teams = 6 students to a team. Alternatively, students may progressively re-allocate players from the larger teams to the smaller teams until they are equally distributed (possibly starting with 5 as a likely ‘near-middle’ value) such as:
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Team
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Number of students
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Comets
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4
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4 + 1 = 5
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5 + 1 = 6
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|
Supers
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7
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7
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7 – 1 = 6
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|
Stars
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5
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5
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5 + 1 = 6
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|
Champs
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11
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11 – ( 1 + 2 ) = 8
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8 – 2 = 6
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|
Kings
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3
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3 + 2 = 5
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5 + 1 = 6
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Question 19 (2008) and Question 26 (2010) require students to find a missing number that is the next value in a sequence. Students are required to identify and apply a simple recursion relation for obtaining the next term in a given sequence from the previous term. For Question 19 (2008), this is a simple linear relation obtained by constant addition of 100 (which is given) to the previous term to obtain the next term: 810 + 100 = 910, 910 + 100 = 1010.
For Question 26 (2009), this is a simple non-linear relation obtained by adding one more than was added to the previous term (which the student must identify) to obtain the next term: 10 + 1 = 11; 11 + 2 = 13; 13 + 3 = 16; 16 + 4 = 20; 20 + 5 = 25; 25 + 6 = 31 and so on.
If students were to complete variations of these questions including exercises involving solving given number sentences with missing numbers they would be working at Level 3 with respect to the key elements of the standard described below.
Elements of standard
(Students) understand the meaning of the ‘=’ in mathematical statements …
(Students) construct number sentences with missing numbers and solve them.
