National Standards – Fractions – After Two Years – End of Year Four
After Two Years / After three years / End of Year 4Standard / use equal sharing and symmetry to find fractions of sets, shapes and quantities / apply basic addition facts and knowledge of place value and symmetry to find fractions of sets, shapes, and quantities / apply basic addition and subtraction facts, simple multiplication facts, and knowledge of place value and symmetry to find fractions of sets, shapes, and quantities
Example / Here is a string of 12 sausages to feed two hungry dogs.
Each dog should get the same number of sausages. How many will each dog get? / Here is a string of 12 sausages to feed 3 hungry dogs.
Each dog should get the same number of sausages. How many will each dog get? / A photo of 4 students
There are 24 marbles in the bag, how many should each student get?
B / If the student solves the problem by one-to-one equal sharing, they don’t meet the expectation.
At / The student uses equal sharing to distribute the sausages between the dogs. This might involve skip-counting, while tracing the count mentally or with fingers, or it might involve halving, that is dividing 12 into 6 and 6 ( Note that 6 and 6 is a symmetrical partitioning of 12) / The student applies basic addition facts to share out the sausages equally between the dogs. Their thinking could be based on doubles or equal dealing e.g. 5+5 + 2 = 12, so 4 + 4+ 4 = 12 (redistributing 1 from each 5) or 6+6 = 12, so 4 + 4 + 4 = 12. / The student applies their knowledge of symmetry or number facts to partition the set of 24 e.g. by using repeated halving or by using repeated halving or by using trial and improvement with addition facts.
Half of 24 is 12, and half of 12 is 6. So one quarter of 24 equals 6
Ab / If they solve the problem using multiplication facts (3x4=12 or
12 ÷ 3 = 4) they exceed the expectation. / If the student knows or derives the fact 4 x 6 = 24, they exceed the expectation.
National Standards – Fractions Year 5 - 8
Standard / apply additive and simple multiplicative strategies and knowledge of symmetry to find fractions of sets, shapes, and quantities / apply additive and simple multiplicative strategies flexibly to find fractions of sets, shapes, and quantities / apply additive and multiplicative strategies flexibly to whole numbers, ratios, and equivalent fractions (including percentages); apply additive strategies to decimals / apply multiplicative strategies flexibly to whole numbers, ratios, and equivalent fractions (including decimals and percentages)
Example / The wheel factory makes toys for children:
Scooters need 2 wheels
Tricycles need 3 wheels
Pushchairs need 4 wheels
Cars with trailers need 6 wheels
Trucks need 8 wheels.
The factory orders 48 wheels. How many of each toy can they make with 48 wheels? / Example 2. 2.
You have 60 jelly beans to decorate the top of the cake. If the jelly beans are spread evenly, how many of them will be on 4 tenths of the cake? / Tauranga – 180kms
The Smith family are one third of the way home.
Westport – 90 kms
The Hohepa family are four sixths of the way home.
The Smith family and Hohepa family are both driving home from their holidays. Which family has travelled the greatest distance?
B / If the student bases their answer on just the amounts (e.g. “The Smiths because 180 is greater than 90”) or just the fractions (e.g. “The Hohepas because four sixths is greater than one third”) they do not meet the expectation.
At / The student used known multiplication facts or builds up answers with addition and multiplication. E.g. To find out how many twos are in 48 for scooters, they may use doubles knowledge (24 +24), they may use addition and multiplication (e.g. 12 x 3 = 36, so 13 x 3 = 36 + 3 = 39, 14 x 3 - / They use division and multiplication to find the number of jelly beans on four tenths of the cake ( e.g. “60 ÷ 10 = 6 jelly beans on one tenth, 4 x 6 = 24 jelly beans) / The student shows that they understand that the value of a fraction of an amount depends on both the fraction and the amount. They do so by calculating the distance each family has travelled, using multiplication and division (e.g. one third of 180 =
180 ÷ 3 = 60). If the student recognises that four sixths is equivalent to two thirds, the second calculation is considerably simplified.
(two thirds of 90 = 90 ÷ 3 x 2 = 60
Ab / If they use properties of multiplication efficiently, they exceed the expectation e.g. 48 ÷ 3 is the same as 30 ÷ 3 = 10 plus 18 ÷ 3 = 6 so 48 ÷ 3 = 16 / If they notice and use the doubling and halving relationship (one third of 180 = four sixths of 90 because four sixths = 2 x one third), they exceed the expectation.