Rivervale Primary School Maths Workshop for P4 to P6 Parents 15 April 2016, Friday 3.00 4.30 pm
P4 P6 Maths Assessment & Exam Paper Format
Assessments: Formative (Assessment for Learning, AfL): Gather feedback about how pupils have learnt, e.g. multiplication of fractions Teacher to reinforce concepts, skills to close any learning gap / stretch pupils learning Workbooks, worksheets, supplementary books The Rivervale School of Excellence, Individuals of Character
Assessments: Summative (Assessment of Learning, AoL): Measure pupil achievement at a particular point in time Covers a range of topics, concepts and skills Used for banding, promotion, awards Exams The Rivervale School of Excellence, Individuals of Character
2016 Assessments: P4, P5: 1 Semestral Continual Assessment (Term 3) 2 Semestral Assessments (Terms 2, 4) P6: 1 Semestral Continual Assessment (Term 1) 1 Semestral Assessment (Term 2) 1 Term Test (Term 3) 1 Preliminary Exam (Term 3) The Rivervale School of Excellence, Individuals of Character
Achievement Bands, Grades / Mark Range: P4: Band Mark Range 1 85-100 2 70-84 3 50-69 4 Below 50 P5, P6: Grade Mark Range A* 91-100 A 75-90 B 60-74 C 50-59 D 35-49 E 20-34 U (Ungraded) Below 20 The Rivervale School of Excellence, Individuals of Character
Exam Formats: P4: 1 h 45 min, 100 marks Section A (MCQ): Section B (SAQ): Section C (LAQ): 15 2-marks 15 2-marks 10 3/4/5-marks The Rivervale School of Excellence, Individuals of Character
Exam Formats: P5, P6: Paper 1: 50 min paper, 40 marks (calculator not allowed) Booklet A (MCQ): Booklet B (SAQ): List of approved calculators for PSLE: http://www.seab.gov.sg/content/calcula tor/guidelinescalculators.pdf 10 1-mark and 5 2-marks 10 1-mark and 5 2-marks Paper 2: 1 h 40 min paper, 60 marks (calculator allowed) SAQ: 5 2-marks LAQ: 13 3/4/5-marks The Rivervale School of Excellence, Individuals of Character
Assessments: P4 Subject-based Banding: Choice to take a combination of standard or foundation subjects Recognise pupils different learning abilities and provide greater flexibility to concentrate on the subjects they are good at Details: https://www.moe.gov.sg/docs/defaultsource/document/education/primary/files/subject-basedbanding2015.pdf The Rivervale School of Excellence, Individuals of Character
Assessments: P6 PSLE: T-score Aggregate score Performance is relative http://nurs8004.wikispaces.com/file/view/no rmal-distributioncurve.jpg/365012298/normal-distributioncurve.jpg The Rivervale School of Excellence, Individuals of Character
P4 & P5 Heuristics Guess-and-check Supposition Method Model Drawing (Before and After) Gaps and Differences (Listing and Model drawing) Grouping v.s Number value The Rivervale School of Excellence, Individuals of Character
Overview of Heuristics from P2 P5 P2 P3 P4 P5 Guess & Check Guess & Check Guess & Check Revision Guess & Check Revision Supposition Mtd (drawing) Supposition Mtd Supposition Mtd Supposition Mtd Revision Units x Values Pattern Pattern Pattern Pattern (grouping) Working Backwards Working Backwards Working Backwards (Model drawing) Working Backwards (Model drawing & listing) Equal Gaps Equal Gaps revision Equal Gaps extension Listing Method (integrated into Fractions) Listing Method (in gaps & diff qn) Listing Method & Model Drawing (in gaps & diff qn)
Guess-and-check A farmer has 15 chickens and rabbits. These animals have 40 feet altogether. How many of each type of animals does the farmer have? No. of rabbits (4 feet) 7 7 x 4 = 28 6 6 x 4 = 24 5 5 x 4 = 20 No. of chickens (2 feet) 8 8 x 2 = 16 9 9 x 2 = 18 10 10 x 2 = 20 FIXED C + R = 15 7 + 8 = 15 6 + 9 = 15 5 + 10 = 15 TARGET Total feet = 40 Check / x 28 + 16 = 44 x 24 + 18 = 42 x 20 + 20 = 40 When From doing the guess 1 st guess, and check/trial we need to and look error at method, the it s encouraged number of feet to guess to see the if number we should of decrease each animal the by number dividing of the rabbits total by or 2 chickens. first. Since rabbits In have this case, more 15 feet, 2 we = 7 should R 1 decrease So 1 type Ans: of 5 animal the rabbits number will and of 7 while rabbits. 10 the chickens other is 8.
Guess-and-check Grace bought 20 blouses and skirts. There were 5 buttons on each blouse and 2 buttons on each skirt. She counted a total of 64 buttons. How many of each type of clothes did she buy? No. of blouses (5 buttons) 10 10 x 5 = 50 9 9 x 5 = 45 8 8 x 5 = 40 No. of skirts (2 buttons) 10 10 x 2 = 20 11 11 x 2 = 22 12 12 x 2 = 24 FIXED B + S = 20 10 + 10 = 20 9 + 11 = 20 8 + 12 = 20 TARGET Total buttons = 64 Check / x 50 + 20 = 70 x 45 + 22 = 67 x 40 + 24 = 64 Ans: 8 blouses and 12 skirts
Supposition Method A farmer has 15 chickens and rabbits. These animals have 40 feet altogether. How many of each type of animals does the farmer have? Determine which animal has fewer legs. Suppose: 15 chickens no. of feet (chickens) = 15 x 2 = 30 excess feet = 40 30 = 10 A rabbit has 2 more feet than a chicken. no. of rabbits= 10 2 = 5 no. of chickens= 15 5 = 10 There are 5 rabbits and 10 chickens.
Supposition Method Grace bought 20 blouses and skirts. There were 5 buttons on each blouse and 2 buttons on each skirt. She counted a total of 64 buttons. How many of each type of clothes did she buy? Determine which piece of clothing has lesser buttons. Suppose: 20 skirts no. of buttons = 20 x 2 = 40 excess buttons= 64 40 24 = A blouse has 3 more buttons than a skirt. no. of skirts = 24 3 = 8 no. of blouses = 20 8 = 12 There are 8 skirts and 12 blouses.
Before and after (equal stage) Daniel and Patrick had an equal number of stickers at first. Daniel then gave away 20 of his stickers to his friend Melvin, and Patrick bought another 12 stickers. In the end, 3 units Patrick had thrice as many stickers as Daniel. Find the number of stickers Daniel had at first. 1 unit Summary Before: After: D 1u 20 D 1u P 1u +12 P 3u Before After D P? D 1u 20 2 units P 1u 20 12 3 units 2 units 20 + 12 = 32 1 unit 32 2 = 16 Daniel at 1st 16 + 20 = 36 Daniel had 36. stickers
Before and after (equal stage) Mrs Tan had a total of 115 oranges and apples. She sold half the apples and 25 oranges. In the end, she had an equal number of oranges and apples left. Find the number of oranges Mrs Tan had at first. Before? Summary O 25 Before: O A A 115 25 half After After: O 1u A 1u O A 115 3 units 115 25 = 90 1 unit 90 3 = 30 Oranges at 1st 30+ 25 = 55 She had. 55 oranges
Before and after (equal stage) 4) Rakesh and Xijie had a total of 216 stickers at first. After Rakesh bought another 24 stickers and Xijie gave away 60 of his stickers to David, both of them had the same number of stickers left. How many stickers did Rakesh have at first? Summary Before: R X After: +24 R 1u 216 60 X 1u Before After R 24 X 24 60 R X? 216 2 units 216 24 60 = 132 1 unit 132 2 = 66 He had. 66 stickers
Gaps and Differences (Listing) A group of children had to share a bag of badges. When each child took 4 badges, there were 5 badges left. When each child took 6 badges instead, 3 more badges were needed. How many badges were there in the bag? No. of children 1 2 3 4 x 4 (x4,+5): 4 9 8 13 12 17 16 21 x 6 (x6, 3): 6 3 12 9 18 15 24 21 Ans: 21 badges
Gaps and Differences (Model) A group of children had to share a bag of badges. When each child took 4 badges, there were 5 badges left. When each child took 6 badges instead, 3 more badges were needed. How many badges were there in the bag? The When difference looking in at badges this type between of question, the 2 we scenarios have to will bear always in mind be 2. that So, in both we have scenarios, to find the how bars many of the sets model of 2 we drawn need should to cover be of equal the gaps length. (8). x4 x6 (because the total amount of items for both cases should be the same). diff 6 4 = 2 gaps 5+ 3 = 8 5 5? 3 No. of sets 8 2 = 4 (children) 1 child 4 4 children 4 x 4 = 16 total badges 16 + 5 = 21 Ans: 21 badges
Gaps and Differences (Model) 1) Sue bought some pencils for her nephews. If she gave them 8 pencils each, she would have 3 pencils left. If she gave them 11 pencils each, she would need 9 more pencils. How many pencils did Sue buy? The When difference looking in at pencils this type between of question, the 2 we scenarios have to will bear always in mind be 3. that So, in both we have scenarios, to find the how bars many of the sets model of 3 we drawn need should to cover be of equal the gaps length. (12). x8 x11 (because the total amount of items for both cases should be the same). 3 3 diff 11 8 = 3 gaps 3+ 9 = 12? 9 No. of sets 12 3 = 4 (nephews) 1 nephew 8 4 nephews 8 x 4 = 32 total pencils 32 + 3 = 35 Ans: 35 pencils
3 things to look out for: 1) no. of units for each item 2) amount/value of each item 3) total amount/value Number value At a school funfair, every girl was given 3 packet drinks while every boy was given 3 units 1 unit 5 packet drinks. There were thrice as many girls as boys at the funfair. Given that a total of 350 packet drinks were given out, how many boys were at the funfair? No. of children (units) x No. of drinks Total unit (packet drinks) girl 3 units x 3 9 packet units boy 1 unit x 5 5 packet units total amount 9 units + 5 units = 14 units 14 units 350 1 unit 350 14 = (boys) 25 There were. 25 boys
Grouping with extra part 5) 30 pupils shared a box of pencils. Each girl received 4 pencils and each boy received 5 pencils. If the girls received 30 more pencils than the boys, how many girls were there? Girls (4 pencils each) Boys (5 pencils each) B + G (30) (G B) 30 pencils /x 15 x 4 = 60 15 x 5 = 75 15 + 15 60-75 x 17 x 4 = 68 13 x 5 = 65 17 + 13 68-65 =3 x 19 x 4 = 76 11 x 5 = 55 19 + 11 76-55 =21 x 20 x 4 = 80 10 x 5 = 50 20 + 10 80-50 =30 Ans: 20 girls
Grouping or Number value? In a farm, there are twice as many chickens as horses. Given that there are 288 legs altogether, how many chickens are there in the farm? Number x value Mrs Lim had a total of $360, consisting of $10 and $2 notes. Given that she had 6 more $10-notes than $2 notes, how many notes did Mrs Lim have? Grouping A sum of $960 is to be divided amongst a group of adults and children. Each adult will receive $7 and each child will receive $4. Given that there are four times as many adults as children, how many adults are there? Number x value
P6 Heuristics and Exam Skills in Mathematics Problem Solving The Rivervale School of Excellence, Individuals of Character
Commonly Used Heuristics Draw a model Look for patterns Act it out Guess and Check Draw a diagram Work backwards Solve part of the problem Before and After Supposition method Examination Skills Scan the whole question Simplify the problem Read the question a few times Underline the key points Metacognitive questioning: - Which is the best method to use? - Have I done a similar question previously? - How does this number help me to solve the problem? - Am I answering to the question?
Working Backwards Start with the end result and work backwards towards the beginning to find a solution. Then check your answer by working forward.
Peter had some money. He spent half of the amount on a pair of jeans. He then spent half of the remainder on a pair of shoes. After that, he was left with $12. How much money did Peter have at first? Solution $12 2 = $24 $24 2 = $48 Ali had $48 at first. jeans? shoes $12 Check $48 2 = $24 (jeans), $48 $24 = $24 $24 2 = $12 (shoes), $24 $2 = $12 ( )
Draw a Model
Ahmad and Benny have a total of 66 toys. Of Ahmad s toys and 2 of Benny s toys are Matchbox cars. If they have 3 an equal number of Matchbox cars, how many toys does each one of them have? 4 5 Ahmad Benny 66 11 units 66 1 unit 66 11 = 6 5 units 6 x 5 = 30 6 units 6 x 6 = 36 Ahmad has 30 toys and Benny has 36 toys
4 2 Ahmad and Benny have a total of 66 toys. of Ahmad s toys and of Benny s 5 3 toys are Matchbox cars. If they have an equal number of Matchbox cars, how many toys does each one of them have? Alternative solution: Ahmad 4 5 Benny = Total number of units = 5 + 6 =11 11 units 66 1 unit 6 5 units 5 x 6 = 30 (Ahmad) 6 units 6 x 6 = 36 (Benny) Ahmad has 30 toys and Benny has 36 toys 2 3 4 6
Common mistakes Reading too quickly and missed out details Misread of questions or numbers 355 as 335 Number left instead of number sold Forget to write the units or the wrong units Incorrect standard unit conversion 1m = 100cm 1 hour = 60 mins
Common mistakes Wrong mathematical statements : Wrong use of equal signs / approximation signs 2 = 40 2 40 5 5 1 = 40 2 = 20 1 40 2 = 20 5 5 Express 3 7 as a percentage, rounded off to the nearest 1 decimal place. 3 7 = 42.9% 3 7 42.9%
Common mistakes Not putting given information into diagrams ABCD is a rhombus and AEFD is a parallelogram.
Answering PSLE Questions & PSLE Exam tips The Rivervale School of Excellence, Individuals of Character
Question 1 PSLE Maths 2015 P2 Q16 Peiyi and Jamal bought potted plants at the prices shown below. Large potted plants 2 for $15 Small potted plants 3 for $10
Question 1 PSLE Maths 2015 P2 Q16 (a) Peiyi bought an equal number of large and small potted plants. She spent $175 more on the large ones. How many potted plants did she buy altogether? (b) Jamal spent an equal amount of money on the large and small potted plants. What fraction of the potted plants he bought were large?
(a) Correct Solution Cost of 6 large potted plants = $15 x 3 = $45 Cost of 6 small potted plants = $10 x 2 = $20 Difference in cost = $45 - $20 = $25 No. of sets of 6 large potted plants she bought = $175 $25 = 7 Number of potted plants she bought altogether = (7 x 6) x 2 = 42 x 2 = 84
(b) Correct Solution No. Of large potted plants he can buy for $30 = 2 2 = 4 No. Of small potted plants he can buy for $30 = 3 3 = 9 Fraction of potted plants he bought that are large = 4 9+4 = 4 13
Question 2 PSLE Maths 2015 P2 Q17 Three girls Amy, Beth and Cindy had the same number of coins. Amy and Beth each had a mix of fifty-cent and ten-cent coins. Amy had 9 ten-cent coins while Beth had 15 ten-cent coins. Cindy had only fifty-cent coins.
(a) Of the three girls, who had the most money and who had the least? (b) What was the difference in the total value of Amy and Beth s coins? (c) Beth used all her fifty-cent coins to buy some food. She then had $10 less in coins than Cindy. How many fifty-cent coins did Cindy have?
(a) Cindy had the most money and Beth had the least money. (b) Difference in value between a 10-cent and a 50- cent coin = $0.50 - $0.10 = $0.40 Difference in total value of Amy and Beth s coins = 6 x $0.40 = $2.40
(c) Original difference in total value of Beth and Cindy s coins = 15 x $0.40 = $6 Number of fifty-cent coins Beth had = ($10 - $6) $0.50 = 8 Number of fifty-cent coins Cindy had = 15 + 8 = 23
Tips For Answering PSLE Questions 1) Be open-minded 2) Think flexibly and divergently 3) Do not always narrow down to 1 method of solution
Tips For Answering PSLE Questions 4)Don t skip too many questions --- attempt all the questions 5) Check your work means physically check, NOT just read 6) Complete the whole PSLE Mathematics book