[ { "id": "smc-2024-pdf", "kind": "past-paper", "title": "SMC 2024 — Primary 4 (Grade 4) Contest Paper", "href": "https://www.dropbox.com/scl/fo/7b136f7n5gx0m5a3cfq1w/ACdfDFukqVp3FM4tIkYn4Ik?rlkey=hyyiq5gbfrzzggekaeg5cpjzi&preview=elsa0x28/papers/2026-08-08-smc-math/2024-SMC-G4%20-%20TS1729.pdf&dl=1", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.compare", "fraction.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.mass", "measurement.money", "measurement.time", "measurement.volume", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "word-problem.model" ], "meta": "the original paper · 2024-SMC-G4 - TS1729.pdf", "blurb": "The original paper as a PDF, with the figures exactly as the publisher printed them — for when the reprinted question is not enough.", "text": "2024-SMC-G4 - TS1729.pdf" }, { "id": "smc-2024-key", "kind": "answer-key", "title": "SMC 2024 — Primary 4 (Grade 4) Contest Paper", "href": "keys/smc-2024-key.html", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.compare", "fraction.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.mass", "measurement.money", "measurement.time", "measurement.volume", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "word-problem.model" ], "meta": "31 questions · every answer, method and trap", "blurb": "Every answer, with the method under it and the mistake to watch for. Best opened after you have had a go. This paper prints NO answers, so these were worked out here and can be argued with.", "text": "What is the value of 20 thousands and 9 tens? Build the number place by place: 20 thousands is 20 000, and 9 tens is 90. Add them: 20 000 + 90 = 20 090. Writing the 9 in the ones place (20 009). \"Tens\" means the second column from the right, so the 9 needs a 0 after it. How many one-thirds are there in 4 wholes? One whole holds 3 thirds, so 4 wholes hold 4 × 3 = 12 thirds. Answering 3 — that is how many thirds are in ONE whole. Or dividing 3 by 4 instead of multiplying. Which of the following is NOT an equivalent fraction of 1/4? Multiply the top and bottom of 1/4 by the same number: × 2 gives 2/8 ✓, × 3 gives 3/12 ✓, × 25 gives 25/100 ✓. For eighths you would need 2/8 to make 1/4, so 5/8 is the odd one out. Reading past the word NOT and picking one that IS equivalent. Underline NOT before looking at the options. A figure folded in half along the dotted line of symmetry is shown: below a horizontal dotted line sit two half-circles, each with one dot and a single stroke leaning to the right. Which of the following figures is the symmetric figure when opened up? The fold line is HORIZONTAL, so opening up reflects everything upwards. The single dot in each half-circle gains a partner directly above it, giving two dots stacked one over the other. The leaning stroke reflects too: a stroke leaning one way, joined to its mirror leaning the other way, makes a curve that bulges out to the right — a ) shape. Only B has both. Picking D, which mirrors the two faces LEFT to RIGHT. The fold line drawn is horizontal, so the mirror is up-and-down, and both faces must end up identical to each other. The table shows the number of pets kept by a group of children. How many children have MORE than 1 pet? \"More than 1 pet\" means 2, 3 or 4 pets — not 1, and not 0. Add those three columns: 5 + 8 + 2 = 15 children. Including the 17 children who have exactly 1 pet (giving 32). \"More than 1\" excludes 1 itself; \"1 or more\" would include it. The figure below is made up of 5 identical squares in an I-shape — two across the top, one in the middle, two across the bottom. The area of each square is 36 cm². What is the perimeter of the figure? Each square has area 36 cm², so its side is 6 cm (because 6 × 6 = 36). Start with the rectangle that just surrounds the figure: it is 2 squares wide and 3 squares tall, so 12 cm by 18 cm, giving 2 × (12 + 18) = 60 cm. Then each notch cut into the waist adds twice its depth, because you walk in and back out: 2 notches, each half a square (3 cm) deep, add 2 × 3 × 2 = 12 cm. Perimeter = 60 + 12 = 72 cm. Taking 36 cm as the SIDE of each square instead of the area, or adding the five squares' perimeters (5 × 24 = 120 cm), which counts the edges where they join. Arrange the following from the greatest to the smallest: 9/20, 0.54, 4.05, 1/2 Turn everything into decimals so they can be compared directly: 9/20 = 45/100 = 0.45, and 1/2 = 0.5. Now the list is 0.45, 0.54, 4.05, 0.5. Greatest to smallest: 4.05, then 0.54, then 0.5 (= 1/2), then 0.45 (= 9/20). Comparing 0.54 and 0.5 by counting digits and deciding 0.54 is \"longer so bigger\" — true here, but it fails for 0.5 versus 0.45. Line up the tenths first, every time. Jason has $20. He bought 2 packets of chips and 6 oranges. Chips are $2.50 a packet and oranges are 3 for $1.20. How much money did he have left? Chips: 2 × $2.50 = $5.00. Oranges: 6 is two lots of 3, so 2 × $1.20 = $2.40. He spent $5.00 + $2.40 = $7.40. Money left = $20 − $7.40 = $12.60. Answering $7.40 — that is what he SPENT. The question asks what is left, so there is one more subtraction to do. Kayden is standing at point K facing the park. He makes a 135° clockwise turn followed by a 315° anticlockwise turn. Where will he be facing now? Combine the two turns first instead of doing them one at a time: 135° clockwise then 315° anticlockwise leaves 315 − 135 = 180° anticlockwise — which is the same as a half turn, so he ends up facing the exact opposite direction. He started facing the Park (north-east), and the opposite of north-east is south-west, which is the Food Centre. Working the two turns separately and losing track. Also note a 180° turn needs no direction — clockwise or anticlockwise both land in the same place. Linda had 3 litres of milk. She spilled 1/6 ℓ of milk and used 3/4 ℓ to bake some cookies. How much milk did she have left? Put everything over twelfths: 1/6 = 2/12 and 3/4 = 9/12, so she lost 2/12 + 9/12 = 11/12 ℓ altogether. Taking that from 3 ℓ: 3 − 11/12 = 2 12/12 − 11/12 = 2 1/12 ℓ. Answering 11/12 — that is how much milk went, not how much is left. Or subtracting only one of the two amounts. A protractor is placed with its centre at Y and its baseline along the bottom. Two lines are drawn from Y: one up to the left towards X, reading 18° above the baseline, and one up to the right towards Z, reading 38° above the baseline. Find the value of ∠XYZ. The whole straight line along the protractor's base is 180°. The angle sits between the two drawn lines, so take off what lies outside it on each side: the 18° between YX and the baseline on the left, and the 38° between YZ and the baseline on the right. ∠XYZ = 180 − 18 − 38 = 124°. Answering 142° by treating YX as lying ALONG the baseline (180 − 38). It does not — it is drawn 18° above it, and that 18° has to come off too. Mrs Bala wanted to sew 6 ribbons on each side of a square handkerchief. There was a ribbon at each corner of the handkerchief. How many ribbons did she sew in total? Counting 6 ribbons on each of the 4 sides gives 4 × 6 = 24 — but each corner ribbon has been counted twice, once for each side it sits on. There are 4 corners, so take 4 off: 24 − 4 = 20 ribbons. Answering 24 and forgetting that a corner ribbon belongs to two sides at once. Mark the corners on a quick sketch before multiplying. Mariam started her piano lesson at the time shown on the clock (the minute hand points at 12 and the hour hand at 1, so 1 o’clock in the afternoon). The duration of the piano lesson is 75 minutes. However, the lesson ended 10 minutes later than that. What time did her lesson end? Give your answer in 24-hour clock format. The clock reads 1 o’clock in the afternoon, which is 13 00. The lesson runs 75 minutes, which is 1 hour 15 minutes, so it was due to end at 13 00 + 1 h 15 min = 14 15. It actually ended 10 minutes after that: 14 15 + 10 min = 14 25. Answering 14 15 and forgetting the extra 10 minutes, or writing 13 65 — there is no such time, because 60 minutes roll over into the next hour. The graph shows the number of watches sold by a shop from January to April: January 38, February 22, March 52, April 16. The total number of watches sold from January to April was 4 times the number of watches sold in May. How many watches were sold in May? Add the four months: 38 + 22 + 52 + 16 = 128 watches. That total is 4 times May's figure, so May = 128 ÷ 4 = 32 watches. Multiplying by 4 instead of dividing. The four months together are the BIG number, so May must be smaller than it. A piece of paper is cut straight down into Square A and Rectangle B. The area of A is twice the area of B. The breadth of Rectangle B is 14 cm. What is the area of Square A? The cut is straight down, so A and B are the same height — call it h. A is a SQUARE, so its width is h too and its area is h × h. B is h tall and 14 cm wide, so its area is 14 × h. Now use \"A is twice B\": h × h = 2 × 14 × h, so h = 28 cm. The area of Square A = 28 × 28 = 784 cm². Answering 784 by luck but 28 by stopping early — 28 is the SIDE, and the question asks for the area. Or using 14 as the square's side. A stall had a promotion: buy 1 muffin for $3 each, or buy 4 muffins at $11. Mandy wanted to buy 50 muffins. What was the least amount of money Mandy had to pay for all the 50 muffins? The 4-for-$11 deal works out at $2.75 a muffin, cheaper than $3, so take as many fours as possible: 50 ÷ 4 = 12 sets of four (48 muffins) with 2 left over. That is 12 × $11 = $132, plus 2 singles at $3 = $6. Total = $138. (Buying 13 sets would give 52 muffins for $143 — more muffins AND more money, so it is not cheaper.) Buying 13 sets of four to avoid paying full price for the last two. Always price the leftover BOTH ways — here the singles win. Ashley is 10 years old and her mother is 46 years old. In how many years’ time will Ashley’s mother be 4 times as old as Ashley? The gap between their ages never changes: 46 − 10 = 36 years, now and always. At the moment we want, Ashley is 1 unit and her mother is 4 units, so the gap is 4 − 1 = 3 units = 36, giving 1 unit = 12. So Ashley will be 12, which is 12 − 10 = 2 years from now. Answering 12 — that is Ashley's AGE at that time, not how many years away it is. Read the last line again before writing. In the figure, ABCD is a rectangle and DEFG is a square, overlapping at D. ∠CDG = 64° and ∠EDH = 33°, where DH is a straight line from D. Find ∠ADH. Use the two right angles the shapes give you. ABCD is a rectangle, so its corner at D is 90°: ∠CDA = 90. Since ∠CDG = 64, what is left is ∠GDA = 90 − 64 = 26°. DEFG is a square, so its corner at D is also 90°: ∠GDE = 90. Going from G round to E the rays pass A then H, so ∠GDA + ∠ADH + ∠HDE = 90, that is 26 + ∠ADH + 33 = 90. So ∠ADH = 90 − 59 = 31°. Adding 64 and 33 and subtracting from 90 straight away. The 64° is measured inside the RECTANGLE's corner and the 33° inside the SQUARE's — two different right angles, so each has to be used on its own shape first. There were 163 ribbons in Box A and 115 ribbons in Box B. Some ribbons were moved from Box A to Box B. In the end, there were 26 more ribbons in Box A than Box B. How many ribbons were there in Box A in the end? Moving ribbons between the boxes does not change the TOTAL: 163 + 115 = 278 ribbons, before and after. At the end A is 26 more than B, so take that 26 off and split the rest equally: (278 − 26) ÷ 2 = 252 ÷ 2 = 126 — that is Box B. Box A = 126 + 26 = 152. Trying to work out how many ribbons were moved first. You never need to know — the unchanged total plus the final difference is enough. Square A and Rectangles B, C and D form the rectangle WXYZ. A sits top-left and is a square of side 7 cm, B is top-right, D is bottom-left and C is bottom-right. The area of Rectangle B is twice the area of Rectangle D. The shaded part (A, B and D together) has an area of 385 cm². Find the area of Rectangle C, in cm². A is a 7 cm square, so the top row is 7 cm tall and the left column is 7 cm wide. Call the right column's width w and the bottom row's height h. Then B = 7 × w, D = 7 × h, and C = w × h. Since B is twice D: 7w = 2 × 7h, so w = 2h. The shaded area is A + B + D = 49 + 7w + 7h = 385, so 7w + 7h = 336 and w + h = 48. Putting w = 2h in: 3h = 48, so h = 16 and w = 32. Area of C = 16 × 32 = 512 cm². Trying to find C by subtracting 385 from the whole rectangle — you do not know the whole rectangle's area yet. Naming the two unknown lengths and using \"B is twice D\" is what unlocks it. Julia had some beads in a jar. 1/4 of the beads were red and the rest were green. After Julia put another 318 red beads into the jar, the fraction of green beads in the jar became 3/7. What was the total number of beads in the jar at first? The GREEN beads never change — only red ones were added. At first green was 3/4 of the jar; afterwards green is 3/7 of the bigger jar. Say the green beads number G. Then at first the jar held G ÷ 3 × 4 beads, and at the end it held G ÷ 3 × 7. The jar grew by exactly the 318 red beads added, so G ÷ 3 × 7 − G ÷ 3 × 4 = 318, that is G ÷ 3 × 3 = 318, so G = 318. The jar at first = 318 ÷ 3 × 4 = 424 beads. Trying to track the red beads, which change. Spotting the quantity that STAYS THE SAME — the green ones — is what makes the two fractions comparable. David paid $7374 for 3 laptops and 2 headphones. Kumar paid $5002 more than David for 5 laptops and 4 headphones. What was the cost of 1 headphone? Kumar paid 7374 + 5002 = $12 376 for 5 laptops and 4 headphones. Now double David's order so the headphones match: 6 laptops and 4 headphones cost 2 × $7374 = $14 748. Comparing that with Kumar's 5 laptops and 4 headphones, the headphones cancel and the difference is exactly 1 laptop: 14 748 − 12 376 = $2372. So 3 laptops cost 3 × 2372 = $7116, leaving 7374 − 7116 = $258 for 2 headphones, and one headphone is $129. Subtracting David's order from Kumar's directly (2 laptops + 2 headphones = $5002) and then guessing. Doubling one order first is what makes a pair of items cancel exactly. The total mass of a box and a watermelon was 7 kg 34 g. When some strawberries were added into the box, the total mass became 8600 g. The watermelon was 3 times as heavy as all the strawberries added. Find the mass of the box, in grams. Work in grams: 7 kg 34 g = 7034 g. Adding the strawberries took the total from 7034 g to 8600 g, so the strawberries weigh 8600 − 7034 = 1566 g. The watermelon is 3 times that: 3 × 1566 = 4698 g. The box and watermelon together were 7034 g, so the box = 7034 − 4698 = 2336 g. Reading 7 kg 34 g as 7340 g. It is 7000 + 34 = 7034 g — the 34 fills the ones and tens columns, not the hundreds. Jamie decorated a square classroom of side 6 m 40 cm. She tied balloons in a repeating pattern on a string and hung the string around the classroom once. In every 80 cm of string the pattern holds 4 small balloons (with big balloons between them). How many small balloons did she use to decorate the 4 sides of the classroom? Find how much string is needed: the classroom is a square of side 6 m 40 cm = 640 cm, so once round is 4 × 640 = 2560 cm. The pattern repeats every 80 cm, so it repeats 2560 ÷ 80 = 32 times. Each repeat carries 4 small balloons, so she used 32 × 4 = 128 small balloons. Working out the balloons for one side and forgetting to multiply by 4, or counting the big balloons too. The question asks only for the SMALL ones. At a fruit stall, mangoes are sold at 2 for $7 and pears are sold at 3 for $5.50. Mdm Fauziah bought the same number of mangoes and pears. She paid $160 in total. How many mangoes did she buy? Choose a number of each that both deals divide into — 6 is the smallest. 6 mangoes are three lots of 2, costing 3 × $7 = $21. 6 pears are two lots of 3, costing 2 × $5.50 = $11. So 6 of each costs $21 + $11 = $32. Then $160 ÷ $32 = 5 such groups, and she bought 5 × 6 = 30 mangoes. Working out the price of one mango and one pear. A pear costs $5.50 ÷ 3, which is not a whole number of cents — buying in sixes keeps every figure exact. Mr Lee had an apple, a papaya and a honeydew. He weighed them in pairs: apple + papaya = 730 g, apple + honeydew = 1960 g, honeydew + papaya = 2470 g. What was the mass of the honeydew, in grams? Add all three weighings: 730 + 1960 + 2470 = 5160 g. Every fruit has been weighed exactly twice, so the three fruits together weigh 5160 ÷ 2 = 2580 g. The honeydew is the one missing from Setup 1, so honeydew = 2580 − 730 = 1850 g. Subtracting two of the weighings and hoping. Adding all three and halving is what turns three pair-weights into the single total, and the rest is one subtraction. Kenny has 3 pieces of rope of length 24 cm, 60 cm and 96 cm. He wants to cut them into smaller pieces of equal length without any leftovers. Each smaller piece must be as long as possible. How many such pieces of equal length can he get? The piece length must divide 24, 60 and 96 exactly, and be as long as possible — so it is the highest common factor. 24 = 12 × 2, 60 = 12 × 5, 96 = 12 × 8, and no bigger number divides all three, so each piece is 12 cm. Counting the pieces: 2 + 5 + 8 = 15 pieces. Answering 12 — that is the LENGTH of each piece, and the paper's answer line unhelpfully prints \"cm\" even though the question asks how many. Read the question, not the blank. Sam used Square X (side 5 cm), Rectangle Y (9 cm by 4 cm) and Rectangle Z (14 cm by 3 cm) — with a second copy of Square X — to form Figure A: Z lies along the bottom, one X sits on Z at the left, Y stands upright beside it, and the other X sits at the right-hand end, level with the bottom of Z. What is the perimeter of Figure A, in cm? Set the bottom-left corner of Z at 0. Z runs from 0 to 14 across and 0 to 3 up. The left X sits on Z, from 0 to 5 across and 3 to 8 up. Y stands next to it, from 5 to 9 across and 3 to 12 up. The right X sits beyond Z, from 14 to 19 across and 0 to 5 up. Now walk the outline once: up the left side 8, across the top of X 5, up Y's side 4, across Y's top 4, down Y's far side 9, along Z's top 5, up the step 2, across the right X's top 5, down its side 5, and back along the bottom 19. Adding: 8 + 5 + 4 + 4 + 9 + 5 + 2 + 5 + 5 + 19 = 66 cm. Adding the pieces' perimeters (5 × 4 + 5 × 4 + 26 + 34 = 100 cm). Every edge where two pieces touch is inside the figure and must not be counted — walking the outline once is the only safe way. A spider was climbing to the top of a garden wall, starting from the bottom. After climbing up 1/3 of the wall, it began to rain. During the rain the spider slipped down 30 cm and stayed there until the rain stopped. Then it climbed up the remaining 5/6 of the height of the wall to reach the top. What was the total height the spider had climbed before and after the rain, in cm? Let the wall be 1 whole. The spider got to 1/3, slipped down 30 cm, then climbed 5/6 of the wall to finish at the top. So (1/3 of the wall) − 30 cm + (5/6 of the wall) = the whole wall. In sixths, 1/3 + 5/6 = 2/6 + 5/6 = 7/6, so 7/6 of the wall minus 30 cm equals 1 wall — meaning the extra 1/6 of the wall is exactly the 30 cm it slipped. So the wall is 6 × 30 = 180 cm. It climbed 1/3 of 180 = 60 cm before the rain and 5/6 of 180 = 150 cm after, a total of 60 + 150 = 210 cm. Answering 180 — that is the height of the WALL. The question asks how far the spider CLIMBED, which is more, because it had to re-climb the 30 cm it slipped. A rectangular park measures 54 m by 28 m. A pedestrian path 2 m wide and a cyclist path 3 m wide are built along two sides of the park — the pedestrian path wraps the top and left of the park, and the cyclist path wraps the top and left of that. It costs $18 to construct each square metre of the cyclist path. How much does it cost to construct the cyclist path? Build outwards from the park. The park is 54 by 28. Adding the 2 m pedestrian path along the top and the left makes that block 54 + 2 = 56 by 28 + 2 = 30. Adding the 3 m cyclist path along the top and left of THAT makes the whole thing 56 + 3 = 59 by 30 + 3 = 33. The cyclist path is the difference between those two: 59 × 33 − 56 × 30 = 1947 − 1680 = 267 m². At $18 a square metre, the cost is 267 × $18 = $4806. Measuring the cyclist path's strips from the PARK's 54 and 28 instead of from the outside of the pedestrian path. The paths are stacked, so the outer one is longer than the inner one. In a library, there were some story books at first. Betty, the librarian, added another 17 story books and removed 36 of them. Then, Betty received 3 times as many new story books as what were left on the shelves. She then arranged all the story books equally between 2 sections. Each section had 458 story books in the end. How many story books were in the library at first? Work backwards. Two sections of 458 make 2 × 458 = 916 books at the end. Just before that, the shelves held some number and Betty received 3 times as many again, so the 916 is 1 + 3 = 4 equal shares: one share = 916 ÷ 4 = 229 books on the shelves. That 229 came after adding 17 and removing 36, a net loss of 19. So at first there were 229 + 19 = 248 books. Reading \"3 times as many new books as what were left\" as making the total 3 times bigger. It ADDS 3 shares to the 1 already there, giving 4 shares in all." }, { "id": "smc-2023-pdf", "kind": "past-paper", "title": "Singapore Math Challenge 2023 — Primary 4 Contest Paper", "href": "https://www.dropbox.com/scl/fo/7b136f7n5gx0m5a3cfq1w/ACdfDFukqVp3FM4tIkYn4Ik?rlkey=hyyiq5gbfrzzggekaeg5cpjzi&preview=elsa0x28/papers/2026-08-08-smc-math/2023%20Singapore%20Math%20Challenge%20Grade%204.pdf&dl=1", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.length", "measurement.money", "measurement.time", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "word-problem.model" ], "meta": "the original paper · 2023 Singapore Math Challenge Grade 4.pdf", "blurb": "The original paper as a PDF, with the figures exactly as the publisher printed them — for when the reprinted question is not enough.", "text": "2023 Singapore Math Challenge Grade 4.pdf" }, { "id": "smc-2023-key", "kind": "answer-key", "title": "Singapore Math Challenge 2023 — Primary 4 Contest Paper", "href": "keys/smc-2023-key.html", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.length", "measurement.money", "measurement.time", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "word-problem.model" ], "meta": "45 questions · every answer, method and trap", "blurb": "Every answer, with the method under it and the mistake to watch for. Best opened after you have had a go. The answers are the published ones; the explanations were written here.", "text": "What is the value of the digit ‘6’ in the numeral 36 574? Read the places from the right: 4 ones, 7 tens, 5 hundreds, 6 thousands, 3 ten-thousands. The 6 sits in the thousands place, so its value is 6 × 1000 = 6000. Answering 6, or naming the place (\"thousands\") instead of the VALUE. The question asks what the digit is worth. A number when rounded to the nearest 100 is 8000. What is the greatest possible number? Rounding to the nearest 100 gives 8000 for anything from 7950 up to 8049. Anything from 8050 upwards rounds to 8100 instead. So the greatest is 8049. Answering 8050 — that is the first number that rounds UP to 8100, not the last that rounds to 8000. Test the boundary by rounding it back. A van can transport 8 passengers. How many passengers can 240 vans transport? 240 × 8 = 1920 passengers. Dividing instead of multiplying. More vans means more passengers, so the answer must be bigger than 240. The figure below is made up of 2 squares of different sizes. AG is 8 cm and DE is 6 cm. What is the length of AD? Give your answer in cm. AG = 8 cm is the side of the big square, so AB = 8 cm too. DE = 6 cm is the side of the small square, so BD (the part of the top line beyond B) = 6 cm. AD runs along the top from A to D: 8 + 6 = 14 cm. Trying to find AD from the slanted-looking picture instead of from the two square sides. In a square all four sides are equal — that is the only fact this question needs. Look at the figure below — shaded squares on a 6 × 6 grid, with a dashed diagonal line running from the bottom-left corner to the top-right. Which one of the options will form a symmetric figure with the above, about that dashed line? The dashed diagonal is the mirror line. Take each shaded square above the line and count its steps to the line; its partner must be the same number of steps on the other side. Only option 1 places every partner correctly, including the half-shaded triangle, which must reflect into a triangle leaning the opposite way. Checking only the big blocks and not the half-squares. The triangles are what separate option 1 from option 3 — they must flip, not just move. Ben was standing at Point X. He turned 225° anti-clockwise and then made another 1/4-turn in the clockwise direction. He ended up facing the Bus Bay. Which location was Ben facing at first? Work out the NET turn first: 225° anti-clockwise, then 90° clockwise, leaves 225 − 90 = 135° anti-clockwise. He finished facing the Bus Bay (north-east), so to find where he began, undo it — turn 135° CLOCKWISE from north-east. North-east → (90°) → south-east → (45° more) → south. South is the Hall. Applying the turns forwards from the Bus Bay instead of backwards. The Bus Bay is where he ENDED, so the working has to run in reverse. Which of the following is an equivalent fraction of 3/4? Multiply the top and bottom of 3/4 by the same number: × 2 gives 6/8. Checking the others: 12/24 simplifies to 1/2, 9/16 and 5/6 do not reduce to 3/4. Picking 9/16 because 3 × 3 = 9 and 4 × 4 = 16. Both parts must be multiplied by the SAME number, not each by itself. Zainal had 5 m of ribbon at first. He used 1.4 m of it to wrap a present and some ribbon to decorate 2 cards. He had 2.6 m of ribbon left. Which of the following is the length of ribbon for each card? Find the ribbon used on the cards: 5 − 1.4 − 2.6 = 1 m for BOTH cards. Then share it: 1 ÷ 2 = 0.5 m for each card. Stopping at 1 m — that is the ribbon for two cards, and the question asks for each one. The table shows the favourite colours of 108 pupils in Primary Four. Some numbers have been blotted out by ink stains. There were twice as many pupils who chose green than yellow. There is a total of 38 pupils who like 2 of the colours. Which of the following options shows the two colours? Green and yellow together = 108 − 22 − 38 = 48. Green is twice yellow, so that is 3 equal shares: yellow = 48 ÷ 3 = 16 and green = 32. Now test the pairs: blue + green = 54, yellow + blue = 16 + 22 = 38 ✓, green + red = 70, red + yellow = 54. Splitting the 48 evenly into 24 and 24. \"Twice as many\" makes THREE shares, not two. Which of the following figures has only ONE line of symmetry? Count the folds for each. A parallelogram has none — no fold makes its halves match. A regular pentagon has five. A rectangle has two, one across and one down. The isosceles trapezium has exactly one, straight down the middle. Choosing the rectangle. It folds down the middle AND across the middle, so it has two, and the question asks for ONLY one. Rectangle PQRS is folded to form the figure below. ∠y is three times ∠x. Find ∠x. The whole corner at Q is 90°, and it is cut into three parts: ∠y, then the two angles the fold makes. A fold reflects, so the crease from Q sits exactly halfway between the edge in its new position and the edge in its old one — those two angles are equal, and each is ∠x. So y + x + x = 90. With y = 3x: 3x + 2x = 90, so 5x = 90 and x = 18°. Writing y + x = 90 and getting 22.5°. The fold makes TWO equal angles below y, not one — that is the whole point of a crease. A 4-digit number is made up of three different odd digits and a zero. What is the smallest possible value of the number? The odd digits are 1, 3, 5, 7, 9 and three of them must be different. To make the number small, put the smallest digit first — but not the 0, because a 4-digit number cannot start with 0. So the thousands digit is 1, the hundreds digit is the 0, and the last two places take the next smallest odd digits: 3 then 5. That gives 1035. Starting with 0 (0135 is a 3-digit number), or forgetting the three odd digits must be DIFFERENT and writing 1011. Study the pattern. What is the missing number in the box? star × star = 49; moon + moon + moon + moon = 100; star × moon = ? From the first line, a number times itself is 49, so the star is 7. From the second, four moons make 100, so the moon is 100 ÷ 4 = 25. Then star × moon = 7 × 25 = 175. Reading the second line as 4 × moon = 100 and then halving instead of dividing by 4, or taking the star as 24.5 (half of 49) instead of the number that multiplies by ITSELF. Alice’s present age is a 2-digit number which is also a multiple of 4. 5 years ago, her age was a factor of 54. What is Alice’s age 5 years later? The factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Adding 5 to each gives the possible present ages: 6, 7, 8, 11, 14, 23, 32, 59. Of those, only 32 is both a 2-digit number and a multiple of 4. So Alice is 32 now, and 5 years later she is 32 + 5 = 37. Answering 32 — that is her age NOW. The question asks for five years later. Colin, Dylan and Eden shared some sweets. Colin took 1/6 of the sweets. Dylan took the remaining sweets and shared it equally with Eden. Eden ate 12 of his sweets and has 18 sweets left. How many sweets were there altogether? Work backwards for Eden: he has 18 left after eating 12, so he had 12 + 18 = 30. Colin took 1/6, leaving 5/6, and Dylan and Eden split that equally — so Eden got half of 5/6, which is 5/12 of the whole. If 5/12 is 30 sweets, then 1/12 is 6, and the whole is 12 × 6 = 72. Taking Eden's share as half of everything (1/2) instead of half of what was LEFT (5/12). The product of 7 and a number is 84. What is the product of 252 and that number? Find the number first: 84 ÷ 7 = 12. Then 252 × 12 = 3024. Multiplying 252 by 84, or by 7. Find the hidden number before using it. A rectangular piece of paper was folded differently on both ends as shown below. What is the area of the rectangular piece of paper before it was folded? Give your answer in cm². Each crease is at 45°, so the slanted edge runs sideways exactly as far as it drops — and that distance is the paper's WIDTH. The 4 cm across the top right is that same distance, so the paper is 4 cm wide. Now unfold each end: the left flap reaches 3 + 4 = 7 cm, and the right flap reaches 1 + 4 = 5 cm. Laid flat, the length is 7 + 9 + 5 = 21 cm. Area = 21 × 4 = 84 cm². Adding up the visible pieces as if nothing were hidden. A fold tucks paper underneath, so the flat sheet is always LONGER than the folded shape. A rectangular piece of paper was cut into 3 smaller rectangles with dimensions 9 cm by 12 cm, 9 cm by 2 cm, and 9 cm by 5 cm. Find the length of the paper before it was cut. Give your answer in cm. Every piece is 9 cm on one side, so 9 cm is the paper's breadth and the cuts ran straight across it. The other sides — 12, 2 and 5 — are the pieces of the length, so the length was 12 + 2 + 5 = 19 cm. Adding all six numbers, or adding the 9s as well. The 9 is the side the three pieces SHARE, so it is counted once and is not part of the length. The table shows the cost of pizza: a Regular is $12 and a Large is $22. Mrs Tan ordered 6 large pizzas and 5 regular pizzas. She gave the delivery man $200. How much change did Mrs Tan receive? 6 large: 6 × $22 = $132. 5 regular: 5 × $12 = $60. Total = $132 + $60 = $192. Change from $200 = 200 − 192 = $8. Answering 192 — that is what she spent, not the change. Or pairing the wrong price to the wrong size. A length of sticky tape is made up of repeated designs: a 5 cm striped band, then a 3 cm checked band, then a 4 cm grid band, repeating. The sticky tape is 80 cm long. How many checked bands are there altogether? One full repeat is 5 + 3 + 4 = 12 cm. In 80 cm: 80 ÷ 12 = 6 repeats with 8 cm left over. Those 6 repeats give 6 checked bands. The leftover 8 cm starts the next repeat: 5 cm of stripes, then 3 cm of checks — exactly enough for one more complete checked band. So 6 + 1 = 7. Answering 6 and ignoring the leftover. Always ask what the remaining centimetres are long enough to reach. Sam spent a total of 3 h 10 min revising for his English, Science and Mathematics tests over the weekend. He spent 50 min on English and another 1 h 25 min on Science. How long did Sam spend revising Mathematics? Express your answer in minutes. Change everything to minutes first. Total = 3 × 60 + 10 = 190 min. English = 50 min. Science = 60 + 25 = 85 min. Mathematics = 190 − 50 − 85 = 55 min. Subtracting hours and minutes separately without converting, which turns 3 h 10 min − 1 h 25 min into a negative number of minutes. The figure below is made up of 7 identical rectangles. The perimeter of the whole figure is 340 cm. Find the breadth of 1 such rectangle. Give your answer in cm. The two rows are the same width, so 2 lengths = 5 breadths. The figure's width is 2 lengths and its height is 1 breadth + 1 length, so the perimeter is 2 × (2 lengths + breadth + length) = 2 × (3 lengths + 1 breadth) = 340, giving 3 lengths + 1 breadth = 170. From 2 lengths = 5 breadths, one length is 2½ breadths, so 3 lengths is 7½ breadths. Then 7½ + 1 = 8½ breadths = 170, and one breadth = 170 ÷ 8.5 = 20 cm. Adding the seven rectangles' perimeters. Only the OUTSIDE edge counts — the lines where rectangles touch are inside the figure. Yiping had 100 fewer stickers than Joanne. Joanne had 4 times as many stickers as Wendy. Wendy had 40 stickers. How many stickers did they have altogether? Start from the one you know. Wendy = 40. Joanne = 4 × 40 = 160. Yiping = 160 − 100 = 60. Altogether = 40 + 160 + 60 = 260. Working from Yiping first, or adding the 100 to Joanne instead of taking it off. \"100 fewer than Joanne\" means Yiping is the smaller one. Star Bookshop is having a promotion: notebooks are $5 each, and for every 2 notebooks bought the third notebook is free (buy 3 for the price of 2). Karen bought 25 notebooks. What is the least amount of money she has spent in total? Group the notebooks in threes, because every group of 3 costs only 2 × $5 = $10. In 25 notebooks there are 25 ÷ 3 = 8 groups of three (24 notebooks) with 1 left over. The 8 groups cost 8 × $10 = $80, and the last notebook has no partners so it costs the full $5. Total = $80 + $5 = $85. Paying full price for all 25 ($125), or counting the free ones as groups of 2. The deal gives 3 notebooks for the price of 2, so it is groups of THREE that matter. Meena left her house to watch a movie at the cinema. Her watch showed 1630 but it was actually 5 minutes slower than the actual time. She took 35 minutes to reach the cinema. In the end, she was 10 minutes early for the movie. What time did the movie begin? Give your answer in 24-hour clock format. Her watch is 5 minutes SLOW, so the real time was 1630 + 5 = 1635. The journey took 35 minutes, so she arrived at 1635 + 35 = 1710. She was 10 minutes early, so the movie began 10 minutes after she arrived: 1710 + 10 = 1720. Taking 5 minutes off instead of adding it. A slow watch shows a time EARLIER than the real one, so the real time is later. Ken had some money. He used 1/2 of it to buy a shirt and 1/5 of it to buy a pair of pants. He had $180 left. How much money did he have at first? He spent 1/2 + 1/5 of his money. Using tenths: 5/10 + 2/10 = 7/10. So what is left is 10/10 − 7/10 = 3/10, and that is $180. One tenth is 180 ÷ 3 = $60, so the whole is 10 × $60 = $600. Adding 1/2 and 1/5 as 2/7 by adding tops and bottoms. Fractions need a common denominator before they can be added. A rectangular piece of paper PQRS was folded as shown below, so that the edge RQ came down onto the line RU. Find ∠TRU. The corner of the rectangle at R is 90°, and it is now split into three: ∠QRT, ∠TRU and ∠URS. The fold carried RQ onto RU, and a crease always sits exactly halfway between where an edge was and where it went — so ∠QRT = ∠TRU. That leaves 90 = ∠TRU + ∠TRU + 52, so 2 × ∠TRU = 38 and ∠TRU = 19°. Halving 52 instead. The 52° is the LEFTOVER angle outside the fold; the two equal angles are what remains of the 90° after taking it off. The line graph shows the earnings at Mr Lim’s chicken rice stall over a 6-month period: January $3000, February $4500, March $5000, April $3500, May $3000, June $2500. Mr Lim charged $5 for every plate of chicken rice. What is the difference in the number of plates of chicken rice sold between the highest earnings month and the lowest earnings month? The highest month is March at $5000 and the lowest is June at $2500. The difference in money is $5000 − $2500 = $2500. Each plate is $5, so that is 2500 ÷ 5 = 500 plates. Working out each month's plates and then subtracting is fine too (1000 − 500 = 500), but dividing the DIFFERENCE once is quicker. The real slip is reading May as the lowest — June is lower. The figure is made up of 2 identical squares overlapping each other. The unshaded rectangle where they overlap has an area of 10 cm². The total area of the shaded parts is 142 cm². Find the length of the square. Give your answer in cm. Each square is made of its shaded part plus its half of the overlap. Both squares contain the whole overlap, so the two squares together = shaded + overlap + overlap = 142 + 10 + 10 = 162 cm². That is two squares, so one square is 162 ÷ 2 = 81 cm². A square of area 81 has side 9 cm, since 9 × 9 = 81. Subtracting the overlap once instead of adding it twice. The overlap belongs to BOTH squares, so it has to be counted twice to rebuild them. A carpet is laid on a rectangular floor measuring 20 m by 13 m, leaving a border of 3 m around it. Find the area of the floor that is not covered by the carpet. Give your answer in square metres. The whole floor is 20 × 13 = 260 m². The border is 3 m on EVERY side, so it takes 3 m off each end of both measurements: the carpet is (20 − 3 − 3) by (13 − 3 − 3) = 14 × 7 = 98 m². Uncovered = 260 − 98 = 162 m². Taking off 3 m once instead of twice, giving a 17 × 10 carpet. The border runs all the way round, so each dimension loses 3 m at BOTH ends. Max bought the same number of books and files. Each book cost $7 and each file cost $2. He paid $243 altogether. How many books and files did he buy altogether? Because the numbers are equal, buy them in pairs: one book and one file cost $7 + $2 = $9. Then $243 ÷ $9 = 27 pairs. Each pair is 2 items, so altogether he bought 27 × 2 = 54 books and files. Answering 27 — that is the number of BOOKS (and of files), but the question asks for both kinds added together. John bought 2 crates of oranges. Crate A contained 200 more oranges than Crate B. After he transferred 50 oranges from Crate B to Crate A, there were twice as many oranges in Crate A than in Crate B. How many oranges were there in Crate A at first? After the move, Crate B has 50 fewer and Crate A has 50 more, so the gap between them grows from 200 to 200 + 50 + 50 = 300. At that point A is twice B, so the gap is exactly one B: B after = 300. So B at first = 300 + 50 = 350, and A at first = 350 + 200 = 550. Thinking the gap stays at 200. Moving oranges from one crate to the other changes BOTH, so the gap shifts by twice the number moved. Some circles and triangles are arranged in a repeating pattern: triangle, circle, circle, circle, triangle, circle, circle, circle, triangle, … How many circles are there if there are 122 shapes? One repeat is a triangle and 3 circles — 4 shapes. In 122 shapes: 122 ÷ 4 = 30 repeats with 2 shapes left over. The 30 repeats give 30 × 3 = 90 circles. The 2 leftover shapes continue the pattern: a triangle, then a circle. So 90 + 1 = 91 circles. Answering 90 and forgetting the leftovers, or counting the leftover 2 shapes as 2 circles. The pattern always restarts with a TRIANGLE. Mr Tan has less than 40 sweets for a group of pupils. If he gives his pupils 4 sweets each, he will be short of 1 sweet. If he gives each pupil 5 sweets each, he will be short of 11 sweets. How many pupils does he have in the group? Compare the two plans. Going from 4 sweets each to 5 sweets each means one extra sweet per pupil, and the shortfall grows from 1 to 11 — that is 10 more sweets needed. So there are 10 pupils. (Check: 10 pupils × 4 = 40, short of 1, so he has 39, which is indeed less than 40.) Trying to find the number of sweets first. The difference between the two plans gives the number of PUPILS straight away, because each pupil accounts for exactly one extra sweet. There are 17 pupils in a class. The teacher gets each pupil to give a hi-five to every classmate once. How many hi-fives are exchanged altogether? Each pupil hi-fives the other 16, which suggests 17 × 16 = 272. But a hi-five needs two people, so every one has been counted twice — once from each side. The real number is 272 ÷ 2 = 136. Answering 272 and forgetting to halve. A hi-five between Amy and Ben is the SAME hi-five, however you look at it. Jean is watching a movie in a hall. The chairs are arranged in rows with the same number, and all are occupied. There are 7 people behind Jean and 4 people in front of her. There are 3 people on her left and 4 people on her right. How many people are watching the movie? Count Jean's own column: 7 behind + 4 in front + Jean herself = 12 rows. Count her row: 3 on the left + 4 on the right + Jean herself = 8 seats across. Every seat is taken, so the hall holds 12 × 8 = 96 people. Forgetting to add Jean herself to each count, giving 11 × 7 = 77. She is in her own row and her own column. Mike had 72 oranges and 45 pears. He packed the fruits such that each bag contained an equal number of oranges and an equal number of pears. He packed the fruits into as many bags as possible. How many fruits were there in each bag? The number of bags must divide both 72 and 45 exactly, and we want as many bags as possible — so it is the largest common factor. 72 = 8 × 9 and 45 = 5 × 9, so the highest common factor is 9. With 9 bags: 72 ÷ 9 = 8 oranges and 45 ÷ 9 = 5 pears in each, giving 8 + 5 = 13 fruits per bag. Answering 9 — that is the number of BAGS. Or using a smaller common factor like 3, which works but does not give as many bags as possible. The figure shows a rectangular garden 12 m by 16 m with a footpath (shaded) running along the top, down the left side, and part-way down the right side. The width of the footpath is 2 m. The area of the garden not covered by the footpath is 124 m². Find the length of AB, the part of the right-hand edge below the footpath. Give your answer in m. The garden is 12 × 16 = 192 m², so the footpath covers 192 − 124 = 68 m². Now build the footpath from its parts, taking care not to count the corners twice. The left strip is 2 × 16 = 32 m². The top strip, not counting the corner already used, is (12 − 2) × 2 = 20 m². That leaves 68 − 32 − 20 = 16 m² for the right-hand strip, which is 2 m wide, so it is 16 ÷ 2 = 8 m long below the top strip — reaching 2 + 8 = 10 m down from the top. AB is the rest of that edge: 16 − 10 = 6 m. Adding the three strips as 2×16 + 2×12 + 2×h and double-counting the two corner squares where they meet. In a General Knowledge quiz, 5 marks were awarded for a correct answer but 2 marks were deducted for an incorrect answer. Mei Ling answered all 50 questions and was awarded a total of 166 marks. How many questions did she answer correctly? Suppose every answer were correct: 50 × 5 = 250 marks. She actually scored 166, which is 84 marks less. Each wrong answer costs 5 marks not gained plus 2 marks deducted — 7 marks in all. So the number wrong is 84 ÷ 7 = 12, and the number correct is 50 − 12 = 38. Counting a wrong answer as costing only 2 marks. It costs 7: the 5 she did not win AND the 2 taken away. Uncle Lee had some pears in his shop. He sold 36 of them in the morning and threw away 5 bad ones. He sold 10 pears in the afternoon and packed half of the remaining pears into 10 bags. Each bag contained 7 pears. How many pears did he have at first? Work backwards. The 10 bags hold 10 × 7 = 70 pears, and that was HALF of what remained, so 140 pears remained after the afternoon sale. Before that he sold 10, so there were 150. Before the morning he had also lost 36 sold and 5 thrown away: 150 + 10 is already counted, so add back 36 + 5 = 41 to the 150, giving 191. Taking the 70 bagged pears as everything that was left. It is half of it, so the remainder is 140, not 70. Mother bought 5 more cups than plates. Each cup cost $8 and each plate cost $3. She paid $183 altogether. How many cups did she buy? Set the 5 extra cups aside first: they cost 5 × $8 = $40. That leaves $183 − $40 = $143 for equal numbers of cups and plates. One cup and one plate together cost $8 + $3 = $11, so there are $143 ÷ $11 = 13 pairs. So there are 13 plates and 13 + 5 = 18 cups. Answering 13 — that is the number of PLATES (and of the matched cups). The five extra cups still have to be added on. Sandy and Tracy had $600 altogether. Sandy gave 1/5 of her money to Tracy and then Tracy gave 1/4 of the money she then had to Sandy. In the end, both of them had the same amount of money. How much money did Sandy have at first? Work backwards from the end: they finished equal, so each had $300. The last move was Tracy giving away 1/4 of her money, keeping 3/4 — and 3/4 of Tracy's money was $300, so Tracy had $400 just before, and gave $100 to Sandy. So before that move Sandy had $300 − $100 = $200. That $200 was Sandy's money after giving away 1/5, so it is 4/5 of what she started with: 1/5 = $50, and Sandy began with 5 × $50 = $250. Taking 1/4 of Tracy's ORIGINAL money instead of the amount she held after Sandy's gift. Each fraction refers to the money at that moment. The total cost of 6 skirts and 3 blouses was $144. Three blouses cost as much as 2 skirts. Find the total cost of one skirt and one blouse. Swap the blouses for skirts: 3 blouses cost the same as 2 skirts, so 6 skirts + 3 blouses costs the same as 6 skirts + 2 skirts = 8 skirts. So 8 skirts = $144, and one skirt = $18. Then 3 blouses = 2 skirts = $36, so one blouse = $12. One skirt and one blouse = $18 + $12 = $30. Assuming a blouse costs the same as a skirt, or dividing $144 by 9 items. The swap is what makes everything one kind of thing. Study the staircase pattern: Figure 1 has 2 squares along the bottom and 1 above the right-hand one (3 squares). Figure 2 has 3 along the bottom, then 2, then 1 (6 squares). Figure 3 has 4, then 3, then 2, then 1 (10 squares). Find the total number of squares needed to form Figure 50. Each figure is a staircase. Figure 1 is 2 + 1 = 3, Figure 2 is 3 + 2 + 1 = 6, Figure 3 is 4 + 3 + 2 + 1 = 10. So Figure n counts down from (n + 1) to 1. Figure 50 is 51 + 50 + 49 + … + 1. Pair the ends: 51 + 1 = 52, 50 + 2 = 52, and so on — there are 51 numbers, giving 51 × 52 ÷ 2 = 1326. Counting down from 50 instead of 51. Figure 1's bottom row has 2 squares, not 1, so the bottom row of Figure 50 has 51. Sam had three coins in his wallet. They could be 10-cent coins, 20-cent coins or 50-cent coins. How many different possible amounts of money could Sam have in his wallet? List the ways to choose 3 coins from the three kinds, working down in an order so none is missed: three 10s = 30c; two 10s and a 20 = 40c; two 10s and a 50 = 70c; one 10 and two 20s = 50c; one of each = 80c; one 10 and two 50s = 110c; three 20s = 60c; two 20s and a 50 = 90c; one 20 and two 50s = 120c; three 50s = 150c. That is 10 combinations, and all 10 totals are different, so there are 10 possible amounts. Counting the ORDER of the coins as different (10-10-20 and 10-20-10 are the same wallet), or stopping before all ten combinations are found. A written-down order is what stops one being missed." }, { "id": "singa-2021-p4-pdf", "kind": "past-paper", "title": "SINGA 2021 Assessment — Primary 4 (SMC Practice Book)", "href": "https://www.dropbox.com/scl/fo/7b136f7n5gx0m5a3cfq1w/ACdfDFukqVp3FM4tIkYn4Ik?rlkey=hyyiq5gbfrzzggekaeg5cpjzi&preview=elsa0x28/papers/2026-08-08-smc-math/SMC%20Primary%204.pdf&dl=1", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.add-sub", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "logic.counting", "logic.reasoning", "measurement.length", "measurement.money", "measurement.time", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "number.remainder", "word-problem.model" ], "meta": "the original paper · SMC Primary 4.pdf", "blurb": "The original paper as a PDF, with the figures exactly as the publisher printed them — for when the reprinted question is not enough.", "text": "SMC Primary 4.pdf" }, { "id": "singa-2021-p4-key", "kind": "answer-key", "title": "SINGA 2021 Assessment — Primary 4 (SMC Practice Book)", "href": "keys/singa-2021-p4-key.html", "created": "2026-09-18", "skills": [ "data.tables-graphs", "decimal.add-sub", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "logic.counting", "logic.reasoning", "measurement.length", "measurement.money", "measurement.time", "number.factors-multiples", "number.four-operations", "number.patterns", "number.place-value", "number.remainder", "word-problem.model" ], "meta": "45 questions · every answer, method and trap", "blurb": "Every answer, with the method under it and the mistake to watch for. Best opened after you have had a go. Part published key, part worked out here — each question says which.", "text": "Write the missing number in the number pattern below: 7648, 7798, 7948, __________, 8248 Find the step first: 7798 − 7648 = 150, and 7948 − 7798 = 150 as well. So the pattern adds 150 each time: 7948 + 150 = 8098. (Check forwards: 8098 + 150 = 8248 ✓) Guessing from the last gap only. Check the step between at least two pairs, then check your answer reaches the number after it. Round the sum of 18.208 and 1.42 to the nearest whole number. Add first, lining up the decimal points: 18.208 + 1.420 = 19.628. Then round to the nearest whole number — the digit after the point is 6, which is 5 or more, so round up to 20. Rounding each number first (18 + 1 = 19). The question says round the SUM, and rounding early loses the 0.628 that pushes it over. Find the sum of all the factors of 81. List the factors in pairs: 1 × 81, 3 × 27, 9 × 9. So the factors are 1, 3, 9, 27 and 81 — the 9 appears only once because it pairs with itself. Sum = 1 + 3 + 9 + 27 + 81 = 121. Writing 9 twice because 9 × 9 = 81. A factor is listed once however many times it appears in a pair. David went to sleep at 21 50. He slept for 9 h 20 min. At what time did he wake up? (Give your answer in 24-hour clock format.) Add the hours first: 21 50 + 9 h = 30 50, which is past midnight, so take off 24 h to get 06 50 the next morning. Then add the 20 min: 06 50 + 20 min = 07 10. Writing 30 50, or adding the minutes as if 50 + 20 made 70. Sixty minutes make an hour, so 50 + 20 rolls over into the next hour. I am a 3-digit odd number. All my three digits are different. All my three digits are multiples of 3. The digit in the ‘hundreds’ place is twice the digit in the ‘ones’ place. What number am I? The digits that are multiples of 3 are 3, 6 and 9. The number is odd, so the ones digit is 3 or 9. If the ones digit were 9, the hundreds digit would be 18 — not a digit. So the ones digit is 3 and the hundreds digit is 2 × 3 = 6. The tens digit must be a different multiple of 3, which leaves 9. The number is 693. Counting 0 as a multiple of 3 and offering 603 as well. In primary work the multiples of 3 are 3, 6, 9, … — they start at 3, not 0. Using a protractor, measure ∠ABC. BA runs horizontally to the left from B; BC rises steeply up and slightly to the right. Put the protractor's centre on B with its baseline along BA, then read where BC crosses the scale. BC leans back past the upright, so the angle is obtuse — a little over 90°, not under it. The reading is 107°. Reading the wrong scale on the protractor and getting 73° instead. The two scales run in opposite directions, so decide FIRST whether the angle is acute or obtuse and pick the reading on that side of 90°. (A reading a degree or two either way is a fair measurement; the published key gives 107.) Mary gave 55 beads to Betty. Then she received 36 beads from Lydia. Mary had 342 beads in the end. How many beads did Mary have at first? Work backwards, undoing each step. Before receiving 36 she had 342 − 36 = 306. Before giving away 55 she had 306 + 55 = 361. Running the story forwards from 342, which adds the 55 and subtracts the 36 the wrong way round. Going backwards means every give becomes a take and every take becomes a give. A man changed $210 into $5 and $2 notes. The number of $5 notes and $2 notes was the same. How many $5 notes did he get? Because the numbers are equal, take one of each together: $5 + $2 = $7 per pair. Then $210 ÷ $7 = 30 pairs, so he got 30 five-dollar notes (and 30 two-dollar notes). Dividing $210 by $5 alone. The pairing is what makes the two equal counts usable in one division. The line graph shows the number of books borrowed by students in a library: March 500, April 600, May 800, June 1100, July 800. How many more books were borrowed in July than in March? Read July off the graph: 800. Read March: 500. \"How many more\" means subtract: 800 − 500 = 300. Reading the peak (June, 1100) instead of July. Follow the month labels along the bottom, not the highest point. Using the same line graph (March 500, April 600, May 800, June 1100, July 800), what was the total number of books borrowed from March to June? Add the four months named — March, April, May and June: 500 + 600 + 800 + 1100 = 3000 books. July is not included, because the question stops at June. READ THIS ONE BEFORE TEACHING IT. The published answer key says 3800, which is the total of all FIVE months including July. For the question as printed — March to June — the answer is 3000. Check the wording with her rather than the key: if she answers 3000 she has read the question correctly. The box below contains a mixture of circles and stars: 4 stars and 6 circles. ?/5 of the shapes in the box are stars. What is the missing number? Count first: 4 stars out of 10 shapes, so the fraction is 4/10. The answer must be written in fifths, so simplify by dividing top and bottom by 2: 4/10 = 2/5. The missing number is 2. Answering 4 — that is the number of stars, not the numerator once the fraction is written in fifths. The figure is made up of 5 identical squares arranged as a cross of diamonds, each outer square joined to the middle one along a full side. The perimeter of the figure is 132 cm. What is the area of each square? Five squares have 5 × 4 = 20 sides altogether. The middle square shares each of its 4 sides with an outer square, so 4 + 4 = 8 of those sides are tucked inside the figure and do not show. That leaves 20 − 8 = 12 sides on the outside. So 12 sides = 132 cm, one side = 132 ÷ 12 = 11 cm, and the area is 11 × 11 = 121 cm². Dividing 132 by 20 (all the sides) or by 4 (one square). Only the sides on the OUTSIDE make up the perimeter — count how many are hidden first. Andy bought 4 identical shirts at $12.80 each and spent the rest of the money on 8 identical pairs of socks. He gave $100 to the cashier and received 80 cents change. How much did each pair of socks cost? He spent $100 − $0.80 = $99.20 in total. The shirts cost 4 × $12.80 = $51.20, so the socks cost $99.20 − $51.20 = $48. There are 8 pairs, so each pair is $48 ÷ 8 = $6. Treating 80 cents as $80, or forgetting to take the change off the $100 before splitting the rest. Siti is 17 years old and Julia is 5 years old. How old will Julia be when she is 3/5 as old as Siti? The gap between their ages never changes: 17 − 5 = 12 years, now and forever. At the moment we want, Julia is 3 units and Siti is 5 units, so the gap is 5 − 3 = 2 units. So 2 units = 12 years, 1 unit = 6 years, and Julia (3 units) is 3 × 6 = 18. Taking 3/5 of 17 straight away and getting 10.2. Both of them get older, so the fraction applies to their ages LATER, not to Siti's age now — the fixed gap is what pins the moment down. The number of girls is 3/7 the number of boys at a National Kids’ Run competition. There are 960 more boys than girls. How many children took part in the competition? Draw boys as 7 units and girls as 3 units. The difference is 7 − 3 = 4 units = 960, so 1 unit = 960 ÷ 4 = 240. Altogether there are 7 + 3 = 10 units, so 10 × 240 = 2400 children. Answering 1680 (the boys) or 720 (the girls). The question asks for everyone, which is all 10 units. Mrs Thompson baked 175 tarts. She sold 3/5 of them. Each tart was sold for $4. How much money did she collect in total? Find how many she sold: 1/5 of 175 is 35, so 3/5 is 3 × 35 = 105 tarts. Each sold for $4, so she collected 105 × $4 = $420. Multiplying all 175 tarts by $4. She only sold three fifths of them — the unsold ones brought in nothing. In the diagram, ABEF is a square and BCDE is a rectangle. The length of AF is twice the length of BC and AB = 56 cm. Find the length of ED. ABEF is a square, so all four of its sides are equal: AF = AB = 56 cm. We are told AF is twice BC, so BC = 56 ÷ 2 = 28 cm. In rectangle BCDE, ED is the side opposite BC, and opposite sides of a rectangle are equal — so ED = 28 cm. Doubling instead of halving, giving 112. \"AF is twice BC\" makes BC the SMALLER one. Using the same diagram (ABEF a square, BCDE a rectangle, with A, B, C on one straight line and F, E, D on another), ∠ACF = 35° and ∠DCE = 18°. Find ∠FCE. At corner C the angle between the top line (towards A) and the side CD is a right angle, 90°. Three angles sit inside it, side by side: ∠ACF = 35°, then ∠FCE, then ∠ECD = 18°. So 35 + ∠FCE + 18 = 90, giving ∠FCE = 90 − 53 = 37°. Adding the two given angles and stopping at 53, or subtracting from 180 instead of 90. The corner of a rectangle is 90°. Jeff is 12 years older than Leon now. 10 years ago, the total age of Jeff and Leon was 78 years. How old is Leon now? Ten years ago both were 10 years younger, so their ages today total 78 + 10 + 10 = 98. Jeff is 12 more than Leon, so take the 12 off and split what is left equally: (98 − 12) ÷ 2 = 86 ÷ 2 = 43. Leon is 43 now. Adding only 10 to the 78 instead of 10 for EACH of them, or answering 55 (Jeff's age). Shirley packed 163 lollipops into plastic bags of 7 lollipops. She used the most number of plastic bags possible. She sold each plastic bag of lollipops at $6 and each remaining lollipop at $2. How much did she get after selling all the lollipops? Divide to find the bags: 163 ÷ 7 = 23 bags with 2 lollipops left over. The bags bring in 23 × $6 = $138, and the 2 loose lollipops bring in 2 × $2 = $4. Total = $138 + $4 = $142. Throwing away the remainder. The 2 left-over lollipops are still sold — at a different price, which is exactly why the remainder matters here. Melody baked some muffins. 2/7 of the muffins were blueberry muffins and the rest were chocolate muffins. There were 72 more chocolate muffins than blueberry muffins. How many muffins did she bake altogether? Blueberry is 2 units out of 7, so chocolate is the other 5 units. The difference is 5 − 2 = 3 units = 72 muffins, so 1 unit = 24. Altogether there are 7 units: 7 × 24 = 168 muffins. Treating 72 as one of the fractions rather than as the DIFFERENCE between them, or answering 120 (the chocolate ones). A glue stick cost $1.80. A pair of scissors cost $3.20. Thomas bought an equal number of glue sticks and scissors. He spent $25 in all. How many glue sticks and scissors did he buy in total? Buy them in pairs, since the numbers are equal: one glue stick and one pair of scissors cost $1.80 + $3.20 = $5. Then $25 ÷ $5 = 5 pairs. Each pair is 2 items, so in total he bought 5 × 2 = 10 items. Answering 5 — that is the number of PAIRS. The question asks for glue sticks and scissors added together. Gupta boarded the bus at 11.40 a.m. The bus ride took 38 minutes. After alighting from the bus, he walked for 28 minutes from the bus stop to his school. What time did he reach his school? (Give your answer in 24-hour clock format.) Add the two journeys to 11.40 a.m. A bus ride of 38 min: 11.40 + 20 min reaches 12.00 noon, and 18 min more gives 12.18 p.m. Then walk 28 min: 12.18 + 28 min = 12.46 p.m. In 24-hour clock that is 1246. Writing 0046 or 1146 by mishandling the crossing of 12 noon, or converting to 24-hour clock by adding 12 to a time that is already past noon. The 8-point compass shows the houses of 8 children around point O: Perry north, Ziming north-east, Bill east, Hadi south-east, Ali south, Sam south-west, Dave west, Shawn north-west. Roy is at point O facing Perry’s house. How many degrees must he turn in a clockwise direction so that he can face Shawn’s house? Each step round an 8-point compass is 360 ÷ 8 = 45°. Starting at Perry (north) and going CLOCKWISE, count the steps to Shawn (north-west): Ziming, Bill, Hadi, Ali, Sam, Dave, Shawn — that is 7 steps. So the turn is 7 × 45 = 315°. Turning the short way and answering 45°. That is anti-clockwise; the question asks for clockwise, which is the long way round. Tania had 156 stickers and Amelia had 114 stickers. After each of them gave away an equal number of stickers, Tania had 4 times as many stickers as Amelia. How many stickers did each of them give away? They give away the same number, so the gap between them never changes: 156 − 114 = 42. At the end Tania is 4 units and Amelia is 1 unit, so the gap is 3 units = 42, giving 1 unit = 14. So Amelia ended with 14 stickers, and she gave away 114 − 14 = 100. Answering 14 (what Amelia has LEFT) instead of what she gave away. The unchanging gap is the key — it is what makes the 42 usable. A rectangular piece of paper is folded as shown below, with both top corners folded down. The flat top that is left is 7 cm across, each folded corner covers 6 cm, and the height of the figure is 23 cm. What is the area of the piece of paper at first? The dashed lines show the paper's original top edge, which runs the full width: 6 + 7 + 6 = 19 cm. The folds only turned the corners down — they did not shorten the sheet — so the height is still 23 cm. Area = 19 × 23 = 437 cm². Using 7 cm as the width because that is the top edge you can see. The folded corners are part of the same sheet, so their 6 cm each still counts. The table shows the boys and girls in four Primary Four classes, with some cells blank: 4A has 22 boys and 18 girls; 4B has 25 girls and 41 students in total; 4C has 13 boys; 4D has 20 girls. There are as many students in Primary 4A as in Primary 4C. How many girls are there in Primary 4C? Fill in 4A first: 22 + 18 = 40 students. We are told 4C has as many students as 4A, so 4C also has 40. Of those, 13 are boys, so the girls number 40 − 13 = 27. Comparing 4A's BOYS with 4C's boys. The sentence says as many STUDENTS, which means the totals match, not the boys. Using the same table, the four classes have a total of 160 students. How many students are from the class that has the least number of students? Three of the classes are now known: 4A = 40, 4B = 41, 4C = 40, which is 121 students. So 4D = 160 − 121 = 39. Comparing 40, 41, 40 and 39, the smallest is 4D with 39 students. Stopping once 4D is found without checking it really is the smallest, or picking 4C's 13 boys as \"the least\". The question asks about whole classes. ABCD is a square of side 8 cm, cut out of a 28 cm by 12 cm rectangle. What is the area of the shaded part (the rectangle outside the square)? Find the whole rectangle: 28 × 12 = 336 cm². Find the square: 8 × 8 = 64 cm². The shaded part is everything except the square, so 336 − 64 = 272 cm². Taking 8 cm as the square's AREA instead of its side, or forgetting to subtract the square at all. The figures show a square and a rectangle. Square X has sides of 10 cm. Rectangle Y is 2 cm wide. The area of Square X is 4 times the area of Rectangle Y. Find the perimeter of Rectangle Y. Square X has area 10 × 10 = 100 cm². Rectangle Y is a quarter of that: 100 ÷ 4 = 25 cm². Its width is 2 cm, so its length is 25 ÷ 2 = 12.5 cm. Perimeter = 2 × (12.5 + 2) = 2 × 14.5 = 29 cm. Multiplying by 4 instead of dividing. Square X is the BIGGER one, so Rectangle Y's area must come out smaller. Find the perimeter of the I-shaped figure below. All lines meet at right angles. The figure is 14 cm across and 16 cm tall; the top and bottom bars are each 5 cm deep, and the waist is set in 3 cm on the left and 4 cm on the right. For a shape like this with no overhangs, start with the rectangle that just surrounds it: 2 × (14 + 16) = 60 cm. Then each notch cut into the side adds twice its depth, because you walk in and back out again: the left notch adds 2 × 3 = 6 cm and the right notch adds 2 × 4 = 8 cm. Perimeter = 60 + 6 + 8 = 74 cm. Assuming the notches make the perimeter SMALLER. Cutting a bite out of the side removes no edge — it replaces one straight edge with three, so the perimeter grows. A bookshop had a total of 160 erasers and pencils. After 28 erasers and 24 pencils were sold, the number of erasers became thrice the number of pencils. How many pencils did the bookshop have at first? After the sale, 160 − 28 − 24 = 108 items were left. At that point erasers were 3 units and pencils 1 unit, so 4 units = 108 and 1 unit = 27 — that is the pencils left. Before the sale the shop had 27 + 24 = 51 pencils. Answering 27, the pencils remaining. The 24 sold pencils have to be added back to reach \"at first\". A road divider is painted in repeating segments: white 2 m, black 5 m, white 2 m, black 5 m, and so on, beginning and ending with white. If the road divider is 149 m long, how many white segments are there? One repeat is a white and a black together: 2 + 5 = 7 m. In 149 m: 149 ÷ 7 = 21 repeats with 2 m left over. The 21 repeats give 21 white segments, and the last 2 m is exactly one more white segment. So 21 + 1 = 22. Answering 21 and ignoring the 2 m remainder — which is precisely the length of one more white segment, and is why the divider ends white. There are 256 balls in a box. 1/4 of the balls are red and 5/8 of them are white. The rest are black. How many more white balls than black balls are there? Red = 1/4 of 256 = 64. White = 5/8 of 256 = 5 × 32 = 160. Black is what is left: 256 − 64 − 160 = 32. So there are 160 − 32 = 128 more white balls than black. Answering 160 (the white balls) instead of the difference, or adding 1/4 and 5/8 as 6/12 by adding tops and bottoms. John spilled some paint on his report card, hiding his Mathematics and Science scores. English was 86, Chinese was 80, and the total of all four subjects was 342. He scored 12 fewer marks in Science than in Mathematics. How many marks did he score for Mathematics? Maths and Science together = 342 − 86 − 80 = 176. Science is 12 fewer than Maths, so take that 12 off and split the rest equally: (176 − 12) ÷ 2 = 164 ÷ 2 = 82 — that is Science. Maths = 82 + 12 = 94. Splitting the 176 evenly into 88 and 88 and forgetting the 12, or adding the 12 to the wrong subject. Science is the SMALLER one. Charles bought 9 bags of sweets. Each bag had 48 sweets. He gave away 5/6 of all the sweets. How many sweets did he have left? Find the total first: 9 × 48 = 432 sweets. He gave away 5/6, so he kept the other 1/6: 432 ÷ 6 = 72 sweets. Working out 5/6 of 432 (= 360) and giving that as the answer. That is what he gave AWAY; the question asks what is left. Study the pattern of hexagons made from sticks: Pattern 1 is one hexagon (6 sticks), Pattern 2 is two hexagons sharing a side (11 sticks), Pattern 3 is three in a row (16 sticks). In a particular pattern, 206 sticks are used. What is the pattern number? The first hexagon takes 6 sticks. Every hexagon after that shares a side with the one before, so it only needs 5 more. Sticks = 6 + 5 × (pattern number − 1). Setting that to 206: 206 − 6 = 200 extra sticks, and 200 ÷ 5 = 40 more hexagons, so the pattern number is 40 + 1 = 41. Dividing 206 by 6 because each hexagon \"has 6 sides\". After the first, each new hexagon adds only 5 sticks — the shared side is already there. The figure is made up of 4 identical rectangles: two standing upright side by side (22 cm tall, 14 cm across the pair) resting centrally on two more lying end to end to form the base. Find the length of XY, the part of the base sticking out to the right of the upright pair. The two uprights together are 14 cm across, so each rectangle is 14 ÷ 2 = 7 cm wide, and each is 22 cm long. The base is two of the same rectangles laid end to end, so it is 22 + 22 = 44 cm long. The upright pair sits centrally, so the base sticks out equally at both ends: (44 − 14) ÷ 2 = 30 ÷ 2 = 15 cm. Forgetting that all four rectangles are IDENTICAL, so the base's length comes from the uprights' 22 cm. Or taking the whole 30 cm overhang instead of halving it between the two sides. Kate and Sam had the same number of books at first. Kate gave away 2/3 of her books while Sam gave away 1/6 of his books. The number of books that Kate gave away was 63 more than Sam. Find the number of books that each of them had at first. They started with the same number, so both fractions are of the same amount. Kate gave 2/3 and Sam gave 1/6. In sixths that is 4/6 and 1/6, so Kate gave 3/6 — that is half — more than Sam. If half the books is 63, the whole is 63 × 2 = 126 books each. Subtracting the fractions as 2/3 − 1/6 = 1/3 by taking the bottoms away too. Rewrite both in sixths before subtracting. Bob wrote letters in a repeating pattern: Z E S T Z E S T Z E S T Z E … How many letters ‘Z’ and ‘T’ are there altogether if there is a total of 87 letters in the whole series? One repeat is Z E S T — 4 letters. In 87 letters: 87 ÷ 4 = 21 repeats with 3 left over. The 21 repeats give 21 Zs and 21 Ts. The 3 leftover letters restart the pattern as Z, E, S — one more Z and no more T. So Z = 22 and T = 21, giving 22 + 21 = 43. Splitting the leftovers evenly, or counting a T in the remainder. The leftover always starts from the BEGINNING of the pattern, so it reaches Z before it reaches T. The mass of a container filled with 2 identical ping pong balls was 80 g. The mass of the same container filled with 3 identical marbles was 220 g. The mass of each marble was 3 times the mass of each ping pong ball. What was the mass of 1 ping pong ball? Swap the marbles for ping pong balls: each marble weighs the same as 3 ping pong balls, so 3 marbles weigh the same as 9 ping pong balls. Now both weighings hold the same container: container + 2 balls = 80 g, and container + 9 balls = 220 g. Subtracting removes the container: 7 balls = 220 − 80 = 140 g, so one ball is 140 ÷ 7 = 20 g. Dividing 80 by 2 and forgetting the container has a mass of its own. Subtracting the two weighings is what makes the container disappear. Mrs Sim wanted to buy 20 glass bowls but she was short of $8. She then bought 19 glass bowls and had $6 left. Each glass bowl cost the same. How much money did Mrs Sim have at first? Compare the two plans. Buying one bowl fewer took her from being $8 short to having $6 spare — a swing of 8 + 6 = $14. That swing is the price of exactly one bowl, so a bowl costs $14. She bought 19 of them and had $6 left, so she started with 19 × $14 + $6 = $266 + $6 = $272. Subtracting the 8 and the 6 instead of adding them. Being short and having spare are on opposite sides of zero, so the gap between the two plans is their sum. Lisa and Max each had a roll of ribbon with the same length. They cut their roll of ribbon into shorter pieces. Each piece of Lisa’s ribbon was 6 cm. Each piece of Max’s ribbon was 10 cm. After cutting, Lisa had 16 more pieces of ribbon than Max. What was the length of each roll of ribbon? Look at a length both cut sizes divide — 30 cm. In every 30 cm, Lisa gets 30 ÷ 6 = 5 pieces and Max gets 30 ÷ 10 = 3 pieces, so Lisa is 2 pieces ahead for every 30 cm. She is 16 pieces ahead in all, which is 16 ÷ 2 = 8 lots of 30 cm. So each roll was 8 × 30 = 240 cm. Multiplying 16 by 6 or by 10. The 16 counts PIECES, not centimetres, so it has to be converted through a length that both cut sizes fit into. A picture frame is made up of 4 identical rectangular pieces joined together as a pinwheel. The area of each rectangular piece is 69 cm² and its breadth is 3 cm. A square picture fits exactly in the centre of the frame. What is the area of the picture? Each frame piece is 69 cm² with breadth 3 cm, so its length is 69 ÷ 3 = 23 cm. In a pinwheel frame each piece lies along one side of the picture, overlapping the next at the corner: the picture's side is the piece's length minus its breadth, 23 − 3 = 20 cm. So the picture's area is 20 × 20 = 400 cm². Taking the picture's side as the full 23 cm. Look at one side of the hole: the neighbouring piece's 3 cm width eats into it at one end. Sally arranges a pattern with sticks and coins. Pattern 1 uses 6 coins and 7 sticks (13 in all), Pattern 2 uses 9 coins and 12 sticks (21 in all), Pattern 3 uses 12 coins and 17 sticks (29 in all). There is a total of 93 coins and sticks in one of the patterns. What is the Pattern Number? Look at the totals: 13, 21, 29. Each one is 8 more than the last. So the total is 13 + 8 × (pattern number − 1). Setting that to 93: 93 − 13 = 80, and 80 ÷ 8 = 10 more steps, so the pattern number is 10 + 1 = 11. Dividing 93 by 8 and answering 11 remainder 5 without checking, or forgetting the +1 because Pattern 1 is the starting point, not a step." }, { "id": "smc-2026-mock-pdf", "kind": "past-paper", "title": "SMC 2026 Mock Exam — Grade 4 (Geniebook)", "href": "https://www.dropbox.com/scl/fo/7b136f7n5gx0m5a3cfq1w/ACdfDFukqVp3FM4tIkYn4Ik?rlkey=hyyiq5gbfrzzggekaeg5cpjzi&preview=elsa0x28/papers/2026-08-08-smc-math/MockTestSMC_G4_2026%20-%20WS1621.pdf&dl=1", "created": "2026-09-17", "skills": [ "data.tables-graphs", "fraction.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.rate", "measurement.time", "number.factors-multiples", "number.four-operations", "number.place-value", "number.remainder", "word-problem.model" ], "meta": "the original paper · MockTestSMC_G4_2026 - WS1621.pdf", "blurb": "The original paper as a PDF, with the figures exactly as the publisher printed them — for when the reprinted question is not enough.", "text": "MockTestSMC_G4_2026 - WS1621.pdf" }, { "id": "smc-2026-mock-key", "kind": "answer-key", "title": "SMC 2026 Mock Exam — Grade 4 (Geniebook)", "href": "keys/smc-2026-mock-key.html", "created": "2026-09-17", "skills": [ "data.tables-graphs", "fraction.add-sub", "fraction.equivalent", "fraction.of-quantity", "geometry.angles", "geometry.area-perimeter", "geometry.symmetry", "logic.counting", "logic.reasoning", "measurement.rate", "measurement.time", "number.factors-multiples", "number.four-operations", "number.place-value", "number.remainder", "word-problem.model" ], "meta": "31 questions · every answer, method and trap", "blurb": "Every answer, with the method under it and the mistake to watch for. Best opened after you have had a go. This paper prints NO answers, so these were worked out here and can be argued with.", "text": "The digit 6 in 63 470 has the same value as __________. Read the places from the RIGHT: 0 ones, 7 tens, 4 hundreds, 3 thousands, 6 ten-thousands. So the 6 is worth 6 × 10 000 = 60 000. Counting places from the left, or counting the digit's position (\"it is the 1st digit\") instead of its place value. How many sixths are there in 2 wholes? One whole holds 6 sixths, so 2 wholes hold 2 × 6 = 12 sixths. Answering 6 — that is how many sixths are in ONE whole, not two. The following numbers are arranged in descending order: A, 17 898, 8929, 879. A is a 5-digit odd number. What is the value of A? Descending means A is the largest, so A must be GREATER than 17 898 — that rules out 17 791. A must also be odd, so it ends in 1, 3, 5, 7 or 9 — that rules out 18 642 and 27 424. Only 71 889 passes both tests. Checking only that the number is odd and picking 17 791, forgetting A also has to be bigger than 17 898. Four different shapes J, K, L and M are shown below. Which of the following figures are symmetrical? Fold each shape and see whether the two halves land on each other. J (isosceles triangle) folds down the middle. K (isosceles trapezium) folds down the middle. L (square) folds four different ways. M is a slanted parallelogram — there is no fold that makes its halves match. So J, K and L only. Thinking a parallelogram is symmetrical because it looks balanced. It has ROTATIONAL symmetry (turn it half-way round and it matches) but no LINE of symmetry — and this question asks about folding. The clock shown below was 45 minutes behind the actual time. To set the clock to the correct time, how many 1/4-turn(s) must the minute hand be moved clockwise? A full turn of the minute hand is 60 minutes, so a 1/4-turn is 60 ÷ 4 = 15 minutes. The clock is 45 minutes behind, so the minute hand must go forward 45 minutes. 45 ÷ 15 = 3 quarter-turns. Moving the hand 45 minutes and answering 45, or working with the hour hand instead of the minute hand. The figure below is made up of 4 identical rectangles. Find the perimeter of the figure. The arrows say each rectangle is 10 cm long and 2 cm wide. The figure is a pinwheel — turn it a quarter turn and it looks exactly the same, so the outline is four identical pieces. One piece is 2 cm (the end of an arm) + 10 cm (the long side of that arm) + 8 cm (the part of the next arm still showing) = 20 cm. Four of them: 4 × 20 = 80 cm. Adding the four rectangles' perimeters (4 × 24 = 96 cm). That counts the edges where the rectangles touch, and those edges are inside the figure, not on its outline. Uncle Jack bought 60 boxes of pens. There were 9 pens in each box. He repacked the pens into smaller packets of 5 each. How many packets are there? First find how many pens there are altogether: 60 × 9 = 540 pens. Then share them into packets of 5: 540 ÷ 5 = 108 packets. Stopping at 540 (that is the number of PENS, not packets), or dividing 60 by 5 and ignoring the 9 pens per box. Which of the following is NOT an equivalent fraction of 1/4? Multiply the top and bottom of 1/4 by the same number: 2/8 ✓, 3/12 ✓, 6/24 ✓. For sixteenths you would need 4/16 to make 1/4, so 5/16 is the one that does not fit. Reading past the word NOT and picking one that IS equivalent. Underline NOT before looking at the options. Helen was at a restaurant in a zoo. At the restaurant, Helen was facing south-west at first. She then made a 3/4-turn clockwise. Which animal would she be facing then? Each 1/4-turn is 90°. Going CLOCKWISE from south-west: 1/4-turn → north-west, 2/4-turn → north-east, 3/4-turn → south-east. On the map the south-east path leads to the Monkey. Turning anticlockwise, which lands on north-west (the Elephant). Check which way the clock hands go before you start counting. At a bakery, muffins are sold at the prices shown. Nick wants to order 57 muffins for a birthday party. What is the least amount of money he will need to pay for the muffins? Work out the value of each deal first: 10 for $8.90 is 89c each, 6 for $5.70 is 95c each, and a single is $1.40. So take as many 10-packs as possible: 5 × 10 = 50 muffins for 5 × $8.90 = $44.50. The remaining 7 are cheapest as one 6-pack plus one single: $5.70 + $1.40 = $7.10. Total = $44.50 + $7.10 = $51.60. Buying six 10-packs (60 muffins for $53.40) or paying for the last 7 as singles ($9.80). Both are more than $51.60 — always price the leftover two ways. Using all the digits 8, 0, 9, 2, form the smallest multiple of 5. A multiple of 5 ends in 0 or 5. There is no 5 among the digits, so the number must end in 0. That leaves 8, 9 and 2 for the first three places, and the smallest arrangement of those is 2, 8, 9. So the number is 2890. Forgetting that ALL four digits must be used, or trying to put the 0 first to make the number small. The table below shows the 3-h PSI readings from 8 am to 12 noon on 22nd February. Which one of the line graphs best represents the information in the table? Describe the numbers before looking at the graphs: 35 → 55 is a sharp RISE, then 55 → 50 → 45 → 40 falls every single step. Only graph (2) starts low, jumps to its highest point at 9 am, and then goes down at every point after. Picking (4), which rises again at 11 am. The table never rises after 9 am — checking the LAST few points is what separates (2) from (4). Shaun ran 3 km on Monday. He ran 2/5 km more on Monday than on Tuesday. How many kilometres did he run on both days? Monday = 3 km. He ran 2/5 km MORE on Monday, so Tuesday is the smaller one: 3 − 2/5 = 2 3/5 km. Both days together = 3 + 2 3/5 = 5 3/5 km. Adding 2/5 to Monday instead of subtracting (giving 6 2/5), or answering 2 3/5 — that is Tuesday alone, and the question asked for BOTH days. Susan and Tina had the same number of beads at first. After Susan threw away 18 beads and Tina bought another 72 beads, Tina had 4 times as many beads as Susan. How many beads did Susan have at first? Draw Susan's beads at the end as 1 unit and Tina's as 4 units. They started equal, then Susan lost 18 and Tina gained 72, so the gap between them grew by 18 + 72 = 90 beads. That gap is 4 units − 1 unit = 3 units, so 1 unit = 90 ÷ 3 = 30 beads. Susan has 30 left, and she threw away 18, so she started with 30 + 18 = 48 beads. Answering 30 — that is what Susan has LEFT at the end, not what she started with. Read the question again before writing the number down. Among Jim and two other friends, John is the lightest and Jack is the heaviest. The table below shows the total weight of two boys weighing themselves each time. Find Jim's weight. The three readings are the three possible pairs. Since John is lightest and Jack is heaviest, the LIGHTEST pair is John + Jim = 54 and the HEAVIEST pair is Jim + Jack = 70, leaving John + Jack = 56. Adding all three readings counts every boy exactly twice: 54 + 56 + 70 = 180, so the three boys together weigh 90 kg. Jim = 90 − (John + Jack) = 90 − 56 = 34 kg. Assuming the readings are listed in a helpful order (that the 1st reading is John + Jim). Sort them smallest to largest first — it is the smallest and largest pairs that the clue pins down. The table shows the number of customers at a restaurant over 4 days; part of it was covered by a stain. The number of customers on Day 3 is equal to the total number of customers on Day 1 and Day 2. The total number of customers on all 4 days is 240 when rounded to the nearest ten. What is the greatest possible number of customers on Day 4? Day 3 = Day 1 + Day 2 = 30 + 45 = 75. So Days 1, 2 and 3 come to 30 + 45 + 75 = 150. The 4-day total rounds to 240 to the nearest ten, so the real total is anywhere from 235 to 244. The question wants the GREATEST Day 4, so take the greatest total: 244 − 150 = 94 customers. Taking the total as exactly 240 and answering 90. \"Rounds to 240\" means a range (235 to 244), and the largest number in that range is what makes Day 4 largest. The figure is made up of 2 identical squares and 3 identical rectangles. What is the length of the unknown side? Look down the right-hand side: 18 cm is the top square's side plus the three stacked rectangles. Now look at the left: the bottom square sits beside those same three rectangles, so the three rectangles stacked are exactly one square's side tall. That makes 18 cm = one square + one square = 2 squares, so each square has side 9 cm. Along the bottom the whole figure is 26 cm, and the top square (9 cm wide) is lined up at the right-hand end, so the marked length = 26 − 9 = 17 cm. Forgetting the two squares are IDENTICAL. That fact is what lets 18 be split as 9 + 9 — without it there is nothing to pin the size down. The table shows the programme for the Children's Day celebration. What was the total duration of the Form Teachers' performances? Pick out only the rows that say Form Teachers' performance. Primary 4: 08 00 to 08 10 = 10 min. Primary 5: 08 10 to 08 25 = 15 min. Primary 6: 08 40 to 08 50 = 10 min. Total = 10 + 15 + 10 = 35 min. Including the Mass Singing from 08 25 to 08 40 because it sits between two performances, giving 50 min. It is not a Form Teachers' performance — the rows must be chosen by NAME, not by position. A farmer planted some trees in a straight row, at an equal distance apart from one another. The distance between the 2nd tree and the 5th tree was 2340 m. What was the distance between the 1st tree and the 10th tree? Count GAPS, not trees. From the 2nd tree to the 5th there are 5 − 2 = 3 gaps, so one gap = 2340 ÷ 3 = 780 m. From the 1st tree to the 10th there are 10 − 1 = 9 gaps, so the distance = 9 × 780 = 7020 m. Counting trees instead of gaps — using 4 trees and 10 trees instead of 3 gaps and 9 gaps. Mark the gaps on a quick sketch before dividing. Barry has a garden with an area of 216 m². It is made up of a rectangle and a square. The area of the rectangle is 5 times the area of the square. Barry wants to build a fence around part of his garden as indicated by the dashes in the figure. Given that the breadth of the rectangle is 9 m, how many metres of fence does he need? The square is 1 share and the rectangle is 5 shares, so the garden is 6 equal shares: 216 ÷ 6 = 36 m² for the square and 5 × 36 = 180 m² for the rectangle. A square of area 36 m² has side 6 m. The rectangle has area 180 m² and breadth 9 m, so its length = 180 ÷ 9 = 20 m. Now walk the outline: 20 (top) + 9 (right side) + 9 (left side) + 14 (the bottom of the rectangle, which is 20 minus the 6 m the square covers) + 6 + 6 + 6 (the square's other three sides) = 70 m. Adding the two shapes' perimeters separately (2×(20+9) + 4×6 = 82 m). Where the square joins the rectangle there is no fence — that edge is inside the garden. At first, Mary had $166 and Larry had $304. Each of them bought 5 similar plates, and each plate was the same price. After buying the plates, Larry had 4 times as much money as Mary had left. What was the price of one plate? They spend exactly the same amount, so the GAP between them never changes: 304 − 166 = $138, before and after. At the end Larry has 4 units and Mary has 1 unit, so the gap is 4 − 1 = 3 units. 3 units = $138, so 1 unit = $46 — that is Mary's money left. Mary spent 166 − 46 = $120 on 5 plates, so one plate costs 120 ÷ 5 = $24. Not noticing that the difference stays the same when both people spend the same amount. That one idea turns a hard problem into two divisions. At a bento shop in Tokyo, three friends ordered: Hana bought 4 bento sets and 2 bottles of tea for $35. Ken bought 2 bento sets, 3 mochi cakes and 1 bottle of tea for $35.50. Miki bought 2 bento sets, 1 mochi cake and 2 bottles of tea for $27.50. 2 mochi cakes cost the same as 3 bottles of tea. What is the cost of 1 bento set? First swap every mochi for tea: 2 mochi = 3 tea, so 1 mochi = 1½ tea. Ken becomes 2 bento + (3 × 1½) tea + 1 tea = 2 bento + 5½ tea = $35.50. Miki becomes 2 bento + 1½ tea + 2 tea = 2 bento + 3½ tea = $27.50. Both have 2 bento sets, so subtracting cancels them: 2 tea = $35.50 − $27.50 = $8, so 1 tea = $4. Put that back into Miki: 2 bento + 3½ × $4 = $27.50 → 2 bento = $27.50 − $14 = $13.50 → 1 bento = $6.75. (Check with Hana: 4 × 6.75 + 2 × 4 = 27 + 8 = $35 ✓) Starting from Hana's line. Ken's and Miki's both contain 2 bento sets, so subtracting THOSE two removes the bento and leaves only tea — choosing which two lines to subtract is the whole trick. Joshua had more than 10 and fewer than 40 candies. When she packed the candies in bags of 7, she was left with 2 candies. When she packed them in bags of 3, she had no candies left. How many candies did Joshua have? \"Bags of 7 with 2 left over\" means the number is 2 more than a multiple of 7: 7+2=9, 14+2=16, 21+2=23, 28+2=30, 35+2=37. Keeping only those between 10 and 40: 16, 23, 30, 37. \"Bags of 3 with none left\" means a multiple of 3, and of those four only 30 is. So Joshua had 30 candies. Listing the multiples of 7 themselves (14, 21, 28, 35) and forgetting to add the remainder of 2. The figure below shows Marcus's backyard with 2 identical 8-m square ponds. What is the remaining area of the empty space around the two ponds in the backyard? Find the whole backyard first: 20 × 15 = 300 m². Each pond is an 8 m SQUARE, so its area is 8 × 8 = 64 m², and there are two: 2 × 64 = 128 m². The empty space is what is left: 300 − 128 = 172 m². Treating \"8-m square pond\" as an area of 8 m² instead of a square with side 8 m, or subtracting only one pond. At a bakery, the price of a cupcake was $2 and the price of a tart was $5. Mrs Lim paid $38 to buy a total of 13 cupcakes and tarts. How many more cupcakes than tarts did Mrs Lim buy? Suppose all 13 were cupcakes: that would cost 13 × $2 = $26. She actually paid $38, which is $12 more. Changing one cupcake into a tart adds $5 − $2 = $3 to the bill, so the number of tarts is 12 ÷ 3 = 4. Then cupcakes = 13 − 4 = 9. The question asks how many MORE: 9 − 4 = 5. Answering 9 — that is the number of cupcakes, not the DIFFERENCE the question asked for. Circle the words \"how many more\" before you start. After a rectangular piece of paper was cut into a maximum of 18 squares of sides 4 cm each, an L-shaped strip of paper was left over as shown. Given that the length of the rectangular piece of paper is 26 cm, what is the perimeter of the rectangular piece of paper before it was cut? Fit the 4 cm squares along the 26 cm length: 26 ÷ 4 = 6 remainder 2, so 6 squares fit across and a 2 cm strip is wasted down one side. With 18 squares in rows of 6, there must be 18 ÷ 6 = 3 rows, using 3 × 4 = 12 cm of the breadth. The leftover strip along the bottom is 1 cm, so the breadth is 12 + 1 = 13 cm. Perimeter = 2 × (26 + 13) = 2 × 39 = 78 cm. Trying to get the breadth by dividing the squares' area (18 × 16 = 288 cm²) by 26. The L-strip is wasted paper, so the squares' area is NOT the paper's area. A tortoise can crawl 8 m in one minute. A rabbit can run at a speed of 40 m in one minute. The two are having a race one day. While the tortoise crawls continually, the rabbit rests for 3 minutes after every 2 minutes' run. How far does the rabbit run when the tortoise crawls 360 m? The tortoise never stops, so use it as the clock: 360 ÷ 8 = 45 minutes for the whole race. The rabbit's pattern repeats every 2 + 3 = 5 minutes, and in each 5-minute cycle it covers 2 × 40 = 80 m. In 45 minutes there are 45 ÷ 5 = 9 complete cycles, so the rabbit runs 9 × 80 = 720 m. Working out 45 minutes × 40 m = 1800 m and forgetting the rabbit spends 3 of every 5 minutes resting. Find the TIME first, then the repeating cycle. The figure is not drawn to scale. It is made up of two identical rectangles overlapping each other, forming Square A. The area of Square A is 9 cm² and the area of each rectangle is 50 cm². The length of the rectangle is twice its breadth. Find the perimeter of the figure. Square A has area 9 cm², so its side is 3 cm. Each rectangle has area 50 cm² with length = 2 × breadth, so breadth × 2 × breadth = 50 → breadth × breadth = 25 → breadth = 5 cm and length = 10 cm. One rectangle's perimeter is 2 × (10 + 5) = 30 cm, so the two separately total 60 cm. But where they overlap, two sides of Square A are hidden inside each rectangle — four 3 cm edges in all. Perimeter of the figure = 60 − 4 × 3 = 60 − 12 = 48 cm. Taking Square A's side as 9 cm instead of 3 cm (9 is the AREA), or answering 60 by forgetting that the overlap hides edges. Mary had 48 stalks of flowers in a basket. 1/3 of them were roses and the rest were lilies and daisies. There were 4 more lilies than daisies. How many daisies should Mary buy such that the number of daisies would be 1/2 of the total number of flowers in the basket? Roses = 1/3 of 48 = 16, so lilies and daisies together = 48 − 16 = 32. There are 4 more lilies than daisies, so take the 4 off and share the rest equally: (32 − 4) ÷ 2 = 14 daisies, and 14 + 4 = 18 lilies. Now she buys more daisies. Wanting daisies to be HALF the total is the same as wanting daisies to equal everything else put together — that is roses + lilies = 16 + 18 = 34. She already has 14, so she must buy 34 − 14 = 20 daisies. (Check: 34 daisies out of 48 + 20 = 68 flowers, and 34 is half of 68 ✓) Taking \"half the total\" as half of 48 (= 24) and answering 10. The total GROWS with every daisy she buys, which is why it is easier to match the daisies against everything else. John divided the corridor of a school building into equal parts of length 4 m, and placed 2 potted plants in each part. For the same corridor, he divided it into equal parts of length 6 m and hung 5 lanterns in each part. If there were 24 more lanterns than potted plants, how long was the corridor? The two patterns use different part lengths, so compare them over a length both divide — 12 m. In 12 m there are 12 ÷ 4 = 3 parts of plants, giving 3 × 2 = 6 plants, and 12 ÷ 6 = 2 parts of lanterns, giving 2 × 5 = 10 lanterns. That is 10 − 6 = 4 more lanterns for every 12 m. We need 24 more, so the corridor is 24 ÷ 4 = 6 lots of 12 m = 72 m. (Check: 72 m gives 18 × 2 = 36 plants and 12 × 5 = 60 lanterns, and 60 − 36 = 24 ✓) Comparing 2 plants against 5 lanterns directly and using a difference of 3. The parts are different LENGTHS, so the counts can only be compared over the same distance. In a restaurant, a rectangular table can seat 10 people and a round table can seat 5 people. During lunch time, all the tables in the restaurant were occupied. The number of each type of table in the restaurant is more than 10 and less than 15. If there were 205 customers, how many rectangular and round tables altogether were there for the 205 customers? \"More than 10 and less than 15\" means each type is 11, 12, 13 or 14. Seats: 10 × rectangular + 5 × round = 205, and dividing everything by 5 gives 2 × rectangular + round = 41. Now test each value: 11 rectangular → round = 41 − 22 = 19 ✗; 12 → 17 ✗; 13 → 15 ✗ (15 is not less than 15); 14 → 41 − 28 = 13 ✓. So 14 rectangular and 13 round, giving 14 + 13 = 27 tables. Reading \"more than 10 and less than 15\" as including 10 and 15, or finding one pair that makes 205 without checking that BOTH counts sit inside the range." }, { "id": "flashcards-02", "kind": "english", "title": "Flashcards — containers, party, verbs, prepositions", "href": "english/flashcards-02.html", "created": "2026-09-17", "skills": [], "meta": "cards to tap through", "blurb": "The second set of cards: what things come in, what you do, and where things are.", "text": "The second set of cards: what things come in, what you do, and where things are." }, { "id": "flashcards-01", "kind": "english", "title": "Flashcards — people, animals, objects, places", "href": "english/flashcards-01.html", "created": "2026-09-10", "skills": [], "meta": "cards to tap through", "blurb": "The A7 picture vocabulary, as cards to tap through. Units 5 to 8.", "text": "The A7 picture vocabulary, as cards to tap through. Units 5 to 8." }, { "id": "mv-u5-8-speaking-01", "kind": "english", "title": "Movers Speaking — Units 5 to 8", "href": "english/mv-u5-8-speaking-01.html", "created": "2026-08-25", "skills": [], "meta": "speaking · 4 parts · 17 marks", "blurb": "Out loud, with a person. The page asks and models; a grown-up marks it from the printed card.", "text": "Movers Speaking — Units 5 to 8 Find the differences Look at the two pictures. They look the same, but five things are different. Tell me what is different. The cat and the kite Look at the four pictures. They tell a story. I will start it, and then you tell me the rest. Which one is different? Look at each row of four things. Tell me which one is different, and why it is different. Now some questions about you I am going to ask you some questions about yourself. Try to answer in a whole sentence, not just one word." }, { "id": "mv-u5-8-listening-01", "kind": "english", "title": "Movers Listening — Units 5 to 8", "href": "english/mv-u5-8-listening-01.html", "created": "2026-08-24", "skills": [], "meta": "listening · 5 parts · 25 marks", "blurb": "Five recordings, played twice, with a sheet to print. The audio is the question, so it is a real recording and not a robot voice.", "text": "Movers Listening — Units 5 to 8 Who is wearing what? Listen and match each name to what that person is wearing. There is one example. The school trip Listen and write. There is one example. You only need one or two words for each answer. The weather last week Charlie is telling Lily about the weather on his holiday. Match each day to the weather. There is one example, and there are two letters you will not need. Listen and choose Listen and choose the best answer. Careful — you will hear all three things mentioned, so wait for the whole conversation before you choose. Listen and colour Look at the picture. Tap a colour, then tap the thing you want to colour. There is one example." }, { "id": "ratio-share", "kind": "guide", "title": "Sharing in a ratio", "href": "rooms/ratio-share.html", "created": "2026-08-22", "skills": [ "word-problem.model", "number.four-operations" ], "meta": "a guide — read it", "blurb": "An takes 3 while Bình takes 4, and they end up with 133 — why the parts trick works, and what changes when the question gives you a gap instead of a total. (#32)", "text": "An takes 3 while Bình takes 4, and they end up with 133 — why the parts trick works, and what changes when the question gives you a gap instead of a total. (#32)" }, { "id": "equal-height-towers", "kind": "guide", "title": "Two towers, the same height", "href": "rooms/equal-height-towers.html", "created": "2026-08-22", "skills": [ "word-problem.model", "measurement.length" ], "meta": "a guide — read it", "blurb": "6 cubes reach as high as 5 cylinders — so why does the working say 5 : 6? Reading a ratio off a picture, and the check that catches you when you read it backwards. (#32)", "text": "6 cubes reach as high as 5 cylinders — so why does the working say 5 : 6? Reading a ratio off a picture, and the check that catches you when you read it backwards. (#32)" }, { "id": "mv-u5-8-progress-01", "kind": "english", "title": "Movers Progress Test — Units 5 to 8", "href": "english/mv-u5-8-progress-01.html", "created": "2026-08-20", "skills": [], "meta": "written paper · 6 parts · 50 marks", "blurb": "The written paper — ninety minutes, printed rather than marked on screen.", "text": "Movers Progress Test — Units 5 to 8 Vocabulary and definitions Look and read. Choose the correct words from the box and write them on the lines. Dialogue and communication Read the conversation between Lily and Charlie. Circle A, B or C. Reading gap fill Read the story. Choose a word from the box and write it on the lines. Grammar and language structure Complete the sentences using the correct word in brackets. Story comprehension and sentence completion Read the text and complete the sentences with 1, 2 or 3 words. Two marks each. Picture writing and story production Read the description of the park scene and write your answers. Two marks each." }, { "id": "smc-p4-drill-03", "kind": "drill", "title": "SMC Primary 4 — Drill 3: counting without losing count", "href": "rooms/smc-p4-drill-03.html", "created": "2026-08-12", "skills": [ "logic.counting" ], "meta": "6 questions · 12 marks · about 12 minutes", "blurb": "Short enough to finish in one sitting — the browser marks it and she sends the result back. Topics: logic · counting.", "text": "How many rectangles are there in the figure below? How many rectangles are there in the figure below? How many ways are there from A to B, moving only up or right along the lines? How many ways are there from A to B, moving only up or right along the lines? How many squares are there in the figure below? In the same 3-by-3 figure as the question above, how many rectangles are there that are NOT squares?" }, { "id": "counting-rectangles", "kind": "guide", "title": "How many rectangles?", "href": "rooms/counting-rectangles.html", "created": "2026-08-12", "skills": [ "logic.counting" ], "meta": "a guide — read it", "blurb": "Counting without losing count — pick two lines, then two more. (#32)", "text": "Counting without losing count — pick two lines, then two more. (#32)" }, { "id": "counting-paths", "kind": "guide", "title": "How many ways from A to B?", "href": "rooms/counting-paths.html", "created": "2026-08-12", "skills": [ "logic.counting" ], "meta": "a guide — read it", "blurb": "The add-as-you-go trick, and the bump in the figure that catches people. (#32)", "text": "The add-as-you-go trick, and the bump in the figure that catches people. (#32)" }, { "id": "smc-p4-drill-02", "kind": "drill", "title": "SMC Primary 4 — Drill 2: bars and units", "href": "rooms/smc-p4-drill-02.html", "created": "2026-08-11", "skills": [ "word-problem.model" ], "meta": "5 questions · 10 marks · about 10 minutes", "blurb": "Short enough to finish in one sitting — the browser marks it and she sends the result back. Topics: word problem · model.", "text": "Mai has 3 times as many stickers as Linh. Altogether they have 24 stickers. How many stickers does Mai have? Duc has 5 more sweets than Linh. Duc then gives Linh 2 sweets. How many more sweets does Duc have than Linh now? Nam has twice as many stamps as Hoa. If Nam gives Hoa 4 stamps, they will have the same number. How many stamps does Nam have at first? An has 3 times as many marbles as Bao. If An gives Bao 10 marbles, they will have the same number. How many marbles does An have at first? Linh has 4 times as many beads as Mai. If Linh gives Mai 9 beads, they will have the same number. How many beads do the two girls have altogether?" }, { "id": "bar-model", "kind": "guide", "title": "The bar model", "href": "rooms/bar-model.html", "created": "2026-08-11", "skills": [ "word-problem.model" ], "meta": "a guide — read it, or tap “Teach me this” and let it explain itself out loud", "blurb": "When someone gives someone else something — the trick behind Drill 1, Question 5. Tap “Teach me this” and it explains itself out loud.", "text": "When someone gives someone else something — the trick behind Drill 1, Question 5. Tap “Teach me this” and it explains itself out loud." }, { "id": "smc-p4-drill-01", "kind": "drill", "title": "SMC Primary 4 — Drill 1", "href": "rooms/smc-p4-drill-01.html", "created": "2026-08-08", "skills": [ "data.tables-graphs", "fraction.of-quantity", "number.factors-multiples", "number.four-operations", "number.place-value", "number.remainder", "word-problem.model" ], "meta": "5 questions · 10 marks · about 10 minutes", "blurb": "Short enough to finish in one sitting — the browser marks it and she sends the result back. Topics: data · tables graphs, fraction · of quantity, number · factors multiples, and more.", "text": "In the number 56 748, add the value of the digit 5 to the value of the digit 4. What do you get? What is the smallest number greater than 1 that leaves a remainder of 1 when it is divided by 2, by 3 and by 4? ELSA had 40 stickers. She gave 3/8 of them to her sister. How many stickers does she have left? The table shows how many cakes a shop sold on four days. How many more cakes were sold on the last two days than on the first two days? Ben has twice as many marbles as Cara. If Ben gives Cara 6 marbles, they will have the same number. How many marbles does Ben have at first?" } ]