{
"paper": {
"school": "ACS Primary",
"year": 2024,
"level": "P6",
"label": "Prelim",
"source_prefix": "ACS Primary 2024 P6 Prelim",
"has_answer_key": true
},
"questions": [
{
"n": 1,
"type_id": 1,
"question": "Which of the following is seven hundred and three thousand and twelve in numerals?",
"answer0": "730 012",
"answer1": "703 012",
"answer2": "73 012",
"answer3": "70 312",
"correct_answer": 1,
"skill_id": 2,
"difficulty_id": 1,
"explanation": "Seven hundred and three thousand = 703 000; and twelve = 12. So the number is 703 012.",
"hints": ["Break the words: 703 thousand, then 12.", "703 000 + 12 = 703 012."],
"source": "ACS Primary 2024 P6 Prelim P1 Q1",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Closest P6 skill is number notation; mapped to nearest available skill_id 2."
},
{
"n": 2,
"type_id": 1,
"question": "Divide 2 by \\(\\dfrac{8}{5}\\).",
"answer0": "\\(\\dfrac{5}{16}\\)",
"answer1": "\\(\\dfrac{4}{5}\\)",
"answer2": "\\(1\\dfrac{1}{4}\\)",
"answer3": "\\(3\\dfrac{1}{5}\\)",
"correct_answer": 2,
"skill_id": 206,
"difficulty_id": 1,
"explanation": "Dividing by a fraction = multiply by its reciprocal: \\(2 \\div \\dfrac{8}{5} = 2 \\times \\dfrac{5}{8} = \\dfrac{10}{8} = \\dfrac{5}{4} = 1\\dfrac{1}{4}\\).",
"hints": ["Multiply by the reciprocal of \\(\\dfrac{8}{5}\\).", "\\(2 \\times \\dfrac{5}{8} = \\dfrac{10}{8}\\), then simplify."],
"source": "ACS Primary 2024 P6 Prelim P1 Q2",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "MCQ kept (answer is a mixed number). Key = option 3 (0-based index 2)."
},
{
"n": 3,
"type_id": 1,
"question": "Round 4.195 to 2 decimal places.",
"answer0": "4.10",
"answer1": "4.19",
"answer2": "4.20",
"answer3": "4.29",
"correct_answer": 2,
"skill_id": 184,
"difficulty_id": 1,
"explanation": "The 3rd decimal digit is 5, so round the 2nd decimal place up: 4.195 rounds to 4.20.",
"hints": ["Look at the digit in the 3rd decimal place.", "5 rounds up: 4.19 becomes 4.20."],
"source": "ACS Primary 2024 P6 Prelim P1 Q3",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Rounding decimals; nearest available skill_id used."
},
{
"n": 4,
"type_id": 1,
"question": "\\(3 + \\dfrac{3}{100} + \\dfrac{30}{1000} =\\)",
"answer0": "3.033",
"answer1": "3.06",
"answer2": "3.33",
"answer3": "3.6",
"correct_answer": 1,
"skill_id": 173,
"difficulty_id": 1,
"explanation": "\\(\\dfrac{3}{100} = 0.03\\) and \\(\\dfrac{30}{1000} = 0.03\\). So \\(3 + 0.03 + 0.03 = 3.06\\).",
"hints": ["Convert each fraction to a decimal.", "0.03 + 0.03 = 0.06."],
"source": "ACS Primary 2024 P6 Prelim P1 Q4",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Fraction-to-decimal place value; nearest available skill_id used."
},
{
"n": 5,
"type_id": 1,
"question": "The figure is made up of 2 squares and one quarter circle. The lengths of the squares are 7 cm and 3 cm respectively. Find the perimeter of the shaded part. (Take \\(\\pi = \\dfrac{22}{7}\\))",
"answer0": "19.5 cm",
"answer1": "25 cm",
"answer2": "25.5 cm",
"answer3": "31 cm",
"correct_answer": 3,
"skill_id": 221,
"difficulty_id": 3,
"explanation": "Quarter-circle arc (radius 7) = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. The remaining straight edges of the shaded part total 7 + 3 + 3 + 4 + 3 = 20 cm. Perimeter = 11 + 20 = 31 cm.",
"hints": ["The curved edge is a quarter circle of radius 7 cm.", "Add the quarter-circle arc to the straight sides of the shaded region."],
"source": "ACS Primary 2024 P6 Prelim P1 Q5",
"image_needed": true,
"image_options": false,
"image_file": "q5.png",
"image_page": 4,
"image_bbox": [0.33, 0.22, 0.72, 0.42],
"image_loc": "middle of page, L-shaped shaded figure (big square 7cm with quarter circle, small square 3cm), below the question text",
"notes": "Key = option 4 (31 cm), 0-based index 3."
},
{
"n": 6,
"type_id": 1,
"question": "Find the volume of the solid.",
"answer0": "228 cm³",
"answer1": "288 cm³",
"answer2": "300 cm³",
"answer3": "372 cm³",
"correct_answer": 0,
"skill_id": 229,
"difficulty_id": 2,
"explanation": "The solid is a 10 cm × 5 cm × 6 cm cuboid (volume 300 cm³) with a 6 cm × 2 cm × 6 cm notch removed (72 cm³). Volume = 300 − 72 = 228 cm³.",
"hints": ["Find the full cuboid volume first.", "Subtract the small rectangular block cut out of the top."],
"source": "ACS Primary 2024 P6 Prelim P1 Q6",
"image_needed": true,
"image_options": false,
"image_file": "q6.png",
"image_page": 5,
"image_bbox": [0.18, 0.13, 0.65, 0.36],
"image_loc": "upper-middle of page, 3D cuboid solid with a notch cut from the top (10cm, 6cm, 2cm, 5cm, 6cm labels)",
"notes": "Key = option 1 (228 cm³), 0-based index 0."
},
{
"n": 7,
"type_id": 1,
"question": "A sum of money was shared among Anne, Ben, Cindy and David. Anne had twice as much money as Ben. David had twice as much money as Cindy. Anne and David had the same amount of money.
Which pie chart represents the information?",
"answer0": "Pie chart 1",
"answer1": "Pie chart 2",
"answer2": "Pie chart 3",
"answer3": "Pie chart 4",
"correct_answer": 2,
"skill_id": 210,
"difficulty_id": 2,
"explanation": "Let Cindy = 1 unit, so David = 2 units. Anne = David = 2 units, so Ben = 1 unit. The shares are Anne 2, Ben 1, Cindy 1, David 2. The correct pie chart shows Anne = David (equal large sectors) and Ben = Cindy (equal small sectors), which is chart 3.",
"hints": ["Set Cindy as 1 unit and work out the others.", "Anne and David are equal and largest; Ben and Cindy are equal and smallest."],
"source": "ACS Primary 2024 P6 Prelim P1 Q7",
"image_needed": true,
"image_options": true,
"image_file": null,
"image_page": 6,
"image_bbox": [0.18, 0.24, 0.72, 0.58],
"image_loc": "four pie-chart options arranged in a 2x2 grid below the question",
"notes": "Image-option MCQ: the four options are pie charts (1)-(4). Crop each option to q7_opt0.png..q7_opt3.png. Key = chart 3 (0-based index 2). Option positions: opt0=top-left(1), opt1=top-right(2), opt2=bottom-left(3), opt3=bottom-right(4)."
},
{
"n": 8,
"type_id": 1,
"question": "The average of three numbers is 40. The sum of another two numbers is 50. What is the average of the 5 numbers?",
"answer0": "13",
"answer1": "18",
"answer2": "34",
"answer3": "65",
"correct_answer": 2,
"skill_id": 205,
"difficulty_id": 2,
"explanation": "Sum of the three numbers = 3 × 40 = 120. Total of all five = 120 + 50 = 170. Average = 170 ÷ 5 = 34.",
"hints": ["Total of the three numbers = 3 × 40.", "Add the other sum, then divide by 5."],
"source": "ACS Primary 2024 P6 Prelim P1 Q8",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = option 3 (34), 0-based index 2."
},
{
"n": 9,
"type_id": 1,
"question": "The figure is made up of four straight lines, ST, TU, XY and YZ.
Students J, K and L wrote these statements: J: \\(\\angle b = \\angle c\\); K: \\(\\angle d = \\angle e\\); L: \\(\\angle a + \\angle c = 180°\\).
Which student(s) is/are correct?",
"answer0": "J only",
"answer1": "K only",
"answer2": "J and L",
"answer3": "K and L",
"correct_answer": 1,
"skill_id": 194,
"difficulty_id": 2,
"explanation": "\\(\\angle d\\) and \\(\\angle e\\) are vertically opposite angles, so \\(\\angle d = \\angle e\\) (K correct). \\(\\angle b\\) and \\(\\angle c\\) are base angles of different positions and need not be equal (J wrong). \\(\\angle a\\) and \\(\\angle c\\) are not supplementary in this figure (L wrong). Only K is correct.",
"hints": ["\\(\\angle d\\) and \\(\\angle e\\) are vertically opposite at the right crossing.", "Check whether the other equalities are forced by the figure."],
"source": "ACS Primary 2024 P6 Prelim P1 Q9",
"image_needed": true,
"image_options": false,
"image_file": "q9.png",
"image_page": 7,
"image_bbox": [0.18, 0.14, 0.74, 0.30],
"image_loc": "top of page, figure of two crossing line pairs forming angles a, b, c, d, e with points S, T, U, X, Y, Z",
"notes": "Key = option 2 (K only), 0-based index 1. The statements table is transcribed into the stem; do not crop the table, only the geometric figure."
},
{
"n": 10,
"type_id": 1,
"question": "ABCD is a square and \\(\\angle AEB = 86°\\). Find \\(\\angle CBE\\).",
"answer0": "41°",
"answer1": "43°",
"answer2": "45°",
"answer3": "49°",
"correct_answer": 0,
"skill_id": 230,
"difficulty_id": 2,
"explanation": "BE and the diagonal AC meet at E. \\(\\angle BEC = 180° − 86° = 94°\\). The diagonal AC of the square makes 45° with side BC, so \\(\\angle BCE = 45°\\). In triangle BEC, \\(\\angle CBE = 180° − 94° − 45° = 41°\\).",
"hints": ["\\(\\angle AEB\\) and \\(\\angle BEC\\) are angles on a straight line.", "A square's diagonal makes a 45° angle with its side."],
"source": "ACS Primary 2024 P6 Prelim P1 Q10",
"image_needed": true,
"image_options": false,
"image_file": "q10.png",
"image_page": 8,
"image_bbox": [0.27, 0.15, 0.62, 0.37],
"image_loc": "upper-middle of page, square ABCD with diagonal AC and line BE meeting at E, marked 86°",
"notes": "Key = option 1 (41°), 0-based index 0."
},
{
"n": 11,
"type_id": 1,
"question": "In a marathon relay race, four runners ran one after another without any stop in between. The second runner started running at the same time the first runner completed running and so on. The chart shows the time each runner started running. The last runner completed the race at 1.30 p.m.
Start times: First 10.30 a.m., Second 11.20 a.m., Third 12.15 p.m., Last 12.55 p.m.
How much more time did the slowest runner take than the fastest runner?",
"answer0": "15 min",
"answer1": "20 min",
"answer2": "35 min",
"answer3": "55 min",
"correct_answer": 1,
"skill_id": 205,
"difficulty_id": 2,
"explanation": "Each runner's time = next start minus own start (last finishes 1.30 p.m.). First = 10.30→11.20 = 50 min; Second = 11.20→12.15 = 55 min; Third = 12.15→12.55 = 40 min; Last = 12.55→1.30 = 35 min. Slowest = 55 min, fastest = 35 min. Difference = 55 − 35 = 20 min.",
"hints": ["Find each runner's duration from one start time to the next.", "Subtract the smallest duration from the largest."],
"source": "ACS Primary 2024 P6 Prelim P1 Q11",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "The start-time table is transcribed into the stem, no figure needed. Key = option 2 (20 min), 0-based index 1."
},
{
"n": 12,
"type_id": 1,
"question": "ABC is an equilateral triangle and DEF is an isosceles triangle. DE = EF. Find \\(\\angle DEF\\).",
"answer0": "102°",
"answer1": "120°",
"answer2": "138°",
"answer3": "141°",
"correct_answer": 2,
"skill_id": 197,
"difficulty_id": 3,
"explanation": "ABC equilateral so each angle = 60°. The 39° marked at C is between line DCA and CB, giving \\(\\angle EDF = 60° − 39° = 21°\\) at D. Since DE = EF, \\(\\angle EDF = \\angle EFD = 21°\\). So \\(\\angle DEF = 180° − 21° − 21° = 138°\\).",
"hints": ["Each angle of the equilateral triangle is 60°.", "Base angles of the isosceles triangle (DE = EF) are equal; angle sum = 180°."],
"source": "ACS Primary 2024 P6 Prelim P1 Q12",
"image_needed": true,
"image_options": false,
"image_file": "q12.png",
"image_page": 10,
"image_bbox": [0.22, 0.17, 0.80, 0.43],
"image_loc": "middle of page, large triangle with apex A and base D-C-B-F, inner isosceles triangle DEF with 39° marked near C",
"notes": "Key = option 3 (138°), 0-based index 2."
},
{
"n": 13,
"type_id": 1,
"question": "The figure shows a cuboid made up of 16 identical small cubes. All the faces of this cuboid including the base are painted green. This cuboid was then cut into 2 big identical cubes, each made up of 8 small cubes. Find the total area of the unpainted faces of these 2 big cubes.",
"answer0": "36 cm²",
"answer1": "72 cm²",
"answer2": "144 cm²",
"answer3": "216 cm²",
"correct_answer": 1,
"skill_id": 229,
"difficulty_id": 3,
"explanation": "Each small cube has edge 3 cm (6 cm ÷ 2). The cuboid (6×6×12) is cut into two 6×6×6 cubes; the cut creates two new square faces of 6×6 = 36 cm² each that were previously internal (unpainted). Total unpainted area = 2 × 36 = 72 cm².",
"hints": ["Each small cube has edge 6 ÷ 2 = 3 cm; each big cube is 6 cm.", "Only the freshly cut faces are unpainted — one new 6×6 face on each big cube."],
"source": "ACS Primary 2024 P6 Prelim P1 Q13",
"image_needed": true,
"image_options": false,
"image_file": "q13.png",
"image_page": 11,
"image_bbox": [0.36, 0.20, 0.72, 0.36],
"image_loc": "middle of page, 3D cuboid drawn as a grid of small cubes (6cm, 6cm, 12cm labels)",
"notes": "Key = option 2 (72 cm²), 0-based index 1."
},
{
"n": 14,
"type_id": 1,
"question": "Aaron had $55 more than Shanti at first. After Aaron used \\(\\dfrac{2}{5}\\) of his money and Shanti used \\(\\dfrac{1}{3}\\) of her money, Aaron had $27 more than Shanti. How much did Shanti have left at the end?",
"answer0": "$27",
"answer1": "$30",
"answer2": "$54",
"answer3": "$60",
"correct_answer": 3,
"skill_id": 209,
"difficulty_id": 3,
"explanation": "Let Shanti = S, Aaron = S + 55. After spending: Aaron has \\(\\dfrac{3}{5}(S+55)\\), Shanti has \\(\\dfrac{2}{3}S\\). Difference: \\(\\dfrac{3}{5}(S+55) − \\dfrac{2}{3}S = 27\\). \\(\\dfrac{3}{5}S + 33 − \\dfrac{2}{3}S = 27\\); \\(-\\dfrac{1}{15}S = -6\\); S = 90. Shanti left = \\(\\dfrac{2}{3} × 90 = 60\\).",
"hints": ["Write Aaron's amount as Shanti's amount + 55.", "Form an equation from the new $27 difference and solve."],
"source": "ACS Primary 2024 P6 Prelim P1 Q14",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = option 4 ($60), 0-based index 3."
},
{
"n": 15,
"type_id": 1,
"question": "A sum of money was shared between John and Ken in the ratio 3 : 4. John used all his money to buy 8 identical toy cars that cost $1.50 each. Ken used some of his money to buy 3 such toy cars. How much money had Ken left?",
"answer0": "$4.50",
"answer1": "$11.50",
"answer2": "$12",
"answer3": "$16",
"correct_answer": 1,
"skill_id": 183,
"difficulty_id": 3,
"explanation": "John spent 8 × $1.50 = $12 (all his share, 3 units). So 1 unit = $4, Ken's share = 4 units = $16. Ken bought 3 cars = 3 × $1.50 = $4.50. Ken left = $16 − $4.50 = $11.50.",
"hints": ["John's 3 units equal what he spent on 8 cars.", "Find 1 unit, then Ken's 4 units, then subtract his 3 cars."],
"source": "ACS Primary 2024 P6 Prelim P1 Q15",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = option 2 ($11.50), 0-based index 1."
},
{
"n": 16,
"type_id": 2,
"question": "Find the value of \\(20 - (4 + 8 \\div 4) + 5\\). [?]",
"answer0": "19",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 206,
"difficulty_id": 1,
"explanation": "Order of operations: \\(8 \\div 4 = 2\\), so brackets \\(= 4 + 2 = 6\\). Then \\(20 - 6 + 5 = 19\\).",
"hints": ["Work out the bracket first, doing division before addition.", "20 − 6 + 5 = 19."],
"source": "ACS Primary 2024 P6 Prelim P1 Q16",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 19. Whole-number answer, FIB. Skill mapped to nearest four-operations track."
},
{
"n": 17,
"type_id": 2,
"question": "Find the value of \\(68.2 \\div 4\\). [?]",
"answer0": "17.05",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 206,
"difficulty_id": 1,
"explanation": "\\(68.2 \\div 4 = 17.05\\).",
"hints": ["Divide 68.2 by 4.", "68 ÷ 4 = 17, and 0.2 ÷ 4 = 0.05."],
"source": "ACS Primary 2024 P6 Prelim P1 Q17",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 17.05. Decimal answer, FIB."
},
{
"n": 18,
"type_id": 2,
"question": "May's clock at home slows down 1 minute after every 6 hours since the battery runs low from 12 midnight. It is now showing 6.00 p.m. What should be the correct time now? Leave your answer in 24-hour clock. [?]",
"answer0": "1803",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 206,
"difficulty_id": 2,
"explanation": "From midnight, the clock shows 6.00 p.m. = 18 hours elapsed on the clock. It loses 1 min per 6 hours, so in 18 hours it has lost 3 minutes. The correct time is 3 minutes ahead: 18:00 + 3 min = 18:03, i.e. 1803.",
"hints": ["18 hours have passed on the slow clock since midnight.", "It loses 1 min every 6 hours: 18 ÷ 6 = 3 min lost, so add 3 min."],
"source": "ACS Primary 2024 P6 Prelim P1 Q18",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 1803 (24-hour clock). Whole-number/time answer entered as 1803."
},
{
"n": 19,
"type_id": 2,
"question": "What is the area of the shaded triangle? [?]",
"answer0": "30",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 186,
"difficulty_id": 2,
"explanation": "The shaded triangle has base 10 cm and height 6 cm (the height is the vertical 6 cm at the right). Area = \\(\\dfrac{1}{2} \\times 10 \\times 6 = 30\\) cm².",
"hints": ["Use \\(\\dfrac{1}{2} \\times \\text{base} \\times \\text{height}\\).", "Base = 10 cm, perpendicular height = 6 cm."],
"source": "ACS Primary 2024 P6 Prelim P1 Q19",
"image_needed": true,
"image_options": false,
"image_file": "q19.png",
"image_page": 15,
"image_bbox": [0.18, 0.12, 0.70, 0.27],
"image_loc": "top of page, shaded slanted triangle with base segments 10 cm and 8 cm and a 6 cm vertical height on the right",
"notes": "Key = 30. Answer units cm² stated by the answer line. FIB whole number."
},
{
"n": 20,
"type_id": 2,
"question": "The amount of water in a beaker at first is shown. After Zoe poured in 50 ml of water into the beaker, what is the total amount of water in the beaker now? [?]",
"answer0": "230",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 192,
"difficulty_id": 2,
"explanation": "The water level reads 180 ml at first. After adding 50 ml, total = 180 + 50 = 230 ml.",
"hints": ["Read the starting water level from the scale.", "Add 50 ml to the starting amount."],
"source": "ACS Primary 2024 P6 Prelim P1 Q20",
"image_needed": true,
"image_options": false,
"image_file": "q20.png",
"image_page": 15,
"image_bbox": [0.36, 0.42, 0.66, 0.66],
"image_loc": "middle of page, beaker with measuring scale (40-240 ml) shaded to about the 180 ml mark",
"notes": "Key = 230. Answer units ml. FIB whole number. The figure (beaker + scale) is needed since the start level is read from it."
},
{
"n": 21,
"type_id": 2,
"question": "ADE is a right-angled isosceles triangle and BE is a straight line. AE = DE and \\(\\angle ACB = 84°\\). Find \\(\\angle ABC\\). [?]",
"answer0": "51",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 197,
"difficulty_id": 3,
"explanation": "In right-angled isosceles triangle ADE, \\(\\angle DAE = \\angle DEA = 45°\\). At A, \\(\\angle DAB\\) sits on line BE; \\(\\angle BAC = 84° − 45° = 39°\\) (angle between AB and AC). In triangle ABC, \\(\\angle ABC = 180° − \\angle ACB − \\angle BAC\\). With \\(\\angle ACB = 84°\\), the printed key obtains \\(\\angle ABC = 51°\\).",
"hints": ["The right-angled isosceles triangle gives two 45° angles.", "Use the angle sum of triangle ABC with the given 84°."],
"source": "ACS Primary 2024 P6 Prelim P1 Q21",
"image_needed": true,
"image_options": false,
"image_file": "q21.png",
"image_page": 16,
"image_bbox": [0.20, 0.24, 0.56, 0.46],
"image_loc": "upper-left area, figure with A and B at top, E and D at bottom, lines crossing at C with 84° and a right angle at E",
"notes": "Key answer = 51° (degrees). FIB whole number; answer line is in degrees. Printed key working is terse but final answer 51° is given."
},
{
"n": 22,
"type_id": 2,
"question": "A baker had three boxes of flour weighing 2.7 kg, 3.4 kg and 5.02 kg. He repacked them equally into four bags. What is the mass of each bag? [?]",
"answer0": "2780",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 184,
"difficulty_id": 2,
"explanation": "Total mass = 2.7 + 3.4 + 5.02 = 11.12 kg = 11120 g. Each bag = 11120 ÷ 4 = 2780 g.",
"hints": ["Add the three masses (convert to grams: 2700 + 3400 + 5020).", "Divide the total by 4."],
"source": "ACS Primary 2024 P6 Prelim P1 Q22",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 2780 g (answer line states grams). FIB whole number."
},
{
"n": 23,
"type_id": 2,
"question": "The table shows the prices of pens and erasers sold at a bookshop: an Eraser costs \\(b\\) cents and a Pen costs \\(2b\\) cents. Mark spent $4.20 on 5 erasers and 1 pen. Find the value of \\(b\\). [?]",
"answer0": "60",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 241,
"difficulty_id": 2,
"explanation": "Cost = 5 erasers + 1 pen = \\(5b + 2b = 7b\\) cents. $4.20 = 420 cents, so \\(7b = 420\\), giving \\(b = 60\\).",
"hints": ["Write the total cost as 5b + 2b cents.", "Convert $4.20 to 420 cents, then divide by 7."],
"source": "ACS Primary 2024 P6 Prelim P1 Q23",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = b = 60 cents. FIB whole number. Price table transcribed into the stem; no figure needed."
},
{
"n": 24,
"type_id": 2,
"question": "The figure shows the net of a cube. Jane wrote the numbers 1, 2, 3, 4, 5 and 6 without repeating on the faces of the cube. The sum of the four numbers added horizontally is the same as the sum of the three numbers when added vertically. When the net is folded to form a cube, '3' is seen on the bottom of the cube. What are the two possible numbers on the top of the cube? [?] and [?]",
"answer0": "6",
"answer1": "2",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 229,
"difficulty_id": 3,
"explanation": "Total of all faces = 21. The horizontal row (4 faces) and vertical column (3 faces) share the centre face '3'. Let horizontal sum = vertical sum = S. The two arms overlap only at the centre 3, and all six numbers are used, so horizontal(4) + vertical(3) − 3 (shared centre) = 21, giving 2S − 3 = 21, S = 12. The face opposite the centre '3' is the top; the printed key gives the two possible numbers as 6 and 2.",
"hints": ["All six faces sum to 21; the centre face 3 is shared by both arms.", "Set the horizontal and vertical sums equal and solve."],
"source": "ACS Primary 2024 P6 Prelim P1 Q24",
"image_needed": true,
"image_options": false,
"image_file": "q24.png",
"image_page": 17,
"image_bbox": [0.40, 0.55, 0.66, 0.78],
"image_loc": "lower-middle of page, cross-shaped cube net with '3' and '1' in the horizontal row, labelled Horizontal and Vertical arrows",
"notes": "Two-blank FIB; key = '6 and 2'. Order not significant; entered answer0=6, answer1=2."
},
{
"n": 25,
"type_id": 2,
"question": "Mrs Goh had 25 apples and some pears at first. She gave away some apples and \\(\\dfrac{1}{4}\\) of the remaining fruits were pears. After buying 8 mangoes, \\(\\dfrac{1}{6}\\) of the total number of fruits were pears. How many apples did Mrs Goh give away? [?]",
"answer0": "13",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 209,
"difficulty_id": 3,
"explanation": "Let pears = p, apples given away = x. After giving away: remaining = (25 − x) + p, with pears = \\(\\dfrac{1}{4}\\): 25 − x + p = 4p, so 25 − x = 3p. After 8 mangoes: 33 − x + p = 6p, so 33 − x = 5p. Subtract: 8 = 2p, p = 4. Then 25 − x = 12, x = 13.",
"hints": ["Let the number of pears be p and apples given away be x.", "Form two equations from the two pear fractions and solve."],
"source": "ACS Primary 2024 P6 Prelim P1 Q25",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 13 apples. FIB whole number."
},
{
"n": 26,
"type_id": 2,
"question": "Two empty tanks with bases measuring 40 cm by 25 cm and 30 cm by 20 cm are filled with a total of 8 litres of water to the same height. What is the height of the water level in both tanks? [?]",
"answer0": "5",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 191,
"difficulty_id": 2,
"explanation": "8 litres = 8000 cm³. Combined base area = (40×25) + (30×20) = 1000 + 600 = 1600 cm². Height = 8000 ÷ 1600 = 5 cm.",
"hints": ["8 litres = 8000 cm³; add the two base areas.", "Height = total volume ÷ total base area."],
"source": "ACS Primary 2024 P6 Prelim P1 Q26",
"image_needed": true,
"image_options": false,
"image_file": "q26.png",
"image_page": 18,
"image_bbox": [0.18, 0.46, 0.88, 0.66],
"image_loc": "middle of page, two open rectangular tanks side by side (40 cm × 25 cm and 30 cm × 20 cm bases)",
"notes": "Key = 5 cm (answer line states cm). FIB whole number."
},
{
"n": 27,
"type_id": 1,
"question": "The table and pie chart show a group of students and the number of digital devices they owned: Group A owns 0 devices (40%), Group B owns 1 device, Group C owns 2 devices (10%), Group D owns more than 2 devices (5%). 80 students did not have any digital devices. What fraction of the total number had only 1 digital device? Give your answer in the simplest form.",
"answer0": "\\(\\dfrac{9}{20}\\)",
"answer1": "\\(\\dfrac{2}{5}\\)",
"answer2": "\\(\\dfrac{1}{10}\\)",
"answer3": "\\(\\dfrac{9}{40}\\)",
"correct_answer": 0,
"skill_id": 210,
"difficulty_id": 2,
"explanation": "Group B (1 device) = 100% − (40% + 10% + 5%) = 45%. As a fraction this is \\(\\dfrac{45}{100} = \\dfrac{9}{20}\\). (Check: 40% = 80 students so total = 200; Group B = 90 students = \\(\\dfrac{90}{200} = \\dfrac{9}{20}\\).)",
"hints": ["Find Group B's percentage: 100% − 40% − 10% − 5%.", "Write 45% as a fraction and simplify."],
"source": "ACS Primary 2024 P6 Prelim P2 Q27",
"image_needed": true,
"image_options": false,
"image_file": "q27.png",
"image_page": 19,
"image_bbox": [0.46, 0.16, 0.80, 0.40],
"image_loc": "right side of page, pie chart with sectors A 40%, B, C 10%, D 5%",
"notes": "Converted to MCQ because the answer is a fraction (9/20). Key fraction = 9/20 (index 0). Distractors are plausible mis-simplifications. The data table is transcribed into the stem; crop only the pie chart figure."
},
{
"n": 28,
"type_id": 2,
"question": "Using the same survey, 80 students did not have any digital devices (Group A = 40%). How many students had more than 1 device? [?]",
"answer0": "30",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 209,
"difficulty_id": 2,
"explanation": "Total students = 80 ÷ 40% = 200. 'More than 1 device' = Group C (2 devices, 10%) + Group D (more than 2, 5%) = 15% of 200 = 30 students.",
"hints": ["Total = 80 ÷ 0.40 = 200 students.", "More than 1 device = Group C (10%) + Group D (5%)."],
"source": "ACS Primary 2024 P6 Prelim P2 Q28",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 30 students. FIB whole number. Refers to the same pie chart as Q27; data needed is given in the stem so no figure required here."
},
{
"n": 29,
"type_id": 2,
"question": "Peter spent $60 of his weekly allowance and saved the rest. During Christmas, his spending increased by 20%. As a result, his savings decreased by 10%. How much was his weekly allowance? [?]",
"answer0": "180",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 209,
"difficulty_id": 3,
"explanation": "Let allowance = A. Original savings = A − 60. New spending = 60 × 1.2 = 72. New savings = 0.9(A − 60). Also new savings = A − 72. So A − 72 = 0.9(A − 60) = 0.9A − 54, giving 0.1A = 18, A = 180.",
"hints": ["New spending = 60 × 1.2 = 72.", "Set 'allowance − new spending' equal to 0.9 × original savings."],
"source": "ACS Primary 2024 P6 Prelim P2 Q29",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = $180 (answer line prefixed $). FIB whole number entered as 180."
},
{
"n": 30,
"type_id": 2,
"question": "ABGH is a parallelogram, CFGH is a rhombus and DEGH is a trapezium. \\(\\angle EDH = 132°\\) and \\(\\angle CHD = 20°\\). Find \\(\\angle BAH\\). [?]",
"answer0": "56",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 200,
"difficulty_id": 3,
"explanation": "In trapezium DEGH, \\(\\angle DHG = 180° − \\angle EDH = 180° − 132° = 48°\\). Then \\(\\angle HGF\\) (the angle inside the rhombus arrangement) = (180° − (48° + 20°)) ÷ 2 = 56°. Since ABGH is a parallelogram, \\(\\angle BAH = \\angle HGF = 56°\\) (using the parallel relationships in the figure, per the printed key).",
"hints": ["Co-interior angles in the trapezium: \\(\\angle DHG = 180° − 132°\\).", "Use the rhombus and parallelogram properties to transfer the angle to BAH."],
"source": "ACS Primary 2024 P6 Prelim P2 Q30",
"image_needed": true,
"image_options": false,
"image_file": "q30.png",
"image_page": 21,
"image_bbox": [0.18, 0.16, 0.80, 0.46],
"image_loc": "upper portion of page, composite figure of parallelogram ABGH, rhombus CFGH and trapezium DEGH with 132° and 20° marked",
"notes": "Key: ∠DHG = 180−132 = 48°; ∠HGF = (180−(48+20))÷2 = 56°. Final answer 56°. FIB whole number, answer line in degrees."
},
{
"n": 31,
"type_id": 1,
"question": "Which column can the number 100 be found in the table? (The table fills the numbers 1, 2, 3, ... row by row across 6 columns A, B, C, D, E, F.)",
"answer0": "Column A",
"answer1": "Column B",
"answer2": "Column D",
"answer3": "Column F",
"correct_answer": 1,
"skill_id": 240,
"difficulty_id": 2,
"explanation": "Numbers fill 6 per row. 100 ÷ 6 = 16 remainder 4, so 100 is the 4th number in its row. The 4th position in the listed pattern lands in column B (the table is read in a boustrophedon/zig-zag pattern as printed, placing 100 in column B).",
"hints": ["Each row holds 6 numbers; find 100's position with 100 ÷ 6.", "Track which column the 4th-in-row position corresponds to."],
"source": "ACS Primary 2024 P6 Prelim P2 Q1",
"image_needed": true,
"image_options": false,
"image_file": "q31.png",
"image_page": 24,
"image_bbox": [0.18, 0.24, 0.66, 0.40],
"image_loc": "upper-middle of page, 6-column table (A-F) with numbers 1-12 arranged in a zig-zag fill pattern",
"notes": "Converted to MCQ (answer is a column letter, not numeric). Printed key Q1 = B (index 1). The figure/table is needed because the fill order is non-obvious."
},
{
"n": 32,
"type_id": 2,
"question": "A 2-m wide path was built around a square garden ABCD. The area of the shaded path is 56 m². What is the area of the square garden ABCD? [?]",
"answer0": "25",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 223,
"difficulty_id": 3,
"explanation": "Let the garden side = s. Outer square side = s + 4 (2 m on each side). Path area = (s+4)² − s² = 8s + 16 = 56, so 8s = 40, s = 5. Garden area = 5 × 5 = 25 m².",
"hints": ["Outer side = garden side + 2 + 2.", "Path area = outer square − garden square."],
"source": "ACS Primary 2024 P6 Prelim P2 Q2",
"image_needed": true,
"image_options": false,
"image_file": "q32.png",
"image_page": 24,
"image_bbox": [0.40, 0.58, 0.66, 0.80],
"image_loc": "lower-middle of page, square garden ABCD with a shaded 2 m border path around it",
"notes": "Key = 25 m². FIB whole number (answer line states m²)."
},
{
"n": 33,
"type_id": 0,
"question": "Ahmad bought some books which cost an average of $26. Two of the books cost $46 and $34 respectively. The average cost of the remaining books was $22. Each statement below is either true, false or not possible to tell. Statement 1: Ahmad bought more than 10 books. Statement 2: Two of the books cost an average of $24.",
"answer0": null,
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 205,
"difficulty_id": 3,
"explanation": "Key: Statement 1 ('Ahmad bought more than 10 books') = False. Statement 2 ('Two of the books cost an average of $24') = Not possible to tell.",
"hints": [],
"source": "ACS Primary 2024 P6 Prelim P2 Q3",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "type_id 0: a true/false/not-possible-to-tell tick-table format that has no app question type. Answers (from key): S1 = False, S2 = Not possible to tell. SKIPPED on insert."
},
{
"n": 34,
"type_id": 2,
"question": "ABC and ABE are identical isosceles triangles. AF is parallel to BE, AB = AE = BC, \\(\\angle BCA = 36°\\) and \\(\\angle AEF = 85°\\). Find \\(\\angle AFE\\). [?]",
"answer0": "59",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 197,
"difficulty_id": 3,
"explanation": "In isosceles triangle ABC, \\(\\angle BAC = \\angle ABC = 72°\\) (since \\(\\angle BCA = 36°\\)). Since AF ∥ BE, \\(\\angle AEF = \\angle EAF = 36°\\) is not directly used; instead in triangle AEF: \\(\\angle EAF = \\angle ABE = 36°\\) (AF ∥ BE), and \\(\\angle AEF = 85°\\), so \\(\\angle AFE = 180° − 85° − 36° = 59°\\).",
"hints": ["Base angles of the isosceles triangle: (180° − 36°) ÷ 2 = 72°.", "Use AF ∥ BE to find the angle at A in triangle AEF, then the angle sum."],
"source": "ACS Primary 2024 P6 Prelim P2 Q4",
"image_needed": true,
"image_options": false,
"image_file": "q34.png",
"image_page": 25,
"image_bbox": [0.16, 0.40, 0.74, 0.74],
"image_loc": "middle/lower portion of page, two isosceles triangles sharing vertex B, with F and E at the bottom, marked 36° at C and 85° at E",
"notes": "Key = 59°. FIB whole number, answer line in degrees."
},
{
"n": 35,
"type_id": 2,
"question": "May had some chocolates and lollipops in the ratio of 3 : 4 at first. After May ate 12 chocolates and bought 4 more lollipops, the ratio became 3 : 8. What was the total number of chocolates and lollipops May had in the end? [?]",
"answer0": "55",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 214,
"difficulty_id": 3,
"explanation": "Let chocolates = 3x, lollipops = 4x. After: \\(\\dfrac{3x-12}{4x+4} = \\dfrac{3}{8}\\). Cross-multiply: 8(3x−12) = 3(4x+4); 24x − 96 = 12x + 12; 12x = 108; x = 9. Final chocolates = 27 − 12 = 15; final lollipops = 36 + 4 = 40. Total = 15 + 40 = 55.",
"hints": ["Let the first amounts be 3x and 4x.", "Form a ratio equation for the new amounts and solve for x."],
"source": "ACS Primary 2024 P6 Prelim P2 Q5",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 55. FIB whole number."
},
{
"n": 36,
"type_id": 2,
"question": "Julie had 5 times as many coins as Gavin at first. She gave \\(\\dfrac{1}{4}\\) of her coins to Gavin. Then Gavin used \\(\\dfrac{1}{3}\\) of his total coins. In the end, Julie had 189 coins more than Gavin. How many coins did Julie and Gavin have at first altogether? [?]",
"answer0": "504",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 206,
"difficulty_id": 3,
"explanation": "Let Gavin = G, Julie = 5G. Julie gives \\(\\dfrac{1}{4}\\)(5G) = \\(\\dfrac{5G}{4}\\), leaving Julie \\(\\dfrac{15G}{4}\\) and Gavin \\(G + \\dfrac{5G}{4} = \\dfrac{9G}{4}\\). Gavin uses \\(\\dfrac{1}{3}\\), leaving \\(\\dfrac{2}{3} \\times \\dfrac{9G}{4} = \\dfrac{3G}{2}\\). Difference: \\(\\dfrac{15G}{4} − \\dfrac{3G}{2} = \\dfrac{9G}{4} = 189\\), so G = 84, Julie = 420. Total at first = 84 + 420 = 504.",
"hints": ["Set Gavin = G and Julie = 5G, then track each transfer.", "Julie's final − Gavin's final = 189; solve for G."],
"source": "ACS Primary 2024 P6 Prelim P2 Q6",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key: Gavin 84, Julie 420; question asks total at first = 504. FIB whole number."
},
{
"n": 37,
"type_id": 2,
"question": "The bar graph shows the number of potted plants owned by different families in a housing estate.
(a) What is the total number of potted plants in the housing estate? [?]
(b) What is the ratio of the number of families with 3 or less potted plants to the number of families with more than 3 potted plants? Give your answer in the simplest form. [?]",
"answer0": "126",
"answer1": "7 : 10",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 461,
"difficulty_id": 2,
"explanation": "From the bar graph, families owning 1,2,3,4,5,6 plants = 2,5,7,10,7,3. (a) Total plants = 2×1 + 5×2 + 7×3 + 10×4 + 7×5 + 3×6 = 2 + 10 + 21 + 40 + 35 + 18 = 126. (b) Families with 3 or less = 2 + 5 + 7 = 14; families with more than 3 = 10 + 7 + 3 = 20; ratio 14 : 20 = 7 : 10.",
"hints": ["Multiply each 'plants owned' value by its family count, then sum.", "For the ratio, group families by 3-or-less vs more-than-3 and simplify."],
"source": "ACS Primary 2024 P6 Prelim P2 Q7",
"image_needed": true,
"image_options": false,
"image_file": "q37.png",
"image_page": 28,
"image_bbox": [0.18, 0.14, 0.84, 0.44],
"image_loc": "top of page, bar graph: x-axis 'Number of potted plants owned' (1-6), y-axis 'Number of families' (0-10)",
"notes": "Two-part FIB. (a) key = 126; (b) key = 7 : 10. answer0 numeric, answer1 is a ratio string (not numeric exact-match). Bar heights read from figure: 2,5,7,10,7,3."
},
{
"n": 38,
"type_id": 2,
"question": "ABE is an equilateral triangle and ADF is a right-angled triangle. \\(\\angle ADF = 20°\\) and \\(\\angle BAF = 98°\\). Find \\(\\angle DCE\\). [?]",
"answer0": "92",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 197,
"difficulty_id": 3,
"explanation": "ABE equilateral so \\(\\angle BAE = 60°\\). \\(\\angle FAE = \\angle BAF − \\angle BAE = 98° − 60° = 38°\\). In right-angled triangle ADF, \\(\\angle DAF = 90° − 20° = 70°\\), so \\(\\angle DAE = \\angle DAF + \\angle FAE = 70° + 38° = 108°\\). Then \\(\\angle DCE = 180° − (180° − \\angle DAE − \\angle AFD) = 180° − (180° − 108° − 60°) = 92°\\) (per printed key).",
"hints": ["Equilateral triangle gives 60° at A.", "Find ∠FAE and ∠DAE, then use the triangle angle sums per the key."],
"source": "ACS Primary 2024 P6 Prelim P2 Q8",
"image_needed": true,
"image_options": false,
"image_file": "q38.png",
"image_page": 29,
"image_bbox": [0.20, 0.13, 0.66, 0.45],
"image_loc": "upper-middle of page, equilateral triangle ABE with right-angled triangle ADF, points B, C, D, A, E, F, marked 98° at A, 20° at D, right angle at F",
"notes": "Key = 92°. FIB whole number, answer line in degrees."
},
{
"n": 39,
"type_id": 2,
"question": "Sean left City A for City B travelling at a constant speed of 80 km/h. At the same time, Yazid left City B for City A travelling at a constant speed of 60 km/h. Sean and Yazid passed each other at 30 km from the midpoint. What is the distance between City A and City B? [?]",
"answer0": "420",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 219,
"difficulty_id": 3,
"explanation": "They travel for the same time, so distances are in the ratio of speeds 80 : 60 = 4 : 3. Let total distance = D, midpoint = D/2. Sean travels D/2 + 30, Yazid travels D/2 − 30. So \\(\\dfrac{D/2 + 30}{D/2 − 30} = \\dfrac{4}{3}\\). Cross-multiply: 3(D/2 + 30) = 4(D/2 − 30); 1.5D + 90 = 2D − 120; 0.5D = 210; D = 420 km.",
"hints": ["Equal time means distance ratio = speed ratio 4 : 3.", "Sean's distance = half + 30; Yazid's = half − 30. Solve for D."],
"source": "ACS Primary 2024 P6 Prelim P2 Q9",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = 420 km. FIB whole number."
},
{
"n": 40,
"type_id": 2,
"question": "Mdm Siti bought \\(4p\\) kg of beef at $7 per kg and \\(6p\\) kg of chicken at $8 per kg for Hari Raya.
(a) Express the total amount she spent in terms of \\(p\\). [?]
(b) If \\(p = 2\\), find the total amount spent. [?]",
"answer0": "76p",
"answer1": "152",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 241,
"difficulty_id": 2,
"explanation": "(a) Beef cost = 4p × 7 = 28p; chicken cost = 6p × 8 = 48p. Total = 28p + 48p = 76p dollars. (b) When p = 2: 76 × 2 = $152.",
"hints": ["Multiply quantity by price per kg for each meat.", "Add the two expressions, then substitute p = 2."],
"source": "ACS Primary 2024 P6 Prelim P2 Q10",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Two-part FIB. (a) algebraic expression '76p' (entered as a literal string); (b) numeric 152. Part (a) is an algebraic-expression answer — grading is exact-match on the string '76p'."
},
{
"n": 41,
"type_id": 2,
"question": "The pie chart represents the number of cars, motorcycles and trucks in a car park. Cars make up \\(\\dfrac{2}{3}\\) of the chart. The ratio of the number of trucks to the number of motorcycles is 5 : 7. There are 48 cars in the car park.
(a) How many trucks are there in the car park? [?]
(b) The cars and trucks have 4 wheels each while the motorcycles have 2 wheels each. Find the total number of wheels represented in the pie chart. [?]",
"answer0": "10",
"answer1": "260",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 213,
"difficulty_id": 3,
"explanation": "Cars = \\(\\dfrac{2}{3}\\) of total = 48, so total = 48 × \\(\\dfrac{3}{2}\\) = 72. Trucks + motorcycles = \\(\\dfrac{1}{3}\\) × 72 = 24, split 5 : 7 (12 parts). (a) Trucks = 24 × \\(\\dfrac{5}{12}\\) = 10. (b) Motorcycles = 24 − 10 = 14. Wheels = cars 48×4 + trucks 10×4 + motorcycles 14×2 = 192 + 40 + 28 = 260.",
"hints": ["Total vehicles = 48 ÷ (2/3) = 72.", "Trucks+motorcycles = 24, divided 5:7; then count wheels."],
"source": "ACS Primary 2024 P6 Prelim P2 Q11",
"image_needed": true,
"image_options": false,
"image_file": "q41.png",
"image_page": 31,
"image_bbox": [0.34, 0.14, 0.70, 0.40],
"image_loc": "upper-middle of page, pie chart with sectors Cars (2/3), Trucks and Motorcycles",
"notes": "Two-part FIB. (a) key = 10 trucks; (b) key = 260 wheels. Both numeric."
},
{
"n": 42,
"type_id": 2,
"question": "At first, Sam had \\(\\dfrac{2}{5}\\) as much money as Paul. Ean had $36 more than Sam. After Paul gave $98 to Sam and some money to Ean, all three boys had the same amount of money. How much did they have altogether? [?]",
"answer0": "810",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 209,
"difficulty_id": 3,
"explanation": "Let Paul = 5u, Sam = 2u, Ean = 2u + 36. The total money is unchanged by transfers. In the end each boy has the same amount M, so 3M = total = 5u + 2u + (2u + 36) = 9u + 36. Sam's final = 2u + 98 = M. From the printed key, solving gives M = 270, so total = 3 × 270 = $810.",
"hints": ["Let Paul = 5 units, Sam = 2 units; Ean = 2 units + 36.", "Sam's final = 2 units + 98 = M; total = 3M (money is conserved)."],
"source": "ACS Primary 2024 P6 Prelim P2 Q12",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key = $810 altogether (each boy ended with $270). FIB whole number."
},
{
"n": 43,
"type_id": 2,
"question": "The figure is made up of a big quarter circle, two small identical quarter circles and a square. The radius of the big quarter circle is 10 cm and the radius of the small quarter circle is 5 cm.
(a) Find the area of the square. [?]
(b) Find the total area of the shaded parts. (Take \\(\\pi = 3.14\\)) [?]",
"answer0": "12.5",
"answer1": "53.5",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 222,
"difficulty_id": 3,
"explanation": "(a) The square's side equals half the small radius arrangement; the printed key gives the square area as \\(\\dfrac{1}{2} × 2.5 × 2.5 × 4 = 12.5\\) cm². (b) Big quarter circle = 3.14 × 10 × 10 × \\(\\dfrac{1}{4}\\) = 78.5 cm². One small quarter circle = 3.14 × 5 × 5 × \\(\\dfrac{1}{4}\\) = 19.625 cm². Following the key: 78.5 − 19.625 = 58.875; 58.875 − 12.5 = 46.375; 19.625 − 12.5 = 7.125; total shaded = 46.375 + 7.125 = 53.5 cm².",
"hints": ["Quarter-circle area = \\(\\dfrac{1}{4}\\pi r^2\\).", "Combine the big quarter circle, small quarter circles and square per the figure to get the shaded total."],
"source": "ACS Primary 2024 P6 Prelim P2 Q13",
"image_needed": true,
"image_options": false,
"image_file": "q43.png",
"image_page": 33,
"image_bbox": [0.36, 0.14, 0.74, 0.40],
"image_loc": "upper-middle of page, composite figure: big quarter circle (radius 10cm), two small quarter circles (radius 5cm) and an unshaded square",
"notes": "Two-part FIB. (a) key = 12.5 cm²; (b) key = 53.5 cm². Both decimal/whole numbers. Answer lines state the values without units forced."
},
{
"n": 44,
"type_id": 2,
"question": "A rectangular tank, 50 cm by 25 cm by 30 cm, is separated into two compartments P and Q by a plastic panel. Compartment P (10 cm wide) holds water to a height of 20 cm and compartment Q holds water to a height of 5 cm.
(a) What is the height of the water in the tank when the plastic panel is removed? [?]
(b) How many more litres of water are needed to fill the tank to the brim after the plastic panel is removed? [?]",
"answer0": "8",
"answer1": "27.5",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 193,
"difficulty_id": 3,
"explanation": "Water in P = 10 × 25 × 20 = 5000 cm³. Water in Q = 40 × 25 × 5 = 5000 cm³. Total water = 10000 cm³. Tank base = 50 × 25 = 1250 cm². (a) Height = 10000 ÷ 1250 = 8 cm. (b) Tank capacity = 50 × 25 × 30 = 37500 cm³. Water needed = 37500 − 10000 = 27500 cm³ = 27.5 litres.",
"hints": ["Find the water volume in each compartment and add them.", "Height = total volume ÷ full base area; then capacity − water = amount to fill."],
"source": "ACS Primary 2024 P6 Prelim P2 Q14",
"image_needed": true,
"image_options": false,
"image_file": "q44.png",
"image_page": 34,
"image_bbox": [0.16, 0.16, 0.84, 0.42],
"image_loc": "upper-middle of page, 3D rectangular tank split by a plastic panel into P (10cm) and Q, with water levels 20cm and 5cm; overall 50cm × 25cm × 30cm",
"notes": "Two-part FIB. (a) key = 8 cm; (b) key = 27.5 litres. Numeric/decimal answers."
},
{
"n": 45,
"type_id": 2,
"question": "Machine A and Machine B had the same mass of biscuits to be packed into identical boxes. The machines started packing at 2 p.m. After 15 minutes, Machine A had packed 250 boxes with 700 g remaining, and Machine B had packed 245 boxes with 1.5 kg remaining.
(a) What was the mass of biscuits in each box? Give your answer in kilograms. [?]
(b) At 2.15 p.m., Machine B broke down and all the remaining biscuits of both machines were packed by Machine A. After packing the maximum boxes possible, how many grams of biscuits were left at the end? [?]",
"answer0": "0.16",
"answer1": "120",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 206,
"difficulty_id": 3,
"explanation": "Both started with equal mass: 250x + 0.7 = 245x + 1.5 (kg), so 5x = 0.8, x = 0.16 kg per box. (b) Remaining biscuits = 0.7 + 1.5 = 2.2 kg. Boxes Machine A can pack = 2.2 ÷ 0.16 = 13.75 → 13 full boxes (2.08 kg). Left = 2.2 − 2.08 = 0.12 kg = 120 g.",
"hints": ["Equal starting mass: 250x + 0.7 = 245x + 1.5.", "For (b), pack the combined remaining mass in whole 0.16 kg boxes; find the leftover."],
"source": "ACS Primary 2024 P6 Prelim P2 Q15",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Two-part FIB. (a) key = 0.16 kg; (b) key = 120 g. Decimal/whole-number answers. The data table is transcribed into the stem; no figure needed."
},
{
"n": 46,
"type_id": 2,
"question": "ACKL is a parallelogram, ACGH is a rectangle and ADE is an isosceles triangle. KIFE is a straight line. AD = AE, \\(\\angle ADE = 75°\\), \\(\\angle EBK = 102°\\) and \\(\\angle CFK = 93°\\).
(a) Find \\(\\angle LAJ\\). [?]
(b) Find \\(\\angle JKI\\). [?]",
"answer0": "42",
"answer1": "45",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 230,
"difficulty_id": 3,
"explanation": "Since AD = AE, \\(\\angle DAE = \\angle DEA = 75°\\), so the apex \\(\\angle ADE\\)-based angle gives \\(\\angle ACB = 180° − (180° − 75° − 75°) − 102° = 180° − 30° − 102° = 48°\\). (a) ACGH is a rectangle so \\(\\angle ACG = 90°\\); \\(\\angle LAJ = 180° − \\angle ACB − 90° = 180° − 48° − 90° = 42°\\). (b) KIFE straight line: \\(\\angle JKI = 180° − \\angle CFK − \\angle LAJ = 180° − 93° − 42° = 45°\\).",
"hints": ["Isosceles triangle: \\(\\angle DAE = \\angle DEA = 75°\\).", "Use the rectangle's 90° and the straight line KIFE summing to 180°."],
"source": "ACS Primary 2024 P6 Prelim P2 Q16",
"image_needed": true,
"image_options": false,
"image_file": "q46.png",
"image_page": 36,
"image_bbox": [0.16, 0.16, 0.82, 0.46],
"image_loc": "upper-middle of page, composite figure with parallelogram ACKL, rectangle ACGH and isosceles triangle ADE; points A,B,C,D,E,F,G,H,I,J,K,L marked with 75°, 102°, 93°",
"notes": "Two-part FIB. (a) key = ∠LAJ = 42°; (b) key = ∠JKI = 45°. Whole numbers, answer lines in degrees."
},
{
"n": 47,
"type_id": 2,
"question": "Andy and Betty had 864 stamps altogether. Andy gave Betty some of his stamps and Betty's stamps increased by 60%. In return, Betty gave Andy some of her stamps and Andy's stamps increased by 30%. Then, Andy gave some of his stamps to Betty and Betty's stamps increased by 80%. In the end, each of them had an equal number of stamps. Find the percentage increase in the number of Betty's stamps. [?]",
"answer0": "80",
"answer1": null,
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 208,
"difficulty_id": 3,
"explanation": "Each of them ends with 432 (half of 864). The final transfer makes Betty's stamps increase by 80% to reach her final equal share. The printed key concludes the percentage increase in Betty's stamps at the final step is 80%.",
"hints": ["Each person ends with 864 ÷ 2 = 432 stamps.", "Track Betty's stamps through each transfer to the final 80% increase."],
"source": "ACS Primary 2024 P6 Prelim P2 Q17",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Key (as printed) = 80%. The printed worked solution is incomplete/hand-waved but states the final answer 80%. FIB whole number (percentage), entered as 80."
}
]
}