{
  "paper": {
    "school": "Anglo-Chinese School (Junior)",
    "year": 2024,
    "level": "P6",
    "label": "Prelim",
    "source_prefix": "ACS 2024 P6 Prelim",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Singapore's population was 6 014 723 last year. Express this number to the nearest thousand.",
      "answer0": "6 000 000",
      "answer1": "6 010 000",
      "answer2": "6 014 000",
      "answer3": "6 015 000",
      "correct_answer": 3,
      "skill_id": 150,
      "difficulty_id": 1,
      "explanation": "Method: round to the nearest thousand by looking at the hundreds digit. 6 014 723 has 7 in the hundreds place, so round up: 6 014 723 → 6 015 000. Final answer: 6 015 000.",
      "hints": ["Look at the hundreds digit (7).", "7 is 5 or more, so round the thousands digit up."],
      "source": "ACS 2024 P6 Prelim Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (4)"
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "In 13.02, which digit is in the tenths place?",
      "answer0": "1",
      "answer1": "2",
      "answer2": "3",
      "answer3": "0",
      "correct_answer": 3,
      "skill_id": 105,
      "difficulty_id": 1,
      "explanation": "Method: identify decimal place values. In 13.02 the first digit after the decimal point is the tenths place, which is 0. Final answer: 0.",
      "hints": ["The first digit after the decimal point is the tenths.", "13.0|2 — the 0 is in the tenths place."],
      "source": "ACS 2024 P6 Prelim Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (4) = 0"
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Rina makes a necklace using 12 pink pearls and 18 white pearls. What fraction of the pearls are white?",
      "answer0": "\\(\\dfrac{2}{5}\\)",
      "answer1": "\\(\\dfrac{3}{5}\\)",
      "answer2": "\\(\\dfrac{2}{3}\\)",
      "answer3": "\\(\\dfrac{3}{2}\\)",
      "correct_answer": 1,
      "skill_id": 212,
      "difficulty_id": 1,
      "explanation": "Method: white ÷ total. Total pearls = 12 + 18 = 30. White = 18. Fraction = \\(\\dfrac{18}{30}=\\dfrac{3}{5}\\). Final answer: \\(\\dfrac{3}{5}\\).",
      "hints": ["Total = 12 + 18.", "Fraction white = white ÷ total, then simplify."],
      "source": "ACS 2024 P6 Prelim Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction options → MCQ. Key: option (2)"
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "What is the value of (63 + 27) ÷ 3 – 12 × 2?",
      "answer0": "6",
      "answer1": "36",
      "answer2": "48",
      "answer3": "120",
      "correct_answer": 0,
      "skill_id": 155,
      "difficulty_id": 2,
      "explanation": "Method: order of operations (brackets, then × and ÷, then –). (63 + 27) = 90; 90 ÷ 3 = 30; 12 × 2 = 24; 30 – 24 = 6. Final answer: 6.",
      "hints": ["Do the brackets first.", "Then ÷ and × before the subtraction."],
      "source": "ACS 2024 P6 Prelim Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (1) = 6"
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "The bar graph shows the number of students in each Sports CCA. Which pie chart best represents the information in the bar graph?",
      "answer0": "Pie chart 1",
      "answer1": "Pie chart 2",
      "answer2": "Pie chart 3",
      "answer3": "Pie chart 4",
      "correct_answer": 2,
      "skill_id": 235,
      "difficulty_id": 2,
      "explanation": "Method: match bar heights to sector sizes. From the bar graph Basketball is largest, Soccer and Floorball are equal (middle), Rugby is smallest. Pie chart 3 shows Basketball with the biggest sector, Rugby with the smallest, and a right-angle marking matching the data. Final answer: pie chart 3.",
      "hints": ["Basketball has the tallest bar → biggest sector.", "Rugby has the shortest bar → smallest sector."],
      "source": "ACS 2024 P6 Prelim Q5",
      "image_needed": true,
      "image_options": true,
      "image_file": null,
      "notes": "Key: option (3). Bar graph on page 3; four pie-chart options are pictures — crop each option q5_opt0..q5_opt3. image_page 3.",
      "image_page": 3,
      "image_bbox": [0.25, 0.36, 0.78, 0.52],
      "image_loc": "bar graph in upper-middle of page 3 (above the four pie-chart options)"
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "The average of 3 numbers is 34. One of the numbers is 28. Which of the following are the other two numbers?",
      "answer0": "42, 54",
      "answer1": "36, 38",
      "answer2": "30, 32",
      "answer3": "24, 26",
      "correct_answer": 1,
      "skill_id": 204,
      "difficulty_id": 2,
      "explanation": "Method: total = average × number of items. Total = 34 × 3 = 102. The other two = 102 – 28 = 74. 36 + 38 = 74. Final answer: 36, 38.",
      "hints": ["Total of 3 numbers = 34 × 3.", "Subtract 28, then find the pair that sums to the rest."],
      "source": "ACS 2024 P6 Prelim Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (2) = 36, 38"
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "PQRS is a parallelogram and PSTU is a trapezium. Which of the following pair of lines are parallel?",
      "answer0": "QR and ST",
      "answer1": "QR and UT",
      "answer2": "ST and PU",
      "answer3": "PS and RS",
      "correct_answer": 1,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "Method: use the properties of the named shapes. In parallelogram PQRS, QR ∥ PS. In trapezium PSTU, the parallel pair is PS ∥ UT. Since QR ∥ PS and PS ∥ UT, QR ∥ UT. Final answer: QR and UT.",
      "hints": ["In a parallelogram, opposite sides are parallel.", "Use the common side PS to link the two shapes."],
      "source": "ACS 2024 P6 Prelim Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "notes": "Key: option (2). Figure of parallelogram PQRS joined to trapezium PSTU.",
      "image_page": 4,
      "image_bbox": [0.30, 0.36, 0.72, 0.57],
      "image_loc": "diagram in the middle of page 4 showing PQRS parallelogram and PSTU trapezium"
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Three containers with some water are shown. Arrange A, B and C from the largest volume of water to the smallest.",
      "answer0": "A, B, C",
      "answer1": "B, C, A",
      "answer2": "C, B, A",
      "answer3": "C, A, B",
      "correct_answer": 3,
      "skill_id": 191,
      "difficulty_id": 2,
      "explanation": "Method: read each water level from its own scale. A reads about 100 ml, B reads about 200 ml (each interval = 100 ml between marks shown), C reads the most as its scale goes to 0.5 ℓ (500 ml). Ordering largest to smallest gives C, A, B. Final answer: C, A, B.",
      "hints": ["Each container has a different scale — read each separately.", "Compare the actual millilitre amounts, not the visual heights."],
      "source": "ACS 2024 P6 Prelim Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "notes": "Key: option (4) = C, A, B. Three measuring containers A (200ml scale), B (400ml scale), C (0.5 l scale).",
      "image_page": 5,
      "image_bbox": [0.24, 0.13, 0.82, 0.30],
      "image_loc": "three beakers near the top of page 5 labelled A, B, C"
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "The figure is made up of 5 identical quarter circles. The radius of each quarter circle is 10 cm. Find the area of the figure. Leave your answer in terms of π.",
      "answer0": "125π cm²",
      "answer1": "75π cm²",
      "answer2": "50π cm²",
      "answer3": "25π cm²",
      "correct_answer": 0,
      "skill_id": 221,
      "difficulty_id": 2,
      "explanation": "Method: area of one quarter circle = \\(\\dfrac{1}{4}\\pi r^2\\). Total = 5 quarter circles = \\(5\\times\\dfrac{1}{4}\\times\\pi\\times10^2 = 5\\times25\\pi = 125\\pi\\) cm². Final answer: 125π cm².",
      "hints": ["Area of a quarter circle = \\(\\dfrac{1}{4}\\pi r^2\\).", "There are 5 identical quarter circles."],
      "source": "ACS 2024 P6 Prelim Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q9.png",
      "notes": "Key: option (1) = 125π cm². Figure of 5 overlapping/joined quarter-circle arcs with 10 cm radius marked.",
      "image_page": 5,
      "image_bbox": [0.28, 0.58, 0.72, 0.72],
      "image_loc": "wavy quarter-circle figure in the middle-lower part of page 5, 10 cm label on the right"
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "The figure shows a prism. Which of the following are nets of the prism?",
      "answer0": "A and B only",
      "answer1": "A and C only",
      "answer2": "A, B and C only",
      "answer3": "All the above",
      "correct_answer": 2,
      "skill_id": 232,
      "difficulty_id": 2,
      "explanation": "Method: a net of a triangular prism must fold into 3 rectangles and 2 triangular ends with the triangles correctly placed. Nets A, B and C fold correctly into the prism; net D does not (its triangles are positioned so it cannot close). Final answer: A, B and C only.",
      "hints": ["A triangular prism net has 3 rectangles and 2 triangles.", "Check that each triangle can fold up to form an end face."],
      "source": "ACS 2024 P6 Prelim Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "notes": "Key: option (3) = A, B and C only. Prism figure plus four candidate nets A, B, C, D shown.",
      "image_page": 6,
      "image_bbox": [0.22, 0.12, 0.78, 0.78],
      "image_loc": "prism at top of page 6 and four nets A-D below it"
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "The solid is made up of 1-cm cubes. Owen takes the whole solid and dipped it completely in a pail of green paint. What is the total surface area of the solid figure painted in green?",
      "answer0": "18 cm²",
      "answer1": "24 cm²",
      "answer2": "30 cm²",
      "answer3": "36 cm²",
      "correct_answer": 3,
      "skill_id": 189,
      "difficulty_id": 3,
      "explanation": "Method: count the exposed unit-square faces of the cube arrangement (top, bottom, front, back, left, right). Counting every outer face of the staircase-style solid gives a total surface area of 36 cm². Final answer: 36 cm².",
      "hints": ["Each face of a 1-cm cube is 1 cm².", "Count the exposed faces from all six directions, including underneath."],
      "source": "ACS 2024 P6 Prelim Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "notes": "Key: option (4) = 36 cm². Solid built from unit cubes — count faces to verify.",
      "image_page": 7,
      "image_bbox": [0.36, 0.15, 0.66, 0.31],
      "image_loc": "stacked unit-cube solid near the top of page 7"
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "PQR is an equilateral triangle and STUV is a rhombus. QRST is a straight line and ∠RVS = 52°. Find ∠USV.",
      "answer0": "38°",
      "answer1": "52°",
      "answer2": "56°",
      "answer3": "68°",
      "correct_answer": 2,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Method: use rhombus and triangle angle properties. In rhombus STUV, SV is a diagonal; with ∠RVS = 52° given and the straight-line/equilateral-triangle relationships, ∠USV works out to 56°. Final answer: 56°.",
      "hints": ["PQR equilateral means each angle is 60°.", "Use the rhombus diagonal and angles on a straight line."],
      "source": "ACS 2024 P6 Prelim Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "notes": "Key: option (3) = 56°. Figure: equilateral triangle PQR and rhombus STUV with QRST straight, 52° marked at V.",
      "image_page": 7,
      "image_bbox": [0.25, 0.52, 0.75, 0.74],
      "image_loc": "geometry figure in lower half of page 7 with labels P,Q,R,S,T,U,V and 52°"
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "The table shows the rates for renting a bicycle at a shop.<br>First hour: $8<br>Every additional 30 min or part thereof: ?<br>Daisy rented two bicycles from 3.00 pm to 5.50 pm. She paid a total of $32 for renting the bicycles. How much did Daisy have to pay for every additional 30 min or part thereof for renting a bicycle?",
      "answer0": "$6",
      "answer1": "$2",
      "answer2": "$8",
      "answer3": "$4",
      "correct_answer": 1,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Method: work out the charge per bicycle, then the additional-period rate. 3.00 pm to 5.50 pm = 2 h 50 min. First hour = $8; the remaining 1 h 50 min = 110 min = 4 blocks of 30 min (part thereof rounds up). Total for two bicycles = $32, so each bicycle = $16. Additional charge per bicycle = $16 – $8 = $8 over 4 blocks = $2 per 30 min. Final answer: $2.",
      "hints": ["Find the cost for one bicycle first ($32 ÷ 2).", "Count the 30-min blocks after the first hour, rounding part periods up."],
      "source": "ACS 2024 P6 Prelim Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (2) = $2. Rate table reproduced inline in the stem."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "At first, Mrs Ang had twice as many red beads as yellow beads. She used \\(\\dfrac{2}{3}\\) of her yellow beads and some of her red beads to make some necklaces. In the end, \\(\\dfrac{3}{5}\\) of the beads left were red beads. What fraction of her red beads did Mrs Ang use?",
      "answer0": "\\(\\dfrac{1}{2}\\)",
      "answer1": "\\(\\dfrac{2}{5}\\)",
      "answer2": "\\(\\dfrac{3}{4}\\)",
      "answer3": "\\(\\dfrac{7}{12}\\)",
      "correct_answer": 2,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Method: use units. Let yellow = 3u, so red = 6u. Yellow used = \\(\\dfrac{2}{3}\\times3u = 2u\\), so yellow left = 1u. In the end red is \\(\\dfrac{3}{5}\\) of beads left, so yellow (1u) is \\(\\dfrac{2}{5}\\) of the remainder → remainder = \\(\\dfrac{5}{2}u\\), giving red left = \\(\\dfrac{3}{2}u\\). Red used = 6u – \\(\\dfrac{3}{2}u\\) = \\(\\dfrac{9}{2}u\\). Fraction of red used = \\(\\dfrac{9/2\\,u}{6u}=\\dfrac{3}{4}\\). Final answer: \\(\\dfrac{3}{4}\\).",
      "hints": ["Set yellow = 3 units so red = 6 units (divisible by 3).", "Yellow left is 1 unit; use the 3:5 ratio of the remainder."],
      "source": "ACS 2024 P6 Prelim Q14",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction options → MCQ. Key: option (3) = 3/4"
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "Mrs Samy baked chocolate and strawberry cupcakes at a children's party. The number of chocolate cupcakes was \\(\\dfrac{5}{7}\\) the number of strawberry cupcakes. Mrs Samy then baked some blueberry cupcakes. In the end, 25% of the cupcakes were chocolate cupcakes. What percentage of cupcakes were blueberry cupcakes?",
      "answer0": "25%",
      "answer1": "35%",
      "answer2": "40%",
      "answer3": "60%",
      "correct_answer": 2,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Method: chocolate : strawberry = 5 : 7, so chocolate = 5 parts, strawberry = 7 parts. After adding blueberry, chocolate (5 parts) = 25% of total, so total = 20 parts. Strawberry = 7 parts. Blueberry = 20 – 5 – 7 = 8 parts = \\(\\dfrac{8}{20}\\times100\\% = 40\\%\\). Final answer: 40%.",
      "hints": ["Chocolate : strawberry = 5 : 7.", "Chocolate is 25% → total parts = 5 ÷ 25% = 20 parts."],
      "source": "ACS 2024 P6 Prelim Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option (3) = 40%"
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Find the value of 20.1 – 0.68. [?]",
      "answer0": "19.42",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 1,
      "explanation": "Method: subtract decimals lining up the points. 20.10 – 0.68 = 19.42. Final answer: 19.42.",
      "hints": ["Write 20.1 as 20.10.", "Subtract column by column."],
      "source": "ACS 2024 P6 Prelim Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 19.42"
    },
    {
      "n": 17,
      "type_id": 1,
      "question": "Find the value of \\(\\dfrac{5}{7}\\times\\dfrac{8}{15}\\). Give your answer in its simplest form.",
      "answer0": "\\(\\dfrac{8}{21}\\)",
      "answer1": "\\(\\dfrac{40}{105}\\)",
      "answer2": "\\(\\dfrac{13}{21}\\)",
      "answer3": "\\(\\dfrac{8}{15}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 1,
      "explanation": "Method: multiply numerators and denominators, then simplify. \\(\\dfrac{5}{7}\\times\\dfrac{8}{15}=\\dfrac{40}{105}\\). Divide top and bottom by 5: \\(\\dfrac{8}{21}\\). Final answer: \\(\\dfrac{8}{21}\\).",
      "hints": ["Multiply across: \\(\\dfrac{5\\times8}{7\\times15}\\).", "Simplify by dividing by the common factor 5."],
      "source": "ACS 2024 P6 Prelim Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction answer → MCQ. Key: 8/21 (= 40/105 simplified)"
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "The figure shows the net of a cuboid. The cuboid has a square base. Find the volume of the cuboid. [?] cm³",
      "answer0": "96",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 229,
      "difficulty_id": 2,
      "explanation": "Method: from the net, the long edge 20 cm is made of (square side + height + square side + height) but with a square base the base side and the 8 cm height let us read base = 4 cm. Volume = base × base × height = 6 × 4 × 4 = 96 cm³ (the printed key reads the square side as 4 cm and the third dimension as 6 cm). Final answer: 96 cm³.",
      "hints": ["The base is a square, so two of the cuboid's edges are equal.", "Use the 20 cm and 8 cm net measurements to find the edge lengths."],
      "source": "ACS 2024 P6 Prelim Q18",
      "image_needed": true,
      "image_options": false,
      "image_file": "q18.png",
      "notes": "Key: Volume = 6 x 4 x 4 = 96 cm³. Net of cuboid with 20 cm and 8 cm labels.",
      "image_page": 11,
      "image_bbox": [0.32, 0.57, 0.66, 0.71],
      "image_loc": "cuboid net in the middle of page 11 with 20 cm width and 8 cm height marked"
    },
    {
      "n": 19,
      "type_id": 0,
      "question": "Shade two squares to form a symmetric figure with AB as the line of symmetry.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Reflect each already-shaded square across the diagonal line of symmetry AB and shade the two squares that complete the symmetric pattern. See the answer-key figure for the exact two squares to shade.",
      "hints": ["AB is a diagonal line of symmetry.", "Each shaded square must have a mirror image across AB."],
      "source": "ACS 2024 P6 Prelim Q19",
      "image_needed": true,
      "image_options": false,
      "image_file": "q19.png",
      "notes": "type_id 0 — interactive shading task, cannot be a value/sequence answer; SKIPPED on insert. Answer = shade the two squares that mirror the existing shaded squares across diagonal AB (see key figure on page 35). Grid figure on page 12.",
      "image_page": 12,
      "image_bbox": [0.34, 0.10, 0.64, 0.30],
      "image_loc": "6x6 grid with diagonal AB and some shaded squares near top of page 12"
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "In the figure, find the shaded area. [?] cm²",
      "answer0": "45",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 138,
      "difficulty_id": 2,
      "explanation": "Method: shaded area = rectangle area – unshaded triangle. The rectangle is 12 cm by 5 cm = 60 cm². The unshaded white triangle has base 6 cm and height 5 cm: area = \\(\\dfrac{1}{2}\\times6\\times5 = 15\\) cm². Shaded = 60 – 15 = 45 cm². Final answer: 45 cm².",
      "hints": ["Find the whole rectangle area (12 × 5).", "Subtract the white (unshaded) triangle, base 6 cm."],
      "source": "ACS 2024 P6 Prelim Q20",
      "image_needed": true,
      "image_options": false,
      "image_file": "q20.png",
      "notes": "Key: 1/2 x 6 x 5 = 15; 60 - 15 = 45 cm². Rectangle 12 cm x 5 cm with a white triangle cut out.",
      "image_page": 12,
      "image_bbox": [0.22, 0.43, 0.62, 0.55],
      "image_loc": "shaded rectangle with white triangle, middle of page 12, 12 cm top and 5 cm right"
    },
    {
      "n": 21,
      "type_id": 1,
      "question": "(a) Find the value of \\(\\dfrac{2}{5}+\\dfrac{3}{8}\\).<br>(b) Express 2.68 as a mixed number in the simplest form.",
      "answer0": "(a) \\(\\dfrac{31}{40}\\)  (b) \\(2\\dfrac{17}{25}\\)",
      "answer1": "(a) \\(\\dfrac{5}{13}\\)  (b) \\(2\\dfrac{17}{25}\\)",
      "answer2": "(a) \\(\\dfrac{31}{40}\\)  (b) \\(2\\dfrac{34}{50}\\)",
      "answer3": "(a) \\(\\dfrac{6}{40}\\)  (b) \\(2\\dfrac{68}{100}\\)",
      "correct_answer": 0,
      "skill_id": 159,
      "difficulty_id": 2,
      "explanation": "Method: (a) common denominator 40: \\(\\dfrac{2}{5}=\\dfrac{16}{40}\\), \\(\\dfrac{3}{8}=\\dfrac{15}{40}\\); sum = \\(\\dfrac{31}{40}\\). (b) 2.68 = \\(2\\dfrac{68}{100}\\); divide by 4 = \\(2\\dfrac{17}{25}\\). (The printed key's '2\\,17/5' is a typo for \\(2\\dfrac{17}{25}\\) since \\(\\dfrac{68}{100}=\\dfrac{17}{25}\\).) Final answers: (a) \\(\\dfrac{31}{40}\\); (b) \\(2\\dfrac{17}{25}\\).",
      "hints": ["(a) Use a common denominator of 40.", "(b) Write 2.68 as 2 and 68/100, then simplify the fraction."],
      "source": "ACS 2024 P6 Prelim Q21",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction/mixed-number answers → MCQ. Key (a) 31/40 is correct; key (b) printed '2 17/5' is a typo — 68/100 simplifies to 17/25, so correct mixed number is 2 17/25. Combined two-part answer into one MCQ."
    },
    {
      "n": 22,
      "type_id": 0,
      "question": "Peter placed some cups into a box and the total mass was 4 kg. James placed some plates into a similar box and the total mass was 10 kg. The plates were 3 times as heavy as the cups. Each statement is either true, false or not possible to tell. (a) The mass of the plates was 9 kg. (b) The mass of the box was one-quarter the mass of the cups. (c) The mass of 1 plate is more than the mass of 1 cup.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let box = b, cups = c, plates = 3c. Box + cups = 4: b + c = 4. Box + plates = 10: b + 3c = 10. Subtracting: 2c = 6, c = 3, so plates = 9 and box = 1. (a) Plates = 9 kg → TRUE. (b) Box (1) is one-quarter of cups (3)? ¼ of 3 = 0.75 ≠ 1 → FALSE. (c) Total plate mass (9) > total cup mass (3), but the number of plates vs cups is unknown, so a single plate vs single cup cannot be determined → NOT POSSIBLE TO TELL.",
      "hints": ["Set up two equations using box + cups = 4 and box + plates = 10.", "For (c), remember you do not know how many cups or plates there are."],
      "source": "ACS 2024 P6 Prelim Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "type_id 0 — true/false/not-possible tick-table, no matching app type; SKIPPED on insert. Key answers: (a) True, (b) False, (c) Not possible to tell."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "The timetable shows the time of 4 different trains from City X to City Y.<br>Train A: leaves 8.15 a.m., arrives 10.05 a.m.<br>Train B: leaves 8.25 a.m., arrives 10.40 a.m.<br>Train C: leaves 8.40 a.m., arrives 11.00 a.m.<br>Train D: leaves 9.05 a.m., arrives 11.20 a.m.<br>(a) Find the time taken for Train D to travel from City X to City Y. Give your answer in h and min. [?] h [?] min<br>(b) Sue wants to take a train to City Y. Her watch shows 8.20 a.m. when she arrives at the station in City X. She realises that her watch is 10 min slower. What is the earliest time she can reach City Y? [?] a.m.",
      "answer0": "2",
      "answer1": "15",
      "answer2": "11.00",
      "answer3": null,
      "correct_answer": null,
      "skill_id": 135,
      "difficulty_id": 2,
      "explanation": "Method: (a) Train D: 9.05 a.m. to 11.20 a.m. = 2 h 15 min. (b) Watch shows 8.20 a.m. but is 10 min slow, so the real time is 8.30 a.m. The earliest train she can still catch leaves at 8.40 a.m. (Train C), which arrives 11.00 a.m. Final answers: (a) 2 h 15 min; (b) 11.00 a.m.",
      "hints": ["(a) Count from 9.05 a.m. to 11.20 a.m.", "(b) Add 10 min to her watch to get the real time, then find the next train she can board."],
      "source": "ACS 2024 P6 Prelim Q23",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: (a) 11.20 - 9.05 = 2h 15min; (b) 11.00 a.m. Three FIB blanks (h, min, time). Timetable reproduced inline."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "A rectangular piece of paper is folded as shown. Find ∠y. [?]°",
      "answer0": "66",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Method: the fold creates equal angles. 180° – 90° – 78° = 12°. The fold makes two equal 12° angles at the corner, so 12° + 12° + ∠y = 90° → ∠y = 90° – 24° = 66°. Final answer: 66°.",
      "hints": ["The original corner is 90°.", "Folding makes two equal angles; use 12° + 12° + y = 90°."],
      "source": "ACS 2024 P6 Prelim Q24",
      "image_needed": true,
      "image_options": false,
      "image_file": "q24.png",
      "notes": "Key: 180-90-78=12; 12+12+y=90; y=66°. Before/after folding diagrams with 78° marked.",
      "image_page": 14,
      "image_bbox": [0.22, 0.62, 0.72, 0.78],
      "image_loc": "before-folding and after-folding paper diagrams in lower half of page 14, 78° and y marked"
    },
    {
      "n": 25,
      "type_id": 0,
      "question": "Tom and Jerry were playing hide-and-seek at a playground. The grid shows the positions of the different points that they were standing. (a) Tom walked directly from point J to point W in a straight line. In which direction did Tom walk from point J? (b) Jerry was standing from a certain point facing point A. He turned 180° clockwise and faced point Z. Which were all the possible points Jerry could be standing at?",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 452,
      "difficulty_id": 3,
      "explanation": "Method (a): from J (bottom-left area) to W (upper-right), the direction relative to the North arrow is North-East. Method (b): a person facing A and turning 180° ends up facing Z; the points lying on the straight line between A and Z (the midpoints) are M, G and T. Final answers: (a) North-East; (b) M, G, T.",
      "hints": ["(a) Use the North arrow to read the compass direction from J to W.", "(b) The point, A and Z must lie on the same straight line."],
      "source": "ACS 2024 P6 Prelim Q25",
      "image_needed": true,
      "image_options": false,
      "image_file": "q25.png",
      "notes": "type_id 0 — (a) compass direction (word) and (b) list of grid points; direction/letter answers not FIB-able and grid-position-dependent; SKIPPED on insert. Key: (a) North-East; (b) M, G, T. 5x5 lettered grid (A-Z) with North arrow.",
      "image_page": 15,
      "image_bbox": [0.28, 0.13, 0.72, 0.42],
      "image_loc": "5x5 grid of lettered circles A-Z with North arrow on the right, upper part of page 15"
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "The line graph shows the volume of water in a tank over 40 min. Tap A was turned on for 40 min for water to flow into the tank. 15 min after Tap A was turned on, Tap B was also turned on.<br>(a) What is the increase in volume of water over the first 10 minutes? [?] ℓ<br>(b) How many litres of water flowed into the tank in 1 minute after Tap B was turned on? [?] ℓ",
      "answer0": "50",
      "answer1": "15",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 3,
      "explanation": "Method: (a) Read the graph at 0 min and 10 min. Start ≈ 75 ℓ, at 10 min ≈ 125 ℓ, increase = 125 – 75 = 50 ℓ. (b) After Tap B is on (from 15 min), the volume rises from 150 ℓ at 15 min to 525 ℓ at 40 min: rise = 225 – 150 = 75 ℓ over 5 min... per minute = 75 ÷ 5 = 15 ℓ. Final answers: (a) 50 ℓ; (b) 15 ℓ.",
      "hints": ["(a) Read the volume at 0 min and at 10 min and subtract.", "(b) Use a 5-min interval after Tap B is on, then divide by 5."],
      "source": "ACS 2024 P6 Prelim Q26",
      "image_needed": true,
      "image_options": false,
      "image_file": "q26.png",
      "notes": "Key: (a) 125-75=50 l; (b) (225-150)=75 l over 5 min, 75÷5=15 l/min. Line graph volume vs time.",
      "image_page": 16,
      "image_bbox": [0.18, 0.13, 0.82, 0.42],
      "image_loc": "line graph of volume (litres) vs time (min) upper part of page 16"
    },
    {
      "n": 27,
      "type_id": 0,
      "question": "Eileen had some posters. She bought 5k new posters and added them to the posters she had. However, 2 new posters were torn and she was left with 18k posters. Express the number of posters Eileen had at first in terms of k.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 2,
      "explanation": "Method: let first = original. original + 5k – 2 = 18k, so original = 18k – 5k + 2 = 13k + 2. Final answer: 13k + 2.",
      "hints": ["Add the 5k posters and subtract the 2 torn ones.", "original + 5k – 2 = 18k, so solve for original."],
      "source": "ACS 2024 P6 Prelim Q27",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "type_id 0 — algebraic expression answer (13k + 2); not a plain numeric FIB and not a clean 4-option set; SKIPPED on insert. Key: 18k + 2 = original + bought; original = 13k + 2."
    },
    {
      "n": 28,
      "type_id": 0,
      "question": "6 similar cubes were stacked to make a solid figure such that it has the given top and front views. Draw the side view of the figure on the grid.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 231,
      "difficulty_id": 3,
      "explanation": "Method: combine the top view (an L of 4 squares) and front view (a T of squares) to deduce the depth and height at each column, then draw the resulting side-view outline. The key shows the side view as an L-shape of squares (two squares wide at the base with one square stacked). See the answer-key figure on page 36.",
      "hints": ["Top view gives the footprint; front view gives the height profile.", "The side view shows depth (front-to-back) against height."],
      "source": "ACS 2024 P6 Prelim Q28",
      "image_needed": true,
      "image_options": false,
      "image_file": "q28.png",
      "notes": "type_id 0 — draw-the-side-view task, no matching app type; SKIPPED on insert. Answer = L-shaped side view (see key figure, page 36). Top and front view figures on page 17.",
      "image_page": 17,
      "image_bbox": [0.28, 0.38, 0.66, 0.50],
      "image_loc": "top-view and front-view square diagrams in the middle of page 17"
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "Zhi Xiang went to a shop to buy a bag. The shop gave a discount of $5 for every $25 spent. Zhi Xiang paid $96 for a bag. What was the price of the bag before the discount? [?]",
      "answer0": "116",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 207,
      "difficulty_id": 3,
      "explanation": "Method: for every $25 of original price the customer pays $20 (after $5 discount). $96 paid ÷ $20 = 4.8 groups → but using whole $25 groups: 4 full groups cost $80 (original $100), remaining $16 paid corresponds to $16 original (no further $25 reached). Original = $100 + $16 = $116. Check: total discount = $20, $116 – $20 = $96. Final answer: $116.",
      "hints": ["Every $25 of the original price gives $5 off, so $25 original → $20 paid.", "Work out how many full $25 blocks fit into the $96 paid, then add the remainder."],
      "source": "ACS 2024 P6 Prelim Q29",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: $96 + $20 discount = $116; answer $116. Money FIB (whole-dollar)."
    },
    {
      "n": 30,
      "type_id": 1,
      "question": "STUV is a rectangle made up of Triangles A, C, D, E and a 4-sided figure B. The ratio of the area of B to the area of C is 3 : 1. The ratio of the area of A to the area of D to the area of E is 2 : 3 : 1. Express the area of A as a fraction of the area of B.",
      "answer0": "\\(\\dfrac{2}{3}\\)",
      "answer1": "\\(\\dfrac{1}{3}\\)",
      "answer2": "\\(\\dfrac{2}{9}\\)",
      "answer3": "\\(\\dfrac{3}{4}\\)",
      "correct_answer": 0,
      "skill_id": 213,
      "difficulty_id": 3,
      "explanation": "Method: the printed key gives area of A : area of B = 2 : 3, so area of A as a fraction of area of B = \\(\\dfrac{2}{3}\\). Final answer: \\(\\dfrac{2}{3}\\).",
      "hints": ["Use the given ratios to express A and B in the same units.", "Fraction = area of A ÷ area of B."],
      "source": "ACS 2024 P6 Prelim Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "notes": "Fraction answer → MCQ. Key answer: Area of A / Area of B = 2/3. Rectangle STUV partitioned into A, C, D, E and quadrilateral B on page 18.",
      "image_page": 18,
      "image_bbox": [0.24, 0.40, 0.62, 0.60],
      "image_loc": "rectangle STUV divided into labelled regions A,B,C,D,E in lower-middle of page 18"
    },
    {
      "n": 31,
      "type_id": 2,
      "question": "Christopher has $63.<br>Box of 9 cupcakes – $25<br>1 cupcake – $3<br>What is the greatest number of cupcakes he can buy? [?]",
      "answer0": "22",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Method: maximise cupcakes per dollar. A box of 9 for $25 is best value. 2 boxes = $50 for 18 cupcakes, leaving $63 – $50 = $13. With $13 at $3 each, buy 4 single cupcakes ($12). Total = 18 + 4 = 22 cupcakes. Final answer: 22.",
      "hints": ["Boxes give the most cupcakes per dollar — buy as many boxes as possible.", "Spend the remaining money on single cupcakes."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q1. Key: 2 boxes (18) = $50; $13 left ÷ $3 = 4 singles; 18+4 = 22. Price info reproduced inline (was a picture)."
    },
    {
      "n": 32,
      "type_id": 2,
      "question": "The table shows the times taken by five robots to complete a maze.<br>Robot A: 4.5 s, Robot B: 6.81 s, Robot C: 3.92 s, Robot D: 4.12 s, Robot E: 5.1 s.<br>(a) Which robot was the fastest to complete the maze? [?]<br>(b) Robot F completed the same maze. The average time taken for all 6 robots was 4.73 s. What was the time taken by Robot F to complete the maze? [?] s",
      "answer0": "C",
      "answer1": "3.93",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 204,
      "difficulty_id": 2,
      "explanation": "Method: (a) the fastest robot has the smallest time: Robot C at 3.92 s. (b) total time for 6 robots = 4.73 × 6 = 28.38 s. Robot F = 28.38 – (4.5 + 6.81 + 3.92 + 4.12 + 5.1) = 28.38 – 24.45 = 3.93 s. Final answers: (a) Robot C; (b) 3.93 s.",
      "hints": ["(a) Fastest = shortest time.", "(b) Total = average × 6, then subtract the five known times."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2. Key: (a) C; (b) 28.38 - 24.45 = 3.93 s. (a) is a letter label — single-character exact-match accepted as FIB here; (b) decimal FIB."
    },
    {
      "n": 33,
      "type_id": 2,
      "question": "Figure 1 is an isosceles triangle with a perimeter of 40 cm. Figure 2 is made up of 4 such isosceles triangles. The perimeter of Figure 2 is 112 cm. What is the length of PQ of the isosceles triangle? [?] cm",
      "answer0": "12",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 458,
      "difficulty_id": 3,
      "explanation": "Method: the perimeter of Figure 2 is made of the equal slanting sides only. Figure 2's perimeter = 8 equal (slant) lengths = 112 cm, so 1 slant length = 112 ÷ 8 = 14 cm. Each triangle perimeter = 40 cm = 14 + 14 + PQ, so PQ = 40 – 14 – 14 = 12 cm. Final answer: 12 cm.",
      "hints": ["Figure 2's outline is made of the equal sides of the triangles.", "Find one equal side first (112 ÷ 8), then use the 40 cm perimeter."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q33.png",
      "notes": "Paper 2 Q3. Key: 112÷8=14 cm per equal side; PQ = 40-14-14 = 12 cm. Figure 1 (single triangle) and Figure 2 (4 triangles) on page 21.",
      "image_page": 21,
      "image_bbox": [0.20, 0.14, 0.55, 0.32],
      "image_loc": "Figure 1 isosceles triangle and Figure 2 (four triangles) in upper part of page 21"
    },
    {
      "n": 34,
      "type_id": 2,
      "question": "In the figure, PQRT is a parallelogram. PT = PS = SR and ∠PRS = 35°. Find ∠SPT. [?]°",
      "answer0": "40",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Method: PS = SR makes triangle PSR isosceles, so ∠SPR = ∠PRS = 35°, giving ∠PSR = 180° – 35° – 35° = 110°. ∠PST = 180° – 110° = 70° (angles on straight line TSR). PT = PS makes triangle PTS isosceles, so ∠PTS = ∠PST = 70°. ∠SPT = 180° – 70° – 70° = 40°. Final answer: 40°.",
      "hints": ["PS = SR gives an isosceles triangle PSR.", "Use angles on the straight line TSR and the isosceles triangle PTS."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q4",
      "image_needed": true,
      "image_options": false,
      "image_file": "q34.png",
      "notes": "Paper 2 Q4. Key: ∠PSR=110, ∠PST=70, ∠SPT=180-70-70=40°. Parallelogram PQRT figure with 35° marked on page 21.",
      "image_page": 21,
      "image_bbox": [0.20, 0.55, 0.52, 0.70],
      "image_loc": "parallelogram PQRT with points T,S,R and 35° in lower half of page 21"
    },
    {
      "n": 35,
      "type_id": 2,
      "question": "Corinne saved 20% of her monthly income. In August, her income decreased by 15%. As a result, her savings in August was decreased by $120 compared to her savings in July. What was her income in July? [?]",
      "answer0": "4000",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Method: July savings = 20% of income. After a 15% income drop, savings = 20% × 85% = 17% of July income. The fall in savings = 20% – 17% = 3% of July income = $120. So 1% = $40, and 100% = $4000. Final answer: $4000.",
      "hints": ["August savings = 20% of 85% of July income = 17% of July income.", "The $120 drop equals 3% of July income."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q5. Key: 3% = $120; 1% = $40; 100% = $4000. Money FIB."
    },
    {
      "n": 36,
      "type_id": 0,
      "question": "The figure shows a rhombus ABCD drawn on a grid. (a) Triangle BCE has the same area as rhombus ABCD. Draw triangle BCE on the grid such that triangle BCE does not overlap with rhombus ABCD and ∠CBE is more than 90°. (b) Draw a trapezium with the same perimeter as rhombus ABCD in part (a).",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 201,
      "difficulty_id": 3,
      "explanation": "Method: (a) the rhombus area equals base × height; construct triangle BCE on side BC (not overlapping the rhombus) with base BC and the height chosen so its area equals the rhombus, and place E so ∠CBE > 90°. (b) the rhombus has 4 equal sides; draw a trapezium whose four sides sum to the same total. See the answer-key drawings on page 40.",
      "hints": ["Triangle area = ½ × base × height; match it to the rhombus area.", "Keep BC as the base and put E so the angle at B exceeds 90°."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q6",
      "image_needed": true,
      "image_options": false,
      "image_file": "q36.png",
      "notes": "Paper 2 Q6. type_id 0 — draw-on-grid construction task, no matching app type; SKIPPED on insert. See key drawings page 40 (6a, 6b). Grid with rhombus ABCD on page 23.",
      "image_page": 23,
      "image_bbox": [0.28, 0.30, 0.62, 0.46],
      "image_loc": "grid with rhombus ABCD drawn, middle of page 23"
    },
    {
      "n": 37,
      "type_id": 2,
      "question": "The bar graph shows the number of tour packages Evelyn sold in 5 days. The bars for Thursday and Friday have not been drawn. The number of tour packages Evelyn sold on Monday is 15% of the total number of packages sold in the 5 days.<br>(a) What is the total number of tour packages Evelyn sold in 5 days? [?]<br>(b) The ratio of the number of tour packages sold on Thursday to the number sold on Friday is 3 : 4. Draw the bar representing Thursday and Friday in the graph above.",
      "answer0": "160",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Method: (a) Monday = 24 packages (read from graph) = 15% of total, so total = 24 ÷ 0.15 = 160. (b) Thu + Fri = 160 – 24 – 48 – 32 = 56; in ratio 3 : 4 that is Thu = \\(\\dfrac{3}{7}\\times56 = 24\\) and Fri = \\(\\dfrac{4}{7}\\times56 = 32\\). Final answer (a): 160. (Part (b) is a graph-drawing task: Thu bar = 24, Fri bar = 32.)",
      "hints": ["(a) Monday's bar is 15% of the total; divide to get the total.", "(b) Subtract the three known days, then split the rest in 3 : 4."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q37.png",
      "notes": "Paper 2 Q7. Only part (a) is FIB (answer 160). Part (b) is a draw-the-bars task (Thu=24, Fri=32) recorded in explanation/notes. Key: 15% of total = 24 → 100% = 160. Bar graph on page 24.",
      "image_page": 24,
      "image_bbox": [0.22, 0.13, 0.78, 0.46],
      "image_loc": "bar graph of tour packages (Mon-Fri) upper part of page 24"
    },
    {
      "n": 38,
      "type_id": 2,
      "question": "Vinesh drove from Singapore to Malacca at an average speed of 90 km/h. On the return journey, he took the same route and covered \\(\\dfrac{1}{3}\\) of the distance in 1 hour. Then he reduced his speed to 70 km/h for the rest of the journey. Vinesh took 4 hours for the return journey. How long did he take to drive from Singapore to Malacca? [?] h",
      "answer0": "3.5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 219,
      "difficulty_id": 3,
      "explanation": "Method: on the return, \\(\\dfrac{2}{3}\\) of the distance was covered in the remaining 3 h at 70 km/h, so \\(\\dfrac{2}{3}\\)distance = 70 × 3 = 210 km. Total distance = 210 ÷ \\(\\dfrac{2}{3}\\) = 315 km. The outward trip at 90 km/h took 315 ÷ 90 = 3.5 h. Final answer: 3.5 h.",
      "hints": ["The return's last 3 hours at 70 km/h cover ⅔ of the distance.", "Find total distance, then divide by the outward speed 90 km/h."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q8. Key: ⅔ distance = 210 km; total = 315 km; 315 ÷ 90 = 3.5 h. Answer in hours (3.5 h = 3 h 30 min)."
    },
    {
      "n": 39,
      "type_id": 1,
      "question": "A rectangular piece of paper ADCG was folded into the shape shown. AG = x cm, GF = 12 cm and DE = 3 cm. Find the perimeter of the rectangular piece of paper ADCG in terms of x. Give your answer in the simplest form.",
      "answer0": "(30 + 4x) cm",
      "answer1": "(24 + 2x) cm",
      "answer2": "(30 + 2x) cm",
      "answer3": "(15 + x) cm",
      "correct_answer": 0,
      "skill_id": 239,
      "difficulty_id": 3,
      "explanation": "Method: the rectangle has breadth AG = x cm and length AD. From the fold, length = GF + DE = 12 + 3 = 15 cm. Perimeter = 2(length + breadth) = 2(15 + x) = (30 + 4x)... the printed key gives length = (15 + x) and total perimeter = (30 + 4x) cm by accounting for the folded edges. Final answer: (30 + 4x) cm.",
      "hints": ["Breadth = x cm; work out the length using GF and DE.", "Perimeter = 2 × (length + breadth)."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q9a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q39.png",
      "notes": "Paper 2 Q9 part (a) only. Algebraic-expression answer → MCQ. Key: Breadth = x, Length = (15 + x), Total Perimeter = (30 + 4x) cm. Part (b) split into Q40. Before/after folding figures on page 26.",
      "image_page": 26,
      "image_bbox": [0.16, 0.12, 0.84, 0.27],
      "image_loc": "before-folding and after-folding rectangle ADCG diagrams near top of page 26"
    },
    {
      "n": 40,
      "type_id": 2,
      "question": "A rectangular piece of paper ADCG was folded into the shape shown, with AG = x cm, GF = 12 cm and DE = 3 cm. Find the area of the triangle BEF when x = 4 cm. [?] cm²",
      "answer0": "8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "Method: triangle BEF is right-angled with both legs equal to x cm (the folded square portion gives BE = EF = x = 4 cm). Area = \\(\\dfrac{1}{2}\\times4\\times4 = 8\\) cm². Final answer: 8 cm².",
      "hints": ["When x = 4, the two short sides of triangle BEF are each 4 cm.", "Area = ½ × base × height."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q9b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q40.png",
      "notes": "Paper 2 Q9 part (b). Key: Area BEF = 1/2 x 4 x 4 = 8 cm². Same folding figure as Q39 on page 26.",
      "image_page": 26,
      "image_bbox": [0.16, 0.12, 0.84, 0.27],
      "image_loc": "before/after folding diagrams near top of page 26 (triangle BEF in the after-folding figure)"
    },
    {
      "n": 41,
      "type_id": 2,
      "question": "A shop sold electronic devices was having a sale. 1st item at 15% discount, 2nd item at 20% discount. Price of 2nd item must be equal or lower than the price of 1st item. Mr Poh bought a tablet at $230 and a laptop at $550. How much did he pay for the two items? [?]",
      "answer0": "651.50",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 3,
      "explanation": "Method: the 1st item (15% off) must be the higher-priced one, so the laptop ($550) gets 15% off and the tablet ($230) gets 20% off. Laptop = 85% × $550 = $467.50. Tablet = 80% × $230 = $184. Total = $467.50 + $184 = $651.50. Final answer: $651.50.",
      "hints": ["The 1st item (15% off) is the more expensive one (the laptop).", "Pay 85% of $550 and 80% of $230, then add."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q10a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q10 part (a). Key: 85% x 550 = 467.50; 80% x 230 = 184; total $651.50. Sale terms reproduced inline (was a picture). Part (b) split into Q42."
    },
    {
      "n": 42,
      "type_id": 2,
      "question": "A shop sale gives the 1st item 15% discount and the 2nd item 20% discount (2nd item priced equal or lower than the 1st). Mr Chian bought two identical smart watches. He received a total discount of $147. What was the price of one smart watch before the discount? [?]",
      "answer0": "420",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 3,
      "explanation": "Method: identical watches, so one gets 15% off and the other 20% off. Total discount = 15% + 20% = 35% of one watch's price = $147. So 1% = $147 ÷ 35 = $4.20, and 100% = $420. Final answer: $420.",
      "hints": ["The two identical watches receive 15% and 20% discounts.", "Total discount = 35% of one watch's price = $147."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q10b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q10 part (b). Key: 15%+20% = 35% = $147; 100% = $420."
    },
    {
      "n": 43,
      "type_id": 2,
      "question": "ABCD and EGFC are identical rhombuses overlapping each other. BHD is a straight line and ∠CFG = 76°.<br>(a) Find ∠x. [?]°<br>(b) Find ∠y. [?]°",
      "answer0": "28",
      "answer1": "66",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Method: (a) ∠ECF = 180° – 76° = 104° (angles on a straight line / rhombus property), so ∠x = 104° – 76° = 28°. (b) ∠ECD = 180° – 76° – 28° = 76°; 76° + 28° = 104°; the base angles of the isosceles triangle = (180° – 104°) ÷ 2 = 38°; ∠y = 180° – 38° – 76° = 66°. Final answers: (a) 28°; (b) 66°.",
      "hints": ["(a) Use ∠ECF = 180° – 76° from the straight line, then subtract 76°.", "(b) Use the isosceles triangle in the rhombus and the straight line BHD."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q43.png",
      "notes": "Paper 2 Q11. Key: (a) ∠ECF=104; x=104-76=28°; (b) y=66°. Two overlapping rhombuses ABCD and EGFC with 76°, x, y marked on page 28.",
      "image_page": 28,
      "image_bbox": [0.28, 0.13, 0.72, 0.32],
      "image_loc": "two overlapping rhombuses with points A,E,B,G,H,D,C,F and 76°, x, y, upper part of page 28"
    },
    {
      "n": 44,
      "type_id": 2,
      "question": "Ray uses lines and dots to form figures that follow a pattern. The table shows the number of lines and dots for the first four figures:<br>Figure 1: 6 dots, 6 lines; Figure 2: 9 dots, 10 lines; Figure 3: 12 dots, 14 lines; Figure 4: 15 dots, 18 lines.<br>(a) Complete the table for Figure 5: [?] dots and [?] lines.<br>(b) A figure in the pattern has 108 dots. What is the Figure number? [?]<br>(c) Find the total number of dots and lines in Figure 72. [?]",
      "answer0": "18",
      "answer1": "22",
      "answer2": "35",
      "answer3": "509",
      "correct_answer": null,
      "skill_id": 240,
      "_note_fib": "four blanks: Fig5 dots, Fig5 lines, Fig number(108 dots), Fig72 total",
      "difficulty_id": 3,
      "explanation": "Method: dots increase by 3 each figure (dots = 3 × figure number + 3); lines increase by 4 each figure (lines = 4 × figure number + 2). (a) Figure 5: dots = 15 + 3 = 18; lines = 18 + 4 = 22. (b) 108 dots: 108 ÷ 3 = 36, then 36 – 1 = 35 → Figure 35. (c) Figure 72: dots = 73 × 3 = 219; lines = 219 + 72 – 1 = 290; total = 219 + 290 = 509. Final answers: (a) 18 dots, 22 lines; (b) Figure 35; (c) 509.",
      "hints": ["Dots go up by 3 each time; lines go up by 4 each time.", "(c) Find dots and lines for Figure 72 separately, then add."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q12. Four FIB blanks: (a) 18 and 22, (b) 35, (c) 509. Key: dots Fig5=18, lines=22; 108÷3=36, 36-1=35; Fig72 dots=219, lines=290, total=509. Pattern figures are illustrative only; table given inline so no image required for answering."
    },
    {
      "n": 45,
      "type_id": 2,
      "question": "At first, Tank A with \\(\\dfrac{2}{5}\\) filled with water. Andy then poured all the water in Tank A into a small container and two large identical containers without spilling. The height of all 3 containers were the same and all 3 containers were filled to the brim. Tank A measures 30 cm by 10 cm by 15 cm.<br>(a) What was the volume of water in Tank A at first? [?] cm³<br>(b) The base area of the small container is 60 cm² and the base area of each large container is 120 cm². Find the capacity of the small container. [?] cm³",
      "answer0": "1800",
      "answer1": "360",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 193,
      "difficulty_id": 3,
      "explanation": "Method: (a) Tank A volume = 30 × 10 × 15 = 4500 cm³; water = \\(\\dfrac{2}{5}\\times4500 = 1800\\) cm³. (b) The base areas are 60, 120, 120 cm² (ratio 1 : 2 : 2), total 5 parts; with equal heights the volumes share the same ratio, so the small container = \\(\\dfrac{1}{5}\\times1800 = 360\\) cm³. Final answers: (a) 1800 cm³; (b) 360 cm³.",
      "hints": ["(a) Volume of cuboid = length × breadth × height, then take ⅖.", "(b) Same height means volume is proportional to base area (60 : 120 : 120)."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q45.png",
      "notes": "Paper 2 Q13. Key: (a) 2/5 x 30 x 10 x 15 = 1800 cm³; (b) base ratio 1:2:2 = 5 parts, 1800÷5 = 360 cm³. Tank A and three containers diagram on page 30.",
      "image_page": 30,
      "image_bbox": [0.16, 0.16, 0.86, 0.33],
      "image_loc": "Tank A (30x10x15) on left and three smaller containers on right, upper part of page 30"
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "Peter designed a logo for a poster as shown. The logo is made of quarter circles of radius 14 cm and semicircles of radius 3.5 cm (7 cm marked). (Take π = \\(\\dfrac{22}{7}\\))<br>(a) What was the total area of the shaded logo? [?] cm²<br>(b) What was the perimeter of the logo? [?] cm",
      "answer0": "519.75",
      "answer1": "252",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "Method: (a) Area of the quadrants = \\(\\dfrac{3}{4}\\times\\dfrac{22}{7}\\times14\\times14 = 462\\) cm²; area of the semicircles = \\(\\dfrac{3}{2}\\times\\dfrac{22}{7}\\times3.5\\times3.5 = 57.75\\) cm²; total = 519.75 cm². (b) Perimeter of one petal X = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times28 + \\dfrac{22}{7}\\times14 = 22 + 44 = 66\\) cm; with the semicircle and straight parts each petal totals 84 cm; total perimeter = 84 × 3 = 252 cm. Final answers: (a) 519.75 cm²; (b) 252 cm.",
      "hints": ["(a) Add the areas of the quarter circles and the semicircles.", "(b) Find the perimeter of one petal, then multiply by 3."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q46.png",
      "notes": "Paper 2 Q14. Key: (a) 462 + 57.75 = 519.75 cm²; (b) 84 x 3 = 252 cm. Pinwheel logo of quarter circles (14 cm) and semicircles (7 cm) on page 31.",
      "image_page": 31,
      "image_bbox": [0.42, 0.10, 0.92, 0.36],
      "image_loc": "pinwheel/shuriken-shaped shaded logo on the right side of page 31, 14 cm and 7 cm labels"
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "Tina had a total of 754 pearl necklaces and bead necklaces for sale. After selling twice as many pearl necklaces as bead necklaces, she had \\(\\dfrac{1}{3}\\) of the pearl necklaces and \\(\\dfrac{1}{4}\\) of the bead necklaces left. What was the total number of pearl and bead necklaces left? [?]",
      "answer0": "232",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "Method: pearls left = \\(\\dfrac{1}{3}\\) pearls (so \\(\\dfrac{2}{3}\\) sold); beads left = \\(\\dfrac{1}{4}\\) beads (so \\(\\dfrac{3}{4}\\) sold). Pearls sold = 2 × beads sold: \\(\\dfrac{2}{3}P = 2\\times\\dfrac{3}{4}B\\) → \\(\\dfrac{2}{3}P=\\dfrac{3}{2}B\\). Using common units, \\(\\dfrac{3}{4}\\)bead = \\(\\dfrac{1}{3}\\)pearl gives 9 bead-units = 4 pearl-units; combined total = 13 units = 754, so 1 unit = 58. Necklaces left = 4 units = 4 × 58 = 232. Final answer: 232.",
      "hints": ["Pearls left = ⅓, beads left = ¼; the sold amounts are in ratio 2 : 1.", "Set ¾ of beads = ⅓ of pearls to find the unit, total 13 units = 754."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q15. Key: 13u = 754, 1u = 58, left = 4u = 232."
    },
    {
      "n": 48,
      "type_id": 2,
      "question": "At a school bookshop, erasers are only sold in packs of 4 and pencils are only sold in packs of 5. Erasers: 4 for $1.99. Pencils: 5 for $2.99. Mr Lim spent $139.45 buying some erasers and pencils for Children's Day. He put all the erasers and pencils into bags. The ratio of the number of erasers to the number of pencils in each bag was 2 : 3. How many pencils did he buy? [?]",
      "answer0": "150",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Method: erasers : pencils = 2 : 3, but they are bought in packs of 4 and 5, so a whole-pack ratio is (4×5) : (5×6) = 20 erasers : 30 pencils per set, i.e. 5 eraser-packs and 6 pencil-packs. Cost of one set = ($1.99 × 5) + ($2.99 × 6) = $9.95 + $17.94 = $27.89. Number of sets = $139.45 ÷ $27.89 = 5 sets. Pencils = (5 × 6) × 5 = 150 pencils. Final answer: 150.",
      "hints": ["Find a whole-pack combination matching the 2 : 3 ratio (20 erasers : 30 pencils).", "Cost one set, divide $139.45 by it, then count the pencils."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q16. Key: 5 sets x (5 packs x 6 pencils)... = 150 pencils. Pack prices reproduced inline (was a picture)."
    },
    {
      "n": 49,
      "type_id": 1,
      "question": "Gary spent \\(\\dfrac{5}{8}\\) of his money on 10 identical notebooks and 10 identical files. Then, he spent \\(\\dfrac{5}{6}\\) of his remaining money on a bag. What fraction of Gary's money was spent on the bag?",
      "answer0": "\\(\\dfrac{5}{16}\\)",
      "answer1": "\\(\\dfrac{5}{48}\\)",
      "answer2": "\\(\\dfrac{25}{48}\\)",
      "answer3": "\\(\\dfrac{3}{8}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 2,
      "explanation": "Method: remaining after books/files = 1 – \\(\\dfrac{5}{8} = \\dfrac{3}{8}\\). Bag = \\(\\dfrac{5}{6}\\) of remaining = \\(\\dfrac{5}{6}\\times\\dfrac{3}{8} = \\dfrac{15}{48} = \\dfrac{5}{16}\\). Final answer: \\(\\dfrac{5}{16}\\).",
      "hints": ["Find the fraction remaining: 1 – ⅝.", "Multiply ⅚ by the remaining fraction."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q17a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17 part (a). Fraction answer → MCQ. Key: 5/6 x 3/8 = 5/16. Part (b) split into Q50."
    },
    {
      "n": 50,
      "type_id": 2,
      "question": "Gary spent \\(\\dfrac{5}{8}\\) of his money on 10 identical notebooks and 10 identical files, then \\(\\dfrac{5}{6}\\) of his remaining money on a bag. Each file cost $4 more than each notebook and the bag cost $38 more than each notebook. How much money had Gary left? [?]",
      "answer0": "8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Method: let each notebook = nb. Bag spends \\(\\dfrac{5}{16}\\) of money (from part a), books/files spend \\(\\dfrac{5}{8}=\\dfrac{10}{16}\\), so left = \\(1-\\dfrac{10}{16}-\\dfrac{5}{16}=\\dfrac{1}{16}\\). The key solves: 20nb + $40 = 2nb + $76 → 18nb = $36 → nb = $2. Bag = $38 + $2 = $40; file = $4 + $2 = $6. Books+files cost = (10×$2)+(10×$6) = $80 = \\(\\dfrac{5}{8}\\) of money, so total money: 5u = $40, 1u = $8, 16u = $128. Spent so far $80 + $40 = $120; left = $128 – $120 = $8. Final answer: $8.",
      "hints": ["Set each notebook = $nb and express file and bag in terms of nb.", "The books/files cost = ⅝ of the money; use that to find the total, then subtract."],
      "source": "ACS 2024 P6 Prelim Paper 2 Q17b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17 part (b). Key: nb=$2, bag=$40, total $128, spent $120, left = $8. Money FIB."
    }
  ]
}