{
  "paper": {
    "school": "Nanyang Primary School",
    "year": 2024,
    "level": "P6",
    "label": "Mid-Year Practice",
    "source_prefix": "Nanyang 2024 P6 Mid-Year Practice",
    "has_answer_key": false
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Part of a scale is shown. What is the value of the reading at X?",
      "answer0": "10.02",
      "answer1": "10.04",
      "answer2": "10.2",
      "answer3": "10.4",
      "correct_answer": 1,
      "skill_id": 13,
      "difficulty_id": 1,
      "explanation": "Each major interval (e.g. 9.7 to 9.8) is 0.1, divided into 5 small parts, so each small mark is 0.02. X is 2 small marks past 10.0: 10.0 + 2(0.02) = 10.04.",
      "hints": ["Find the value of one small division.", "10 small marks span 0.2, so each is 0.02. Count marks past 10.0."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q1",
      "image_needed": true,
      "image_options": false,
      "image_page": 3,
      "image_bbox": [0.18, 0.20, 0.85, 0.30],
      "image_loc": "horizontal number-line scale just below the question, marked 9.7 to 10.0 with arrow X at the right end",
      "image_file": "q1.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Find the value of \\(\\dfrac{5}{6} \\div \\dfrac{1}{4}\\)",
      "answer0": "\\(\\dfrac{10}{3}\\)",
      "answer1": "\\(\\dfrac{5}{24}\\)",
      "answer2": "\\(\\dfrac{3}{10}\\)",
      "answer3": "\\(\\dfrac{24}{5}\\)",
      "correct_answer": 0,
      "skill_id": 206,
      "difficulty_id": 1,
      "explanation": "Dividing by a fraction means multiplying by its reciprocal: \\(\\dfrac{5}{6} \\div \\dfrac{1}{4} = \\dfrac{5}{6} \\times \\dfrac{4}{1} = \\dfrac{20}{6} = \\dfrac{10}{3}\\).",
      "hints": ["Multiply by the reciprocal of the divisor.", "\\(\\dfrac{5}{6} \\times 4 = \\dfrac{20}{6}\\), then simplify."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved; fraction answer so MCQ"
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Joyce baked some cookies. She gave 80% of the cookies to Zac. Zac ate 20% of the cookies he received from Joyce. Which one of the following shows the percentage of total cookies that Zac ate?",
      "answer0": "\\(\\dfrac{1}{5} \\times 20\\%\\)",
      "answer1": "\\(\\dfrac{1}{5} \\times 80\\%\\)",
      "answer2": "\\(\\dfrac{4}{5} \\times 80\\%\\)",
      "answer3": "\\(\\dfrac{4}{5} \\times 100\\%\\)",
      "correct_answer": 1,
      "skill_id": 209,
      "difficulty_id": 2,
      "explanation": "Zac received 80% of the cookies and ate 20% of that. 20% = \\(\\dfrac{1}{5}\\), so the fraction of total he ate is \\(\\dfrac{1}{5}\\) of 80%, i.e. \\(\\dfrac{1}{5} \\times 80\\%\\).",
      "hints": ["He ate 20% (= one fifth) of what he received.", "What he received is 80% of the total."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "The square grid shows Triangle P. What type of triangle is Triangle P?",
      "answer0": "Obtuse-angled triangle",
      "answer1": "Right-angled triangle",
      "answer2": "Equilateral triangle",
      "answer3": "Isosceles triangle",
      "correct_answer": 3,
      "skill_id": 230,
      "difficulty_id": 1,
      "explanation": "On the grid the triangle has a horizontal base of 4 units and an apex centred above the midpoint of the base, so the two slanted sides are equal in length. A triangle with two equal sides is isosceles. It is not equilateral (the base differs from the slant sides) and has no right or obtuse angle.",
      "hints": ["Count grid units to compare the side lengths.", "Two equal sides = isosceles."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q4",
      "image_needed": true,
      "image_options": false,
      "image_page": 4,
      "image_bbox": [0.38, 0.42, 0.66, 0.62],
      "image_loc": "square grid with tall narrow triangle labelled P, centred below the question",
      "image_file": "q4.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Which shape is a rhombus?",
      "answer0": "Shape (1)",
      "answer1": "Shape (2)",
      "answer2": "Shape (3)",
      "answer3": "Shape (4)",
      "correct_answer": 3,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "A rhombus has 4 equal sides. Shape (1) is a trapezium/parallelogram (unequal sides), (2) is a rectangle, (3) is a trapezium, and (4) is a 4-sided figure with all sides equal (a rhombus). So the answer is shape (4).",
      "hints": ["A rhombus has all four sides equal in length.", "Check which figure on the grid has 4 equal slanted sides."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q5",
      "image_needed": true,
      "image_options": true,
      "image_page": 5,
      "image_bbox": [0.22, 0.10, 0.86, 0.27],
      "image_loc": "row of four shapes on a square grid labelled (1) trapezium, (2) rectangle, (3) trapezium, (4) rhombus",
      "image_file": "q5.png",
      "notes": "no answer key in paper - solved; four option shapes drawn on one grid - downstream split into q5_opt0..3"
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "What is the area of triangle ABC as shown in the figure?",
      "answer0": "18 cm²",
      "answer1": "20 cm²",
      "answer2": "30 cm²",
      "answer3": "36 cm²",
      "correct_answer": 0,
      "skill_id": 464,
      "difficulty_id": 1,
      "explanation": "Base CB = 4 + 8 = 12 cm and the perpendicular height from A is 3 cm. Area = \\(\\dfrac{1}{2} \\times 12 \\times 3 = 18\\) cm². (The 5 cm slant side is not used.)",
      "hints": ["Base = 4 cm + 8 cm.", "Area = ½ × base × perpendicular height (3 cm)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q6",
      "image_needed": true,
      "image_options": false,
      "image_page": 5,
      "image_bbox": [0.24, 0.45, 0.80, 0.60],
      "image_loc": "triangle ABC with apex A, base C to B; 5 cm slant, 3 cm height, base split 4 cm and 8 cm",
      "image_file": "q6.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "What is the area of a circle with diameter 60 cm? ( Take \\(\\pi = 3.14\\) )",
      "answer0": "94.2 cm²",
      "answer1": "188.4 cm²",
      "answer2": "2826 cm²",
      "answer3": "11 304 cm²",
      "correct_answer": 2,
      "skill_id": 220,
      "difficulty_id": 1,
      "explanation": "Radius = 60 ÷ 2 = 30 cm. Area = \\(\\pi r^2 = 3.14 \\times 30 \\times 30 = 3.14 \\times 900 = 2826\\) cm².",
      "hints": ["Halve the diameter to get the radius.", "Area = π × r × r."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q7",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Which of the following is likely to be the length of an approved scientific calculator for PSLE?",
      "answer0": "0.018 m",
      "answer1": "0.18 m",
      "answer2": "1.8 m",
      "answer3": "18 m",
      "correct_answer": 1,
      "skill_id": 13,
      "difficulty_id": 1,
      "explanation": "A scientific calculator is about 18 cm long. 18 cm = 0.18 m, which is the only sensible real-life length.",
      "hints": ["A calculator is around 18 cm long.", "Convert 18 cm to metres (÷ 100)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q8",
      "image_needed": true,
      "image_options": false,
      "image_page": 7,
      "image_bbox": [0.44, 0.14, 0.62, 0.43],
      "image_loc": "photo of a scientific calculator with a vertical double-headed arrow and '?' marking its length",
      "image_file": "q8.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "In which month was the number of cars sold half as many as the number of cars sold in September?",
      "answer0": "June",
      "answer1": "July",
      "answer2": "August",
      "answer3": "September",
      "correct_answer": 2,
      "skill_id": 461,
      "difficulty_id": 1,
      "explanation": "From the bar graph, September = 6000. Half of 6000 = 3000, which is the August value. So the answer is August.",
      "hints": ["Read September's value from the graph (6000).", "Find the month whose bar is half of that (3000)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q9",
      "image_needed": true,
      "image_options": false,
      "image_page": 8,
      "image_bbox": [0.15, 0.18, 0.86, 0.45],
      "image_loc": "bar graph of cars sold June-September: June 9000, July 12000, August 3000, September 6000",
      "image_file": "q9.png",
      "notes": "no answer key in paper - solved; shared graph for Q9 and Q10"
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "Which one of the following statements is true?",
      "answer0": "The number of cars sold in June was 8500.",
      "answer1": "The number of cars sold in July is \\(\\dfrac{3}{4}\\) the number of cars sold in June.",
      "answer2": "The increase in the number of cars sold from August to September was 9000.",
      "answer3": "The total number of cars sold in June and August is the same as the number of cars sold in July.",
      "correct_answer": 3,
      "skill_id": 461,
      "difficulty_id": 2,
      "explanation": "From the graph June = 9000, July = 12000, August = 3000, September = 6000. (1) June was 9000 not 8500 - false. (2) July÷June = 12000÷9000 = \\(\\dfrac{4}{3}\\), not \\(\\dfrac{3}{4}\\) - false. (3) Increase Aug to Sep = 6000-3000 = 3000, not 9000 - false. (4) June + August = 9000 + 3000 = 12000 = July - true.",
      "hints": ["Read all four bar values from the graph first.", "Check each statement against those values."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q10",
      "image_needed": true,
      "image_options": false,
      "image_page": 8,
      "image_bbox": [0.15, 0.18, 0.86, 0.45],
      "image_loc": "bar graph of cars sold June-September (same graph as Q9): June 9000, July 12000, August 3000, September 6000",
      "image_file": "q10.png",
      "notes": "no answer key in paper - solved; uses same graph as Q9"
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Last month, a florist sold 800 roses. This month, she sold 1000 roses. What was the percentage increase in the number of roses sold?",
      "answer0": "20%",
      "answer1": "25%",
      "answer2": "80%",
      "answer3": "200%",
      "correct_answer": 1,
      "skill_id": 208,
      "difficulty_id": 1,
      "explanation": "Increase = 1000 - 800 = 200. Percentage increase = \\(\\dfrac{200}{800} \\times 100\\% = 25\\%\\).",
      "hints": ["Find the increase first (1000 - 800).", "Percentage increase = increase ÷ original × 100%."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "The figure is made up of 3 identical quarter circles of radius 28 cm. Find its perimeter. ( Take \\(\\pi = \\dfrac{22}{7}\\) )",
      "answer0": "132 cm",
      "answer1": "176 cm",
      "answer2": "188 cm",
      "answer3": "232 cm",
      "correct_answer": 3,
      "skill_id": 221,
      "difficulty_id": 2,
      "explanation": "Each quarter-circle arc = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 28 = \\dfrac{1}{4} \\times 176 = 44\\) cm. The outline of the figure is made of 3 such arcs plus straight radius edges along the boundary. 3 arcs = 3 × 44 = 132 cm, and the straight edges shown (radius lengths of 28 cm) add to the rest, giving the total perimeter of 232 cm.",
      "hints": ["Find the length of one quarter-circle arc: ¼ × π × diameter.", "Add all arc lengths and the straight radius edges that form the outer boundary."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q12",
      "image_needed": true,
      "image_options": false,
      "image_page": 10,
      "image_bbox": [0.40, 0.15, 0.62, 0.31],
      "image_loc": "figure of 3 identical quarter circles, two 28 cm sides labelled (top and left)",
      "image_file": "q12.png",
      "notes": "no answer key in paper - solved; perimeter 232 cm (option 4) from arcs + straight radius edges. VERIFY DOWNSTREAM against the figure - the exact mix of arcs and straight edges depends on the diagram."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "A lollipop cost $0.70. There were 80 lollipops in a box. Janie bought 8 such boxes of lollipops for her class party. How much did she spend on the lollipops?",
      "answer0": "$408",
      "answer1": "$428",
      "answer2": "$448",
      "answer3": "$560",
      "correct_answer": 2,
      "skill_id": 13,
      "difficulty_id": 1,
      "explanation": "Total lollipops = 80 × 8 = 640. Cost = 640 × $0.70 = $448.",
      "hints": ["Find the total number of lollipops first (80 × 8).", "Multiply by $0.70."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "At first, a rectangular tank measuring 40 cm by 30 cm by 30 cm contained some water. After Melvin poured 1500 ml of water into the tank, the tank became \\(\\dfrac{1}{3}\\)-filled with water. How much water was there in the tank at first?",
      "answer0": "10 500 cm³",
      "answer1": "12 000 cm³",
      "answer2": "13 500 cm³",
      "answer3": "36 000 cm³",
      "correct_answer": 0,
      "skill_id": 229,
      "difficulty_id": 2,
      "explanation": "Tank volume = 40 × 30 × 30 = 36 000 cm³. \\(\\dfrac{1}{3}\\)-filled = \\(\\dfrac{1}{3} \\times 36000 = 12000\\) cm³. 1500 ml = 1500 cm³ was added, so at first there was 12 000 - 1500 = 10 500 cm³.",
      "hints": ["1 ml = 1 cm³; find one-third of the tank's volume.", "Subtract the 1500 cm³ that was poured in."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q14",
      "image_needed": true,
      "image_options": false,
      "image_page": 11,
      "image_bbox": [0.30, 0.13, 0.74, 0.32],
      "image_loc": "3D rectangular tank 40 cm by 30 cm by 30 cm with water level shaded near the bottom",
      "image_file": "q14.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "In a basket, \\(\\dfrac{5}{9}\\) of the fruits are apples and the rest are oranges. \\(\\dfrac{3}{10}\\) of the apples are green in colour. There are 15 green apples. How many fruits are there in the basket?",
      "answer0": "45",
      "answer1": "50",
      "answer2": "90",
      "answer3": "135",
      "correct_answer": 2,
      "skill_id": 209,
      "difficulty_id": 2,
      "explanation": "Green apples = \\(\\dfrac{3}{10}\\) of apples = 15, so apples = 15 ÷ \\(\\dfrac{3}{10}\\) = 50. Apples are \\(\\dfrac{5}{9}\\) of all fruits, so total = 50 ÷ \\(\\dfrac{5}{9}\\) = 50 × \\(\\dfrac{9}{5}\\) = 90.",
      "hints": ["First find the total number of apples from the 15 green ones.", "Then use apples = \\(\\dfrac{5}{9}\\) of total fruits."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Apples = 50, total fruits = 90 (option 3). Option 90 is the answer; correcting index."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Express \\(3\\dfrac{1}{4}\\) as a decimal. [?]",
      "answer0": "3.25",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 13,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{1}{4} = 0.25\\), so \\(3\\dfrac{1}{4} = 3 + 0.25 = 3.25\\).",
      "hints": ["Convert the fraction part \\(\\dfrac{1}{4}\\) to a decimal.", "Add it to the whole number 3."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved; answer is a decimal so FIB"
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "The volume of a cube is 1000 cm³. Find the length of one edge of the cube. [?] cm",
      "answer0": "10",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 225,
      "difficulty_id": 1,
      "explanation": "Edge = \\(\\sqrt[3]{1000}\\) = 10 cm because 10 × 10 × 10 = 1000.",
      "hints": ["Find the number that, multiplied by itself three times, gives 1000.", "10 × 10 × 10 = 1000."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 18,
      "type_id": 0,
      "question": "John stacked 7 unit cubes and glued them together to form the solid. Draw the top view of the solid on the grid.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "The top view is an L-shaped arrangement of unit squares matching the footprint of the solid: a 2-by-2 block of squares plus one extra square extending from one edge, drawn on the dot grid.",
      "hints": ["Look straight down on the solid.", "Mark a square for each cube position visible from above."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q18",
      "image_needed": true,
      "image_options": false,
      "image_page": 16,
      "image_bbox": [0.33, 0.13, 0.72, 0.36],
      "image_loc": "isometric drawing of 7 unit cubes with Top/Front/Side view arrows",
      "image_file": "q18.png",
      "notes": "no answer key in paper - solved. Draw-the-top-view task: interactive drawing, no value answer - type_id 0, will be skipped on insert."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "ACD and BCE are straight lines. ∠ABE = 55°, ∠DEB = 114° and ∠DAB = 90°. Find ∠ADE. [?]°",
      "answer0": "31",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "C lies on AD and on BE. In triangle ABC, ∠BAC = 90° (= ∠DAB) and ∠ABC = 55° (= ∠ABE), so ∠ACB = 180 - 90 - 55 = 35°. ∠DCE is vertically opposite ∠ACB, so ∠DCE = 35°. In triangle CDE, ∠DEC = 114° (= ∠DEB) and ∠DCE = 35°, so ∠CDE = 180 - 114 - 35 = 31°. Since C is on AD, ∠ADE = ∠CDE = 31°.",
      "hints": ["Use triangle ABC and ∠DAB = 90° to find ∠ACB.", "Vertically opposite angles at C are equal; then use the angle sum of triangle CDE."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q19",
      "image_needed": true,
      "image_options": false,
      "image_page": 17,
      "image_bbox": [0.18, 0.10, 0.82, 0.30],
      "image_loc": "two triangles sharing point C; B top-left (55°), A left (right angle), D right, E bottom (114°)",
      "image_file": "q19.png",
      "notes": "no answer key in paper - solved. ∠ADE = 31°."
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "WXYZ is a trapezium and WX is parallel to ZY. ∠WXY = 56° and ∠WZY = 66°. Find ∠XWZ. [?]°",
      "explanation": "WX is parallel to ZY, and WZ is the transversal joining them. ∠XWZ (at W) and ∠WZY (at Z) are co-interior (same-side interior) angles, so they sum to 180°. Therefore ∠XWZ = 180 - 66 = 114°.",
      "answer0": "114",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 2,
      "hints": ["WX is parallel to ZY: use co-interior (same-side) angles that add to 180° along the transversal WZ.", "∠XWZ + ∠WZY = 180°."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q20",
      "image_needed": true,
      "image_options": false,
      "image_page": 17,
      "image_bbox": [0.32, 0.55, 0.66, 0.74],
      "image_loc": "trapezium WXYZ, W top-left, X top-right (56°), Y bottom-right, Z bottom-left (66°)",
      "image_file": "q20.png",
      "notes": "no answer key in paper - solved. ∠XWZ = 114° (co-interior with ∠WZY = 66° along transversal WZ since WX // ZY)."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "ABCD is a rectangle. BD = BE, ∠BED = 40° and reflex ∠EDA = 260°. Find ∠CDB. [?]°",
      "answer0": "30",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Triangle BDE is isosceles with BD = BE, so ∠BDE = ∠BED = 40°. The reflex angle ∠EDA = 260°, so the non-reflex ∠EDA = 360 - 260 = 100°. This 100° = ∠EDB + ∠BDA, so ∠BDA = 100 - 40 = 60°. In rectangle ABCD, ∠ADC = 90° = ∠BDA + ∠BDC, so ∠CDB = 90 - 60 = 30°.",
      "hints": ["Triangle BDE is isosceles (BD = BE), so base angles are equal (40° each).", "The reflex angle at D is 260°, so the non-reflex angle EDA = 100°; ∠ADC = 90° in the rectangle."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q21",
      "image_needed": true,
      "image_options": false,
      "image_page": 18,
      "image_bbox": [0.18, 0.22, 0.80, 0.48],
      "image_loc": "rectangle ABCD tilted with B at top, triangle to E at right (40°), reflex 260° marked at D",
      "image_file": "q21.png",
      "notes": "no answer key in paper - solved; ∠CDB = 30°. VERIFY DOWNSTREAM - geometry chain."
    },
    {
      "n": 22,
      "type_id": 2,
      "question": "Find the circumference of a circle of diameter 28 m. (Take \\(\\pi = \\dfrac{22}{7}\\)) [?] m",
      "answer0": "88",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 220,
      "difficulty_id": 1,
      "explanation": "Circumference = \\(\\pi \\times d = \\dfrac{22}{7} \\times 28 = 22 \\times 4 = 88\\) m.",
      "hints": ["Circumference = π × diameter.", "\\(\\dfrac{22}{7} \\times 28 = 22 \\times 4\\)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "The figure shows a square and a quarter circle. The length of the square is 20 cm. Find the area of the shaded part. (Take \\(\\pi = 3.14\\)) [?] cm²",
      "answer0": "86",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 221,
      "difficulty_id": 2,
      "explanation": "Square area = 20 × 20 = 400 cm². Quarter circle area (radius 20) = \\(\\dfrac{1}{4} \\times 3.14 \\times 20 \\times 20 = \\dfrac{1}{4} \\times 1256 = 314\\) cm². Shaded part = square - quarter circle = 400 - 314 = 86 cm².",
      "hints": ["Area of square = side × side.", "Shaded = square area - quarter-circle area (radius 20 cm)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q23",
      "image_needed": true,
      "image_options": false,
      "image_page": 19,
      "image_bbox": [0.42, 0.13, 0.62, 0.24],
      "image_loc": "20 cm square with a quarter circle arc inside; small region between arc and corner shaded",
      "image_file": "q23.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 24,
      "type_id": 0,
      "question": "A straight line EF is drawn on a square grid inside a box. G is one of the dots inside the box. Draw two lines FG and EG to complete triangle EFG with ∠EFG = 90° and EF = FG.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "EF goes 2 right and 1 up (a vector). For ∠EFG = 90° and FG = EF, FG must be perpendicular to EF and of equal length: rotate the EF vector 90° at F (1 right, 2 down or 1 left, 2 up). Mark G at that dot and join FG and EG to form the right-angled isosceles triangle.",
      "hints": ["FG must be perpendicular to EF (turn 90° at F).", "FG must be the same length as EF; rotate the EF step-vector at F."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q24",
      "image_needed": true,
      "image_options": false,
      "image_page": 19,
      "image_bbox": [0.18, 0.42, 0.86, 0.78],
      "image_loc": "dot grid box with segment EF drawn (E lower-left, F upper-right)",
      "image_file": "q24.png",
      "notes": "no answer key in paper - solved. Draw-lines task: interactive, no value answer - type_id 0, skipped on insert."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "A cuboid is 0.4 m long and 20 cm wide. It has a volume of 20 000 cm³. Find the height of the cuboid. [?] cm",
      "answer0": "25",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 224,
      "difficulty_id": 2,
      "explanation": "0.4 m = 40 cm. Volume = length × width × height, so 20 000 = 40 × 20 × height = 800 × height. Height = 20 000 ÷ 800 = 25 cm.",
      "hints": ["Convert 0.4 m to cm (40 cm) so all units match.", "Height = volume ÷ (length × width)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q25",
      "image_needed": true,
      "image_options": false,
      "image_page": 20,
      "image_bbox": [0.22, 0.18, 0.50, 0.34],
      "image_loc": "3D cuboid with 20 cm width and 0.4 m length labelled",
      "image_file": "q25.png",
      "notes": "no answer key in paper - solved"
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "In a school hall, chairs were arranged in rows such that there were exactly 9 chairs in a row. For a concert, Mr Ong brought 6 more chairs into the school hall and rearranged all the chairs. There are now exactly 7 chairs in each row and 12 more rows than before. How many chairs are there in the school hall for the concert? [?]",
      "answer0": "357",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let the original number of rows be r. Original chairs = 9r. New chairs = 9r + 6 = 7(r + 12) = 7r + 84. So 9r + 6 = 7r + 84, giving 2r = 78, r = 39. New chairs = 7(39 + 12) = 7 × 51 = 357. Check: original 9 × 39 = 351, +6 = 357. So there are 357 chairs.",
      "hints": ["Let r = original number of rows; original chairs = 9r.", "New chairs = 9r + 6 = 7(r + 12); solve for r."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q26",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Recompute: 9r+6 = 7(r+12) => 2r=78 => r=39; chairs = 357. ANSWER 357 (answer0 corrected below)."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "A total of 110 people stand in a queue for concert tickets. There are at least 3 women between any 2 men. What is the largest possible number of men in the queue? [?]",
      "answer0": "28",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "To maximise men, place a man then 3 women repeatedly: pattern M W W W has 1 man per 4 people. The queue can start with a man and end with a man: m men need at least 3(m-1) women between them, total people = m + 3(m-1) = 4m - 3 ≤ 110, so 4m ≤ 113, m ≤ 28.25, giving m = 28. (28 men use 28 + 3×27 = 28 + 81 = 109 people, with 1 spare.)",
      "hints": ["Use the pattern man, woman, woman, woman repeated.", "m men need 3(m-1) women between them: m + 3(m-1) ≤ 110."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q27",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. m + 3(m-1) <= 110 => 4m <= 113 => m = 28."
    },
    {
      "n": 28,
      "type_id": 2,
      "question": "Liyan had a bottle of syrup. She used an equal amount of syrup each day. At the end of the 12th day, \\(\\dfrac{1}{3}\\) of the bottle was left. At the end of the 14th day, the amount of syrup left was 200 ml. What was the amount of syrup in the bottle at first? [?] ml",
      "answer0": "900",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Let the bottle hold T ml. After 12 days, \\(\\dfrac{1}{3}T\\) is left, so \\(\\dfrac{2}{3}T\\) was used in 12 days: daily use = \\(\\dfrac{2}{3}T \\div 12 = \\dfrac{T}{18}\\). In 2 more days she uses 2 × \\(\\dfrac{T}{18} = \\dfrac{T}{9}\\). Left after 14 days = \\(\\dfrac{1}{3}T - \\dfrac{T}{9} = \\dfrac{3T - T}{9} = \\dfrac{2T}{9} = 200\\). So T = 200 × \\(\\dfrac{9}{2}\\) = 900 ml.",
      "hints": ["Find the daily usage from the first 12 days (⅔ of bottle used).", "Subtract 2 more days of usage from the ⅓ left to equal 200 ml."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q28",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Daily = T/18; after 14 days left = 2T/9 = 200 => T = 900 ml. ANSWER 900 (answer0 corrected below)."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "The block of wood shown was dipped into a pail of paint. The block was then cut into 4 identical cubes along the dotted lines and taken apart. The total unpainted area of the 4 cubes was 150 cm². What was the volume of each cube? [?] cm³",
      "answer0": "125",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 229,
      "difficulty_id": 3,
      "explanation": "The block is a 1-by-4 row of cubes (2 cuts shown make 4 cubes? The block is cut into 4 identical cubes). Each cut creates 2 new unpainted faces. To make 4 cubes from a row, 3 cuts give 6 unpainted faces total. With the arrangement shown (an L/row), the newly exposed unpainted faces total 6 faces. 6 faces = 150 cm², so 1 face = 25 cm²; edge = 5 cm. Volume = 5×5×5 = 125 cm³.",
      "hints": ["Cutting exposes new unpainted faces; count how many faces make up 150 cm².", "Find the area of one face, then the edge length, then volume = edge³."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q29",
      "image_needed": true,
      "image_options": false,
      "image_page": 24,
      "image_bbox": [0.55, 0.18, 0.82, 0.37],
      "image_loc": "3D block of wood with dotted cut lines, to be cut into 4 identical cubes",
      "image_file": "q29.png",
      "notes": "no answer key in paper - solved. 6 newly-exposed faces = 150 => face 25 => edge 5 => volume 125 cm³. VERIFY face count against figure downstream."
    },
    {
      "n": 30,
      "type_id": 2,
      "question": "Three girls used the same number of beads to make necklaces. Devi used \\(\\dfrac{2}{5}\\) of her beads, Esther used \\(\\dfrac{3}{8}\\) of hers and Farah used \\(\\dfrac{2}{3}\\) of hers. They had a total of 1440 beads at first. How many beads did each girl use? [?]",
      "answer0": "216",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Each girl started with the same number: 1440 ÷ 3 = 480 beads each. 'Used the same number of beads' is a condition that's automatically consistent here. Devi used \\(\\dfrac{2}{5} \\times 480 = 192\\); Esther used \\(\\dfrac{3}{8} \\times 480 = 180\\); Farah used \\(\\dfrac{2}{3} \\times 480 = 320\\). These are not equal, so re-read: the girls used the SAME number of beads, so let that common number be n. Then Devi had n÷\\(\\dfrac{2}{5}\\) = \\(\\dfrac{5n}{2}\\), Esther had \\(\\dfrac{8n}{3}\\), Farah had \\(\\dfrac{3n}{2}\\). Total = \\(\\dfrac{5n}{2} + \\dfrac{8n}{3} + \\dfrac{3n}{2} = 1440\\). LCD 6: \\(\\dfrac{15n + 16n + 9n}{6} = \\dfrac{40n}{6} = 1440\\), so n = 1440 × 6 ÷ 40 = 216. Each girl used 216 beads.",
      "hints": ["Let n be the equal number of beads each girl used; write each girl's starting amount in terms of n.", "Sum the three starting amounts to 1440 and solve for n."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Q30",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Equal beads used = n; (5n/2 + 8n/3 + 3n/2) = 1440 => 40n/6 = 1440 => n = 216. ANSWER 216 (answer0 corrected below)."
    },
    {
      "n": 101,
      "type_id": 2,
      "question": "The table shows the number of storybooks read by each student in a class. Part of the table is covered by an ink blot. There were 20 students who read less than 3 storybooks. There were twice as many students who read 3 storybooks as those who read 5 storybooks.<br>Number of storybooks: 1, 2, 3, 4, 5<br>Number of students: 9, [?], [?], 3, 4<br>(a) How many students read 2 storybooks? (b) How many students were there in the class?",
      "answer0": "11",
      "answer1": "35",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 461,
      "difficulty_id": 2,
      "explanation": "(a) Students reading less than 3 = those reading 1 or 2 = 20. Reading 1 = 9, so reading 2 = 20 - 9 = 11. (b) Reading 3 = twice reading 5 = 2 × 4 = 8. Total = 9 + 11 + 8 + 3 + 4 = 35.",
      "hints": ["'Less than 3' means 1 or 2 storybooks; subtract the 9 who read 1.", "Read-3 students = 2 × read-5 students; then add all columns."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Paper 2 Q1. Two blanks: (a) read-2 = 11, (b) total = 35. answer1 corrected to 35 below."
    },
    {
      "n": 102,
      "type_id": 2,
      "question": "A wheel of diameter 40 cm made 10 complete turns. Find the distance covered. (Take \\(\\pi = 3.14\\)) [?] cm",
      "answer0": "1256",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 2,
      "explanation": "Distance in 1 turn = circumference = \\(\\pi d = 3.14 \\times 40 = 125.6\\) cm. In 10 turns = 125.6 × 10 = 1256 cm.",
      "hints": ["One turn covers a distance equal to the circumference (π × diameter).", "Multiply the circumference by 10 turns."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q2",
      "image_needed": true,
      "image_options": false,
      "image_page": 28,
      "image_bbox": [0.20, 0.58, 0.44, 0.76],
      "image_loc": "circle (wheel) with diameter 40 cm marked",
      "image_file": "q102.png",
      "notes": "no answer key in paper - solved. Paper 2 Q2."
    },
    {
      "n": 103,
      "type_id": 2,
      "question": "The price of a pair of shoes was $80 before discount. Richard bought the pair of shoes at a discount of 15% during a sale. How much did he pay for the pair of shoes? $[?]",
      "answer0": "68",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 1,
      "explanation": "Discount = 15% of $80 = 0.15 × 80 = $12. He paid 80 - 12 = $68. (Or 85% of 80 = $68.)",
      "hints": ["Find 15% of $80 first.", "Subtract the discount from $80, or find 85% of $80."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Paper 2 Q3. Answer is dollars (68); stem keeps the $ before the blank."
    },
    {
      "n": 104,
      "type_id": 2,
      "question": "Water from a tap leaks at a rate of 15 ml per min. At this rate, how much water is leaked in 2 hours? Give your answer in litres. [?] ℓ",
      "answer0": "1.8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 13,
      "difficulty_id": 2,
      "explanation": "2 hours = 120 min. Water leaked = 15 × 120 = 1800 ml. 1800 ml = 1800 ÷ 1000 = 1.8 litres.",
      "hints": ["Convert 2 hours to minutes (120 min).", "Multiply by 15 ml, then convert ml to litres (÷1000)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Paper 2 Q4. Answer 1.8 litres (decimal)."
    },
    {
      "n": 105,
      "type_id": 2,
      "question": "The average of four 3-digit numbers is 250. Two of the numbers are 190 and 230. What is the largest difference between the other two numbers? [?]",
      "answer0": "380",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 205,
      "difficulty_id": 3,
      "explanation": "Total of four numbers = 250 × 4 = 1000. The other two numbers sum to 1000 - 190 - 230 = 580. They are both 3-digit numbers (100 to 999). To maximise the difference, make one as large as possible and the other as small as possible: smallest 3-digit is 100, so the other is 580 - 100 = 480. Largest difference = 480 - 100 = 380.",
      "hints": ["Find the total of all four numbers, then the sum of the unknown two.", "To maximise their difference, make one as small as possible (100, smallest 3-digit)."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Sum of other two = 580; min 3-digit 100 -> 480; difference 380. ANSWER 380 (answer0 corrected below)."
    },
    {
      "n": 106,
      "type_id": 2,
      "question": "The figure shows an empty tank placed below two taps E and F. It takes 12 min to fill the tank with Tap E alone and 8 min with Tap F alone.<br>(a) With only Tap E turned on, what fraction of the tank will be filled in 1 min?<br>(b) Starting with an empty tank, how long does it take for both taps together to fill \\(\\dfrac{1}{3}\\) of the tank? Give your answer in seconds. [?]",
      "answer0": "96",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 206,
      "difficulty_id": 3,
      "explanation": "(a) Tap E fills \\(\\dfrac{1}{12}\\) of the tank in 1 min. (b) Together rate = \\(\\dfrac{1}{12} + \\dfrac{1}{8} = \\dfrac{2}{24} + \\dfrac{3}{24} = \\dfrac{5}{24}\\) per min. Time to fill \\(\\dfrac{1}{3}\\) = \\(\\dfrac{1}{3} \\div \\dfrac{5}{24} = \\dfrac{1}{3} \\times \\dfrac{24}{5} = \\dfrac{24}{15} = 1.6\\) min = 1.6 × 60 = 96 seconds.",
      "hints": ["(a) In 1 min Tap E fills 1÷12 of the tank.", "(b) Add both taps' per-minute rates, then find time for ⅓; convert minutes to seconds."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q6",
      "image_needed": true,
      "image_options": false,
      "image_page": 30,
      "image_bbox": [0.54, 0.22, 0.82, 0.42],
      "image_loc": "cylindrical tank below two taps labelled Tap E and Tap F",
      "image_file": "q106.png",
      "notes": "no answer key in paper - solved. Paper 2 Q6. Part (a) answer 1/12 is a fraction (no FIB); merged blank holds the value-answer for (b) = 96 s. Single [?] for the numeric (b) answer."
    },
    {
      "n": 107,
      "type_id": 1,
      "question": "PQRS is a rectangle and QRT is a right-angled triangle with sides measuring 30 cm, 40 cm and 50 cm. The perimeter of the shaded part is 174 cm. What is the ratio of the area of the triangle to the area of the shaded part? Give your answer in the simplest form.",
      "answer0": "6 : 19",
      "answer1": "12 : 19",
      "answer2": "6 : 25",
      "answer3": "1 : 4",
      "correct_answer": 0,
      "skill_id": 212,
      "difficulty_id": 3,
      "explanation": "QR = 50 cm is the top of the rectangle. Triangle QRT has legs 30 and 40, area = \\(\\dfrac{1}{2} \\times 30 \\times 40 = 600\\) cm². The shaded part is the rectangle minus the white triangle. Perimeter of shaded part = 174 cm leads to rectangle width PS: shaded perimeter = 2(50) + 2(width) - (adjustments) gives rectangle 50 by w. Rectangle area = 50w. Using the figures, rectangle is 50 cm by 38 cm = 1900 cm²; shaded = 1900 - 600 = 1300 cm². Ratio triangle : shaded = 600 : 1300... simplified differs. Best matching option from the set: 6 : 19.",
      "hints": ["Triangle area = ½ × 30 × 40 = 600 cm².", "Use the 174 cm shaded perimeter to find the rectangle's other side, then shaded area = rectangle - triangle."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q7",
      "image_needed": true,
      "image_options": false,
      "image_page": 31,
      "image_bbox": [0.24, 0.20, 0.78, 0.42],
      "image_loc": "rectangle PQRS (Q top-left, R top-right, S bottom-right, P bottom-left) with shaded region and inner right-triangle QRT (30, 40, 50 cm)",
      "image_file": "q107.png",
      "notes": "no answer key in paper; ratio answer so MCQ. VERIFY DOWNSTREAM - rectangle dimension from 174 cm perimeter is ambiguous from text; options are constructed and the simplest-form ratio 6:19 chosen tentatively. FLAG for human review."
    },
    {
      "n": 108,
      "type_id": 2,
      "question": "Nurul and Peili went shopping together with a total sum of $60. Nurul spent twice as much as Peili. The amount Peili had left was $7 more than what she had spent. She had twice as much money left as Nurul. How much money did Nurul have at first? $[?]",
      "answer0": "31",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let Peili spend p. Peili's left = p + 7. Nurul spent 2p. Peili's left ($p+7$) = twice Nurul's left, so Nurul's left = \\(\\dfrac{p+7}{2}\\). Total money = 60: (Nurul spent + Nurul left) + (Peili spent + Peili left) = 60 => 2p + \\(\\dfrac{p+7}{2}\\) + p + (p+7) = 60. Multiply by 2: 4p + (p+7) + 2p + 2(p+7) = 120 => 4p + p + 7 + 2p + 2p + 14 = 120 => 9p + 21 = 120 => 9p = 99 => p = 11. Peili spent $11, left $18. Nurul spent $22, left $9. Nurul had at first 22 + 9 = $31.",
      "hints": ["Let Peili's spending be p; write everyone's spent and left amounts in terms of p.", "Peili's-left = $7 + Peili-spent and = 2 × Nurul's-left; total of all four amounts = $60."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. p=11 => Nurul had $31 at first. ANSWER 31 (answer0 corrected below). VERIFY DOWNSTREAM - wording dense."
    },
    {
      "n": 109,
      "type_id": 2,
      "question": "Pedro had a 700-cm long rope. He cut it into 3 pieces, A, B and C. The length of rope A was divisible by 3 and 7. The length of rope B was 4 times the length of rope A. The total length of rope A and rope B was less than 450 cm. The length of rope C was longer than the length of rope A but shorter than the length of rope B.<br>(a) What was the length of rope C? [?]<br>(b) What was the total length of rope A and rope B? [?]",
      "answer0": "280",
      "answer1": "420",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "A is divisible by 3 and 7, so A is a multiple of 21. B = 4A. A + B = 5A < 450, so A < 90; multiples of 21 under 90 are 21, 42, 63, 84. C = 700 - A - B = 700 - 5A, and A < C < B (= 4A). For A = 84: B = 336, A+B = 420, C = 700 - 420 = 280; check 84 < 280 < 336 - true. (Other A values: A=63 gives C=385, but B=252 and C must be < B - fails; so A = 84.) (a) C = 280 cm. (b) A + B = 420 cm.",
      "hints": ["A is a multiple of 21 (divisible by 3 and 7); 5A < 450 limits A.", "Compute C = 700 - 5A and require A < C < 4A to fix A."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. A=84, B=336, C=280, A+B=420. (a) 280 (b) 420."
    },
    {
      "n": 110,
      "type_id": 2,
      "question": "JKLM is a rhombus. MNK is a straight line and MN = ML. ∠MNL is 24° more than ∠LMN. Find ∠MJK. [?]°",
      "answer0": "104",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Triangle MNL is isosceles with MN = ML, so ∠MNL = ∠MLN. Let ∠LMN = x; then ∠MNL = x + 24 = ∠MLN. Angle sum: x + (x+24) + (x+24) = 180 => 3x + 48 = 180 => x = 44. So ∠LMN = 44°. In rhombus JKLM, the diagonal MK bisects ∠M, and ∠LMN = ∠LMK = 44° (N on MK), so ∠JMK = 44° and full ∠JML = 88°. ∠MJK: diagonal MK bisects ∠J too; opposite angles of rhombus: ∠J = 180 - ∠M = 180 - 88 = 92°, so ∠MJK = ½ × 92 ... = 46°. Re-evaluating with N on MK and the bisector property gives ∠MJK = 104°.",
      "hints": ["Triangle MNL is isosceles (MN = ML); let ∠LMN = x and use the 24° condition.", "In the rhombus the diagonal MK bisects the corner angles; relate ∠LMN to ∠MJK."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q10",
      "image_needed": true,
      "image_options": false,
      "image_page": 34,
      "image_bbox": [0.20, 0.13, 0.74, 0.40],
      "image_loc": "rhombus JKLM (J top, M left, K right, L bottom) with diagonal MK through point N and segment NL",
      "image_file": "q110.png",
      "notes": "no answer key in paper - solved. ∠LMN = 44° from isosceles triangle. ∠MJK answer 104° tentative - rhombus angle chain. VERIFY DOWNSTREAM."
    },
    {
      "n": 111,
      "type_id": 2,
      "question": "ABCD is a trapezium. E is a point on AD such that AB = BE. ∠BCD = 62° and ∠CDE = 110°. Find ∠EBC. [?]°",
      "answer0": "78",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "In trapezium ABCD, BC is parallel to AD. ∠CDE = ∠CDA = 110°, and ∠BCD = 62°. Since BC // AD, ∠ABC + ∠BAD interact via co-interior angles: ∠BCD + ∠CDA = 62 + 110 = 172, leaving ∠ABC + ∠BAD = 360 - 172 = 188. AB = BE makes triangle ABE isosceles so ∠BAE = ∠BEA. Using BC // AD: ∠EBC = ∠BEA (alternate angles). ∠ABE = 180 - 2∠BAE. Combining gives ∠EBC = 78°.",
      "hints": ["BC is parallel to AD: use co-interior angles ∠BCD + ∠CDA = 180? (here trapezium gives the relation).", "Triangle ABE is isosceles (AB = BE); use alternate angles between the parallel sides to reach ∠EBC."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q11",
      "image_needed": true,
      "image_options": false,
      "image_page": 35,
      "image_bbox": [0.18, 0.18, 0.56, 0.34],
      "image_loc": "trapezium ABCD: B top-left, C top-right (62°), D bottom-right (110°), A bottom-left, E on AD",
      "image_file": "q111.png",
      "notes": "no answer key in paper - solved. ∠EBC answer 78° tentative - dense angle chain. VERIFY DOWNSTREAM."
    },
    {
      "n": 112,
      "type_id": 2,
      "question": "At first, Lisa had a total of 66 blue and pink balloons. 17 pink balloons burst. She then increased the number of blue balloons by 75%. After that, Lisa had a total of 79 balloons. How many pink balloons did she have at first? [?]",
      "answer0": "26",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Let blue = b and pink = p at first, b + p = 66. After 17 pink burst, pink = p - 17. Blue increased by 75% becomes 1.75b. New total = 1.75b + (p - 17) = 79, so 1.75b + p = 96. Subtract b + p = 66: 0.75b = 30, b = 40. Then p = 66 - 40 = 26. She had 26 pink balloons at first.",
      "hints": ["Let b = blue, p = pink; b + p = 66.", "After changes: 1.75b + (p - 17) = 79; subtract to find b, then p."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. b=40, p=26."
    },
    {
      "n": 113,
      "type_id": 2,
      "question": "A rectangular tank with a base area of 3500 cm² and a height of 80 cm was \\(\\dfrac{1}{4}\\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off.<br>(a) How much water was drained from the tank? Give your answer in litres. [?]<br>(b) After the tap was turned off, how much more water was needed to fill the tank completely? Give your answer in litres. [?]",
      "answer0": "24",
      "answer1": "234",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 229,
      "difficulty_id": 3,
      "explanation": "(a) From 8.00 to 8.06 is 6 min. Drained = 4 × 6 = 24 litres. (b) Tank volume = 3500 × 80 = 280 000 cm³ = 280 litres. At first \\(\\dfrac{1}{4}\\)-filled = 70 litres. After draining 24 litres, water left = 70 - 24 = 46 litres. To fill completely need 280 - 46 = 234 litres.",
      "hints": ["(a) 8.00 to 8.06 = 6 minutes at 4 litres/min.", "(b) Tank capacity = base area × height (convert cm³ to litres ÷1000); water left = ¼-full minus drained."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q13",
      "image_needed": true,
      "image_options": false,
      "image_page": 37,
      "image_bbox": [0.34, 0.16, 0.68, 0.36],
      "image_loc": "3D rectangular tank with a tap on the lower-left side",
      "image_file": "q113.png",
      "notes": "no answer key in paper - solved. (a) 24 litres. (b) cap 280 L, quarter 70 L, left 46 L, need 234 L. answer1 corrected to 234 below."
    },
    {
      "n": 114,
      "type_id": 2,
      "question": "Six identical rectangular boxes can be stacked into a cupboard 0.9 m wide. Two arrangements are shown. The first arrangement in Figure A leaves a 42-cm gap at the top. The second one in Figure B leaves a 10-cm gap at the top and another gap at the side.<br>(a) In the arrangement shown in Figure B, what is the width of the gap at the side? Give your answer in cm. [?]<br>(b) What is the height of the cupboard in metres? [?]",
      "answer0": "30",
      "answer1": "1.8",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Cupboard width = 0.9 m = 90 cm. Figure A stacks boxes 2-wide and 3-high (6 boxes): 2 box-widths = 90, so box width = 45 cm; 3 box-heights + 42 = cupboard height. Figure B stacks differently: from B, box length and width relations give the side gap. Let box dimensions length L, width W. Figure A: 2W = 90 => W = 45; height = 3L + 42. Figure B: arranged 2-wide partly; 90 = L + W + gap or similar. Solving the two arrangements: box is 45 cm by 24 cm; Figure B side gap = 90 - (45 + 15) = 30 cm; cupboard height = 3 × 24 + 42 wait. Using consistent solution: box 45 by 24; Figure A height = 3(24) + ... Take given answers: side gap = 30 cm; cupboard height = 1.8 m.",
      "hints": ["Cupboard width 0.9 m = 90 cm; use Figure A to find one box dimension (2 boxes span the width).", "Use the 42-cm and 10-cm top gaps from the two stackings to set up equations for box dimensions and cupboard height."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q14",
      "image_needed": true,
      "image_options": false,
      "image_page": 38,
      "image_bbox": [0.16, 0.16, 0.84, 0.40],
      "image_loc": "two stacking diagrams: Figure A (boxes with 42 cm top gap) and Figure B (10 cm top gap and side gap '?')",
      "image_file": "q114.png",
      "notes": "no answer key in paper - solved TENTATIVELY. Box dimensions depend on exact box orientation per figure; side gap 30 cm and height 1.8 m are best estimates. VERIFY DOWNSTREAM against the figure - FLAG for human review."
    },
    {
      "n": 115,
      "type_id": 2,
      "question": "Kai Li spent \\(\\dfrac{1}{3}\\) of her money on 5 magnets and 11 postcards. The cost of each magnet is 3 times the cost of each postcard. She bought some more magnets with \\(\\dfrac{3}{4}\\) of her remaining money. How many magnets did Kai Li buy altogether? [?]",
      "answer0": "18",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let a postcard cost 1 unit, so a magnet costs 3 units. First purchase: 5 magnets + 11 postcards = 5(3) + 11(1) = 15 + 11 = 26 units = \\(\\dfrac{1}{3}\\) of her money. So total money = 3 × 26 = 78 units. Remaining = 78 - 26 = 52 units. She spent \\(\\dfrac{3}{4}\\) of remaining on more magnets = \\(\\dfrac{3}{4} \\times 52 = 39\\) units. Magnets at 3 units each: 39 ÷ 3 = 13 more magnets. Altogether = 5 + 13 = 18 magnets.",
      "hints": ["Let a postcard = 1 unit and a magnet = 3 units; cost of first purchase = ⅓ of money.", "Find total money, then ¾ of the remaining money buys more magnets (3 units each); add to the first 5."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. Total 78u, remaining 52u, 3/4 = 39u, 39/3 = 13 more, +5 = 18 altogether. ANSWER 18 (answer0 corrected below)."
    },
    {
      "n": 116,
      "type_id": 2,
      "question": "A symmetric figure is drawn on a rectangular piece of paper 20 cm by 15 cm. Its outline consists of a large semicircle, 2 smaller semicircles and 2 straight lines. (Take \\(\\pi = 3.14\\))<br>(a) What is the area of the figure? Give your answer in cm². [?]<br>(b) What is its perimeter? Give your answer in cm. [?]",
      "answer0": "78.5",
      "answer1": "62.8",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 221,
      "difficulty_id": 3,
      "explanation": "The large semicircle has diameter 20 cm (radius 10). The 2 smaller semicircles have diameter 10 cm each (radius 5), cut out from the bottom. (a) Area = large semicircle - 2 small semicircles = \\(\\dfrac{1}{2}\\pi(10)^2 - 2 \\times \\dfrac{1}{2}\\pi(5)^2 = \\dfrac{1}{2}(3.14)(100) - (3.14)(25) = 157 - 78.5 = 78.5\\) cm². (b) Perimeter = large semicircle arc + 2 small semicircle arcs + ... = π(10) + 2(π × 5) wait: large arc = \\(\\pi r = 3.14 \\times 10 = 31.4\\); two small arcs = 2 × π × 5 = 31.4; total curved = 62.8 cm; plus any straight edges. Perimeter = 62.8 cm (the two straight lines are zero-length at the base join) - tentative.",
      "hints": ["Large semicircle diameter = 20 cm (r = 10); small semicircles r = 5 cm.", "(a) Area = large semicircle - the cut-out semicircles. (b) Perimeter = sum of all the arc lengths plus the straight edges."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q16",
      "image_needed": true,
      "image_options": false,
      "image_page": 40,
      "image_bbox": [0.32, 0.13, 0.66, 0.32],
      "image_loc": "symmetric shaded figure: large semicircle on top, 2 smaller semicircular bites at the bottom, inside a 20 cm by 15 cm dashed rectangle",
      "image_file": "q116.png",
      "notes": "no answer key in paper - solved TENTATIVELY. (a) area = 157 - 78.5 = 78.5 cm2; (b) perimeter = 62.8 cm. answer0 should be 78.5 (corrected below). Geometry of the 2 small semicircles vs straight lines is uncertain - FLAG for human review."
    },
    {
      "n": 117,
      "type_id": 2,
      "question": "Two pouches, Y and Z, contained some gold tokens and silver tokens at first. In Pouch Y, the ratio of gold tokens to silver tokens was 3 : 1. In Pouch Z, the ratio of gold tokens to silver tokens was 1 : 4. Pouch Z had 5 times as many tokens as Pouch Y.<br>(a) What was the ratio of the number of gold tokens in Pouch Y to the number of silver tokens in Pouch Z? [?]<br>(b) After 24 gold tokens and 24 silver tokens were transferred from Pouch Z to Pouch Y, the ratio of gold tokens to silver tokens in Pouch Y became 9 : 5. What was the total number of tokens in Pouch Y in the end? [?]<br>(c) What was the total number of tokens in both pouches, Y and Z, at first? [?]",
      "answer0": "3 : 16",
      "answer1": "112",
      "answer2": "384",
      "answer3": null,
      "correct_answer": null,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "Let Pouch Y have 4 units (gold 3, silver 1). Pouch Z has 5 times as many tokens = 20 units, split 1 : 4 = gold 4, silver 16. (a) Gold in Y : silver in Z = 3 : 16 (in unit terms). As a ratio it is 3 : 16. (b) After transferring 24 gold and 24 silver from Z to Y: Y gold = 3u + 24, Y silver = 1u + 24, ratio (3u+24):(u+24) = 9:5 => 5(3u+24) = 9(u+24) => 15u + 120 = 9u + 216 => 6u = 96 => u = 16. Y in end = (3u+24)+(u+24) = (48+24)+(16+24) = 72 + 40 = 112. (c) At first total = Y + Z = 4u + 20u = 24u = 24 × 16 = 384 tokens.",
      "hints": ["Let Pouch Y = 4 units (3:1); Pouch Z = 20 units (1:4) since Z has 5× as many tokens.", "(b) Set (3u+24):(u+24) = 9:5 and solve for u. (c) Total at first = 24u."],
      "source": "Nanyang 2024 P6 Mid-Year Practice Paper 2 Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "no answer key in paper - solved. u=16. (a) ratio gold-Y : silver-Z = 3:16; (b) 112; (c) 384. answer0 corrected to '3 : 16' and answer2 to '384' below."
    }
  ]
}