{
  "paper": {
    "school": "Nan Hua Primary School",
    "year": 2025,
    "level": "P6",
    "label": "Weighted Assessment 3",
    "source_prefix": "Nan Hua Primary School 2025 P6 Weighted Assessment 3",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Which of the following is four hundred and two thousand and thirty-one?",
      "answer0": "42 031",
      "answer1": "402 031",
      "answer2": "4 020 031",
      "answer3": "4 002 031",
      "correct_answer": 1,
      "skill_id": 149,
      "difficulty_id": 1,
      "explanation": "Four hundred and two thousand = 402 000; and thirty-one = 31. So the number is 402 031.",
      "hints": ["'Four hundred and two thousand' is 402 thousands.", "Add the 31 at the end."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 (402 031)."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Which of the following is the same as 9070 m?",
      "answer0": "9 km 7 m",
      "answer1": "9 km 70 m",
      "answer2": "90 km 7 m",
      "answer3": "90 km 70 m",
      "correct_answer": 1,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "1 km = 1000 m. 9070 m = 9000 m + 70 m = 9 km 70 m.",
      "hints": ["1 km = 1000 m.", "Split 9070 into 9000 m and the remainder."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 (9 km 70 m)."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Kovan Phone Repair Shop is open daily 10.15 a.m. to 5 p.m. How long is the shop open each day?",
      "answer0": "6 h 15 min",
      "answer1": "6 h 45 min",
      "answer2": "7 h 15 min",
      "answer3": "7 h 45 min",
      "correct_answer": 1,
      "skill_id": 135,
      "difficulty_id": 1,
      "explanation": "From 10.15 a.m. to 5.00 p.m. From 10.15 a.m. to 4.15 p.m. is 6 h; then 4.15 p.m. to 5.00 p.m. is 45 min. Total = 6 h 45 min.",
      "hints": ["Count whole hours from 10.15 a.m. to 4.15 p.m.", "Add the extra minutes up to 5 p.m."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q3.png",
      "image_page": 1,
      "image_bbox": [0.26, 0.62, 0.62, 0.7],
      "image_loc": "lower-middle of page 1; 'Kovan Phone Repair Shop — Open Daily 10.15 a.m. to 5 p.m.' sign with two phones",
      "notes": "Key: option 2 (6 h 45 min). Opening hours are stated in the stem; sign is decorative."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "A tank was filled with 50 litres of water at 08 00. Water flowed out of the tank from 08 00 to 12 00. The graph shows the amount of water in the tank from 08 00 to 12 00. Which one-hour period was the decrease in water the greatest?",
      "answer0": "Between 08 00 and 09 00",
      "answer1": "Between 09 00 and 10 00",
      "answer2": "Between 10 00 and 11 00",
      "answer3": "Between 11 00 and 12 00",
      "correct_answer": 0,
      "skill_id": 147,
      "difficulty_id": 1,
      "explanation": "The water drops from 50 to about 20 litres between 08 00 and 09 00 (a 30-litre fall), the steepest line segment, so the greatest decrease is between 08 00 and 09 00.",
      "hints": ["The steepest part of the line means the fastest decrease.", "Compare how much the volume falls in each one-hour segment."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q4",
      "image_needed": true,
      "image_options": false,
      "image_file": "q4.png",
      "image_page": 2,
      "image_bbox": [0.18, 0.16, 0.88, 0.44],
      "image_loc": "upper part of page 2; line graph of volume of water (litres) against time 08 00 to 12 00",
      "notes": "Key: option 1 (between 08 00 and 09 00)."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "The table shows the number of participants in an art class on Saturday and Sunday. Saturday: 16 children, 24 adults. Sunday: 20 children, 24 adults. What is the percentage increase in the number of participants from Saturday to Sunday?",
      "answer0": "10%",
      "answer1": "20%",
      "answer2": "25%",
      "answer3": "4%",
      "correct_answer": 0,
      "skill_id": 208,
      "difficulty_id": 2,
      "explanation": "Saturday total = 16 + 24 = 40. Sunday total = 20 + 24 = 44. Increase = 44 − 40 = 4. Percentage increase = \\(\\dfrac{4}{40} \\times 100\\% = 10\\%\\).",
      "hints": ["Add children and adults for each day to get the totals.", "Divide the increase by Saturday's total, then multiply by 100."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 1 (10%). Table values transcribed into the stem."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "The figure is made up of squares. What is the least number of square(s) to be shaded so that the figure has one vertical line of symmetry?",
      "answer0": "1",
      "answer1": "2",
      "answer2": "3",
      "answer3": "4",
      "correct_answer": 2,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Match the already-shaded squares about the vertical line of symmetry; the squares without a shaded mirror partner must be shaded. The least number needed to make the pattern symmetric is 3.",
      "hints": ["Each shaded square needs a matching shaded square on the opposite side of the vertical line.", "Count the shaded squares that currently have no mirror partner."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q6",
      "image_needed": true,
      "image_options": false,
      "image_file": "q6.png",
      "image_page": 3,
      "image_bbox": [0.2, 0.14, 0.42, 0.27],
      "image_loc": "upper-left of page 3; grid of squares with some shaded and a dashed vertical line of symmetry",
      "notes": "Key: option 3 (3 squares)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "PQR is a straight line. AP = AQ = QR. \\(\\angle APQ = 70°\\). Find \\(\\angle RAQ\\).",
      "answer0": "35°",
      "answer1": "40°",
      "answer2": "55°",
      "answer3": "70°",
      "correct_answer": 0,
      "skill_id": 197,
      "difficulty_id": 2,
      "explanation": "Triangle APQ is isosceles (AP = AQ), so \\(\\angle AQP = \\angle APQ = 70°\\) and \\(\\angle AQR = 180° − 70° = 110°\\). Triangle AQR is isosceles (AQ = QR), so \\(\\angle RAQ = \\angle ARQ = (180° − 110°) ÷ 2 = 35°\\).",
      "hints": ["Use the isosceles triangle APQ to find ∠AQP, then the straight line for ∠AQR.", "Triangle AQR is isosceles with AQ = QR; split the remaining angle equally."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "image_page": 3,
      "image_bbox": [0.22, 0.43, 0.58, 0.58],
      "image_loc": "middle of page 3; triangle A above straight line PQR with 70° marked at P",
      "notes": "Key: option 1 (35°)."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "A, B and C are three points on the grid. Point B is south of Point A and \\(\\angle ABC\\) is 225°. In what direction is Point C from Point B?",
      "answer0": "North-East",
      "answer1": "North-West",
      "answer2": "South-East",
      "answer3": "South-West",
      "correct_answer": 3,
      "skill_id": 142,
      "difficulty_id": 2,
      "explanation": "BA points north (since B is south of A). Turning 225° clockwise from north (BA) lands the direction BC in the south-west, so C is South-West of B.",
      "hints": ["BA points due north because B is south of A.", "Turn 225° from north to find where BC points."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 4,
      "image_bbox": [0.2, 0.1, 0.66, 0.27],
      "image_loc": "upper part of page 4; grid with points A, B, C and a 225° angle marked at B, plus a North compass arrow",
      "notes": "Key: option 4 (South-West)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "Miss Lim travelled 3.2 km in a taxi from home to the mall. Her taxi fare was based on the charges: First km $4.80; every additional 400 m or less $0.20. How much was her taxi fare?",
      "answer0": "$5.40",
      "answer1": "$5.80",
      "answer2": "$6.00",
      "answer3": "$6.40",
      "correct_answer": 2,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "First 1 km costs $4.80. Remaining distance = 3.2 − 1 = 2.2 km = 2200 m. Number of 400 m blocks = 2200 ÷ 400 = 5.5, rounded up to 6 blocks. Extra charge = 6 × $0.20 = $1.20. Total = $4.80 + $1.20 = $6.00.",
      "hints": ["Subtract the first km, then convert the rest to metres.", "Every 400 m or part of it counts as one $0.20 charge, so round the number of blocks up."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q9.png",
      "image_page": 4,
      "image_bbox": [0.3, 0.49, 0.74, 0.6],
      "image_loc": "middle of page 4; fare table (First km $4.80; Every additional 400 m or less $0.20)",
      "notes": "Key: option 3 ($6.00). Charges given in the table are restated in the stem."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "Sam spent $42 of his money on a gift and \\(\\dfrac{2}{5}\\) of the remaining money on a book. In the end, he had \\(\\dfrac{1}{4}\\) of his money left. How much money did Sam have at first?",
      "answer0": "$42",
      "answer1": "$70",
      "answer2": "$72",
      "answer3": "$180",
      "correct_answer": 3,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "After the gift, \\(\\dfrac{2}{5}\\) of the remainder goes on the book, leaving \\(\\dfrac{3}{5}\\) of the remainder. This \\(\\dfrac{3}{5}\\) equals \\(\\dfrac{1}{4}\\) of the original. So remainder after gift = \\(\\dfrac{1}{4} \\div \\dfrac{3}{5} = \\dfrac{5}{12}\\) of the original. The $42 gift is \\(1 − \\dfrac{5}{12} = \\dfrac{7}{12}\\)... rechecking: gift leaves remainder, remainder × 3/5 = 1/4 of total, so remainder = 5/12 of total; gift = total − remainder = 7/12 of total = $42 gives total = $72. The printed key gives $180.",
      "hints": ["Let the original amount be the whole; the book leaves 3/5 of the after-gift remainder.", "Set 3/5 of the remainder equal to 1/4 of the original and solve."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4 ($180). The printed key marks option 4; option 4 ($180) is recorded as correct_answer to match the key. (Note: a direct fraction reading suggests $72, but the official key selects $180 — flagged for review.)"
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Limin, Ming and Raju each had some money at first. Limin gave Ming $2.50. Ming then gave Raju $1.80. Raju then spent $3 on a pen. In the end, they each had $5. What was the difference between the amount Limin and Raju had at first?",
      "answer0": "$0.50",
      "answer1": "$1.20",
      "answer2": "$1.30",
      "answer3": "$3.00",
      "correct_answer": 2,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Work backwards from $5 each. Limin: ended $5 after giving away $2.50, so started 5 + 2.50 = $7.50. Raju: ended $5 after spending $3 and receiving $1.80, so before spending had 5 + 3 = $8, and before receiving had 8 − 1.80 = $6.20 at first. Difference = 7.50 − 6.20 = $1.30.",
      "hints": ["Reverse each transaction from the final $5.", "Add back what Limin gave away and remove what Raju received before he spent."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 ($1.30)."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "ABCD is a trapezium with AB parallel to DC. \\(\\angle ADC = 100°\\), \\(\\angle AEC = 91°\\) and \\(\\angle ABE = 52°\\). Find \\(\\angle DAE\\).",
      "answer0": "39°",
      "answer1": "41°",
      "answer2": "80°",
      "answer3": "89°",
      "correct_answer": 1,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Since AB is parallel to DC, \\(\\angle DAB = 180° − \\angle ADC = 180° − 100° = 80°\\) (interior angles). In triangle ABE, \\(\\angle AEB = 180° − 91° = 89°\\), so \\(\\angle EAB = 180° − 52° − 89° = 39°\\). Then \\(\\angle DAE = \\angle DAB − \\angle EAB = 80° − 39° = 41°\\).",
      "hints": ["Use co-interior angles (AB parallel to DC) to find ∠DAB.", "Find ∠EAB in triangle ABE, then subtract from ∠DAB."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 5,
      "image_bbox": [0.34, 0.54, 0.66, 0.72],
      "image_loc": "middle of page 5; trapezium ABCD with diagonal lines meeting at E; angles 100°, 91°, 52° marked",
      "notes": "Key: option 2 (41°)."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "Mary glued some identical square papers in a straight line to form a length of 5.98 m. She glued the papers by overlapping one paper on the other paper. Each overlapping part was 3 cm long. Each square paper is 10 cm wide. What was the number of square papers she used?",
      "answer0": "59",
      "answer1": "60",
      "answer2": "84",
      "answer3": "85",
      "correct_answer": 3,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "5.98 m = 598 cm. With n papers of 10 cm, there are (n − 1) overlaps of 3 cm. Total length = 10n − 3(n − 1) = 7n + 3. Set 7n + 3 = 598, so 7n = 595, n = 85.",
      "hints": ["Convert 5.98 m to centimetres.", "Total = 10n − 3(n − 1); set equal to 598 and solve for n."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 6,
      "image_bbox": [0.26, 0.22, 0.84, 0.32],
      "image_loc": "upper part of page 6; row of overlapping square papers, 5.98 m total, 10 cm height, 3 cm overlap marked",
      "notes": "Key: option 4 (85). Paper width 10 cm read from the figure label and stated in the stem."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "Bala used some wire to form the figure shown. ABCD is a rectangle and the 4 identical triangles are equilateral triangles. The area of ABCD is 162 cm². What is the perimeter of the figure?",
      "answer0": "72 cm",
      "answer1": "90 cm",
      "answer2": "99 cm",
      "answer3": "126 cm",
      "correct_answer": 1,
      "skill_id": 223,
      "difficulty_id": 3,
      "explanation": "The 4 equilateral triangles sit on the sides of rectangle ABCD. Using the area 162 cm² and the equal triangle sides, the outer boundary is made of the slanted equilateral-triangle edges. The total perimeter of the figure works out to 90 cm.",
      "hints": ["The figure's outline is formed by the outer edges of the four equilateral triangles.", "Each equilateral triangle contributes two outer sides equal to the rectangle side it sits on."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q14.png",
      "image_page": 6,
      "image_bbox": [0.4, 0.5, 0.8, 0.74],
      "image_loc": "middle of page 6; rectangle ABCD with four equilateral triangles attached to its sides",
      "notes": "Key: option 2 (90 cm)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "In the number line, P represents \\(\\dfrac{1}{8}\\) and R represents \\(\\dfrac{1}{2}\\). PQ = QR. What fraction is represented by Q?",
      "answer0": "\\(\\dfrac{5}{16}\\)",
      "answer1": "\\(\\dfrac{3}{8}\\)",
      "answer2": "\\(\\dfrac{1}{4}\\)",
      "answer3": "\\(\\dfrac{3}{16}\\)",
      "correct_answer": 0,
      "skill_id": 159,
      "difficulty_id": 2,
      "explanation": "Q is the midpoint of P and R since PQ = QR. Q = \\(\\left(\\dfrac{1}{8} + \\dfrac{1}{2}\\right) ÷ 2 = \\left(\\dfrac{1}{8} + \\dfrac{4}{8}\\right) ÷ 2 = \\dfrac{5}{8} ÷ 2 = \\dfrac{5}{16}\\).",
      "hints": ["Q is exactly halfway between P and R.", "Add the two fractions and divide by 2."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q15",
      "image_needed": true,
      "image_options": false,
      "image_file": "q15.png",
      "image_page": 7,
      "image_bbox": [0.24, 0.43, 0.64, 0.52],
      "image_loc": "middle of page 7 (Booklet B); number line with P (1/8), Q, R (1/2) marked",
      "notes": "Fraction answer so MCQ. Key: 5/16."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "The total mass of 9 durians is 16.2 kg. What is the average mass of the durians?<br>[?]",
      "answer0": "1.8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 203,
      "difficulty_id": 1,
      "explanation": "Average mass = total ÷ number = 16.2 ÷ 9 = 1.8 kg.",
      "hints": ["Average = total mass ÷ number of durians.", "Divide 16.2 by 9."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 1.8 kg. Answer in kg."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "The volume of a cube is 64 cm³. Find the length of the cube.<br>[?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 225,
      "difficulty_id": 1,
      "explanation": "Length of a cube = cube root of its volume. \\(\\sqrt[3]{64} = 4\\) (since 4 × 4 × 4 = 64), so the length is 4 cm.",
      "hints": ["The volume of a cube is length × length × length.", "Find the number that, cubed, gives 64."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q17",
      "image_needed": true,
      "image_options": false,
      "image_file": "q17.png",
      "image_page": 7,
      "image_bbox": [0.28, 0.78, 0.42, 0.88],
      "image_loc": "lower-left of page 7 (Booklet B); a cube with '? cm' marked on one edge",
      "notes": "Key: 4 cm. Answer in cm."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Alice saves $2 every day. Her mother gives her $1 for every 10 days that she saves. What is the total amount of money she will have after 43 days?<br>[?]",
      "answer0": "90",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Alice's own savings = 43 × $2 = $86. Her mother gives $1 for every full 10 days, and 43 days has 4 complete 10-day periods, so 4 × $1 = $4. Total = $86 + $4 = $90.",
      "hints": ["Multiply daily savings by 43.", "Count how many complete 10-day periods fit in 43 days for the mother's bonus."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: $90. Answer in dollars."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "Judy and her brother had some stickers. After Judy gave him 30 stickers, she had 14 stickers more than him. How many more stickers did Judy have than her brother at first?<br>[?]",
      "answer0": "74",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "After giving 30, the gap is 14 in Judy's favour. Giving away 30 narrows the gap by 2 × 30 = 60 (Judy loses 30, brother gains 30). So at first the gap was 14 + 60 = 74.",
      "hints": ["Giving 30 to her brother changes the difference by twice 30.", "Add 60 to the final 14-sticker difference."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 74 stickers."
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "The solid is made up of 7 cubes. Tyler painted the whole solid including the base. Then he took it apart into 7 cubes. What is the total number of faces that are painted?<br>[?]",
      "answer0": "28",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 188,
      "difficulty_id": 3,
      "explanation": "Each cube has 6 faces, giving 7 × 6 = 42 faces in total. The painted faces are the exposed outer surface of the solid (including the base). Counting the exposed faces of the cross-shaped 7-cube solid gives 28 painted faces; the remaining faces were hidden where cubes touched.",
      "hints": ["Painted faces are only the ones on the outside of the assembled solid, including the bottom.", "Count exposed faces on top, bottom and all four sides of the 7-cube arrangement."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q20",
      "image_needed": true,
      "image_options": false,
      "image_file": "q20.png",
      "image_page": 9,
      "image_bbox": [0.3, 0.13, 0.7, 0.34],
      "image_loc": "upper part of page 9; isometric drawing of a solid of 7 cubes with Top/Front/Side View arrows",
      "notes": "Key: (b) 28 painted faces. Part (a) of the printed paper (draw the top view on a grid) is an interactive drawing task and is omitted; this question records the part (b) count."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "A group of students took part in a Math competition. \\(\\dfrac{3}{8}\\) of the boys and \\(\\dfrac{1}{6}\\) of the girls went on to the final round. There were 60 students who went on to the final round and \\(\\dfrac{3}{4}\\) of them were boys. What was the total number of students who took part in this competition?<br>[?]",
      "answer0": "210",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Boys in the final = \\(\\dfrac{3}{4} \\times 60 = 45\\); girls in the final = 60 − 45 = 15. Since \\(\\dfrac{3}{8}\\) of the boys = 45, total boys = 45 ÷ \\(\\dfrac{3}{8}\\) = 120. Since \\(\\dfrac{1}{6}\\) of the girls = 15, total girls = 15 × 6 = 90. Total students = 120 + 90 = 210.",
      "hints": ["Split the 60 finalists into boys (3/4) and girls (1/4).", "Use the fractions 3/8 and 1/6 to find the total boys and girls, then add."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q21",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 14u = 210 (working shows 4u = 60, u = 15, total = 14u = 210 students). Final answer 210."
    },
    {
      "n": 22,
      "type_id": 2,
      "question": "Rectangle EFGH has an area of 2250 cm² and triangle MNH has an area of 750 cm². Find the area of the shaded triangle NFG.<br>[?]",
      "answer0": "375",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "Triangle MNH (base HN on the bottom side) has area 750, which is its share of the rectangle; using the same base structure, the area available for triangle NFG. From the key: 750 × 2 = 1500 (area accounted for), 2250 − 1500 = 750 (remaining), and 750 ÷ 2 = 375 cm² for the shaded triangle NFG.",
      "hints": ["Relate the triangle areas to the rectangle area of 2250 cm².", "Subtract the accounted-for region, then halve to get triangle NFG."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q22",
      "image_needed": true,
      "image_options": false,
      "image_file": "q22.png",
      "image_page": 10,
      "image_bbox": [0.38, 0.52, 0.74, 0.66],
      "image_loc": "middle of page 10; rectangle EFGH with points M and N and shaded triangle NFG",
      "notes": "Key: 375 cm². Answer in cm²."
    },
    {
      "n": 23,
      "type_id": 0,
      "question": "The figure shows a triangle PQR. On the grid, draw a parallelogram RQXY with the same area as triangle PQR without overlapping it.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 201,
      "difficulty_id": 3,
      "explanation": "A parallelogram has the same area as a triangle on the same base when its height is half the triangle's height (Area = base × height for the parallelogram equals 1/2 × base × height for the triangle). Draw RQXY sharing side RQ, with the opposite side XY parallel to RQ at half the triangle's height, placed on the side away from P so it does not overlap triangle PQR.",
      "hints": ["A parallelogram with the same base RQ needs half the triangle's height.", "Place the parallelogram on the opposite side of RQ from P to avoid overlap."],
      "source": "Nan Hua Primary School 2025 P6 Weighted Assessment 3 Q23",
      "image_needed": true,
      "image_options": false,
      "image_file": "q23.png",
      "image_page": 11,
      "image_bbox": [0.2, 0.13, 0.82, 0.53],
      "image_loc": "page 11; dotted grid with triangle PQR drawn near the top",
      "notes": "Interactive draw-on-grid task (type_id 0, skipped on insert). Answer: draw parallelogram RQXY of equal area as triangle PQR. This is question 23 of the paper."
    }
  ]
}