{
  "paper": {
    "school": "Methodist Girls' School (Primary)",
    "year": 2024,
    "level": "P6",
    "label": "Weighted Assessment 1",
    "source_prefix": "MGS 2024 P6 Weighted Assessment 1",
    "has_answer_key": false,
    "notes": "NO printed answer key in the PDF -- every answer solved by the agent. Single paper, 16 questions: Q1-5 MCQ, Q6-14 short answer, Q15-16 structured. A few multi-part questions whose parts include a name/ratio/fraction are rendered as MCQ per spec."
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Which of the following fractions is greater than \\(\\dfrac{2}{3}\\)?",
      "answer0": "\\(\\dfrac{3}{5}\\)",
      "answer1": "\\(\\dfrac{5}{7}\\)",
      "answer2": "\\(\\dfrac{6}{9}\\)",
      "answer3": "\\(\\dfrac{7}{11}\\)",
      "correct_answer": 1,
      "skill_id": 158,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{2}{3}\\approx 0.667\\). \\(\\dfrac{3}{5}=0.6\\), \\(\\dfrac{5}{7}\\approx 0.714\\), \\(\\dfrac{6}{9}=0.667\\) (= \\(\\dfrac{2}{3}\\), not greater), \\(\\dfrac{7}{11}\\approx 0.636\\). Only \\(\\dfrac{5}{7}\\) is greater.",
      "hints": ["Convert each fraction to a decimal.", "Compare with 0.667."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Find \\(\\angle y\\).",
      "answer0": "58°",
      "answer1": "90°",
      "answer2": "87°",
      "answer3": "122°",
      "correct_answer": 2,
      "skill_id": 194,
      "difficulty_id": 2,
      "explanation": "The marked angles 58°, 90° (right angle), 125° and \\(\\angle y\\) meet at the point and make a full turn: 58° + 90° + 125° + \\(\\angle y\\) = 360°, so \\(\\angle y = 360° - 273° = 87°\\).",
      "hints": ["Angles around a point add up to 360°.", "Include the right angle (90°) shown."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q2",
      "image_needed": true,
      "image_page": 2,
      "image_bbox": [0.28, 0.1, 0.62, 0.26],
      "image_loc": "rays meeting at a point with 58°, right angle, 125° and y marked",
      "image_options": false,
      "image_file": "q2.png",
      "notes": "No answer key -- solved; figure-dependent. Angles around a point = 360° gives 87°; verify the marked angles against the figure."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "ABCD is a rhombus. E and F are midpoints of the 2 sides of the rhombus. What fraction of the rhombus is shaded?",
      "answer0": "\\(\\dfrac{1}{3}\\)",
      "answer1": "\\(\\dfrac{1}{4}\\)",
      "answer2": "\\(\\dfrac{1}{6}\\)",
      "answer3": "\\(\\dfrac{1}{8}\\)",
      "correct_answer": 3,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "E and F are midpoints of DA and AB, so triangle AEF is similar to triangle ADB with linear scale \\(\\dfrac{1}{2}\\), giving area \\(\\dfrac{1}{4}\\) of triangle ADB. Triangle ADB is half the rhombus, so the shaded triangle AEF = \\(\\dfrac{1}{4}\\times\\dfrac{1}{2} = \\dfrac{1}{8}\\) of the rhombus.",
      "hints": ["The small triangle is half the size (linearly) of the top half-triangle.", "Area scales by the square of the length ratio."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q3",
      "image_needed": true,
      "image_page": 2,
      "image_bbox": [0.24, 0.5, 0.62, 0.72],
      "image_loc": "rhombus ABCD with midpoints E, F and shaded top triangle",
      "image_options": false,
      "image_file": "q3.png",
      "notes": "No answer key -- solved. Fraction answer -> MCQ."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "Mr Lim had a sum of money. He gave \\(\\dfrac{2}{5}\\) of the money to his son and shared the remainder equally between his 2 daughters. What fraction of the sum of money did each of his 2 daughters receive?",
      "answer0": "\\(\\dfrac{3}{10}\\)",
      "answer1": "\\(\\dfrac{1}{5}\\)",
      "answer2": "\\(\\dfrac{3}{5}\\)",
      "answer3": "\\(\\dfrac{5}{6}\\)",
      "correct_answer": 0,
      "skill_id": 165,
      "difficulty_id": 2,
      "explanation": "Son got \\(\\dfrac{2}{5}\\), so remainder = \\(\\dfrac{3}{5}\\). Shared between 2 daughters: each gets \\(\\dfrac{3}{5}\\div 2 = \\dfrac{3}{10}\\).",
      "hints": ["Find the remainder after the son's share.", "Divide the remainder by 2."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. Fraction answer -> MCQ."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "WXYZ is a rhombus. \\(\\angle PWZ\\) is \\(\\dfrac{5}{7}\\) of \\(\\angle PZW\\). Find \\(\\angle WPZ\\).",
      "answer0": "90°",
      "answer1": "96°",
      "answer2": "110°",
      "answer3": "120°",
      "correct_answer": 1,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "\\(\\angle WXY = 110°\\), so \\(\\angle XWZ = 70°\\). The diagonal WY bisects this angle, so \\(\\angle PWZ = 35°\\). Then \\(\\angle PWZ = \\dfrac{5}{7}\\angle PZW\\) gives \\(\\angle PZW = 49°\\). In triangle WPZ: \\(\\angle WPZ = 180° - 35° - 49° = 96°\\).",
      "hints": ["A rhombus diagonal bisects the corner angle.", "Use the angle sum of triangle WPZ."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q5",
      "image_needed": true,
      "image_page": 3,
      "image_bbox": [0.22, 0.4, 0.7, 0.66],
      "image_loc": "rhombus WXYZ with diagonal and point P, 110° marked at X",
      "image_options": false,
      "image_file": "q5.png",
      "notes": "No answer key -- solved; figure-dependent. Verify the 110° and that P lies on diagonal WY."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "Arrange the following from the smallest to the largest.<br>\\(1\\dfrac{1}{6}\\), 1.2, \\(\\dfrac{8}{7}\\)",
      "answer0": "\\(\\dfrac{8}{7}\\), \\(1\\dfrac{1}{6}\\), 1.2",
      "answer1": "\\(1\\dfrac{1}{6}\\), \\(\\dfrac{8}{7}\\), 1.2",
      "answer2": "1.2, \\(\\dfrac{8}{7}\\), \\(1\\dfrac{1}{6}\\)",
      "answer3": "\\(\\dfrac{8}{7}\\), 1.2, \\(1\\dfrac{1}{6}\\)",
      "correct_answer": 0,
      "skill_id": 158,
      "difficulty_id": 2,
      "explanation": "Convert to decimals: \\(1\\dfrac{1}{6}\\approx 1.167\\), 1.2, \\(\\dfrac{8}{7}\\approx 1.143\\). Smallest to largest: \\(\\dfrac{8}{7}\\), \\(1\\dfrac{1}{6}\\), 1.2.",
      "hints": ["Convert every value to a decimal.", "Then order them."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. Ordering of fractions -> MCQ (sequence options contain fractions)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "AC is \\(\\dfrac{2}{3}\\) of CB. The number line shows A at \\(\\dfrac{1}{3}\\) and B at \\(\\dfrac{3}{4}\\). What fraction is represented at C?",
      "answer0": "\\(\\dfrac{1}{2}\\)",
      "answer1": "\\(\\dfrac{5}{12}\\)",
      "answer2": "\\(\\dfrac{7}{12}\\)",
      "answer3": "\\(\\dfrac{1}{4}\\)",
      "correct_answer": 0,
      "skill_id": 159,
      "difficulty_id": 3,
      "explanation": "AB = \\(\\dfrac{3}{4} - \\dfrac{1}{3} = \\dfrac{5}{12}\\). AC = \\(\\dfrac{2}{3}\\)CB and AC + CB = AB, so \\(\\dfrac{2}{3}\\)CB + CB = \\(\\dfrac{5}{3}\\)CB = \\(\\dfrac{5}{12}\\) → CB = \\(\\dfrac{1}{4}\\), AC = \\(\\dfrac{1}{6}\\). C = A + AC = \\(\\dfrac{1}{3} + \\dfrac{1}{6} = \\dfrac{1}{2}\\).",
      "hints": ["Find the length AB first.", "Split AB so that AC : CB = 2 : 3."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q7",
      "image_needed": true,
      "image_page": 4,
      "image_bbox": [0.16, 0.5, 0.78, 0.64],
      "image_loc": "number line with A at 1/3, B at 3/4, C marked with '?'",
      "image_options": false,
      "image_file": "q7.png",
      "notes": "No answer key -- solved. Fraction answer -> MCQ."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "In a class library, \\(\\dfrac{3}{5}\\) are fiction books and the rest are non-fiction books. \\(\\dfrac{1}{3}\\) of the non-fiction books are magazines. The rest of the non-fiction books are historical books. What fraction of the class library books are historical books?",
      "answer0": "\\(\\dfrac{4}{15}\\)",
      "answer1": "\\(\\dfrac{2}{15}\\)",
      "answer2": "\\(\\dfrac{2}{5}\\)",
      "answer3": "\\(\\dfrac{1}{3}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 3,
      "explanation": "Non-fiction = \\(\\dfrac{2}{5}\\). Historical = \\(\\dfrac{2}{3}\\) of non-fiction (since \\(\\dfrac{1}{3}\\) are magazines) = \\(\\dfrac{2}{3}\\times\\dfrac{2}{5} = \\dfrac{4}{15}\\).",
      "hints": ["Non-fiction = 1 − fiction.", "Historical = \\(\\dfrac{2}{3}\\) of the non-fiction fraction."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. Fraction answer -> MCQ."
    },
    {
      "n": 9,
      "type_id": 2,
      "question": "Mary bought \\(\\dfrac{9}{10}\\) m of ribbon. She used \\(\\dfrac{5}{6}\\) of it to tie a present. With the remaining ribbon, she used it to make 3 bows. How much ribbon was used to make 1 bow? [?]",
      "answer0": "0.05",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 164,
      "difficulty_id": 3,
      "explanation": "Remaining ribbon = \\(\\dfrac{1}{6}\\times\\dfrac{9}{10} = \\dfrac{9}{60} = \\dfrac{3}{20}\\) m. For 3 bows: \\(\\dfrac{3}{20}\\div 3 = \\dfrac{1}{20}\\) m = 0.05 m per bow.",
      "hints": ["Find the ribbon left after tying the present.", "Divide the remaining ribbon by 3."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. \\(\\dfrac{1}{20}\\) m = 0.05 m, so a decimal FIB is used. Answer in metres."
    },
    {
      "n": 10,
      "type_id": 2,
      "question": "Mrs Wong bought \\(\\dfrac{7}{8}\\) kg of flour to bake some cakes. She needs 0.25 kg of flour for 1 cake. After baking as many cakes as possible, what is the mass of flour left? [?]",
      "answer0": "0.125",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "\\(\\dfrac{7}{8}\\) kg = 0.875 kg. Each cake uses 0.25 kg, so she can make 3 cakes (0.875 ÷ 0.25 = 3.5 → 3 whole cakes) using 0.75 kg. Flour left = 0.875 − 0.75 = 0.125 kg.",
      "hints": ["Convert \\(\\dfrac{7}{8}\\) kg to a decimal.", "Use only whole cakes, then subtract the flour used."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. Answer in kg."
    },
    {
      "n": 11,
      "type_id": 2,
      "question": "Andrew, Chris, Jeremy and Glen sold some funfair tickets. Both Andrew and Glen sold \\(\\dfrac{1}{5}\\) of the tickets each. Chris sold 12 tickets more than Andrew and Jeremy sold 30 tickets. How many tickets did they sell altogether? [?]",
      "answer0": "105",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Let total = T. Andrew = Glen = \\(\\dfrac{T}{5}\\); Chris = \\(\\dfrac{T}{5}\\) + 12; Jeremy = 30. Sum: \\(\\dfrac{T}{5}+\\dfrac{T}{5}+\\dfrac{T}{5}+12+30 = T\\) → \\(\\dfrac{3T}{5}+42 = T\\) → \\(42 = \\dfrac{2T}{5}\\) → T = 105.",
      "hints": ["Write Andrew, Glen and Chris as fractions of T.", "Add all four and set equal to T."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved."
    },
    {
      "n": 12,
      "type_id": 2,
      "question": "PTR and QTS are straight lines. PQ = QR = RS. Find \\(\\angle TRS\\). [?]",
      "answer0": "84",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "PQ = QR so triangle PQR is isosceles: \\(\\angle QPR = \\angle QRP = 34°\\), and \\(\\angle PQR = 112°\\). At T, \\(\\angle PTQ = 180° - 115° = 65°\\), so in triangle PQT \\(\\angle PQT = 180° - 34° - 65° = 81°\\); thus \\(\\angle TQR = 112° - 81° = 31°\\). QR = RS so triangle QRS is isosceles with \\(\\angle RQS = \\angle RSQ = 31°\\) and \\(\\angle QRS = 118°\\). Since T lies on PR, \\(\\angle QRT = 34°\\), giving \\(\\angle TRS = 118° - 34° = 84°\\).",
      "hints": ["Use the isosceles triangles PQR and QRS.", "Track the angles at T (115° marked) and at R."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q12",
      "image_needed": true,
      "image_page": 6,
      "image_bbox": [0.28, 0.45, 0.7, 0.78],
      "image_loc": "figure with straight lines PTR and QTS, 34° at P and 115° at T",
      "image_options": false,
      "image_file": "q12.png",
      "notes": "No answer key -- solved; multi-step geometry. Answer 84° -- verify against the figure (which angle the 115° marks). Answer in degrees."
    },
    {
      "n": 13,
      "type_id": 2,
      "question": "EFGH is a square. EF = FD and \\(\\angle DEF = 83°\\). Find \\(\\angle FGD\\). [?]",
      "answer0": "52",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Triangle EFD has EF = FD, so \\(\\angle FED = \\angle FDE = 83°\\) and \\(\\angle EFD = 180° - 166° = 14°\\). Since EF and FG are equal sides of the square, FD = FG, and \\(\\angle GFD = \\angle EFG - \\angle EFD = 90° - 14° = 76°\\). Triangle FGD is isosceles (FG = FD): \\(\\angle FGD = \\dfrac{180° - 76°}{2} = 52°\\).",
      "hints": ["Find \\(\\angle EFD\\) from the isosceles triangle EFD.", "FD = FG (both equal the square's side); use the isosceles triangle FGD."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q13",
      "image_needed": true,
      "image_page": 7,
      "image_bbox": [0.26, 0.12, 0.66, 0.4],
      "image_loc": "square EFGH with point D and 83° marked at E",
      "image_options": false,
      "image_file": "q13.png",
      "notes": "No answer key -- solved. Answer in degrees."
    },
    {
      "n": 14,
      "type_id": 0,
      "question": "In the square grid, AB and BC form 2 sides of a trapezium. BC is parallel to AD and \\(\\angle BCD\\) is a right angle. Complete the drawing of the trapezium ABCD in the grid.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 201,
      "difficulty_id": 2,
      "explanation": "Draw AD parallel to BC and draw CD so that \\(\\angle BCD\\) is 90°, completing trapezium ABCD on the grid.",
      "hints": ["AD must be parallel to BC.", "Make the angle at C a right angle."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q14",
      "image_needed": true,
      "image_page": 8,
      "image_bbox": [0.16, 0.16, 0.74, 0.42],
      "image_loc": "square grid with points A, B, C marked",
      "image_options": false,
      "image_file": "q14.png",
      "notes": "No answer key -- solved. Draw/construct task (interactive) -> type_id 0 (skipped on insert)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "CD is parallel to GE and GD = GE. CGF is a straight line.<br>(a) Name a trapezium.<br>(b) Name a pair of angles in the figure that add up to 180°.<br>(c) Find \\(\\angle CDE\\).",
      "answer0": "(a) CDEG   (b) \\(\\angle DCG\\) and \\(\\angle CGE\\)   (c) 115°",
      "answer1": "(a) CDEG   (b) \\(\\angle GDE\\) and \\(\\angle DEG\\)   (c) 137.5°",
      "answer2": "(a) GDEF   (b) \\(\\angle DCG\\) and \\(\\angle CGE\\)   (c) 95°",
      "answer3": "(a) CDEG   (b) \\(\\angle DCG\\) and \\(\\angle DGE\\)   (c) 130°",
      "correct_answer": 0,
      "skill_id": 199,
      "difficulty_id": 3,
      "explanation": "(a) CD ∥ GE, so CDEG is a trapezium. (b) CD ∥ GE cut by transversal CG gives co-interior angles \\(\\angle DCG + \\angle CGE = 180°\\). (c) Taking \\(\\angle CGD = 95°\\) and \\(\\angle DCG = 35°\\): triangle CDG gives \\(\\angle CDG = 180° - 95° - 35° = 50°\\). CD ∥ GE gives \\(\\angle DGE = \\angle CDG = 50°\\) (alternate). GD = GE so \\(\\angle GDE = \\angle GED = \\dfrac{180° - 50°}{2} = 65°\\). Hence \\(\\angle CDE = \\angle CDG + \\angle GDE = 50° + 65° = 115°\\).",
      "hints": ["A trapezium has one pair of parallel sides (CD ∥ GE).", "Use alternate angles and the isosceles triangle GDE for part (c)."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q15",
      "image_needed": true,
      "image_page": 9,
      "image_bbox": [0.24, 0.16, 0.66, 0.4],
      "image_loc": "figure CDEGF with CD ∥ GE, 35° and 95° marked",
      "image_options": false,
      "image_file": "q15.png",
      "notes": "No answer key -- solved. Multi-part (a) name, (b) name, (c) angle -> rendered as MCQ. Part (c) = 115° assumes the 95° marks angle CGD; figure-dependent -- verify."
    },
    {
      "n": 16,
      "type_id": 1,
      "question": "Ali, Bala and Charles had 60 marbles altogether. Ali gave \\(\\dfrac{3}{10}\\) of his marbles to Bala and \\(\\dfrac{1}{5}\\) of his marbles to Charles. In the end, all 3 boys had the same number of marbles.<br>(a) Who had more marbles at first, Bala or Charles? How many more?<br>(b) What fraction of the total number of marbles did Bala have at first?",
      "answer0": "(a) Charles, 4 more   (b) \\(\\dfrac{2}{15}\\)",
      "answer1": "(a) Bala, 4 more   (b) \\(\\dfrac{2}{15}\\)",
      "answer2": "(a) Charles, 4 more   (b) \\(\\dfrac{1}{5}\\)",
      "answer3": "(a) Bala, 2 more   (b) \\(\\dfrac{2}{15}\\)",
      "correct_answer": 0,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Each boy ends with 60 ÷ 3 = 20. Ali keeps \\(1 - \\dfrac{3}{10} - \\dfrac{1}{5} = \\dfrac{1}{2}\\) of his marbles = 20, so Ali = 40. Bala + \\(\\dfrac{3}{10}\\times 40\\) = Bala + 12 = 20 → Bala = 8. Charles + \\(\\dfrac{1}{5}\\times 40\\) = Charles + 8 = 20 → Charles = 12. (a) Charles had more, by 12 − 8 = 4. (b) Bala's fraction = \\(\\dfrac{8}{60} = \\dfrac{2}{15}\\).",
      "hints": ["Each boy ends with one-third of 60.", "Ali keeps half his marbles after giving the rest away."],
      "source": "MGS 2024 P6 Weighted Assessment 1 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "No answer key -- solved. Multi-part (a) name + count, (b) fraction -> rendered as MCQ. (a) Charles, 4 more; (b) 2/15."
    }
  ]
}