{
  "paper": {
    "school": "Catholic High",
    "year": 2025,
    "level": "P6",
    "label": "WA2",
    "source_prefix": "Catholic High 2025 P6 WA2",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 2,
      "question": "A bowl of noodles at a restaurant cost $27.25 including 9% GST. Find the cost of the bowl of noodles before GST.<br>$[?]",
      "answer0": "25",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "The price with GST is 109% of the original. 109% = $27.25, so 1% = $27.25 ÷ 109 = $0.25, and 100% = $0.25 x 100 = $25. Answer: $25.",
      "hints": ["Price including 9% GST is 109% of the pre-GST cost.", "$27.25 ÷ 109 x 100 = $25."],
      "source": "Catholic High 2025 P6 WA2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q1 = $25. [2 marks]"
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "The radius of a circle is 21 cm. Find the circumference of the circle. Leave your answer in terms of \\(\\pi\\).",
      "answer0": "\\(42\\pi\\) cm",
      "answer1": "\\(21\\pi\\) cm",
      "answer2": "\\(84\\pi\\) cm",
      "answer3": "\\(441\\pi\\) cm",
      "correct_answer": 0,
      "skill_id": 220,
      "difficulty_id": 2,
      "explanation": "Circumference = \\(2\\pi r = 2 \\times \\pi \\times 21 = 42\\pi\\) cm. Answer: \\(42\\pi\\) cm.",
      "hints": ["Circumference = \\(2\\pi r\\).", "\\(2 \\times 21 = 42\\), so the circumference is \\(42\\pi\\) cm."],
      "source": "Catholic High 2025 P6 WA2 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q2 = 42π cm. Answer is in terms of π (not a plain decimal), so served as MCQ. Distractor options added. [2 marks]"
    },
    {
      "n": 3,
      "type_id": 2,
      "question": "A rectangular piece of paper is folded along the dotted line as shown. Find \\(\\angle p\\).<br>[?]°",
      "answer0": "34",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "The fold creates a right-angled corner. The angle adjacent to the 62° is \\(180° - 90° - 62° = 28°\\). The fold reflects this, so \\(\\angle p = 90° - 28° - 28° = 34°\\). Answer: 34°.",
      "hints": ["The folded corner keeps a right angle; find the 28° angle first.", "\\(90° - 28° - 28° = 34°\\)."],
      "source": "Catholic High 2025 P6 WA2 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q3.png",
      "image_page": 2,
      "image_bbox": [0.18, 0.14, 0.68, 0.27],
      "image_loc": "rectangle before folding and after folding showing angles p and 62°, top of page",
      "notes": "Key: Q3 = 34°. [2 marks]"
    },
    {
      "n": 4,
      "type_id": 2,
      "question": "The usual price of a watch is $320. After a 20% discount, how much does the watch cost?<br>$[?]",
      "answer0": "256",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "After a 20% discount, the watch costs 100% - 20% = 80% of $320. \\(\\dfrac{80}{100} \\times 320 = 256\\). Answer: $256.",
      "hints": ["A 20% discount means paying 80% of the usual price.", "0.80 x $320 = $256."],
      "source": "Catholic High 2025 P6 WA2 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q4 = $256. [2 marks]"
    },
    {
      "n": 5,
      "type_id": 2,
      "question": "ABCD is a square. CEFD is a rhombus. AGF and CGD are straight lines. \\(\\angle GDF = 44°\\). Find \\(\\angle GFD\\).<br>[?]°",
      "answer0": "23",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "In rhombus CEFD, DF = FC so triangle GFD relationships apply. With \\(\\angle GDF = 44°\\) and the rhombus/square geometry, the key computes \\(\\angle GFD = \\dfrac{180° - 134°}{2} = 23°\\). Answer: 23°.",
      "hints": ["Use the equal sides of the rhombus to form an isosceles triangle.", "\\((180° - 134°) ÷ 2 = 23°\\)."],
      "source": "Catholic High 2025 P6 WA2 Q5",
      "image_needed": true,
      "image_options": false,
      "image_file": "q5.png",
      "image_page": 3,
      "image_bbox": [0.28, 0.13, 0.66, 0.31],
      "image_loc": "square ABCD joined to rhombus CEFD with straight lines AGF and CGD, 44° marked at D, top of page",
      "notes": "Key: Q5 = 23°. [2 marks]"
    },
    {
      "n": 6,
      "type_id": 2,
      "question": "The diameter of a circle is 70 cm. Find the area of the circle. Take \\(\\pi = \\dfrac{22}{7}\\).<br>[?] cm²",
      "answer0": "3850",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 220,
      "difficulty_id": 2,
      "explanation": "Radius = 70 ÷ 2 = 35 cm. Area = \\(\\pi r^2 = \\dfrac{22}{7} \\times 35 \\times 35 = 3850\\) cm². Answer: 3850 cm².",
      "hints": ["Find the radius first: 70 ÷ 2 = 35 cm.", "Area = \\(\\dfrac{22}{7} \\times 35 \\times 35 = 3850\\) cm²."],
      "source": "Catholic High 2025 P6 WA2 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q6 = 3850 cm². [2 marks]"
    },
    {
      "n": 7,
      "type_id": 2,
      "question": "The figure is made up of a quarter circle and a square. Find the area of the figure. Take \\(\\pi = 3.14\\).<br>[?] cm²",
      "answer0": "64.26",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 2,
      "explanation": "The square has side 6 cm, area = 6 x 6 = 36 cm². The quarter circle has radius 6 cm, area = \\(\\dfrac{1}{4} \\times 3.14 \\times 6 \\times 6 = 28.26\\) cm². Total area = 28.26 + 36 = 64.26 cm². Answer: 64.26 cm².",
      "hints": ["Add the square's area to the quarter-circle's area.", "\\(\\dfrac{1}{4} \\times 3.14 \\times 6^2 + 6^2 = 28.26 + 36 = 64.26\\) cm²."],
      "source": "Catholic High 2025 P6 WA2 Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "image_page": 4,
      "image_bbox": [0.18, 0.26, 0.44, 0.36],
      "image_loc": "quarter circle joined to a square with 6 cm marked across the top, upper-left of page",
      "notes": "Key: Q7 = 64.26 cm². [3 marks]"
    },
    {
      "n": 8,
      "type_id": 2,
      "question": "EFGH is a parallelogram and EBHA is a trapezium. AE is parallel to HB. \\(\\angle AEH = 34°\\), \\(\\angle EBH = 76°\\), \\(\\angle FGH = 108°\\). Find \\(\\angle BEF\\).<br>[?]°",
      "answer0": "38",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "In triangle EBH (or using AE // HB), \\(\\angle AEB + \\angle EBH = 180°\\) relationships give \\(\\angle HEB = 180° - 76° - 34° = 70°\\). In parallelogram EFGH, \\(\\angle FEH = \\angle FGH = 108°\\). So \\(\\angle BEF = 108° - 70° = 38°\\). Answer: 38°.",
      "hints": ["Use AE // HB and the angle sum to find \\(\\angle HEB = 70°\\).", "\\(\\angle FEH = 108°\\) (parallelogram), so \\(\\angle BEF = 108° - 70° = 38°\\)."],
      "source": "Catholic High 2025 P6 WA2 Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 5,
      "image_bbox": [0.18, 0.13, 0.62, 0.31],
      "image_loc": "parallelogram EFGH with trapezium EBHA, angles 34°, 76°, 108° marked, top of page",
      "notes": "Key: Q8 = 38°. [3 marks]"
    },
    {
      "n": 9,
      "type_id": 2,
      "question": "The figure shows a rectangle and two identical semicircles. The length of the rectangle is 24 m and the radius of each semicircle is 4 m. Find the perimeter of the figure. Take \\(\\pi = 3.14\\).<br>[?] m",
      "answer0": "73.12",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "The two semicircles (radius 4 m) together form one full circle: circumference = \\(2 \\times 3.14 \\times 4 = 25.12\\) m (or \\(8\\pi\\)). The two straight rectangle sides of length 24 m contribute 2 x 24 = 48 m. Perimeter = 25.12 + 48 = 73.12 m. Answer: 73.12 m.",
      "hints": ["The two semicircles combine to one full circle of diameter 8 m.", "Perimeter = circle circumference + two long sides = 25.12 + 48 = 73.12 m."],
      "source": "Catholic High 2025 P6 WA2 Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q9.png",
      "image_page": 6,
      "image_bbox": [0.20, 0.18, 0.66, 0.32],
      "image_loc": "stadium shape: rectangle 24 m long with a semicircle (radius 4 m) on each end, top of page",
      "notes": "Key: Q9 = 73.12 m (8π + 48). [3 marks]"
    },
    {
      "n": 10,
      "type_id": 2,
      "question": "The number of people who visited a museum in February increased by 20% when compared to January. The number of people who visited the same museum in March decreased by 5% when compared to February. 855 people visited the museum in March. How many people visited the museum in January?<br>[?]",
      "answer0": "750",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 208,
      "difficulty_id": 3,
      "explanation": "March = 95% of February, so February = 855 ÷ 95% = 855 ÷ 0.95 = 900. February = 120% of January, so January = 900 ÷ 120% = 900 ÷ 1.20 = 750. Answer: 750.",
      "hints": ["Work backwards: March is 95% of February; February is 120% of January.", "855 ÷ 0.95 = 900; 900 ÷ 1.20 = 750."],
      "source": "Catholic High 2025 P6 WA2 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q10 = 750. [4 marks]"
    },
    {
      "n": 11,
      "type_id": 2,
      "question": "There were 140 pupils in an Art Club at first. 40% of the pupils were girls and the rest were boys. After some girls left the Art Club, 20% of the number of pupils who remained in the Art Club were girls. (a) How many pupils in the Art Club were boys? (b) How many girls left the Art Club?<br>(a) [?] (b) [?]",
      "answer0": "84",
      "answer1": "35",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 175,
      "difficulty_id": 3,
      "explanation": "(a) Boys = 60% of 140 = \\(\\dfrac{60}{100} \\times 140 = 84\\) boys. (b) The number of boys stays 84. After some girls left, girls are 20% and boys are 80% of those remaining: 80% (= 80u) = 84, so 1u = 84 ÷ 80 = 1.05; girls remaining = 20u = 1.05 x 20 = 21. Girls at first = 140 - 84 = 56. Girls who left = 56 - 21 = 35. Answers: (a) 84 boys, (b) 35 girls.",
      "hints": ["(a) Boys are 60% of 140.", "(b) Boys (84) are 80% of those who remained; find remaining girls (21), then 56 - 21 = 35."],
      "source": "Catholic High 2025 P6 WA2 Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q11(a) = 84 boys [1 mark], (b) = 35 girls [4 marks]. END OF PAPER."
    }
  ]
}