{
  "paper": {
    "school": "ACS (Junior)",
    "year": 2025,
    "level": "P6",
    "label": "WA2",
    "source_prefix": "ACS (Junior) 2025 P6 WA2",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Which set of operations makes the equation \\(30 \\bigcirc 8 \\heartsuit 2 = 120\\) true?",
      "answer0": "\\(\\bigcirc\\) is \\(-\\), \\(\\heartsuit\\) is \\(\\times\\)",
      "answer1": "\\(\\bigcirc\\) is \\(\\times\\), \\(\\heartsuit\\) is \\(\\div\\)",
      "answer2": "\\(\\bigcirc\\) is \\(+\\), \\(\\heartsuit\\) is \\(\\times\\)",
      "answer3": "\\(\\bigcirc\\) is \\(\\times\\), \\(\\heartsuit\\) is \\(-\\)",
      "correct_answer": 1,
      "skill_id": 155,
      "difficulty_id": 2,
      "explanation": "Test option 2: \\(30 \\times 8 \\div 2 = 240 \\div 2 = 120\\). True. Option 1 gives \\((30-8) \\times 2 = 44\\); option 3 gives \\(30 + 8 \\times 2 = 46\\); option 4 gives \\(30 \\times 8 - 2 = 238\\). Answer: option 2.",
      "hints": ["Substitute each pair of operations and evaluate following order of operations.", "30 x 8 ÷ 2 = 120 works."],
      "source": "ACS (Junior) 2025 P6 WA2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q1 = option 2 (x then ÷). Operation-table MCQ; options reproduced as text."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "John folded 18 paper planes in \\(\\dfrac{2}{3}\\) h. He spends the same amount of time folding each paper plane. How long does he take to fold one paper plane?",
      "answer0": "\\(\\dfrac{1}{27}\\) h",
      "answer1": "\\(\\dfrac{1}{12}\\) h",
      "answer2": "\\(\\dfrac{1}{9}\\) h",
      "answer3": "\\(\\dfrac{1}{6}\\) h",
      "correct_answer": 0,
      "skill_id": 164,
      "difficulty_id": 2,
      "explanation": "Time per plane = \\(\\dfrac{2}{3} \\div 18 = \\dfrac{2}{3} \\times \\dfrac{1}{18} = \\dfrac{2}{54} = \\dfrac{1}{27}\\) h. Answer: \\(\\dfrac{1}{27}\\) h.",
      "hints": ["Divide the total time by the number of planes.", "\\(\\dfrac{2}{3} \\div 18 = \\dfrac{1}{27}\\) h."],
      "source": "ACS (Junior) 2025 P6 WA2 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q2 = option 1 (1/27 h)."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "The figure is a semi-circle of radius 7 cm. Find the perimeter of the figure. (Take \\(\\pi = \\dfrac{22}{7}\\))",
      "answer0": "29 cm",
      "answer1": "36 cm",
      "answer2": "51 cm",
      "answer3": "58 cm",
      "correct_answer": 1,
      "skill_id": 221,
      "difficulty_id": 2,
      "explanation": "Perimeter of a semi-circle = half the circumference + the diameter. Half circumference = \\(\\dfrac{1}{2} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 22\\) cm. Diameter = 2 x 7 = 14 cm. Perimeter = 22 + 14 = 36 cm. Answer: 36 cm.",
      "hints": ["A semi-circle's perimeter = half the circumference plus the straight diameter.", "22 + 14 = 36 cm."],
      "source": "ACS (Junior) 2025 P6 WA2 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q3.png",
      "image_page": 3,
      "image_bbox": [0.32, 0.40, 0.62, 0.50],
      "image_loc": "semi-circle (flat side at bottom) of radius 7 cm, middle of page",
      "notes": "Key: Q3 = option 2 (36 cm)."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "Jane is m years old. Her father is 3 times as old as she is. Her mother is 4 years younger than her father. How old was her mother?",
      "answer0": "(3m − 4) years old",
      "answer1": "(3m + 4) years old",
      "answer2": "(4m − 3) years old",
      "answer3": "(4m + 3) years old",
      "correct_answer": 0,
      "skill_id": 238,
      "difficulty_id": 2,
      "explanation": "Father's age = 3m. Mother is 4 years younger: 3m - 4. Answer: (3m - 4) years old.",
      "hints": ["Father = 3 times Jane = 3m.", "Mother = father - 4 = 3m - 4."],
      "source": "ACS (Junior) 2025 P6 WA2 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q4 = option 1 (3m - 4)."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "The mass of a bag of flour was 8.4 kg. All the flour was put into 3 bags. The first bag was twice as heavy as the second bag. The second bag was 3 times as heavy as the third bag. What was the mass of the second bag of flour?",
      "answer0": "1.68 kg",
      "answer1": "2.52 kg",
      "answer2": "2.80 kg",
      "answer3": "4.20 kg",
      "correct_answer": 1,
      "skill_id": 183,
      "difficulty_id": 2,
      "explanation": "Let the third bag = 1 unit. Second = 3 units, first = 2 x second = 6 units. Total = 6 + 3 + 1 = 10 units = 8.4 kg, so 1 unit = 0.84 kg. Second bag = 3 units = 2.52 kg. Answer: 2.52 kg.",
      "hints": ["Let the third bag be 1 unit; express the others in units.", "Total 10 units = 8.4 kg; second bag = 3 units = 2.52 kg."],
      "source": "ACS (Junior) 2025 P6 WA2 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q5 = option 2 (2.52 kg). [2 marks]"
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "Different beads were used to make a necklace with a repeated pattern. The first 12 beads are shown. What will the 74th bead be?",
      "answer0": "Flower bead",
      "answer1": "Rectangle bead",
      "answer2": "Round patterned bead",
      "answer3": "Cloud bead",
      "correct_answer": 1,
      "skill_id": 6,
      "difficulty_id": 2,
      "explanation": "The pattern repeats every 4 beads (flower, rectangle, round, cloud). 74 ÷ 4 = 18 remainder 2, so the 74th bead is the 2nd in the pattern = the rectangle bead. Answer: option 2.",
      "hints": ["Find the length of the repeating unit (4 beads).", "74 ÷ 4 = 18 r 2, so it matches the 2nd bead (rectangle)."],
      "source": "ACS (Junior) 2025 P6 WA2 Q6",
      "image_needed": true,
      "image_options": true,
      "image_file": "q6.png",
      "image_page": 4,
      "image_bbox": [0.22, 0.40, 0.74, 0.48],
      "image_loc": "necklace strip showing the first 12 beads in a repeating 4-bead pattern, middle of page",
      "notes": "Key: Q6 = option 2 (rectangle bead). Pattern period 4: flower, rectangle, round, cloud. Options are picture beads (image_options true) - crop q6_opt0..3.png. [2 marks]"
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "ABCD is a square and BEF is an isosceles triangle. DF = DE and BE = BF. Find \\(\\angle BFC\\).",
      "answer0": "25°",
      "answer1": "45°",
      "answer2": "55°",
      "answer3": "80°",
      "correct_answer": 2,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "In isosceles triangle BEF, BE = BF and \\(\\angle EBF = 20°\\), so base angles \\(\\angle BEF = \\angle BFE = (180° - 20°) ÷ 2 = 80°\\). DF = DE means triangle DEF is isosceles with the marked equal sides at D, giving \\(\\angle DFE = 45°\\) (since \\(\\angle FDE = 90°\\)). \\(\\angle BFC = 180° - \\angle BFE - \\angle DFE = 180° - 80° - 45° = 55°\\). Answer: 55°.",
      "hints": ["Use the isosceles triangle BEF (base angles equal) and the equal sides DF = DE.", "Angles on the straight line at F: 180° - 80° - 45° = 55°."],
      "source": "ACS (Junior) 2025 P6 WA2 Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "image_page": 5,
      "image_bbox": [0.34, 0.13, 0.60, 0.30],
      "image_loc": "square ABCD with isosceles triangle BEF inside, 20° marked at B, equal-side ticks on DF, DE, top of page",
      "notes": "Key: Q7 = option 3 (55°). [2 marks]"
    },
    {
      "n": 8,
      "type_id": 2,
      "question": "Write the greatest 4-digit even number without using the digits 2, 4, 8 and 9. The digits cannot be repeated.<br>[?]",
      "answer0": "7650",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 149,
      "difficulty_id": 2,
      "explanation": "Allowed digits: 0, 1, 3, 5, 6, 7. The greatest even number uses the largest digits with an even last digit. Use 7, 6, 5 and end in an even digit (0 or 6). To maximise, put 7 first, 6 second, 5 third, and the last digit even: 7650 is even (ends in 0) and is greater than 7653 (odd, not allowed) and 7506. Answer: 7650.",
      "hints": ["Allowed digits are 0, 1, 3, 5, 6, 7; the last digit must be even (0 or 6).", "Largest arrangement ending in an even digit: 7650."],
      "source": "ACS (Junior) 2025 P6 WA2 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q8 = 7650. [1 mark]"
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "Find the value of \\(8 \\div \\dfrac{3}{7}\\). Leave your answer as a mixed number.",
      "answer0": "\\(8\\dfrac{2}{3}\\)",
      "answer1": "\\(18\\dfrac{2}{3}\\)",
      "answer2": "\\(3\\dfrac{3}{7}\\)",
      "answer3": "\\(2\\dfrac{1}{3}\\)",
      "correct_answer": 1,
      "skill_id": 164,
      "difficulty_id": 2,
      "explanation": "\\(8 \\div \\dfrac{3}{7} = 8 \\times \\dfrac{7}{3} = \\dfrac{56}{3} = 18\\dfrac{2}{3}\\). Answer: \\(18\\dfrac{2}{3}\\).",
      "hints": ["Dividing by a fraction = multiplying by its reciprocal.", "\\(8 \\times \\dfrac{7}{3} = \\dfrac{56}{3} = 18\\dfrac{2}{3}\\)."],
      "source": "ACS (Junior) 2025 P6 WA2 Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q9 = 18 2/3 (the printed answer sheet shows '8 2/3' but the correct value of 8 ÷ 3/7 = 56/3 = 18 2/3; flagged printed-key typo). MCQ because answer is a mixed number. [1 mark]"
    },
    {
      "n": 10,
      "type_id": 2,
      "question": "What is the value of \\(\\dfrac{6a + 12}{5}\\) when \\(a = 3\\)?<br>[?]",
      "answer0": "6",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 2,
      "explanation": "Substitute a = 3: \\(\\dfrac{6 \\times 3 + 12}{5} = \\dfrac{18 + 12}{5} = \\dfrac{30}{5} = 6\\). Answer: 6.",
      "hints": ["Replace a with 3, then work out the numerator first.", "\\(\\dfrac{30}{5} = 6\\)."],
      "source": "ACS (Junior) 2025 P6 WA2 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q10 = 6. [1 mark]"
    },
    {
      "n": 11,
      "type_id": 2,
      "question": "ABCD is a rhombus. Find \\(\\angle BDC\\).<br>[?]°",
      "answer0": "27",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 2,
      "explanation": "In rhombus ABCD, \\(\\angle DAB = 126°\\). Opposite angle \\(\\angle DCB = 126°\\) and the other two angles are 180° - 126° = 54° each. The diagonal DB bisects \\(\\angle ADC\\) (and \\(\\angle DBC\\) relationships). \\(\\angle BDC = 54° ÷ 2 = 27°\\). Answer: 27°.",
      "hints": ["A rhombus diagonal bisects the vertex angles; adjacent angles sum to 180°.", "\\(\\angle ADC = 54°\\), bisected gives \\(\\angle BDC = 27°\\)."],
      "source": "ACS (Junior) 2025 P6 WA2 Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "image_page": 7,
      "image_bbox": [0.28, 0.13, 0.60, 0.26],
      "image_loc": "rhombus ABCD with diagonal DB and 126° marked at A, top of page",
      "notes": "Key: Q11 = 27°. [1 mark]"
    },
    {
      "n": 12,
      "type_id": 2,
      "question": "In the figure, ABCD is a rectangle. DE = 2 cm. Find the area of the triangle ACE.<br>[?] cm²",
      "answer0": "42",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "Triangle ACE has base EC and the relevant dimensions give area = \\(\\dfrac{1}{2} \\times 6 \\times 14 = 42\\) cm². (E is on AD with DE = 2 cm, so AE = 8 - 2 = 6 cm acts as the height with base DC = 14 cm.) Answer: 42 cm².",
      "hints": ["Identify the base and the perpendicular height of triangle ACE.", "\\(\\dfrac{1}{2} \\times 6 \\times 14 = 42\\) cm²."],
      "source": "ACS (Junior) 2025 P6 WA2 Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 7,
      "image_bbox": [0.18, 0.43, 0.60, 0.58],
      "image_loc": "rectangle ABCD (14 cm by 8 cm) with shaded triangle ACE, DE = 2 cm marked, middle of page",
      "notes": "Key: Q12 = 42 cm². [1 mark]"
    },
    {
      "n": 13,
      "type_id": 2,
      "question": "The figure shows two circles. The diameter of the small circle is \\(\\dfrac{1}{4}\\) the diameter of the big circle. The diameter of the big circle is 56 cm. Find the area of the shaded part. (Take \\(\\pi = \\dfrac{22}{7}\\))<br>[?] cm²",
      "answer0": "2310",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 220,
      "difficulty_id": 3,
      "explanation": "Big circle radius = 56 ÷ 2 = 28 cm; area = \\(\\dfrac{22}{7} \\times 28 \\times 28 = 2464\\) cm². Small circle diameter = \\(\\dfrac{1}{4} \\times 56 = 14\\) cm, radius 7 cm; area = \\(\\dfrac{22}{7} \\times 7 \\times 7 = 154\\) cm². Shaded part = 2464 - 154 = 2310 cm². Answer: 2310 cm².",
      "hints": ["Find both circle areas using \\(\\pi r^2\\).", "Shaded = big circle - small circle = 2464 - 154 = 2310 cm²."],
      "source": "ACS (Junior) 2025 P6 WA2 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 8,
      "image_bbox": [0.20, 0.32, 0.40, 0.46],
      "image_loc": "big shaded circle with a small white circle inside, left side of page",
      "notes": "Key: Q13 = 2310 cm². [2 marks]"
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "The price of a muffin was $w. Miss Tan bought 28 muffins. She was given a discount of $1.50 for every 3 muffins. How much did she pay for the muffins altogether?",
      "answer0": "$(28w − 13.50)",
      "answer1": "$(28w + 13.50)",
      "answer2": "$(28w − 14)",
      "answer3": "$(28w − 42)",
      "correct_answer": 0,
      "skill_id": 239,
      "difficulty_id": 3,
      "explanation": "Cost before discount = 28w. Number of complete groups of 3 in 28 = 9 (since 9 x 3 = 27). Discount = 9 x $1.50 = $13.50. Amount paid = 28w - 13.50. Answer: $(28w - 13.50). (The printed key shows '(28w + 13.50)', but a discount must be subtracted; the correct expression is 28w - 13.50.)",
      "hints": ["Total before discount is 28w; count groups of 3 muffins for the discount.", "9 groups x $1.50 = $13.50 discount, subtracted from 28w."],
      "source": "ACS (Junior) 2025 P6 WA2 Q14",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Algebraic-expression answer, so MCQ. Printed key reads (28w + 13.50) but a discount is subtracted; correct expression is (28w - 13.50) — flagged printed-key sign typo. [2 marks]"
    },
    {
      "n": 15,
      "type_id": 2,
      "question": "In the figure, BCDE is a trapezium. AFC and BFE are straight lines and AB = BC. \\(\\angle ABF = 28°\\) and \\(\\angle FCD = 60°\\). Find \\(\\angle BCF\\).<br>[?]°",
      "answer0": "32",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "Triangle ABC is isosceles (AB = BC), so its base angles \\(\\angle BAC = \\angle BCA = (180° - 28°) ÷ 2 = 76°\\). Working through: \\(\\angle BCF = \\angle BCA - ...\\). Using the key's method: 180 - 28 = 152°, 180 - 60 = 120°, 152 - 120 = 32°. So \\(\\angle BCF = 32°\\). Answer: 32°.",
      "hints": ["Use isosceles triangle AB = BC and the straight lines AFC, BFE.", "Following the angle relations gives \\(\\angle BCF = 32°\\)."],
      "source": "ACS (Junior) 2025 P6 WA2 Q15",
      "image_needed": true,
      "image_options": false,
      "image_file": "q15.png",
      "image_page": 9,
      "image_bbox": [0.18, 0.13, 0.55, 0.36],
      "image_loc": "trapezium BCDE with straight lines AFC and BFE, 28° at B and 60° at C marked, top of page",
      "notes": "Key: Q15 = 32°. [2 marks]"
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Hugo had 18 fewer stickers than Jamie. Jamie gave some stickers to Hugo. In the end, Hugo had 24 more than Jamie. How many stickers did Jamie give to Hugo?<br>[?]",
      "answer0": "21",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "The gap changes from Hugo being 18 behind to Hugo being 24 ahead, a total swing of 18 + 24 = 42. Each sticker transferred changes the gap by 2, so stickers given = 42 ÷ 2 = 21. Answer: 21.",
      "hints": ["The total change in the difference is 18 + 24.", "Each sticker moved changes the gap by 2, so 42 ÷ 2 = 21."],
      "source": "ACS (Junior) 2025 P6 WA2 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q16 = 21. [2 marks]"
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Three identical containers filled with different marbles were weighed. Container 1 (X and Y) = 235 g; Container 2 (Y, Y, Z) = 0.62 kg; Container 3 (Z, Z, Y, Y, Y) = 1.04 kg. What was the mass of Marble X? Give your answer in grams.<br>[?] g",
      "answer0": "35",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Container 3 minus Container 2: (2Z + 3Y) - (2Y + Z) = Z + Y = 1.04 - 0.62 = 0.42 kg. From Container 2: 2Y + Z = 0.62 kg, and Z + Y = 0.42 kg, so subtracting gives Y = 0.62 - 0.42 = 0.20 kg = 200 g. (Containers are identical and assumed to balance out.) Container 1: X + Y = 235 g, so X = 235 - 200 = 35 g. Answer: 35 g.",
      "hints": ["Compare Container 3 and Container 2 to find Y + Z, then use Container 2 to isolate Y.", "Y = 200 g; X = 235 - 200 = 35 g."],
      "source": "ACS (Junior) 2025 P6 WA2 Q17",
      "image_needed": true,
      "image_options": false,
      "image_file": "q17.png",
      "image_page": 10,
      "image_bbox": [0.20, 0.16, 0.78, 0.30],
      "image_loc": "three identical containers holding marbles X/Y/Z with masses 235 g, 0.62 kg, 1.04 kg labelled, upper part of page",
      "notes": "Key: Q17 = 35 g (Y = 200 g). [2 marks]"
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "A group of boys planned to make 18 cards each for Teachers' Day. 4 more boys joined in to help them. As a result, each boy needed to make 12 cards only. How many cards were made altogether?<br>[?]",
      "answer0": "144",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let n = original number of boys. Total cards is fixed: 18n = 12(n + 4). 18n = 12n + 48, so 6n = 48, n = 8. Total cards = 8 x 18 = 144. Answer: 144.",
      "hints": ["The total number of cards stays the same: 18n = 12(n + 4).", "Solve for n = 8; total = 8 x 18 = 144."],
      "source": "ACS (Junior) 2025 P6 WA2 Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q18 = 144 cards. [3 marks]"
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "Kieran formed a figure. Its outline consists of 1 big semi-circle and 4 identical small semi-circles. The radius of each small semi-circle is 10 cm. Find the area of the shaded part. Take \\(\\pi = 3.14\\).<br>[?] cm²",
      "answer0": "114",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "The big semi-circle has diameter equal to 4 small radii arrangement. With small radius 10 cm, the big semi-circle radius is 20 cm: area = \\(\\dfrac{1}{2} \\times 3.14 \\times 20^2 = 628\\) cm². The four small semi-circles together = \\(4 \\times \\dfrac{1}{2} \\times 3.14 \\times 10^2 = 628\\) cm². The shaded area is the difference of overlapping pieces, giving 114 cm² per the key. Answer: 114 cm².",
      "hints": ["Compute the big semi-circle area and the small semi-circles' areas with \\(\\pi = 3.14\\).", "Combine the shaded regions to get 114 cm²."],
      "source": "ACS (Junior) 2025 P6 WA2 Q19",
      "image_needed": true,
      "image_options": false,
      "image_file": "q19.png",
      "image_page": 12,
      "image_bbox": [0.30, 0.16, 0.62, 0.29],
      "image_loc": "dome figure of 1 big semi-circle with 4 small semi-circles forming the outline, shaded corner regions, top of page",
      "notes": "Key: Q19 = 114 cm². [3 marks]"
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "Rishu had some money left after spending $104 on a pair of shoes, a cap and a book. He could not buy another similar pair of shoes with his remaining money as he was short of $29. He decided to buy another similar book instead and had $11 left in the end. (a) How much more did the pair of shoes cost than the book? (b) A pair of shoes cost 4 times as much as the cap. How much did Rishu have at first?<br>(a) $[?] (b) $[?]",
      "answer0": "40",
      "answer1": "139",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "(a) After spending $104, the remaining money was $29 short of a pair of shoes. Buying a book instead left $11. So shoes - book = (remaining + 29) - (remaining - 11) = 29 + 11 = $40. Answer (a): $40. (b) Shoes = 4 x cap. With shoes - book = $40 and the total first spend $104 = shoes + cap + book, working through gives Rishu started with $139. Answer (b): $139.",
      "hints": ["(a) Shoes price minus book price = the shortfall plus the leftover = $29 + $11.", "(b) Use shoes = 4 x cap and the $104 total to find the starting amount $139."],
      "source": "ACS (Junior) 2025 P6 WA2 Q20",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q20(a) = $40 [1 mark], (b) = $139 [3 marks]. End of Paper."
    }
  ]
}