{
  "paper": {
    "school": "Henry Park",
    "year": 2025,
    "level": "P5",
    "label": "Weighted Assessment 1",
    "source_prefix": "Henry Park 2025 P5 Weighted Assessment 1",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Nine million, fifty thousand, one hundred and seventy-three in numerals is [?].",
      "answer0": "950 173",
      "answer1": "9 005 173",
      "answer2": "9 050 173",
      "answer3": "9 500 173",
      "correct_answer": 2,
      "skill_id": 149,
      "difficulty_id": 1,
      "explanation": "Nine million = 9 000 000; fifty thousand = 50 000; one hundred and seventy-three = 173. Total = 9 050 173.",
      "hints": ["Build the number place by place: millions, thousands, ones.", "Fifty thousand fills the 050 in the thousands group."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (9 050 173)."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "How many one-fifths are there in \\(2\\dfrac{2}{5}\\)?",
      "answer0": "10",
      "answer1": "11",
      "answer2": "12",
      "answer3": "17",
      "correct_answer": 2,
      "skill_id": 158,
      "difficulty_id": 2,
      "explanation": "\\(2\\dfrac{2}{5} = \\dfrac{12}{5}\\), so there are 12 one-fifths.",
      "hints": ["Convert the mixed number to fifths.", "2 wholes = 10 fifths, plus 2 more fifths."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (12)."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "In the figure, ABC is a triangle. What is the height of triangle ABC when its base is AC?",
      "answer0": "AE",
      "answer1": "BA",
      "answer2": "BC",
      "answer3": "BD",
      "correct_answer": 3,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "The height to base AC is the perpendicular distance from the opposite vertex B to the line AC. BD is drawn perpendicular to AC (right angle at D), so BD is the height when AC is the base.",
      "hints": ["The height must be perpendicular to the chosen base AC.", "Find the line from B that meets AC at a right angle."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q3.png",
      "image_page": 2,
      "image_bbox": [0.28, 0.15, 0.78, 0.34],
      "image_loc": "upper part of page; triangle ABC with perpendiculars BD (to AC at D) and AE (to BC at E), right angles marked at D and E",
      "notes": "Key: option 4 (BD)."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "What is the product of \\(\\dfrac{8}{3}\\) and 24?",
      "answer0": "1",
      "answer1": "9",
      "answer2": "56",
      "answer3": "64",
      "correct_answer": 3,
      "skill_id": 163,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{8}{3} \\times 24 = 8 \\times (24 \\div 3) = 8 \\times 8 = 64\\).",
      "hints": ["Divide 24 by 3 first.", "Then multiply by 8."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4 (64)."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Which one of the following is the second common multiple of 6 and 9?",
      "answer0": "3",
      "answer1": "18",
      "answer2": "36",
      "answer3": "54",
      "correct_answer": 2,
      "skill_id": 115,
      "difficulty_id": 2,
      "explanation": "Common multiples of 6 and 9 are multiples of their LCM, 18: 18, 36, 54 … The second one is 36.",
      "hints": ["Find the LCM of 6 and 9 (the first common multiple).", "Double it for the second common multiple."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (36)."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "Arrange the numbers \\(\\dfrac{8}{5}\\), 6 and \\(\\dfrac{3}{4}\\) from the largest to the smallest.",
      "answer0": "6, \\(\\dfrac{8}{5}\\), \\(\\dfrac{3}{4}\\)",
      "answer1": "6, \\(\\dfrac{3}{4}\\), \\(\\dfrac{8}{5}\\)",
      "answer2": "\\(\\dfrac{3}{4}\\), \\(\\dfrac{8}{5}\\), 6",
      "answer3": "\\(\\dfrac{8}{5}\\), \\(\\dfrac{3}{4}\\), 6",
      "correct_answer": 0,
      "skill_id": 124,
      "difficulty_id": 2,
      "explanation": "\\(\\dfrac{8}{5} = 1.6\\), 6 = 6, \\(\\dfrac{3}{4} = 0.75\\). Largest to smallest: 6, \\(\\dfrac{8}{5}\\), \\(\\dfrac{3}{4}\\).",
      "hints": ["Convert each to a decimal to compare.", "6 is the biggest; 3/4 is less than 1."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Has fraction terms so kept MCQ. Key: option 1 (6, 8/5, 3/4)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Find the value of 5 × 28 + (100 − 76) ÷ 4.",
      "answer0": "41",
      "answer1": "65",
      "answer2": "146",
      "answer3": "170",
      "correct_answer": 2,
      "skill_id": 155,
      "difficulty_id": 2,
      "explanation": "Brackets: 100 − 76 = 24. Then 24 ÷ 4 = 6 and 5 × 28 = 140. So 140 + 6 = 146.",
      "hints": ["Work out the brackets, then multiply and divide.", "5 × 28 = 140; (100 − 76) ÷ 4 = 6."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q7",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (146)."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Mrs Tan bought \\(1\\dfrac{1}{2}\\) kg of sugar and Mrs Lee bought \\(\\dfrac{1}{3}\\) of what Mrs Tan bought. What was the total amount of sugar bought by Mrs Tan and Mrs Lee?",
      "answer0": "\\(1\\dfrac{2}{5}\\) kg",
      "answer1": "\\(1\\dfrac{5}{6}\\) kg",
      "answer2": "2 kg",
      "answer3": "\\(3\\dfrac{1}{2}\\) kg",
      "correct_answer": 2,
      "skill_id": 163,
      "difficulty_id": 2,
      "explanation": "Mrs Lee = \\(\\dfrac{1}{3} \\times 1\\dfrac{1}{2} = \\dfrac{1}{3} \\times \\dfrac{3}{2} = \\dfrac{1}{2}\\) kg. Total = \\(1\\dfrac{1}{2} + \\dfrac{1}{2} = 2\\) kg.",
      "hints": ["Find Mrs Lee's amount (1/3 of 1½ kg).", "Add it to Mrs Tan's 1½ kg."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (2 kg)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "Polly and Queenie had the same number of stickers at first. After Polly gave away 28 stickers and Queenie gave away 76 stickers, Polly had 4 times as many stickers as Queenie. How many stickers did Polly have at first?",
      "answer0": "44",
      "answer1": "48",
      "answer2": "88",
      "answer3": "92",
      "correct_answer": 3,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Both started equal. Difference in what they gave away = 76 − 28 = 48, so Polly has 48 more than Queenie at the end. Polly = 4 × Queenie, so 48 = 3 units → 1 unit (Queenie) = 16. Polly at end = 4 × 16 = 64. Polly at first = 64 + 28 = 92.",
      "hints": ["The end gap of 48 = 3 of Queenie's parts (since Polly = 4× Queenie).", "Add back the 28 Polly gave away to her ending amount."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4 (92)."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "Jin Yong spent \\(\\dfrac{1}{3}\\) of his money on a pen and \\(\\dfrac{3}{4}\\) of the remaining money on a book. In the end, he had $16 left. How much money, in total, did he spend on the pen and the book?",
      "answer0": "$32",
      "answer1": "$48",
      "answer2": "$80",
      "answer3": "$96",
      "correct_answer": 2,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Remaining after the pen = \\(\\dfrac{2}{3}\\). Book = \\(\\dfrac{3}{4}\\) of that, so left = \\(\\dfrac{1}{4} \\times \\dfrac{2}{3} = \\dfrac{1}{6}\\) of total = $16. So total = 6 × $16 = $96. Spent = $96 − $16 = $80.",
      "hints": ["The $16 left is 1/4 of the remainder after the pen, i.e. 1/6 of the total.", "Find the total, then subtract the $16 left."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 ($80)."
    },
    {
      "n": 11,
      "type_id": 2,
      "question": "The figure is a 4 by 4 grid of squares. How many more squares must be shaded so that only \\(\\dfrac{3}{8}\\) of the figure is left unshaded?<br>[?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 159,
      "difficulty_id": 2,
      "explanation": "The grid has 16 squares. If \\(\\dfrac{3}{8}\\) are unshaded, that is \\(\\dfrac{3}{8} \\times 16 = 6\\) unshaded squares, so 16 − 6 = 10 must be shaded. The figure already has 6 shaded, so 10 − 6 = 4 more squares must be shaded.",
      "hints": ["Find how many squares should remain unshaded (3/8 of 16).", "Subtract the already-shaded squares from the required shaded total."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "image_page": 6,
      "image_bbox": [0.45, 0.27, 0.65, 0.43],
      "image_loc": "middle of page; 4 by 4 grid of squares with some squares shaded",
      "notes": "Key: 4. Need to read the count of already-shaded squares from the grid when cropping (6 shaded gives 4 more)."
    },
    {
      "n": 12,
      "type_id": 2,
      "question": "The figure is made up of rectangle ABCD and triangle BDE. AB = 19 cm, BC = 10 cm and EC = 5 cm. Find the area of the shaded triangle BDE.<br>[?]",
      "answer0": "70",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 138,
      "difficulty_id": 3,
      "explanation": "Rectangle area = 19 × 10 = 190 cm². Triangle ABD = \\(\\dfrac{1}{2}\\) of the rectangle = 95 cm². Triangle BEC has base EC = 5 cm and height BC = 10 cm, area = \\(\\dfrac{1}{2} \\times 10 \\times 5 = 25\\) cm². Shaded BDE = 190 − 95 − 25 = 70 cm².",
      "hints": ["Find the rectangle's area, then remove triangle ABD (half the rectangle).", "Also remove triangle BEC; what remains is triangle BDE."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 6,
      "image_bbox": [0.27, 0.55, 0.6, 0.72],
      "image_loc": "lower-middle of page; rectangle ABCD (19 cm by 10 cm) with shaded triangle BDE, E on DC with EC = 5 cm",
      "notes": "Key: 70 cm². Answer in cm². (Key working 190 − 95 − 25 = 70.)"
    },
    {
      "n": 13,
      "type_id": 2,
      "question": "The solid is made up of 1-cm cubes. What is the volume of the solid?<br>[?]",
      "answer0": "8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 188,
      "difficulty_id": 2,
      "explanation": "Counting the 1-cm cubes in the solid gives 8 cubes, so the volume is 8 cm³.",
      "hints": ["Count every 1-cm cube, including any hidden ones.", "Each cube has volume 1 cm³."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 7,
      "image_bbox": [0.27, 0.22, 0.5, 0.38],
      "image_loc": "upper part of page; solid built from 1-cm cubes",
      "notes": "Key: 8 cm³ (count = 8 cubes). Answer in cm³. Verify the cube count against the key when cropping."
    },
    {
      "n": 14,
      "type_id": 2,
      "question": "The table shows museum admission prices: Adult $15, Child $7, Package A (1 adult and 1 child) $20, Package B (2 adults and 1 child) $35. Mr Liew bought tickets for 4 adults and 3 children. What was the least amount that he could have paid?<br>$[?]",
      "answer0": "75",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Buy one Package B (2 adults + 1 child) = $35, leaving 2 adults + 2 children. Buy two Package A (each 1 adult + 1 child) = 2 × $20 = $40. Total = $35 + $40 = $75, which is cheaper than buying singles.",
      "hints": ["Use packages to bundle adults with children cheaply.", "1 Package B + 2 Package A covers 4 adults and 3 children."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q14.png",
      "image_page": 7,
      "image_bbox": [0.35, 0.47, 0.78, 0.64],
      "image_loc": "middle of page; price table of Adult, Child, Package A and Package B tickets",
      "notes": "Key: $75 (key working 35 + 20 × 2). Price table given in stem so figure optional."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "There are 2 boxes A and B in a cupboard. Box A contained twice as many balls as box B. In box A, \\(\\dfrac{1}{4}\\) of the balls are blue. In box B, \\(\\dfrac{1}{2}\\) of the balls are blue. What fraction of the balls in the cupboard are blue? Express your answer in the simplest form.",
      "answer0": "\\(\\dfrac{1}{3}\\)",
      "answer1": "\\(\\dfrac{3}{8}\\)",
      "answer2": "\\(\\dfrac{3}{4}\\)",
      "answer3": "\\(\\dfrac{1}{4}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 3,
      "explanation": "Let box B = 2 balls, box A = 4 balls (twice). Blue in A = \\(\\dfrac{1}{4} \\times 4 = 1\\); blue in B = \\(\\dfrac{1}{2} \\times 2 = 1\\). Total balls = 6, total blue = 2, so fraction blue = \\(\\dfrac{2}{6} = \\dfrac{1}{3}\\).",
      "hints": ["Use convenient numbers: box B = 2 balls, box A = 4 balls.", "Count blue balls in each box, then over the total."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction answer so converted from FIB to MCQ. Key: 1/3."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "The figure is a rectangle with a length of 18 cm and a breadth of 13 cm, made up of shaded and unshaded parts. Find the area of the unshaded parts.<br>[?]",
      "answer0": "117",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 138,
      "difficulty_id": 3,
      "explanation": "Rectangle area = 18 × 13 = 234 cm². The shaded triangles together total half the rectangle, so the unshaded parts are the other half: 234 ÷ 2 = 117 cm².",
      "hints": ["Find the whole rectangle's area first.", "The shaded triangles make up exactly half, so the unshaded parts are the other half."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q16",
      "image_needed": true,
      "image_options": false,
      "image_file": "q16.png",
      "image_page": 8,
      "image_bbox": [0.28, 0.5, 0.62, 0.7],
      "image_loc": "middle of page; rectangle 18 cm by 13 cm with shaded triangles (top and two lower triangles) and unshaded parts",
      "notes": "Key: 117 cm² (key working 18 × 13 = 234, 234 ÷ 2 = 117). Answer in cm²."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Kai Lun made some bookmarks on Tuesday. He made 8 more bookmarks on Wednesday than on Tuesday and twice as many bookmarks on Thursday as on Wednesday. On Friday, he made 14 fewer bookmarks than on Thursday. Kai Lun made a total of 200 bookmarks over the four days. How many bookmarks did he make on Tuesday?<br>[?]",
      "answer0": "29",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Let Tuesday = u. Wednesday = u + 8. Thursday = 2(u + 8) = 2u + 16. Friday = (2u + 16) − 14 = 2u + 2. Total = u + (u + 8) + (2u + 16) + (2u + 2) = 6u + 26 = 200, so 6u = 174 and u = 29.",
      "hints": ["Write each day in terms of Tuesday (u).", "Add the four expressions to 200 and solve for u."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 29 (key working 200 − 26 = 174, 174 ÷ 6 = 29)."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "A repeated pattern is formed using the numbers 2 and 0. The first 18 numbers are: 2, 0, 2, 0, 2, 2, 0, 2, 0, 2, 2, 0, 2, 0, 2, 2 … (the unit '2, 0, 2, 0, 2, 2' repeats). What is the sum of the first 100 numbers?<br>[?]",
      "answer0": "132",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "The repeating block of 6 numbers (2, 0, 2, 0, 2, 2) sums to 8. 100 ÷ 6 = 16 complete blocks with remainder 4. Sixteen blocks contribute 16 × 8 = 128. The next 4 numbers are 2, 0, 2, 0, but per the key the remaining contribution is 2 + 2 = 4, giving 128 + 4 = 132.",
      "hints": ["The pattern repeats every 6 numbers; find the sum of one block.", "100 = 16 blocks + 4 extra numbers; add the extras."],
      "source": "Henry Park 2025 P5 Weighted Assessment 1 Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 132 (key working 100 ÷ 6 = 16 R4, 8 × 16 = 128, 128 + 2 + 2 = 132)."
    }
  ]
}