{
  "paper": {
    "school": "Henry Park",
    "year": 2024,
    "level": "P6",
    "label": "Prelim",
    "source_prefix": "Henry Park 2024 P6 Prelim",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "$80\\,000 + 4\\,000 + 300 + 7 = \\boxed{?}$<br>What is the missing number in the box?",
      "answer0": "80 437",
      "answer1": "84 037",
      "answer2": "84 307",
      "answer3": "84 370",
      "correct_answer": 2,
      "skill_id": 105,
      "difficulty_id": 1,
      "explanation": "Add the place values: 80 000 + 4 000 + 300 + 7 = 84 307. Answer: 84 307.",
      "hints": ["Line up the digits by place value: ten-thousands, thousands, hundreds, tens, ones.", "There are no tens, so put a 0 in the tens place: 84 307."],
      "source": "Henry Park 2024 P6 Prelim P1 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q1 = option 3 (84 307)."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "The mass of a table is 11 kg when rounded to the nearest kilogramme. Which of the following cannot be the mass of the table?",
      "answer0": "10.49 kg",
      "answer1": "10.55 kg",
      "answer2": "11.08 kg",
      "answer3": "11.46 kg",
      "correct_answer": 0,
      "skill_id": 128,
      "difficulty_id": 1,
      "explanation": "A mass rounds to 11 kg if it lies in 10.5 kg to 11.49 kg (rounding to the nearest whole kg). 10.49 kg rounds to 10 kg, not 11 kg, so it cannot be the mass. Answer: 10.49 kg.",
      "hints": ["To round to the nearest whole kg, look at the first decimal digit.", "10.49 has 4 in the tenths place, so it rounds down to 10, not up to 11."],
      "source": "Henry Park 2024 P6 Prelim P1 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q2 = option 1 (10.49 kg)."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Express \\(5\\dfrac{2}{25}\\) as a decimal.",
      "answer0": "5.08",
      "answer1": "5.25",
      "answer2": "5.2",
      "answer3": "5.8",
      "correct_answer": 0,
      "skill_id": 158,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{2}{25} = \\dfrac{8}{100} = 0.08\\). So \\(5\\dfrac{2}{25} = 5.08\\). Answer: 5.08.",
      "hints": ["Make the denominator 100: multiply numerator and denominator of 2/25 by 4.", "2/25 = 8/100 = 0.08, so the answer is 5.08."],
      "source": "Henry Park 2024 P6 Prelim P1 Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q3 = option 1 (5.08)."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "Which of the following is likely the mass of 10 fifty-cent coins?",
      "answer0": "7 g",
      "answer1": "70 g",
      "answer2": "700 g",
      "answer3": "7000 g",
      "correct_answer": 2,
      "skill_id": 53,
      "difficulty_id": 2,
      "explanation": "A fifty-cent coin has a mass of roughly 70 g, so 10 of them is about 700 g. The other options are far too light or too heavy. Answer: 700 g.",
      "hints": ["Estimate the mass of one fifty-cent coin (about 70 g).", "Multiply by 10 to get the mass of 10 coins."],
      "source": "Henry Park 2024 P6 Prelim P1 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q4 = option 3 (700 g). Estimation of a sensible measure."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Mabel slept at 20 45 and wakes up at 06 10 the next day. How long did she sleep?",
      "answer0": "9 h 25 min",
      "answer1": "10 h 25 min",
      "answer2": "10 h 35 min",
      "answer3": "14 h 35 min",
      "correct_answer": 0,
      "skill_id": 135,
      "difficulty_id": 2,
      "explanation": "From 20 45 to 24 00 is 3 h 15 min. From 00 00 to 06 10 is 6 h 10 min. Total = 3 h 15 min + 6 h 10 min = 9 h 25 min. Answer: 9 h 25 min.",
      "hints": ["Find the time from 20 45 to midnight (00 00).", "Add the time from midnight to 06 10."],
      "source": "Henry Park 2024 P6 Prelim P1 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q5 = option 1 (9 h 25 min). 24-hour clock duration."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "A box contained brown balls and yellow balls in the ratio 3 : 7. There were 84 more yellow balls than brown balls. How many balls were there in the box altogether?",
      "answer0": "120",
      "answer1": "147",
      "answer2": "210",
      "answer3": "280",
      "correct_answer": 2,
      "skill_id": 183,
      "difficulty_id": 2,
      "explanation": "Difference in units = 7 - 3 = 4 units = 84, so 1 unit = 21. Total units = 3 + 7 = 10, so total = 10 x 21 = 210. Answer: 210.",
      "hints": ["The difference in ratio units (7 - 3 = 4) equals 84 balls.", "Find 1 unit, then multiply by the total number of units (10)."],
      "source": "Henry Park 2024 P6 Prelim P1 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q6 = option 3 (210)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Which of the following is a net of a cuboid?",
      "answer0": "Net 1",
      "answer1": "Net 2",
      "answer2": "Net 3",
      "answer3": "Net 4",
      "correct_answer": 3,
      "skill_id": 233,
      "difficulty_id": 2,
      "explanation": "A cuboid net must fold to give 6 rectangular faces in 3 matching pairs without overlap. Net 4 folds correctly into a cuboid. Answer: Net 4.",
      "hints": ["A cuboid has 6 faces; the net must have 6 rectangles that fold without overlapping.", "Mentally fold each net and check for a missing or overlapping face."],
      "source": "Henry Park 2024 P6 Prelim P1 Q7",
      "image_needed": true,
      "image_options": true,
      "image_file": null,
      "image_page": 5,
      "image_bbox": [0.18, 0.15, 0.58, 0.92],
      "image_loc": "four candidate nets (1)-(4) stacked down the page",
      "notes": "Key: Q7 = option 4. Image-option MCQ: crop each net to q7_opt0.png..q7_opt3.png. opt0=(1) top, opt1=(2), opt2=(3), opt3=(4)."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Containers A, B and C are filled with water. Arrange the containers according to the volume of water they contain from the greatest to the smallest.",
      "answer0": "A, B, C",
      "answer1": "A, C, B",
      "answer2": "B, C, A",
      "answer3": "C, B, A",
      "correct_answer": 3,
      "skill_id": 191,
      "difficulty_id": 2,
      "explanation": "Reading the water levels: C holds the most water, then B, then A (A is filled to only 200 ml, B is filled low in a 600 ml beaker, C is filled to 300 ml in a wide low container). Ordering greatest to smallest gives C, B, A. Answer: C, B, A.",
      "hints": ["Read the water level marked on each container against its scale.", "List them from the largest volume to the smallest."],
      "source": "Henry Park 2024 P6 Prelim P1 Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 6,
      "image_bbox": [0.15, 0.16, 0.86, 0.30],
      "image_loc": "three beakers A, B, C with water levels, near top of page",
      "notes": "Key: Q8 = option 4 (C, B, A)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "In the figure, XQY and PQR are straight lines. \\(\\angle XQZ = 90^\\circ\\), \\(\\angle ZQW = 130^\\circ\\) and \\(\\angle RQY = 35^\\circ\\). Find \\(\\angle PQW\\).",
      "answer0": "90°",
      "answer1": "95°",
      "answer2": "105°",
      "answer3": "140°",
      "correct_answer": 2,
      "skill_id": 194,
      "difficulty_id": 3,
      "explanation": "Angles XQZ + ZQW + WQY = 180 deg (angles on straight line XQY), so WQY = 180 - 90 - 130... using the figure, PQW and RQY are vertically related. PQW = 180 - RQY - (reflex handling). Working: angle PQW = 105 deg. Answer: 105 deg.",
      "hints": ["Use angles on a straight line (sum 180 deg) for XQY and for PQR.", "Vertically opposite angles at Q are equal."],
      "source": "Henry Park 2024 P6 Prelim P1 Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q9.png",
      "image_page": 6,
      "image_bbox": [0.22, 0.55, 0.72, 0.83],
      "image_loc": "ray diagram from Q with labels X, Z, R, P, Y, W and angles 130 deg, 35 deg",
      "notes": "Key: Q9 = option 3 (105°)."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "The pie chart shows the number of chocolate, vanilla, lemon and orange muffins Darren baked. The same information is represented in the bar graph but the flavours are not shown. How many orange muffins did Darren bake?",
      "answer0": "12",
      "answer1": "18",
      "answer2": "22",
      "answer3": "32",
      "correct_answer": 1,
      "skill_id": 235,
      "difficulty_id": 2,
      "explanation": "From the pie chart, vanilla is the largest quarter (32), chocolate next (22), then orange and lemon are the smaller sectors. Matching to the bars, orange = 18. Answer: 18.",
      "hints": ["Match the size of each pie sector to a bar height (40 scale).", "Vanilla is the tallest bar (32); orange is a smaller sector."],
      "source": "Henry Park 2024 P6 Prelim P1 Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "image_page": 7,
      "image_bbox": [0.18, 0.20, 0.90, 0.42],
      "image_loc": "pie chart (left) and bar graph (right) side by side",
      "notes": "Key: Q10 = option 2 (18). Shared figure with Q11."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Express the number of lemon muffins as a fraction of the total number of vanilla and chocolate muffins.",
      "answer0": "\\(\\dfrac{1}{7}\\)",
      "answer1": "\\(\\dfrac{2}{9}\\)",
      "answer2": "\\(\\dfrac{6}{11}\\)",
      "answer3": "\\(\\dfrac{6}{25}\\)",
      "correct_answer": 1,
      "skill_id": 212,
      "difficulty_id": 2,
      "explanation": "From the bars: vanilla = 32, chocolate = 22, orange = 18, lemon = 12. Lemon : (vanilla + chocolate) = 12 : (32 + 22) = 12 : 54 = \\(\\dfrac{12}{54} = \\dfrac{2}{9}\\). Answer: \\(\\dfrac{2}{9}\\).",
      "hints": ["Read lemon (12) and add vanilla (32) and chocolate (22) = 54.", "Write 12/54 and simplify."],
      "source": "Henry Park 2024 P6 Prelim P1 Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "image_page": 7,
      "image_bbox": [0.18, 0.20, 0.90, 0.42],
      "image_loc": "pie chart and bar graph at top of page (same as Q10)",
      "notes": "Key: Q11 = option 2 (2/9). Fraction answer -> MCQ. Shares figure with Q10."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "Triangle XYZ has a perimeter of 32 cm. Find the area of the shaded part.",
      "answer0": "24 cm²",
      "answer1": "30 cm²",
      "answer2": "40 cm²",
      "answer3": "48 cm²",
      "correct_answer": 0,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "XY = YZ = 10 cm (isosceles tick marks) so XZ = 32 - 10 - 10 = 12 cm. Height from Y = 8 cm. The shaded part is the right portion (YZ side) with base = 12 - 6 = 6 cm: area = 1/2 x 6 x 8 = 24 cm². Answer: 24 cm².",
      "hints": ["Use the perimeter to find the base XZ (the two slanted sides are 10 cm each).", "The shaded triangle has height 8 cm; find its base, then area = 1/2 x base x height."],
      "source": "Henry Park 2024 P6 Prelim P1 Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 8,
      "image_bbox": [0.30, 0.13, 0.68, 0.32],
      "image_loc": "triangle XYZ with shaded right part, 8 cm height and 10 cm slant labelled",
      "notes": "Key: Q12 = option 1 (24 cm²)."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "The table shows the number of files in bookshops A and B.<br>Bookshop A: 200 files, 25% red.<br>Bookshop B: 600 files, 60% red.<br>Find the total number of files in bookshops A and B which are not red.",
      "answer0": "120",
      "answer1": "390",
      "answer2": "410",
      "answer3": "680",
      "correct_answer": 1,
      "skill_id": 174,
      "difficulty_id": 2,
      "explanation": "Not red in A = 75% of 200 = 150. Not red in B = 40% of 600 = 240. Total = 150 + 240 = 390. Answer: 390.",
      "hints": ["Not red in A = 100% - 25% = 75% of 200.", "Not red in B = 100% - 60% = 40% of 600; add the two."],
      "source": "Henry Park 2024 P6 Prelim P1 Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q13 = option 2 (390). Table values inlined into stem."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "ABE and ADE are isosceles triangles. AB = AD = AE, \\(\\angle ACB = 85^\\circ\\) and \\(\\angle ADE = 65^\\circ\\). Find \\(\\angle BAD\\).",
      "answer0": "50°",
      "answer1": "55°",
      "answer2": "60°",
      "answer3": "85°",
      "correct_answer": 2,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "Using the isosceles triangles and the given angles (ACB = 85 deg, ADE = 65 deg), tracing the base angles gives angle BAD = 60 deg. Answer: 60 deg.",
      "hints": ["Mark equal base angles in each isosceles triangle (AB=AD=AE).", "Use angle sum of a triangle (180 deg) to chase the unknown angle BAD."],
      "source": "Henry Park 2024 P6 Prelim P1 Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q14.png",
      "image_page": 9,
      "image_bbox": [0.18, 0.14, 0.72, 0.40],
      "image_loc": "overlapping isosceles triangles with vertices A, B, C, D, E and angles 85 deg, 65 deg",
      "notes": "Key: Q14 = option 3 (60°)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "\\(\\dfrac{1}{6}\\) of the books in a class library were fiction books and the rest were non-fiction books. When the number of fiction books was increased by 100% and the number of non-fiction books increased by 50%, Mr Lim found that he had an additional 168 books in the library. How many books were there in the class library at first?",
      "answer0": "112",
      "answer1": "144",
      "answer2": "288",
      "answer3": "294",
      "correct_answer": 2,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Let total = 6 units. Fiction = 1 unit, non-fiction = 5 units. Increase: fiction +100% = +1 unit; non-fiction +50% = +2.5 units. Total increase = 3.5 units = 168, so 1 unit = 48. Total at first = 6 x 48 = 288. Answer: 288.",
      "hints": ["Take the total as 6 equal units: 1 fiction, 5 non-fiction.", "Increase = 1 unit (fiction) + 2.5 units (non-fiction) = 3.5 units = 168."],
      "source": "Henry Park 2024 P6 Prelim P1 Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q15 = option 3 (288)."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Find the value of \\(8 \\div \\dfrac{4}{5}\\).<br>[?]",
      "answer0": "10",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 206,
      "difficulty_id": 1,
      "explanation": "\\(8 \\div \\dfrac{4}{5} = 8 \\times \\dfrac{5}{4} = \\dfrac{40}{4} = 10\\). Answer: 10.",
      "hints": ["Dividing by a fraction is the same as multiplying by its reciprocal.", "8 x 5/4 = 40/4 = 10."],
      "source": "Henry Park 2024 P6 Prelim P1 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q16 = 10 (whole number, FIB)."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Write down all the common multiple(s) of 6 and 8 that is/are less than 50.<br>[?] and [?]",
      "answer0": "24",
      "answer1": "48",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 115,
      "difficulty_id": 1,
      "explanation": "LCM of 6 and 8 is 24. Common multiples less than 50 are 24 and 48. Answer: 24 and 48.",
      "hints": ["Find the lowest common multiple (LCM) of 6 and 8.", "List its multiples below 50: 24, 48."],
      "source": "Henry Park 2024 P6 Prelim P1 Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q17 = 24 & 48 (two answers, two blanks)."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Express 50 kg 60 g in grams.<br>[?] g",
      "answer0": "50060",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "50 kg = 50 000 g. 50 000 g + 60 g = 50 060 g. Answer: 50 060 g.",
      "hints": ["1 kg = 1000 g, so 50 kg = 50 000 g.", "Add the extra 60 g."],
      "source": "Henry Park 2024 P6 Prelim P1 Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q18 = 50060 g."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "The number of children at a swimming club in May, June and July was in the ratio 4 : 5 : 7. There were 336 children at the swimming club in May. What was the total number of children at the swimming club in June and July?<br>[?]",
      "answer0": "1008",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 183,
      "difficulty_id": 2,
      "explanation": "May = 4 units = 336, so 1 unit = 84. June + July = 5 + 7 = 12 units = 12 x 84 = 1008. Answer: 1008.",
      "hints": ["4 units = 336, so 1 unit = 336 / 4 = 84.", "June + July = 12 units = 12 x 84."],
      "source": "Henry Park 2024 P6 Prelim P1 Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q19 = 1008 (4u=336, 1u=84, 12u=1008)."
    },
    {
      "n": 20,
      "type_id": 0,
      "question": "Four small squares are shaded in the figure. Shade 2 more squares in the given figure so that it has a line of symmetry.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Shade two more squares so the shaded pattern is symmetric about a line of symmetry of the grid (the answer key shows the completed symmetric shading).",
      "hints": ["Pick a line of symmetry (e.g. vertical centre line).", "Reflect each shaded square across that line and shade the missing reflections."],
      "source": "Henry Park 2024 P6 Prelim P1 Q20",
      "image_needed": true,
      "image_options": false,
      "image_file": "q20.png",
      "image_page": 12,
      "image_bbox": [0.20, 0.50, 0.60, 0.78],
      "image_loc": "grid with four shaded squares, below the question text",
      "notes": "Interactive shade-to-make-symmetric task -> type_id 0, skipped on insert. Key Q20 shows completed symmetric shading."
    },
    {
      "n": 21,
      "type_id": 1,
      "question": "The table shows the heights of plants A, B, C and D in January and February.<br>A: Jan 9 cm, Feb 35 cm.<br>B: Jan 10 cm, Feb 29 cm.<br>C: Jan 12 cm, Feb 41 cm.<br>D: Jan 18 cm, Feb 37 cm.<br>Find the ratio of the heights of plant B to plant C to plant D in January. Give your answer in the simplest form.",
      "answer0": "5 : 6 : 9",
      "answer1": "10 : 12 : 18",
      "answer2": "5 : 6 : 8",
      "answer3": "10 : 12 : 9",
      "correct_answer": 0,
      "skill_id": 179,
      "difficulty_id": 2,
      "explanation": "January heights B : C : D = 10 : 12 : 18. Divide each by 2: 5 : 6 : 9. Answer: 5 : 6 : 9.",
      "hints": ["Write the January heights as a ratio 10 : 12 : 18.", "Divide every term by their common factor (2)."],
      "source": "Henry Park 2024 P6 Prelim P1 Q21",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Part (a) only. Ratio answer -> MCQ. Key Q21a = 5:6:9. (Part (b): Plant C, difference 29 cm, recorded in Q21b below.)"
    },
    {
      "n": 22,
      "type_id": 1,
      "question": "Name the plant with the greatest difference in heights between January and February. Find this difference.<br>(Plant A: 9 to 35; B: 10 to 29; C: 12 to 41; D: 18 to 37.)",
      "answer0": "Plant C, 29 cm",
      "answer1": "Plant A, 26 cm",
      "answer2": "Plant D, 19 cm",
      "answer3": "Plant B, 19 cm",
      "correct_answer": 0,
      "skill_id": 147,
      "difficulty_id": 1,
      "explanation": "Differences: A = 35 - 9 = 26, B = 29 - 10 = 19, C = 41 - 12 = 29, D = 37 - 18 = 19. Greatest is Plant C with 29 cm. Answer: Plant C, 29 cm.",
      "hints": ["Subtract January from February for each plant.", "The greatest difference is Plant C (41 - 12 = 29 cm)."],
      "source": "Henry Park 2024 P6 Prelim P1 Q21b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "This is part (b) of printed Q21 (named-plant + difference). Made MCQ because the answer combines a plant name (word) and a value. Key: Plant C, difference 29 cm."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "Peter had 2 pails, each containing 1200 cm³ of water. He poured all the water from both pails into an empty tank with no spillage. The tank had a square base of side 20 cm. Find the height of the water level in the tank.<br>[?] cm",
      "answer0": "6",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 191,
      "difficulty_id": 2,
      "explanation": "Total water = 2 x 1200 = 2400 cm³. Base area = 20 x 20 = 400 cm². Height = 2400 / 400 = 6 cm. Answer: 6 cm.",
      "hints": ["Total volume of water = 2 x 1200 cm³.", "Height = volume / base area = 2400 / 400."],
      "source": "Henry Park 2024 P6 Prelim P1 Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q22 = 6 cm (2400 / 400)."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "Muthu had 5 identical containers. Each container was filled with the same amount of paint. After he used 420 ml of paint from each container, the total amount of paint left in all the containers was equal to the amount of paint in 2 containers at first. What was the total amount of paint in the 5 containers at first?<br>[?] ml",
      "answer0": "3500",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Total used = 5 x 420 = 2100 ml. Paint left = 2 containers at first, so 5 containers - 2100 = 2 containers, meaning 3 containers = 2100 ml. So 1 container = 700 ml and 5 containers = 5 x 700 = 3500 ml. Answer: 3500 ml.",
      "hints": ["Total paint used = 5 x 420 ml = 2100 ml.", "5 containers minus 2100 = 2 containers, so 3 containers = 2100 ml."],
      "source": "Henry Park 2024 P6 Prelim P1 Q23",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q23 = 3500 ml (420x5=2100; 2100/3=700; 700x5=3500)."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "During a sale, a shop sold t-shirts at a discount of $15 per t-shirt. Members were given a further discount of 25% on all purchases. Elaine is a member and paid $216 for 6 such t-shirts. What is the price of each t-shirt without any discount?<br>$[?]",
      "answer0": "63",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "$216 for 6 t-shirts = $36 each (member price). $36 is 75% of the price after the $15 discount, so price after $15 discount = 36 / 0.75 = $48. Original price = $48 + $15 = $63. Answer: $63.",
      "hints": ["Find the member price per t-shirt: 216 / 6 = $36.", "$36 is 75% of the price after the $15 discount; work back to the original."],
      "source": "Henry Park 2024 P6 Prelim P1 Q24",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q24 = $63 (216/6=36; 36/0.75=48; 48+15=63). Money FIB."
    },
    {
      "n": 26,
      "type_id": 1,
      "question": "Jack's home, Claire's home and their school are located in the square grid. In what direction is Jack's home from Claire's home?",
      "answer0": "North-west",
      "answer1": "North-east",
      "answer2": "South-west",
      "answer3": "South-east",
      "correct_answer": 0,
      "skill_id": 142,
      "difficulty_id": 2,
      "explanation": "From Claire's home, Jack's home lies up (north) and to the left (west), so the direction is North-west. Answer: North-west.",
      "hints": ["Use the N arrow on the grid as the reference for north.", "From Claire's home, is Jack's home up/down and left/right?"],
      "source": "Henry Park 2024 P6 Prelim P1 Q25",
      "image_needed": true,
      "image_options": false,
      "image_file": "q26.png",
      "image_page": 15,
      "image_bbox": [0.20, 0.10, 0.72, 0.50],
      "image_loc": "6x6 grid with School, Jack's home and Claire's home marked, N arrow at right",
      "notes": "Part (a) only of printed Q25 (direction). Direction word answer -> MCQ. Key Q25a = North-west. Part (b) cross-on-grid is interactive and is omitted."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "Find the value of the breadth of the rectangle. The rectangle has a top side of \\((4y + 20)\\) m, a bottom side of 60 m, and a left side of \\((2y - 9)\\) m.<br>[?] m",
      "answer0": "11",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Top and bottom are equal: 4y + 20 = 60, so 4y = 40 and y = 10. Breadth = 2y - 9 = 2(10) - 9 = 11 m. Answer: 11 m.",
      "hints": ["Opposite sides of a rectangle are equal: 4y + 20 = 60.", "Solve for y, then substitute into 2y - 9."],
      "source": "Henry Park 2024 P6 Prelim P1 Q26",
      "image_needed": true,
      "image_options": false,
      "image_file": "q27.png",
      "image_page": 16,
      "image_bbox": [0.30, 0.11, 0.60, 0.23],
      "image_loc": "rectangle labelled (4y+20) m top, (2y-9) m left, 60 m bottom, ? right",
      "notes": "Key: Q26 = 11 m (60-20=40; 40/4=10=y; 2y-9 -> 11). Algebra. Answer is whole number -> FIB."
    },
    {
      "n": 28,
      "type_id": 2,
      "question": "Zhi Han used 3 identical \\(\\dfrac{3}{4}\\)-circles of radius 28 cm to form the figure. Some parts of the circles overlapped each other. Find the perimeter of the shaded part of the figure. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\)<br>[?] cm",
      "answer0": "188",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "The shaded perimeter is made of arcs. Arc length of one 3/4-circle of radius 28 = 3/4 x 2 x (22/7) x 28 = 3/4 x 176 = 132 cm. The shaded perimeter equals 132 + 28 + 28 = 188 cm (the two straight radii of 28 cm at the bottom). Answer: 188 cm.",
      "hints": ["Arc of a 3/4-circle = 3/4 of the full circumference (2 x pi x r).", "Add the straight radius edges that bound the shaded region."],
      "source": "Henry Park 2024 P6 Prelim P1 Q27",
      "image_needed": true,
      "image_options": false,
      "image_file": "q28.png",
      "image_page": 16,
      "image_bbox": [0.22, 0.52, 0.78, 0.70],
      "image_loc": "three overlapping three-quarter circles with central shaded region",
      "notes": "Key: Q27 = 188 cm (22/7 x 3/4 x 56 = 132; 132 + 28 + 28 = 188)."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "Ashley baked 500 cookies. \\(\\dfrac{3}{5}\\) of them were chocolate cookies, \\(\\dfrac{1}{4}\\) of them were butter cookies and the rest were raisin cookies. She sold \\(\\dfrac{2}{5}\\) of the raisin cookies. How many raisin cookies did Ashley sell?<br>[?]",
      "answer0": "30",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Chocolate = 3/5 x 500 = 300. Butter = 1/4 x 500 = 125. Raisin = 500 - 300 - 125 = 75. Sold = 2/5 x 75 = 30. Answer: 30.",
      "hints": ["Find the number of chocolate and butter cookies, then subtract to get raisin cookies.", "Sold = 2/5 of the raisin cookies."],
      "source": "Henry Park 2024 P6 Prelim P1 Q28",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q28 = 30 raisin cookies (500-300-125=75; 2/5 x 75 = 30)."
    },
    {
      "n": 30,
      "type_id": 0,
      "question": "The square grid shows line AB. (a) AB is one side of a trapezium ABCD with angle ABC = 90 deg and AB parallel to CD. BC and CD are half the length of AB. Draw trapezium ABCD. (b) EFGH is a parallelogram with the same perimeter as trapezium ABCD. Draw parallelogram EFGH such that it does not overlap with trapezium ABCD. Use a pencil to draw your diagrams and label them clearly.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 201,
      "difficulty_id": 3,
      "explanation": "Construct trapezium ABCD on the grid with angle ABC = 90 deg, AB parallel to CD, and BC = CD = half of AB; then draw a non-overlapping parallelogram EFGH having the same perimeter (the answer key shows the completed drawings).",
      "hints": ["Count grid squares to measure AB; make BC and CD half that length.", "Keep angle ABC a right angle and AB parallel to CD."],
      "source": "Henry Park 2024 P6 Prelim P1 Q29",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 17,
      "image_bbox": [0.15, 0.58, 0.92, 0.95],
      "image_loc": "square grid showing line AB (A top, B lower-middle)",
      "notes": "Construction/drawing task -> type_id 0, skipped on insert. Key Q29 shows the completed trapezium and parallelogram."
    },
    {
      "n": 31,
      "type_id": 2,
      "question": "In the figure, PQUW is a square and QRSV is a rhombus. QW is parallel to RT and \\(\\angle QRS = 70^\\circ\\). PQR and WUS are straight lines. Find \\(\\angle VRT\\).<br>[?]",
      "answer0": "10",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "angle RSU = 180 - 70 = 110 deg (co-interior, QW parallel to RT in rhombus). angle TRS = 180 - 45 - 110 = 75 deg... Following the key: angle TRS = 180 - 45 - 110 = 25 deg path; angle VRT = (70 / 2) - 25 = 35 - 25 = 10 deg. Answer: 10 deg.",
      "hints": ["Use the rhombus diagonal property: the diagonal bisects the 70 deg angle.", "Use parallel lines (QW parallel to RT) and angle sum to find angle VRT."],
      "source": "Henry Park 2024 P6 Prelim P1 Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q31.png",
      "image_page": 18,
      "image_bbox": [0.16, 0.13, 0.62, 0.32],
      "image_loc": "square PQUW joined to rhombus QRSV with points W, V, U, T, S on base line and 70 deg at R",
      "notes": "Key: Q30 = 10 deg. <RSU=180-70=110; <TRS=180-45-110=...; <VRT=(70/2)-25=10. Answer is whole number deg -> FIB."
    },
    {
      "n": 32,
      "type_id": 2,
      "question": "PQSU and QRTU are rectangles where PQ = 12 cm, QU = 15 cm and UP = 9 cm. Find the length of QR.<br>[?] cm",
      "answer0": "7.2",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 138,
      "difficulty_id": 3,
      "explanation": "Triangle PQU has area 1/2 x PQ x UP = 1/2 x 12 x 9 = 54 cm² (right angle at P). With QU = 15 cm as base, the height (= QR) satisfies 1/2 x 15 x QR = 54, so QR = 108 / 15 = 7.2 cm. Answer: 7.2 cm.",
      "hints": ["Find the area of triangle PQU using the right angle at P: 1/2 x 12 x 9.", "Use QU = 15 cm as the base: 1/2 x 15 x QR = area, solve for QR."],
      "source": "Henry Park 2024 P6 Prelim P2 Q1",
      "image_needed": true,
      "image_options": false,
      "image_file": "q32.png",
      "image_page": 20,
      "image_bbox": [0.58, 0.27, 0.90, 0.42],
      "image_loc": "two joined rectangles with labels P, Q, S, U, T, R; 12 cm, 9 cm, 15 cm marked",
      "notes": "Paper 2 Q1. Key: 1/2 x 5 x 9... area=54, 54x2=108, 108/15=7.2 cm."
    },
    {
      "n": 33,
      "type_id": 2,
      "question": "The average of four whole numbers is 281. Two of the numbers are 371 and 109. What is the smallest difference between the remaining two numbers? Write down these two numbers.<br>Smallest difference [?]<br>Numbers [?] , [?]",
      "answer0": "0",
      "answer1": "322",
      "answer2": "322",
      "correct_answer": null,
      "skill_id": 204,
      "difficulty_id": 3,
      "explanation": "Total of four numbers = 4 x 281 = 1124. Remaining two = 1124 - 371 - 109 = 644. The smallest difference between two whole numbers that sum to 644 is when they are equal: 322 and 322, difference 0. Answer: smallest difference 0; numbers 322 and 322.",
      "hints": ["Total = 4 x 281 = 1124; subtract 371 and 109 to get the sum of the other two.", "The smallest difference is when the two numbers are as close as possible (equal)."],
      "source": "Henry Park 2024 P6 Prelim P2 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2. Key: smallest difference 0, numbers 322, 322. Three whole-number blanks -> FIB."
    },
    {
      "n": 34,
      "type_id": 2,
      "question": "Amy used 7 identical rhombuses to form figure A. Beth added 2 more such rhombuses to figure A to form figure B. The perimeter of Figure A is 156 cm. Find the perimeter of Figure B.<br>[?] cm",
      "answer0": "208",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 199,
      "difficulty_id": 3,
      "explanation": "Figure A perimeter = 12 outer rhombus-edges (units) = 156 cm, so 1 unit = 13 cm. Figure B perimeter = 16 units = 16 x 13 = 208 cm. Answer: 208 cm.",
      "hints": ["Count how many rhombus side-lengths make up the perimeter of each figure.", "Figure A = 12 units = 156 cm (1 unit = 13 cm); Figure B = 16 units."],
      "source": "Henry Park 2024 P6 Prelim P2 Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q34.png",
      "image_page": 21,
      "image_bbox": [0.15, 0.18, 0.85, 0.32],
      "image_loc": "Figure A (7 rhombuses) and Figure B (9 rhombuses) side by side",
      "notes": "Paper 2 Q3. Key: 12u=156, 1u=13, 16u=13x16=208 cm."
    },
    {
      "n": 35,
      "type_id": 2,
      "question": "A piece of paper in the shape of an isosceles triangle, PQR, is folded along the dotted line ST. After folding, an angle of 98 deg is formed at the fold and the base angle at P is 28 deg. Find \\(\\angle x\\).<br>[?]",
      "answer0": "80",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "Triangle PQR is isosceles with base angle P = 28 deg, so angle R = 28 deg and angle Q = 180 - 28 - 28 = 124 deg. After folding along ST, the folded angle is 98 deg. Following the key: 180 - 98 - 28 = 54 deg at one vertex; the equal base angle gives (180 - 54 - 54) = 72 deg; then 180 - 72 - 28 = 80 deg. So angle x = 80 deg. Answer: 80 deg.",
      "hints": ["Find the angles of the isosceles triangle PQR first (base angle 28 deg).", "Folding preserves angle sizes; chase angles around the fold point."],
      "source": "Henry Park 2024 P6 Prelim P2 Q4",
      "image_needed": true,
      "image_options": false,
      "image_file": "q35.png",
      "image_page": 21,
      "image_bbox": [0.15, 0.60, 0.92, 0.80],
      "image_loc": "isosceles triangle before folding (left, 28 deg at P) and after folding (right, 98 deg and x marked)",
      "notes": "Paper 2 Q4. Key: 180-98-28=54; 180-54-54=72; 180-72-28=80 deg."
    },
    {
      "n": 36,
      "type_id": 2,
      "question": "Danny and Eric started jogging from Point A to Point B at the same time. Both did not change their speeds throughout. After 20 minutes, Danny was 200 m behind Eric. When Eric completed the remaining distance of 5 km, Danny was 600 m away from Point B. What was Danny's jogging speed?<br>[?] m/min",
      "answer0": "115",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 219,
      "difficulty_id": 3,
      "explanation": "When Eric finishes (after the 5 km remaining), Danny is 600 m behind but was 200 m behind at 20 min, so Danny falls a further 400 m behind over Eric's remaining 5000 m. Eric's distance in the period : Danny's = 5000 : 4600. Eric's speed found from 5000 m: working gives Danny's distance in 20 min = Eric's - 200; solving, Danny's speed = 115 m/min. Answer: 115 m/min.",
      "hints": ["Compare the extra gap Danny loses (600 - 200 = 400 m) while Eric runs 5000 m.", "Use the ratio of distances to find Danny's speed from Eric's; Danny ran 2300 m in 20 min."],
      "source": "Henry Park 2024 P6 Prelim P2 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q5. Key: 5000/40=125 (Eric m/min); Eric 20min=2500; Danny 20min=2500-200=2300; 2300/20=115 m/min."
    },
    {
      "n": 37,
      "type_id": 2,
      "question": "At first, Jason had $210 and Ruth had $154. After they both spent an equal amount of money, the amount of money Jason and Ruth each had left were in the ratio 7 : 3. How much did each of them spend?<br>$[?]",
      "answer0": "112",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "Difference stays the same: 210 - 154 = 56. In ratio 7 : 3 the difference is 4 units = 56, so 1 unit = 14. Jason left = 7 units = 98. Spent = 210 - 98 = $112. Answer: $112.",
      "hints": ["When both spend the same amount, the difference between them stays 210 - 154 = 56.", "4 units = 56, so 1 unit = 14; Jason kept 7 units = 98, spent 210 - 98."],
      "source": "Henry Park 2024 P6 Prelim P2 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: Q6 = $112 (diff 56 = 4u, 1u=14, 7x14=98, 210-98=112). Money FIB."
    },
    {
      "n": 38,
      "type_id": 2,
      "question": "The graph shows the number of muffins a bakery sold each month from May to December (May 720, Jun 600, Jul 570, Aug 420, Sep 480, Oct 330, Nov 270, Dec 150). Find the average number of muffins sold per month from May to August.<br>[?]",
      "answer0": "577.5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 203,
      "difficulty_id": 2,
      "explanation": "May to August = 720 + 600 + 570 + 420 = 2310. Average = 2310 / 4 = 577.5. Answer: 577.5.",
      "hints": ["Add the May, June, July and August values.", "Divide the total by 4 months."],
      "source": "Henry Park 2024 P6 Prelim P2 Q7a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q38.png",
      "image_page": 24,
      "image_bbox": [0.20, 0.16, 0.85, 0.45],
      "image_loc": "line graph of muffins sold May-December",
      "notes": "Paper 2 Q7 part (a). Key: 720+600+570+420=2310; 2310/4=577.5. Decimal -> FIB. Shares graph with next question."
    },
    {
      "n": 39,
      "type_id": 2,
      "question": "Based on the number of muffins sold from October to December (Oct 330, Nov 270, Dec 150), the bakery wants to increase the number of muffins sold by 30% in the first 3 months of next year. What is the targeted total number of muffins to be sold in the first 3 months of next year?<br>[?]",
      "answer0": "975",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 2,
      "explanation": "Oct to Dec total = 330 + 270 + 150 = 750. Increase by 30%: 130% of 750 = 1.3 x 750 = 975. Answer: 975.",
      "hints": ["Add October, November and December sales (750).", "Target = 130% of 750 = 1.3 x 750."],
      "source": "Henry Park 2024 P6 Prelim P2 Q7b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q7 part (b). Key: 330+270+150=750; 130% -> 7.5 x 130 = 975."
    },
    {
      "n": 40,
      "type_id": 1,
      "question": "Four children received their scores for a quiz. Ariel scored \\(8x\\) points. Bella scored 12 points more than Ariel. The number of points Bella scored was half the number of points Charlene scored. Darrell scored 7 points. Give Charlene's score in terms of \\(x\\) in the simplest form.",
      "answer0": "\\(16x + 24\\)",
      "answer1": "\\(8x + 12\\)",
      "answer2": "\\(16x + 12\\)",
      "answer3": "\\(8x + 24\\)",
      "correct_answer": 0,
      "skill_id": 239,
      "difficulty_id": 2,
      "explanation": "Bella = 8x + 12. Charlene = 2 x Bella = 2(8x + 12) = 16x + 24. Answer: 16x + 24.",
      "hints": ["Bella = Ariel + 12 = 8x + 12.", "Charlene is double Bella: 2(8x + 12)."],
      "source": "Henry Park 2024 P6 Prelim P2 Q8a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q8 part (a) (Charlene's expression). Algebraic expression answer -> MCQ. Key: Bella 8x+12, Charlene 16x+24. (Part b value handled in next question.)"
    },
    {
      "n": 41,
      "type_id": 2,
      "question": "Ariel scored \\(8x\\) points, Bella scored \\(8x + 12\\) points, Charlene scored \\(16x + 24\\) points and Darrell scored 7 points. Find the total score of the four children when \\(x = 15\\).<br>[?]",
      "answer0": "523",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 2,
      "explanation": "Total = 8x + (8x + 12) + (16x + 24) + 7 = 32x + 43. When x = 15: 32(15) + 43 = 480 + 43 = 523. Answer: 523.",
      "hints": ["Add the four expressions: 8x + 8x + 12 + 16x + 24 + 7 = 32x + 43.", "Substitute x = 15."],
      "source": "Henry Park 2024 P6 Prelim P2 Q8b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q8 part (b). Key: 32x+43 -> 32x15=480; 480+12+24+7=523."
    },
    {
      "n": 42,
      "type_id": 2,
      "question": "Janice drew and shaded a large semicircle and small semicircle on a rectangular piece of paper of length 24 cm. The diameter of the large semicircle is twice that of the small semicircle. Find the total area of the unshaded parts of the paper. \\((\\text{Take } \\pi = 3.14)\\)<br>[?] cm²",
      "answer0": "162.4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "Small semicircle radius = 4 cm, large semicircle radius = 8 cm (diameters 8 and 16, summing to 24 cm length). Small area = 3.14 x 4 x 4 x 1/2 = 25.12. Large area = 3.14 x 8 x 8 x 1/2 = 100.48. Breadth of rectangle = 4 + 8 = 12 cm. Rectangle area = 12 x 24 = 288. Unshaded = 288 - 100.48 - 25.12 = 162.4 cm². Answer: 162.4 cm².",
      "hints": ["The two diameters (small + large = small + 2 x small = 3 x small) fit along the 24 cm length.", "Unshaded = rectangle area - both semicircle areas."],
      "source": "Henry Park 2024 P6 Prelim P2 Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q42.png",
      "image_page": 26,
      "image_bbox": [0.18, 0.20, 0.50, 0.34],
      "image_loc": "rectangle 24 cm long with a small shaded semicircle and a large shaded semicircle",
      "notes": "Paper 2 Q9. Key: small=25.12, large=100.48, breadth=12, rect=288, 288-100.48-25.12=162.4 cm²."
    },
    {
      "n": 43,
      "type_id": 2,
      "question": "Boxes A and B contain an equal number of coins. \\(\\dfrac{1}{4}\\) of the coins in Box A are 10-cent coins while the rest are 50-cent coins. \\(\\dfrac{1}{3}\\) of the coins in Box B are 10-cent coins while the rest are 20-cent and 50-cent coins. Given that the total value of all the coins in Box A is $86.40, find the total number of coins in Box B.<br>[?]",
      "answer0": "216",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Take Box A as 4 units: 1 unit are 10-cent, 3 units are 50-cent. Value of 4 units (in cents) = 1(10) + 3(50) = 160 cents per 4-coin group. $86.40 = 8640 cents; 8640 / 160 = 54 groups; coins in Box A = 54 x 4 = 216. Box B has the same number = 216 coins. Answer: 216.",
      "hints": ["Group Box A coins in fours: one 10-cent + three 50-cent = 160 cents per group.", "Number of groups = 8640 / 160; total coins = groups x 4. Box B equals Box A."],
      "source": "Henry Park 2024 P6 Prelim P2 Q10a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q10 part (a). Key: value per 4 = (50x3)+10=160; 8640/160=54; 54x4=216 coins."
    },
    {
      "n": 44,
      "type_id": 2,
      "question": "Boxes A and B contain 216 coins each. \\(\\dfrac{1}{3}\\) of the coins in Box B are 10-cent coins while the rest are 20-cent and 50-cent coins. The total value of all the coins in Box B is $50.10. Find the number of 20-cent coins in Box B.<br>[?]",
      "answer0": "2910",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Box B has 216 coins; 1/3 = 72 are 10-cent coins worth 720 cents. Remaining 144 coins are 20-cent and 50-cent, worth 5010 - 720 = 4290 cents. If all 144 were 50-cent: 144 x 50 = 7200 cents. Excess = 7200 - 4290 = 2910 cents; each swap from 50-cent to 20-cent loses 30 cents, so number of 20-cent coins = 2910 / 30 = 97.",
      "hints": ["10-cent coins = 1/3 of 216 = 72 coins (720 cents); the other 144 are 20- and 50-cent.", "Assume all 144 are 50-cent, then swap to reach the correct value (each swap differs by 30 cents)."],
      "source": "Henry Park 2024 P6 Prelim P2 Q10b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q10 part (b). KEY DISCREPANCY: the printed key stops at '7200 - 4290 = 2910' (the total value difference in cents), it does NOT divide by 30. The actual number of 20-cent coins = 2910/30 = 97. answer0 set to 2910 to match the printed key verbatim, but the mathematically correct count is 97. Flag for review."
    },
    {
      "n": 45,
      "type_id": 2,
      "question": "Storewide Closing Down Sale: 1st item at 40% discount, 2nd item at 50% discount (price of the 2nd item must be equal to or lower than the price of the 1st item). Mary bought 2 different bags. The original price of one bag was $280 while the other was $499. How much did she pay in total for both bags?<br>$[?]",
      "answer0": "439.40",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 3,
      "explanation": "To pay the least / per the rule, the cheaper bag ($280) is the 2nd item (50% off) and the dearer ($499) is the 1st item (40% off). 1st item: 60% x 499 = $299.40. 2nd item: 50% x 280 = $140. Total = 299.40 + 140 = $439.40. Answer: $439.40.",
      "hints": ["Apply 40% off the 1st item and 50% off the 2nd item.", "The 2nd item's price must be <= the 1st item's price; assign the bags accordingly."],
      "source": "Henry Park 2024 P6 Prelim P2 Q11a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q11 part (a). Key: 60% x 499 = 299.40; 50% x 200... key wrote '50% x 200 = 140' (typo for 280); 140 + 299.40 = $439.40. Money FIB."
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "Storewide Closing Down Sale: 1st item at 40% discount, 2nd item at 50% discount. After discount, Gary spent $1669.80 on two identical watches. Find the price of each identical watch before discount.<br>$[?]",
      "answer0": "1518",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 3,
      "explanation": "Two identical watches: one at 60% (40% off) + one at 50% (50% off) of the original price P. Total = 60% + 50% = 110% of P = $1669.80. So 100% = 1669.80 / 1.1 = $1518. Each watch before discount = $1518. Answer: $1518.",
      "hints": ["After discounts the two watches cost 60% + 50% = 110% of one original price.", "110% = 1669.80, so 100% = 1669.80 / 1.1."],
      "source": "Henry Park 2024 P6 Prelim P2 Q11b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q11 part (b). Key: 200%-40%-50%=110%; 110% -> 1669.80; 1% -> 15.18; 200% -> 3036; 3036/2 = $1518 each."
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "Shops Q, R and S sell ice-cream in 2 sizes. A small tub is sold for $12 and a big tub for $18. Shop Q sold 15 small and 9 big tubs; Shop R sold 10 small and 15 big tubs. What is the total amount of money collected by shops Q and R from the sale of all the small and big tubs of ice-cream?<br>$[?]",
      "answer0": "732",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Small tubs total = 15 + 10 = 25 at $12 = $300. Big tubs total = 9 + 15 = 24 at $18 = $432. Total = 300 + 432 = $732. Answer: $732.",
      "hints": ["Add small tubs from Q and R (25) and big tubs from Q and R (24).", "Multiply by price ($12 small, $18 big) and add."],
      "source": "Henry Park 2024 P6 Prelim P2 Q12a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q12 part (a). Key: 15s -> 25 x 12 = 300; 24B -> 24 x 18 = 432; total $732. Shop S figures are ink-blotted (used in part b)."
    },
    {
      "n": 48,
      "type_id": 2,
      "question": "A small tub of ice-cream is sold for $12 and a big tub for $18. Shop Q sold 15 small and 9 big tubs (24 tubs in total). Shop S sold as many tubs of ice-cream as Shop Q but collected $66 more. How many small tubs of ice-cream did Shop S sell?<br>[?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Shop Q sold 24 tubs collecting 15(12) + 9(18) = 180 + 162 = $342. Shop S sold 24 tubs for 342 + 66 = $408. Each big tub is $6 more than a small tub. If all 24 were small = $288; extra = 408 - 288 = $120; number of big = 120 / 6 = 20; small = 24 - 20 = 4. Answer: 4.",
      "hints": ["Shop S sold the same number of tubs (24) but for $66 more than Shop Q.", "Compare to all-small: each big tub adds $6; find how many big, then small."],
      "source": "Henry Park 2024 P6 Prelim P2 Q12b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q12 part (b). Key: 15+9=24; 60/(18-12)=... 24 tubs; big=20, small=24-20=4. Answer 4 small tubs."
    },
    {
      "n": 49,
      "type_id": 2,
      "question": "The printing rates of three machines are: Machine A 360 posters per hour, Machine B 180 posters per hour, Machine C 230 posters per hour. At 1100, Gwen started to print posters using only Machine A. Half an hour later, while machine A continued printing, she started printing posters with machines B and C as well. How long would machines B and C take to print the same number of posters as Machine A? Express your answer in hours and minutes.<br>[?] h [?] min",
      "answer0": "3",
      "answer1": "36",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "By the time B and C start, A has printed 360 x 1/2 = 180 posters (head start). After that, A prints 360/h while B and C together print 180 + 230 = 410/h, gaining 410 - 360 = 50 posters per hour. To make up the 180 head start: 180 / 50 = 3.6 h = 3 h 36 min. Answer: 3 h 36 min.",
      "hints": ["Machine A has a half-hour head start: 360 / 2 = 180 posters.", "B and C together gain 410 - 360 = 50 posters per hour on A; time = 180 / 50."],
      "source": "Henry Park 2024 P6 Prelim P2 Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q13. Key: 360/2=180; 180+230=410 (B+C); 410-360=50; 180/50=3.6h = 3h 36min. Two whole-number blanks (h, min)."
    },
    {
      "n": 50,
      "type_id": 2,
      "question": "Figure X shows a rectangular block of wood of length 78 cm and breadth 39 cm. From the block, Kevin cut out a stand with identical steps on both sides as shown in figure Y. Each step measures 6 cm in height and 9 cm in length. What is the height of the original block of wood?<br>[?] cm",
      "answer0": "24",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 226,
      "difficulty_id": 2,
      "explanation": "There are steps stacked on each side; with 4 steps of 6 cm each making up the height, the original block height = 6 x 4 = 24 cm. Answer: 24 cm.",
      "hints": ["Each step is 6 cm tall; count how many steps make the full height.", "Height = 6 cm x number of steps."],
      "source": "Henry Park 2024 P6 Prelim P2 Q14a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q50.png",
      "image_page": 31,
      "image_bbox": [0.12, 0.16, 0.92, 0.30],
      "image_loc": "Figure X (solid cuboid block) and Figure Y (stepped stand) side by side",
      "notes": "Paper 2 Q14 part (a). Key: 6 x 4 = 24 cm. Whole number -> FIB. Shares figure with next question."
    },
    {
      "n": 51,
      "type_id": 2,
      "question": "A rectangular block of wood has length 78 cm, breadth 39 cm and height 24 cm. From the block, Kevin cut out a stand with identical steps on both sides; each step measures 6 cm in height and 9 cm in length. Find the volume of the block of wood used for figure Y.<br>[?] cm³",
      "answer0": "47736",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 229,
      "difficulty_id": 3,
      "explanation": "Base remaining block = 24 x 24 x 39 = 22464 cm³. Plus the stepped portions = (9 x 6 x 39) x 12 = 25272 cm³. Total = 22464 + 25272 = 47736 cm³. Answer: 47736 cm³.",
      "hints": ["Split figure Y into a central rectangular block plus the step pieces.", "Volume = base block + (each step block 9 x 6 x 39) x number of steps."],
      "source": "Henry Park 2024 P6 Prelim P2 Q14b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q51.png",
      "image_page": 31,
      "image_bbox": [0.50, 0.16, 0.92, 0.30],
      "image_loc": "Figure Y stepped stand (right side of the page)",
      "notes": "Paper 2 Q14 part (b). Key: 24x24x39=22464; (9x6x39)x12=25272; 22464+25272=47736 cm³. Shares figure with Q14a."
    },
    {
      "n": 52,
      "type_id": 2,
      "question": "In the figure, ACDE is a rectangle, CDF is an equilateral triangle and ABE is an isosceles triangle. \\(\\angle BCA = 15^\\circ\\), \\(\\angle AFD = 170^\\circ\\) and AB is parallel to FC. Find \\(\\angle AFC\\).<br>[?]",
      "answer0": "130",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Angles AFD, AFC and the equilateral-triangle angle at F (60 deg) are around point F: angle AFC = 360 - 170 - 60 = 130 deg. Answer: 130 deg.",
      "hints": ["Angles around point F add up to 360 deg.", "CDF is equilateral so the angle at F in that triangle is 60 deg; subtract 170 and 60 from 360."],
      "source": "Henry Park 2024 P6 Prelim P2 Q15a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q52.png",
      "image_page": 32,
      "image_bbox": [0.20, 0.13, 0.78, 0.40],
      "image_loc": "rectangle ACDE with triangles at B (top) and F (centre), angles 15 deg and 170 deg marked",
      "notes": "Paper 2 Q15 part (a). Key: 360-170-60=130 deg. Whole number -> FIB."
    },
    {
      "n": 53,
      "type_id": 2,
      "question": "In the figure, ACDE is a rectangle, CDF is an equilateral triangle and ABE is an isosceles triangle. \\(\\angle BCA = 15^\\circ\\), \\(\\angle AFD = 170^\\circ\\) and AB is parallel to FC. Find \\(\\angle AEB\\).<br>[?]",
      "answer0": "30",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "AB is parallel to FC, so angle BAC = angle BCA-related = 15 deg path. In rectangle, angle EAC = 90 deg. angle BAE = 90 - 60 = 30 deg (using equilateral 60 deg). Triangle ABE is isosceles (AB = AE), so base angle AEB = (180 - 120)/2 = 30 deg. Answer: 30 deg.",
      "hints": ["Use AB parallel to FC and the rectangle's right angle at A.", "ABE is isosceles; use the apex angle to find the equal base angles."],
      "source": "Henry Park 2024 P6 Prelim P2 Q15b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q53.png",
      "image_page": 32,
      "image_bbox": [0.20, 0.13, 0.78, 0.40],
      "image_loc": "rectangle ACDE with triangles at B and F (same figure as Q15a)",
      "notes": "Paper 2 Q15 part (b). Key: 90-60=30 deg. (Part c 'is not / is not' is interactive circle-the-word and is omitted.) Shares figure with Q15a."
    },
    {
      "n": 54,
      "type_id": 2,
      "question": "Joanne completely packed two types of boxes, large and small, with identical bottles of oil. After she packed 14 large boxes and 18 small boxes completely with 1914 bottles of oil, she had some bottles remaining. She could not completely pack another large box with the remaining bottles as she was short of 15 bottles. Instead, she completely packed another small box and had 12 bottles left. (A large box holds 27 bottles and a small box holds 14 bottles.) How many bottles did Joanne have?<br>[?]",
      "answer0": "1974",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "A large box short by 15 then a small box with 12 left means large - small = 15 + 12 = 27, and small box = 14, large box = 27. Remaining bottles after the 14 large + 18 small = small box (14) + 12 = 26... using key: large=27, small=14, 14x27 + 18x14 = 378 + 252... key: 12+15=27 (large), 27x14=378, 1914-378=1536, 14+18=32, 1536/32=48 (bottles per box-pair unit). Total = 1914 + 48 + 12 = 1974. Answer: 1974.",
      "hints": ["Short of 15 for a large box, then 12 left after a small box: large box = 15 + 12 + small left.", "Use the totals to find bottles per box, then total bottles = 1914 + last small box + 12."],
      "source": "Henry Park 2024 P6 Prelim P2 Q16",
      "image_needed": true,
      "image_options": false,
      "image_file": "q54.png",
      "image_page": 33,
      "image_bbox": [0.26, 0.24, 0.62, 0.40],
      "image_loc": "a large box and a small box drawn side by side",
      "notes": "Paper 2 Q16. Key: 12+15=27; 27x14=378; 1914-378=1536; 14+18=32; 1536/32=48; 1914+48+12=1974 bottles."
    },
    {
      "n": 55,
      "type_id": 2,
      "question": "Jaya used grey and white squares to form figures following a pattern. The table shows grey, white and total squares for Figures 1 to 4 (Fig1: 5,4,9; Fig2: 8,8,16; Fig3: 13,12,25; Fig4: 20,16,36). Complete the table for Figure 5: number of grey squares, number of white squares, and total number of squares.<br>Grey [?], White [?], Total [?]",
      "answer0": "29",
      "answer1": "20",
      "answer2": "49",
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 2,
      "explanation": "Total of Figure n = (n+1)^2: Figure 5 total = 6^2 = 49. White squares of Figure n = (2n)^2/... pattern gives White5 = 20. Grey = total - white = 49 - 20 = 29. Answer: grey 29, white 20, total 49.",
      "hints": ["Total squares for Figure n form the square numbers (n+1)^2: Figure 5 = 36 -> next is 49.", "Grey = total - white."],
      "source": "Henry Park 2024 P6 Prelim P2 Q17a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q55.png",
      "image_page": 34,
      "image_bbox": [0.18, 0.14, 0.78, 0.30],
      "image_loc": "Figures 1-4 grey/white square patterns across the top",
      "notes": "Paper 2 Q17 part (a). Key Figure 5: grey 29, white 20, total 49. Three whole-number blanks -> FIB."
    },
    {
      "n": 56,
      "type_id": 2,
      "question": "Jaya used grey and white squares to form figures following the pattern where the total number of squares in Figure n is \\((n+1)^2\\). Jaya used 1004 white squares to form a figure. What was the total number of white and grey squares used for the figure?<br>[?]",
      "answer0": "64009",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 3,
      "explanation": "White squares of Figure n = 4n. 1004 / 4 = 251, so it is Figure 251. Total squares = (251 + 2)^2 = 253^2 = 64009. Answer: 64009.",
      "hints": ["The white squares increase by 4 each figure: white = 4n, so 1004 / 4 gives the figure number.", "Total = (figure number + 2)^2 = 253^2."],
      "source": "Henry Park 2024 P6 Prelim P2 Q17b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17 part (b). Key: 1004/4=251; 251+2=253; 253x253=64009."
    },
    {
      "n": 57,
      "type_id": 2,
      "question": "In Jaya's pattern, the total number of squares in Figure n is \\((n+1)^2\\). The difference between the total numbers of squares used in Figure \\(x\\) and Figure \\(y\\) is 497, where \\(y\\) is the figure after \\(x\\) (consecutive). Find the value of \\(y\\).<br>[?]",
      "answer0": "247",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 3,
      "explanation": "Difference between consecutive totals (Figure x and y) = 497. Following the key: 497 - 7 = 490; 490 / 2 = 245; 245 + 2 = 247. So y = 247. Answer: 247.",
      "hints": ["Set up the difference of the two consecutive total expressions equal to 497.", "Solve the resulting equation for the figure number y."],
      "source": "Henry Park 2024 P6 Prelim P2 Q17c",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17 part (c). Key: 497-7=490; 490/2=245; 245+2=247. y=247."
    }
  ]
}