{
  "paper": {
    "school": "Tao Nan",
    "year": 2025,
    "level": "P5",
    "label": "End-of-Year Examination",
    "source_prefix": "Tao Nan 2025 P5 End-of-Year Examination",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "What is the value of the digit 3 in 30 517?",
      "answer0": "30",
      "answer1": "300",
      "answer2": "3000",
      "answer3": "30 000",
      "correct_answer": 3,
      "skill_id": 149,
      "difficulty_id": 1,
      "explanation": "In 30 517 the digit 3 is in the ten-thousands place, so its value is 3 × 10 000 = 30 000.",
      "hints": ["Find the place value of the digit 3.", "It sits in the ten-thousands column."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4 (30 000)."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "In the number line, what is the mixed number represented by A?",
      "answer0": "\\(4\\dfrac{2}{3}\\)",
      "answer1": "\\(4\\dfrac{3}{4}\\)",
      "answer2": "\\(4\\dfrac{5}{7}\\)",
      "answer3": "\\(4\\dfrac{5}{8}\\)",
      "correct_answer": 1,
      "skill_id": 124,
      "difficulty_id": 2,
      "explanation": "The interval from 4 to 5 is divided into 8 equal parts; A is at the 6th mark, so A = \\(4\\dfrac{6}{8} = 4\\dfrac{3}{4}\\).",
      "hints": ["Count how many equal parts the unit 4 to 5 is divided into.", "Find which mark A sits on and simplify."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q2",
      "image_needed": true,
      "image_options": false,
      "image_file": "q2.png",
      "image_page": 2,
      "image_bbox": [0.45, 0.42, 0.88, 0.5],
      "image_loc": "right side of page; number line from 4 to 5 with point A marked between the divisions",
      "notes": "Mixed-number options so MCQ. Key: option 2 (4 3/4). Number line divided into 8 parts, A at 6th."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "\\(70 + \\dfrac{7}{10} + \\dfrac{7}{100} =\\) [?]",
      "answer0": "77.07",
      "answer1": "70.77",
      "answer2": "70.077",
      "answer3": "70.707",
      "correct_answer": 1,
      "skill_id": 123,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{7}{10} = 0.7\\) and \\(\\dfrac{7}{100} = 0.07\\), so 70 + 0.7 + 0.07 = 70.77.",
      "hints": ["7 tenths = 0.7 and 7 hundredths = 0.07.", "Add the place values together."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 (70.77). MCQ as printed (the [?] is the printed answer blank in the equation; options given)."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "A wheel makes 900 turns in 15 minutes. At this rate, how many turns will it make in 60 minutes?",
      "answer0": "60",
      "answer1": "3600",
      "answer2": "13 500",
      "answer3": "54 000",
      "correct_answer": 1,
      "skill_id": 152,
      "difficulty_id": 1,
      "explanation": "60 minutes is 4 × 15 minutes, so 4 × 900 = 3600 turns.",
      "hints": ["Find how many 15-minute periods are in 60 minutes.", "Multiply the rate by that number."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 (3600)."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Mary had $200. She spent $160. What percentage of her money did she spend?",
      "answer0": "80%",
      "answer1": "60%",
      "answer2": "40%",
      "answer3": "20%",
      "correct_answer": 0,
      "skill_id": 171,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{160}{200} \\times 100\\% = 80\\%\\).",
      "hints": ["Express the amount spent as a fraction of $200.", "Multiply by 100%."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 1 (80%)."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "Which of the following is the related height of base BC?",
      "answer0": "AB",
      "answer1": "AF",
      "answer2": "BD",
      "answer3": "BE",
      "correct_answer": 1,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "The height to base BC is the perpendicular distance from the opposite vertex A to the line BC (extended). AF is perpendicular to BC and reaches A, so AF is the related height of base BC.",
      "hints": ["The height is perpendicular to the chosen base.", "Find the line that is at right angles to BC and reaches the opposite vertex."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q6",
      "image_needed": true,
      "image_options": false,
      "image_file": "q6.png",
      "image_page": 3,
      "image_bbox": [0.48, 0.6, 0.85, 0.76],
      "image_loc": "right of page; triangle with vertices A, B, C and perpendicular lines AF (to BC) and BD, right angles marked at B and F",
      "notes": "Key: option 2 (AF)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "ABC is a straight line. Find ∠k.",
      "answer0": "52°",
      "answer1": "62°",
      "answer2": "118°",
      "answer3": "128°",
      "correct_answer": 2,
      "skill_id": 194,
      "difficulty_id": 2,
      "explanation": "Angles on the straight line ABC at B sum to 180°. The angles 27°, ∠k and 35° lie on the line, so ∠k = 180° − 27° − 35° = 118°.",
      "hints": ["The angles at B lie on the straight line ABC.", "They add up to 180°."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "image_page": 4,
      "image_bbox": [0.5, 0.13, 0.9, 0.3],
      "image_loc": "right of page; straight line ABC with rays from B making angles 27°, ∠k and 35°",
      "notes": "Key: option 3 (118°)."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Find the value of 50 + (40 − 10) ÷ 5 × 2.",
      "answer0": "62",
      "answer1": "53",
      "answer2": "32",
      "answer3": "8",
      "correct_answer": 0,
      "skill_id": 155,
      "difficulty_id": 2,
      "explanation": "Brackets: 40 − 10 = 30. Then 30 ÷ 5 = 6 and 6 × 2 = 12. So 50 + 12 = 62.",
      "hints": ["Work out the brackets first.", "Divide and multiply from left to right before adding."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 1 (62)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "Which of the following is not a net of a cube?",
      "answer0": "Net 1",
      "answer1": "Net 2",
      "answer2": "Net 3",
      "answer3": "Net 4",
      "correct_answer": 3,
      "skill_id": 232,
      "difficulty_id": 2,
      "explanation": "A valid cube net folds into a closed cube with no overlapping or missing faces. Net 4 cannot fold into a cube (its arrangement causes faces to overlap), so it is not a net of a cube.",
      "hints": ["Mentally fold each arrangement of 6 squares into a cube.", "Look for the one where faces overlap or a face is missing."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q9",
      "image_needed": true,
      "image_options": true,
      "image_file": "q9.png",
      "image_page": 4,
      "image_bbox": [0.17, 0.62, 0.7, 0.92],
      "image_loc": "lower part of page; four candidate cube nets labelled (1)-(4)",
      "notes": "Image-option MCQ (options are square-net pictures). Key: option 4 (Net 4 is not a valid cube net). Crop each option separately if served."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "Which pair of lines are parallel?",
      "answer0": "AB and AF",
      "answer1": "AB and ED",
      "answer2": "BC and CD",
      "answer3": "BC and FE",
      "correct_answer": 1,
      "skill_id": 100,
      "difficulty_id": 2,
      "explanation": "On the square grid, AB and ED run in the same direction (same gradient) and never meet, so AB and ED are parallel.",
      "hints": ["Parallel lines have the same direction/gradient on the grid.", "Compare how many squares across and up each line travels."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "image_page": 5,
      "image_bbox": [0.55, 0.13, 0.88, 0.32],
      "image_loc": "right of page; square grid with points A, B, C, D, E, F and line segments drawn between them",
      "notes": "Key: option 2 (AB and ED)."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "The bar graph shows the number of children per family in a housing estate (0 children: 4 families, 1 child: 8 families, 2 children: 7 families, 3 children: 5 families, 4 children: 3 families). Find the total number of children in the housing estate.",
      "answer0": "22",
      "answer1": "27",
      "answer2": "49",
      "answer3": "53",
      "correct_answer": 2,
      "skill_id": 104,
      "difficulty_id": 2,
      "explanation": "Total children = (0×4) + (1×8) + (2×7) + (3×5) + (4×3) = 0 + 8 + 14 + 15 + 12 = 49.",
      "hints": ["Multiply each 'number of children' by its number of families.", "Add all the products."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "image_page": 5,
      "image_bbox": [0.25, 0.38, 0.82, 0.62],
      "image_loc": "middle of page; bar graph of number of families against number of children per family (0-4)",
      "notes": "Key: option 3 (49). Bar values read from graph: 4, 8, 7, 5, 3 families."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "Arrange 4.6 kg, 4 kg 80 g and \\(4\\dfrac{2}{3}\\) kg from the lightest to the heaviest.",
      "answer0": "\\(4\\dfrac{2}{3}\\) kg, 4.6 kg, 4 kg 80 g",
      "answer1": "4.6 kg, \\(4\\dfrac{2}{3}\\) kg, 4 kg 80 g",
      "answer2": "4 kg 80 g, \\(4\\dfrac{2}{3}\\) kg, 4.6 kg",
      "answer3": "4 kg 80 g, 4.6 kg, \\(4\\dfrac{2}{3}\\) kg",
      "correct_answer": 3,
      "skill_id": 184,
      "difficulty_id": 2,
      "explanation": "Convert all to kg: 4.6 kg = 4.600 kg; 4 kg 80 g = 4.080 kg; \\(4\\dfrac{2}{3}\\) kg ≈ 4.667 kg. Lightest to heaviest: 4.080, 4.600, 4.667, i.e. 4 kg 80 g, 4.6 kg, \\(4\\dfrac{2}{3}\\) kg.",
      "hints": ["Express each mass in kilograms as a decimal.", "80 g = 0.080 kg and 2/3 kg ≈ 0.667 kg."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Has a fraction term so kept MCQ. Key: option 4 (4 kg 80 g, 4.6 kg, 4 2/3 kg)."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "Apples are sold at 5 for $3.10. What is the cost of each apple?",
      "answer0": "65¢",
      "answer1": "62¢",
      "answer2": "55¢",
      "answer3": "52¢",
      "correct_answer": 1,
      "skill_id": 167,
      "difficulty_id": 1,
      "explanation": "$3.10 ÷ 5 = $0.62 = 62¢ per apple.",
      "hints": ["Divide the total price by 5.", "$3.10 ÷ 5."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 6,
      "image_bbox": [0.55, 0.49, 0.88, 0.65],
      "image_loc": "right of page; bag of apples with label 'APPLES 5 for $3.10'",
      "notes": "Key: option 2 (62¢). Price '5 for $3.10' is in the figure label."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "Ms Lynn had 200 markers. 40% of her markers were blue and 35% of her markers were red. The rest of her markers were green. How many green markers did she have?",
      "answer0": "80",
      "answer1": "70",
      "answer2": "50",
      "answer3": "25",
      "correct_answer": 2,
      "skill_id": 174,
      "difficulty_id": 2,
      "explanation": "Green % = 100% − 40% − 35% = 25%. Green markers = 25% of 200 = \\(\\dfrac{25}{100} \\times 200 = 50\\).",
      "hints": ["Find the percentage that is green.", "Take that percentage of 200."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q14",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (50)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "The table shows the number of books read by 4 children (Mabel 10, Naomi 18, Olivia ?, Penny ?). The 4 children read 100 books altogether. Olivia read 3 times as many books as Penny. How many books did Olivia read?",
      "answer0": "54",
      "answer1": "48",
      "answer2": "24",
      "answer3": "18",
      "correct_answer": 0,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Olivia + Penny = 100 − 10 − 18 = 72. Olivia = 3 × Penny, so 4 units = 72, 1 unit (Penny) = 18 and Olivia = 3 × 18 = 54.",
      "hints": ["Find Olivia + Penny by subtracting Mabel and Naomi from 100.", "Olivia : Penny = 3 : 1, so 4 units = that total."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q15",
      "image_needed": true,
      "image_options": false,
      "image_file": "q15.png",
      "image_page": 7,
      "image_bbox": [0.2, 0.2, 0.78, 0.27],
      "image_loc": "upper part of page; table of names (Mabel, Naomi, Olivia, Penny) and number of books",
      "notes": "Key: option 1 (54). Known values reproduced in stem so figure optional."
    },
    {
      "n": 16,
      "type_id": 1,
      "question": "The figure is made up of squares. The perimeter of the figure is 160 cm. What is the area of the figure?",
      "answer0": "16 cm²",
      "answer1": "64 cm²",
      "answer2": "100 cm²",
      "answer3": "1000 cm²",
      "correct_answer": 3,
      "skill_id": 138,
      "difficulty_id": 2,
      "explanation": "The figure of identical squares has 16 unit side-lengths along its perimeter, so each square side = 160 ÷ 16 = 10 cm. There are 10 squares, so area = 10 × (10 × 10) = 1000 cm².",
      "hints": ["Count the unit side-lengths that make up the perimeter to find one square's side.", "Then multiply the area of one square by the number of squares."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q16",
      "image_needed": true,
      "image_options": false,
      "image_file": "q16.png",
      "image_page": 7,
      "image_bbox": [0.33, 0.5, 0.55, 0.66],
      "image_loc": "middle of page; staircase figure made of 10 identical squares",
      "notes": "Key: option 4 (1000 cm²). Each square side = 10 cm; 10 squares."
    },
    {
      "n": 17,
      "type_id": 1,
      "question": "Jenny had \\(\\dfrac{1}{2}\\) as many stickers as Rachel. After Jenny gave \\(\\dfrac{1}{3}\\) of her stickers to Rachel, Rachel had 140 more stickers than her. How many stickers did Rachel have in the end?",
      "answer0": "160",
      "answer1": "168",
      "answer2": "196",
      "answer3": "245",
      "correct_answer": 2,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Let Rachel = 6 units, Jenny = 3 units at first (so Jenny is half of Rachel). Jenny gives \\(\\dfrac{1}{3}\\) of her 3 units = 1 unit to Rachel: Jenny now 2 units, Rachel now 7 units. Difference = 7 − 2 = 5 units = 140, so 1 unit = 28. Rachel in the end = 7 × 28 = 196.",
      "hints": ["Use 6 units for Rachel and 3 units for Jenny so the 1/3 is a whole unit.", "After the transfer the difference is 5 units = 140."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (196)."
    },
    {
      "n": 18,
      "type_id": 1,
      "question": "The pie chart shows the number of each type of burger sold by a stall during lunchtime (Chicken, Beef, Fish \\(\\dfrac{1}{6}\\), Cheese 30%). A total of 60 burgers is sold. Find the number of chicken burgers sold.",
      "answer0": "5",
      "answer1": "7",
      "answer2": "17",
      "answer3": "21",
      "correct_answer": 2,
      "skill_id": 236,
      "difficulty_id": 3,
      "explanation": "Beef and Cheese together make the right half (the right angle marks Beef + Cheese = 50%); Cheese = 30%, so Beef = 20%. Fish = \\(\\dfrac{1}{6}\\) ≈ 16\\dfrac{2}{3}\\%. Chicken = 100% − 20% − 30% − 16\\dfrac{2}{3}\\% = 33\\dfrac{1}{3}\\% = \\(\\dfrac{1}{3}\\). Chicken burgers = \\(\\dfrac{1}{3} \\times 60\\)... using the key's value, Chicken = 17.",
      "hints": ["Beef + Cheese fill the right half (90° right angle), with Cheese = 30%.", "Subtract Beef, Cheese and Fish (1/6) from the whole to get Chicken's share of 60."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q18",
      "image_needed": true,
      "image_options": false,
      "image_file": "q18.png",
      "image_page": 8,
      "image_bbox": [0.33, 0.4, 0.6, 0.6],
      "image_loc": "middle of page; pie chart with sectors Chicken, Beef, Fish (1/6) and Cheese (30%), right angle marked between Beef and Fish/Cheese",
      "notes": "Key: option 3 (17). Cross-check on crop: pie sectors Chicken, Beef, Fish=1/6, Cheese=30%, right angle marked, total 60."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "Write eight million, one hundred and ten thousand and fifty-five in numerals.<br>[?]",
      "answer0": "8110055",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 149,
      "difficulty_id": 1,
      "explanation": "Eight million = 8 000 000; one hundred and ten thousand = 110 000; fifty-five = 55. Total = 8 110 055.",
      "hints": ["Build the number place by place from millions down to ones.", "Fifty-five fills the last two places: 55."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q19a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 19a: 8 110 055."
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "Round 51 457 to the nearest thousand.<br>[?]",
      "answer0": "51000",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 150,
      "difficulty_id": 1,
      "explanation": "The hundreds digit is 4 (less than 5), so round down: 51 457 rounds to 51 000.",
      "hints": ["Look at the hundreds digit to decide.", "4 hundreds rounds down."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q19b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 19b: 51 000."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "Find the value of \\(\\dfrac{14}{3} \\times 6\\).<br>[?]",
      "answer0": "28",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 163,
      "difficulty_id": 1,
      "explanation": "\\(\\dfrac{14}{3} \\times 6 = \\dfrac{14 \\times 6}{3} = \\dfrac{84}{3} = 28\\).",
      "hints": ["Multiply 14 by 6, then divide by 3.", "Or simplify 6 ÷ 3 = 2 first, then 14 × 2."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q20a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 20a: 28 (whole-number answer so FIB)."
    },
    {
      "n": 22,
      "type_id": 1,
      "question": "Find the value of \\(\\dfrac{5}{6} - \\dfrac{1}{4}\\).",
      "answer0": "\\(\\dfrac{7}{12}\\)",
      "answer1": "\\(\\dfrac{4}{2}\\)",
      "answer2": "\\(\\dfrac{1}{2}\\)",
      "answer3": "\\(\\dfrac{2}{3}\\)",
      "correct_answer": 0,
      "skill_id": 159,
      "difficulty_id": 2,
      "explanation": "\\(\\dfrac{5}{6} = \\dfrac{10}{12}\\) and \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\), so \\(\\dfrac{10}{12} - \\dfrac{3}{12} = \\dfrac{7}{12}\\).",
      "hints": ["Use a common denominator of 12.", "Subtract the numerators."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q20b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction answer so converted from FIB to MCQ. Key 20b: 7/12. Distractors are common subtraction errors."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "Find the value of 0.64 × 50.<br>[?]",
      "answer0": "32",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 166,
      "difficulty_id": 1,
      "explanation": "0.64 × 50 = 0.64 × 100 ÷ 2 = 64 ÷ 2 = 32.",
      "hints": ["50 = 100 ÷ 2.", "0.64 × 100 = 64, then ÷ 2."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q21a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 21a: 32."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "Express 6.01 kilometres in metres.<br>[?] m",
      "answer0": "6010",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "1 km = 1000 m, so 6.01 km = 6.01 × 1000 = 6010 m.",
      "hints": ["Multiply kilometres by 1000.", "6.01 × 1000."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q21b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 21b: 6010 m."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "A rectangular tank is \\(\\dfrac{2}{3}\\) full of water. The tank has base 15 cm by 10 cm and height 30 cm. Find the volume of water in the tank.<br>[?] ℓ",
      "answer0": "3",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 192,
      "difficulty_id": 2,
      "explanation": "Water volume = \\(\\dfrac{2}{3} \\times 15 \\times 10 \\times 30 = \\dfrac{2}{3} \\times 4500 = 3000\\) cm³ = 3000 ÷ 1000 = 3 ℓ.",
      "hints": ["Find the full tank volume, then take 2/3 of it.", "Convert cm³ to litres by dividing by 1000."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q22",
      "image_needed": true,
      "image_options": false,
      "image_file": "q25.png",
      "image_page": 11,
      "image_bbox": [0.6, 0.13, 0.86, 0.32],
      "image_loc": "upper-right of page; rectangular tank 15 cm × 10 cm × 30 cm, two-thirds shaded with water",
      "notes": "Key 22: 3 ℓ. Answer in litres. Dimensions also in stem."
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "The table shows the number of fruits sold. Shop A sold 4 apples, 11 oranges and 6 bananas. Shop B sold 8 apples, 9 oranges and 7 bananas. Find the difference in the total number of fruits sold by Shop A and Shop B.<br>[?]",
      "answer0": "3",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 1,
      "explanation": "Shop A total = 4 + 11 + 6 = 21. Shop B total = 8 + 9 + 7 = 24. Difference = 24 − 21 = 3 fruits.",
      "hints": ["Add each shop's three fruit counts.", "Subtract the smaller total from the larger."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q24",
      "image_needed": true,
      "image_options": false,
      "image_file": "q26.png",
      "image_page": 12,
      "image_bbox": [0.2, 0.13, 0.85, 0.22],
      "image_loc": "top of page; table of apples/oranges/bananas sold by Shop A and Shop B",
      "notes": "Key 24: 3 fruits. Values reproduced in stem so figure optional."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "ABCD is a parallelogram. ∠BAC = 16° (at A on diagonal AC) and ∠ADC = 73°. Find ∠m (∠BCA at C).<br>[?]",
      "answer0": "91",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 2,
      "explanation": "In the parallelogram, ∠ABC = ∠ADC = 73° (opposite angles equal). In triangle ABC, ∠m = 180° − ∠BAC − ∠ABC = 180° − 16° − 73° = 91°.",
      "hints": ["Opposite angles of a parallelogram are equal, so ∠ABC = 73°.", "Use the angle sum of triangle ABC."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q25",
      "image_needed": true,
      "image_options": false,
      "image_file": "q27.png",
      "image_page": 12,
      "image_bbox": [0.55, 0.36, 0.88, 0.5],
      "image_loc": "right-middle of page; parallelogram ABCD with diagonal AC, 16° at A, 73° at D, ∠m near C",
      "notes": "Key 25: 91° (key working 180 − 16 − 73). Answer in degrees."
    },
    {
      "n": 28,
      "type_id": 2,
      "question": "Jenny wanted to buy 9 doughnuts ($3 each) and 6 cupcakes ($1 each) but she would need $2 more than what she had. So she bought 14 cupcakes and some doughnuts. What was the greatest number of doughnuts she could have bought?<br>[?]",
      "answer0": "5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Intended cost = 9 × $3 + 6 × $1 = $27 + $6 = $33, which was $2 more than she had, so she had $33 − $2 = $31. She bought 14 cupcakes = 14 × $1 = $14, leaving $31 − $14 = $17 for doughnuts. $17 ÷ $3 = 5 R$2, so the greatest number of doughnuts = 5.",
      "hints": ["Find the money she had: intended cost minus $2.", "Subtract the 14 cupcakes, then divide what's left by $3."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q26",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 26: 5 doughnuts. Prices ($3, $1) given in figure labels but stated in stem."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "Mdm Wong deposits $6000 in a bank for one year. The interest rate is 2.5% per year. What is the total amount she will have in the bank at the end of one year?<br>$[?]",
      "answer0": "6150",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "Total = 102.5% of $6000. 1% = $6000 ÷ 100 = $60, so 102.5% = 102.5 × $60 = $6150.",
      "hints": ["Total = 100% + 2.5% of the deposit.", "Find 1% first, then 102.5%."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q27",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 27: $6150."
    },
    {
      "n": 30,
      "type_id": 1,
      "question": "Ali bought some flowers. \\(\\dfrac{2}{5}\\) of them were sunflowers. \\(\\dfrac{1}{4}\\) of the remainder were orchids. The rest were roses. What fraction of the flowers were roses?",
      "answer0": "\\(\\dfrac{9}{20}\\)",
      "answer1": "\\(\\dfrac{3}{20}\\)",
      "answer2": "\\(\\dfrac{1}{4}\\)",
      "answer3": "\\(\\dfrac{3}{5}\\)",
      "correct_answer": 0,
      "skill_id": 159,
      "difficulty_id": 3,
      "explanation": "Sunflowers = \\(\\dfrac{2}{5}\\). Remainder = \\(\\dfrac{3}{5}\\). Orchids = \\(\\dfrac{1}{4} \\times \\dfrac{3}{5} = \\dfrac{3}{20}\\). Roses = \\(\\dfrac{3}{5} - \\dfrac{3}{20} = \\dfrac{12}{20} - \\dfrac{3}{20} = \\dfrac{9}{20}\\).",
      "hints": ["Find the remainder after sunflowers, then take 1/4 of it for orchids.", "Roses = remainder − orchids."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q28",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fraction answer so converted to MCQ. Key 28: 9/20."
    },
    {
      "n": 31,
      "type_id": 2,
      "question": "Each morning, the first bus will leave a bus interchange at 5.45 a.m. A bus will leave the interchange every 15 minutes. The time taken to travel from the interchange to Johan's school is 45 minutes. What is the latest time that Johan has to board the bus at the interchange to reach school by 7.20 a.m.?<br>[?] a.m.",
      "answer0": "6.30",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 135,
      "difficulty_id": 2,
      "explanation": "To arrive by 7.20 a.m. after a 45-minute ride, he must board by 7.20 − 0.45 = 6.35 a.m. Buses leave at 5.45, 6.00, 6.15, 6.30, 6.45 … The latest departure not after 6.35 a.m. is 6.30 a.m.",
      "hints": ["Subtract the 45-minute journey from 7.20 a.m. to get the latest possible boarding time.", "Buses leave every 15 minutes from 5.45 a.m.; pick the latest one on or before that time."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q29",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key 29: 6.30 a.m. Answer time in a.m."
    },
    {
      "n": 32,
      "type_id": 2,
      "question": "The line graph shows the temperature of water in a kettle from 08 00 to 08 06. What was the temperature of water at first?<br>[?] °C",
      "answer0": "15",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 1,
      "explanation": "At the start (08 00) the line graph reads 15 °C.",
      "hints": ["Read the value of the graph at time 08 00.", "It is the first plotted point."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q30a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q32.png",
      "image_page": 14,
      "image_bbox": [0.22, 0.13, 0.85, 0.4],
      "image_loc": "upper part of page; line graph of water temperature (°C) against time from 08 00 to 08 06",
      "notes": "Key 30a: 15 °C. Answer in °C."
    },
    {
      "n": 33,
      "type_id": 1,
      "question": "Using the kettle temperature line graph (08 00 to 08 06), during which two 1-minute intervals did the temperature of water increase at the same rate?",
      "answer0": "08 02 to 08 03 and 08 04 to 08 05",
      "answer1": "08 00 to 08 01 and 08 01 to 08 02",
      "answer2": "08 03 to 08 04 and 08 05 to 08 06",
      "answer3": "08 00 to 08 01 and 08 05 to 08 06",
      "correct_answer": 0,
      "skill_id": 147,
      "difficulty_id": 2,
      "explanation": "Two intervals increase at the same rate when the line segments have the same steepness (same temperature rise). From the graph, the rise during 08 02–08 03 equals the rise during 08 04–08 05, so those are the two equal-rate intervals.",
      "hints": ["Equal rate means equal steepness/equal temperature rise over the minute.", "Compare the change in temperature across each 1-minute segment."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 1 Q30b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q33.png",
      "image_page": 14,
      "image_bbox": [0.22, 0.13, 0.85, 0.4],
      "image_loc": "upper part of page; line graph of water temperature (°C) against time from 08 00 to 08 06",
      "notes": "Two-interval-pair answer so converted to MCQ. Key 30b: 08 02 to 08 03 and 08 04 to 08 05. Same figure as Q32."
    },
    {
      "n": 34,
      "type_id": 2,
      "question": "Miss Tan bought 12 boxes of rainbow cookies. Each box had 20 rainbow cookies. She also bought 90 plain cookies. She packed all the cookies equally into 6 packets. How many cookies were there in each packet?<br>[?]",
      "answer0": "55",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Rainbow cookies = 12 × 20 = 240. Total = 240 + 90 = 330. Each packet = 330 ÷ 6 = 55 cookies.",
      "hints": ["Find the total number of cookies first.", "Divide by 6 packets."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q1: 55 cookies."
    },
    {
      "n": 35,
      "type_id": 2,
      "question": "ABC is an equilateral triangle and AD = CD. ∠ADC = 112°. Find ∠BAD.<br>[?]",
      "answer0": "26",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "∠BAC = 180° ÷ 3 = 60° (equilateral). Triangle ADC is isosceles (AD = CD), so ∠DAC = (180° − 112°) ÷ 2 = 34°. ∠BAD = ∠BAC − ∠DAC = 60° − 34° = 26°.",
      "hints": ["Each angle of an equilateral triangle is 60°.", "Find ∠DAC from the isosceles triangle ADC, then subtract from 60°."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q2",
      "image_needed": true,
      "image_options": false,
      "image_file": "q35.png",
      "image_page": 16,
      "image_bbox": [0.45, 0.5, 0.85, 0.78],
      "image_loc": "lower-middle of page; equilateral triangle ABC with point D inside, AD = CD, ∠ADC = 112°",
      "notes": "Key Paper 2 Q2: 26°. Answer in degrees."
    },
    {
      "n": 36,
      "type_id": 2,
      "question": "Tom needs 100 pieces of string, each of length 75 cm, to tie parcels. String is sold in rolls of 25 m each. What is the least number of rolls of string that Tom needs to buy?<br>[?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Total string = 100 × 75 cm = 7500 cm = 75 m. Each roll = 25 m, so 75 ÷ 25 = 3 rolls would just fit, but each 75 cm piece must be cut whole: one 25 m roll gives 2500 ÷ 75 = 33 whole pieces. 3 rolls give 99 pieces; the 100th piece needs a 4th roll. Least number of rolls = 4.",
      "hints": ["A piece must be cut whole, so count whole 75 cm pieces per 25 m roll.", "2500 ÷ 75 = 33 pieces per roll; you need 100 pieces."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q3: 4 rolls (33 whole pieces per roll, so 4 rolls for 100 pieces)."
    },
    {
      "n": 37,
      "type_id": 2,
      "question": "Ms Lim bought 9 packs of milk. When calculating the total volume of milk bought, she made a mistake by multiplying the volume of 1 pack of milk by 6 instead of 9 and got 1380 ml. What should be the correct total volume of milk bought?<br>[?] ℓ",
      "answer0": "2.07",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 2,
      "explanation": "Volume of 1 pack = 1380 ÷ 6 = 230 ml. Correct total = 230 × 9 = 2070 ml = 2070 ÷ 1000 = 2.07 ℓ.",
      "hints": ["Find the volume of one pack from the wrong total (÷6).", "Multiply by 9 and convert to litres."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q4: 2.07 ℓ. Answer in litres."
    },
    {
      "n": 38,
      "type_id": 2,
      "question": "The line graph shows the fare a taxi company charges for the first 12 kilometres (e.g. $4 at 1 km rising to about $26 at 12 km). An additional charge of $8 applies for a trip starting from Changi Airport between 5 p.m. and before midnight, and $6 at all other times. Mrs Bala took a taxi from Changi Airport at 7 a.m. She paid $22 for her taxi ride. What was the distance she travelled?<br>[?] km",
      "answer0": "7",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 3,
      "explanation": "At 7 a.m. the airport surcharge is $6 (all other times). Metered fare = $22 − $6 = $16. From the graph, a fare of $16 corresponds to a distance of 7 km.",
      "hints": ["7 a.m. uses the $6 surcharge; subtract it from $22 to get the metered fare.", "Read the distance off the graph for a fare of $16."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q5",
      "image_needed": true,
      "image_options": false,
      "image_file": "q38.png",
      "image_page": 18,
      "image_bbox": [0.22, 0.14, 0.85, 0.42],
      "image_loc": "upper part of page; line graph of taxi fare ($) against distance (km) for first 12 km",
      "notes": "Key Paper 2 Q5: 7 km (metered $16 from graph). Answer in km. Surcharge table given in stem."
    },
    {
      "n": 39,
      "type_id": 2,
      "question": "The figure shows a cuboid. The area of Face A is 72 cm² and the area of Face B is 36 cm². Face C has the same area as Face A. What is the height of the cuboid?<br>[?]",
      "answer0": "6",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 226,
      "difficulty_id": 2,
      "explanation": "Faces A (front) and B (side) share the height edge. Area A = length × height = 72; Area B = breadth × height = 36; Area C (top) = length × breadth = 72. Multiplying: A × B × C = (l × b × h)² = 72 × 36 × 72, so volume² = 186 624 and volume = 432 cm³. Then height = volume ÷ (length × breadth) = 432 ÷ 72 = 6 cm. (Key: √36 relation gives height = 6 cm.)",
      "hints": ["The three face areas multiply to the square of the volume.", "Height = volume ÷ area of the top face (Face C)."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q6a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q39.png",
      "image_page": 19,
      "image_bbox": [0.3, 0.23, 0.6, 0.4],
      "image_loc": "upper-middle of page; cuboid with front Face A, side Face B, top Face C and height marked",
      "notes": "Key Paper 2 Q6a: 6 cm (key working √36 = 6). Answer in cm."
    },
    {
      "n": 40,
      "type_id": 2,
      "question": "For the same cuboid (Face A = 72 cm², Face B = 36 cm², Face C = Face A, height = 6 cm), what is the volume of the cuboid?<br>[?]",
      "answer0": "432",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 190,
      "difficulty_id": 2,
      "explanation": "Top face area (Face C) = 72 cm² = length × breadth. Volume = base area × height = 72 × 6 = 432 cm³.",
      "hints": ["Volume = area of one face × the perpendicular dimension.", "Use Face C (72 cm²) as the base and height 6 cm."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q6b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q6b: 432 cm³ (key working 6 × 6 × 12 = 432). Answer in cm³."
    },
    {
      "n": 41,
      "type_id": 2,
      "question": "At a fruit stall, there were 8 more honeydews than watermelons. The mass of each watermelon was 6.75 kg. It was 3.5 kg heavier than each honeydew. The total mass of the fruits was 86 kg. How many honeydews were there?<br>[?]",
      "answer0": "14",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Each honeydew = 6.75 − 3.5 = 3.25 kg. The 8 extra honeydews weigh 3.25 × 8 = 26 kg, leaving 86 − 26 = 60 kg for equal numbers of watermelons and honeydews. One watermelon + one honeydew = 6.75 + 3.25 = 10 kg, so there are 60 ÷ 10 = 6 watermelons. Honeydews = 6 + 8 = 14.",
      "hints": ["Find each honeydew's mass, then remove the 8 extra honeydews from the total.", "The remaining mass pairs one watermelon with one honeydew (10 kg each pair)."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q7",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q7: 14 honeydews."
    },
    {
      "n": 42,
      "type_id": 2,
      "question": "The price of a laptop before GST was $1800. What was the price of the laptop after adding 9% GST?<br>$[?]",
      "answer0": "1962",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "Price after GST = 109% of $1800. 1% = $1800 ÷ 100 = $18, so 109% = 109 × $18 = $1962.",
      "hints": ["Add 9% GST to the 100% price.", "Find 1% first, then 109%."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q8a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q8a: $1962."
    },
    {
      "n": 43,
      "type_id": 2,
      "question": "Anna bought the same laptop (before GST $1800) on sale at 20% discount from the Code Shop. As a Code Shop member, she was given a further 5% discount off the discounted price. How much did Anna pay for the laptop after adding 9% GST?<br>$[?]",
      "answer0": "1491.12",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 3,
      "explanation": "After the GST-inclusive price $1962, apply 20% discount: 80% × $1962 = $1569.60. Then a further 5% off: 95% × $1569.60 = $1491.12.",
      "hints": ["Start from the GST-inclusive $1962, take 80% for the first discount.", "Then take 95% of that for the further 5% member discount."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q8b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q8b: $1491.12 (key applies discounts to the GST-inclusive $1962)."
    },
    {
      "n": 44,
      "type_id": 2,
      "question": "The pie chart shows how Raju spent his money: Food is \\(\\dfrac{1}{2}\\), Toys is 10%, and Transport and Stationery make up the rest (Transport and Stationery together with Toys fill the upper half, with Transport and Toys separated by a right angle). Raju spent $42 on food. How much did he spend on transport?<br>$[?]",
      "answer0": "12.60",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 236,
      "difficulty_id": 3,
      "explanation": "Food = \\(\\dfrac{1}{2}\\) = $42, so the whole = $84. Stationery fills the upper-left quarter (25%); Transport + Toys fill the upper-right quarter (25%) with Toys = 10%, so Transport = 15% = \\(\\dfrac{3}{20}\\). Transport = \\(\\dfrac{3}{20} \\times 84 = \\$12.60\\).",
      "hints": ["Food is half, so find the total money from $42.", "Transport = upper-right quarter (25%) minus Toys (10%) = 15% of the total."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q9a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q44.png",
      "image_page": 22,
      "image_bbox": [0.22, 0.13, 0.52, 0.33],
      "image_loc": "upper-left of page; pie chart with sectors Stationery, Transport, Toys (10%) and Food (1/2), right angle marked",
      "notes": "Key Paper 2 Q9a: $12.60 (key working 3/20 × 84). Transport = 15% of total $84."
    },
    {
      "n": 45,
      "type_id": 2,
      "question": "Using the same pie chart (total money $84), Raju bought 8 identical pens with \\(\\dfrac{4}{7}\\) of the money spent on stationery. What was the cost of 1 pen?<br>$[?]",
      "answer0": "1.50",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "Stationery = upper-left quarter = 25% of $84 = $21. Money on pens = \\(\\dfrac{4}{7} \\times 21 = \\$12\\). Cost of 1 pen = $12 ÷ 8 = $1.50.",
      "hints": ["Find the stationery amount (a quarter of the total $84).", "Take 4/7 of it for the pens, then divide by 8."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q9b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q9b: $1.50 (key working 42÷2=21 stationery, 4/7×21=12, 12÷8=1.50)."
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "At first, Farah had 91 star stickers and 78 heart stickers. Then she bought an equal number of boxes of star stickers and heart stickers. Each box of star stickers contained 5 stickers and each box of heart stickers contained 7 stickers. In the end, there were 15 more heart stickers than star stickers. How many boxes of star stickers did she buy?<br>[?]",
      "answer0": "14",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Star stickers started 13 more than hearts (91 − 78 = 13). Each box adds 7 hearts but only 5 stars, so each box closes the gap by 7 − 5 = 2 and then overtakes. To go from stars being 13 ahead to hearts being 15 ahead is a swing of 13 + 15 = 28, so number of boxes = 28 ÷ 2 = 14.",
      "hints": ["Stars begin 13 ahead; each box of stickers shifts the difference by 7 − 5 = 2 toward hearts.", "Total swing = 13 + 15 = 28; divide by 2."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q10: 14 boxes."
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "Tap A and tap B were turned on to fill two identical empty containers X and Y at different rates. Water flowed out of tap B at a rate that was 3 times as fast as tap A. Tap A filled container X with some water for 12 minutes. How long did it take tap B to fill container Y with the same amount of water?<br>[?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Tap B is 3 times as fast, so it fills the same amount in \\(\\dfrac{1}{3}\\) of the time: 12 ÷ 3 = 4 minutes.",
      "hints": ["A faster tap needs less time for the same amount.", "3 times as fast means one-third of the time."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q11a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 24,
      "image_bbox": [0.3, 0.16, 0.65, 0.32],
      "image_loc": "upper part of page; Tap A over Container X and Tap B over Container Y",
      "notes": "Key Paper 2 Q11a: 4 min. Answer in minutes."
    },
    {
      "n": 48,
      "type_id": 2,
      "question": "Mary Ling and Qi Fang... (Tap A/Tap B set-up, part b is a true/false/not-possible-to-tell tick table). [SKIPPED — interactive tick-table format.]",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Statement 1 (more water in X than Y after another 5 min): False. Statement 2 (15 ℓ of water in X when half filled): Not possible to tell. Statement 3 (tap B took 20 min to fill Y to the brim): False.",
      "hints": [],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q11b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "type_id 0 — interactive True/False/Not-possible-to-tell tick table, no app question type. Key 11b: False / Not possible to tell / False."
    },
    {
      "n": 49,
      "type_id": 2,
      "question": "A box contained the same number of red, blue and green toy blocks at first. After 44 green blocks, some red blocks and blue blocks were removed, there were 122 blocks left. There were twice as many red blocks as blue blocks left. The number of green blocks left was 18 fewer than the number of red blocks left.<br>(a) How many more green blocks than blue blocks were left? [?]<br>(b) How many blocks were in the box at first? [?]",
      "answer0": "10",
      "answer1": "246",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Let blue left = 1 unit, red left = 2 units, green left = 2 units − 18. Total left: 1 + 2 + (2 − 18-unit-adjust) ... using the key: red − 18 = green, red − green = 18 → 122 + 18 = 140, 140 ÷ 5 = 28 (blue), red = 56, green = 38, green − blue = 38 − 28 = 10. (b) At first each colour was equal: green at first = green left + 44 = 38 + 44 = 82, so each colour = 82 and total = 82 × 3 = 246.",
      "hints": ["Use 1 unit blue, 2 units red, and red − 18 for green; total left = 122.", "Each colour started equal; green started = green left + 44, then × 3 colours."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q12a: 10 more; Q12b: 246 blocks. Two answer blanks for (a) and (b)."
    },
    {
      "n": 50,
      "type_id": 2,
      "question": "The figure shows rectangle KLMN and triangle MOL. The area of MNPL is \\(\\dfrac{7}{16}\\) of the area of MOL and 7 times the area of PKL. ON = 20 cm and ML = 24 cm. What is the area of MNPL?<br>[?]",
      "answer0": "105",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "Triangle MOL has base ML = 24 cm and height ON = 20 cm (with N on ML), so area MOL = \\(\\dfrac{1}{2} \\times 24 \\times 20 = 240\\) cm². Area MNPL = \\(\\dfrac{7}{16} \\times 240 = 105\\) cm².",
      "hints": ["Find the area of triangle MOL using base ML and height ON.", "MNPL is 7/16 of that area."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q13a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q50.png",
      "image_page": 26,
      "image_bbox": [0.28, 0.18, 0.65, 0.4],
      "image_loc": "middle of page; rectangle KLMN with triangle MOL, O above N (ON = 20 cm), base ML = 24 cm, shaded region MNPL",
      "notes": "Key Paper 2 Q13a: 105 cm² (key working ½ × 24 × 20 = 240, 240 × 7/16 = 105). Answer in cm²."
    },
    {
      "n": 51,
      "type_id": 2,
      "question": "Using the same figure (rectangle KLMN, triangle MOL, MNPL = 105 cm² which is 7 times the area of PKL, ML = 24 cm), what is the length of PK?<br>[?]",
      "answer0": "6",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 136,
      "difficulty_id": 3,
      "explanation": "Area PKL = MNPL ÷ 7 = 105 ÷ 7 = 15 cm². Area of rectangle KLMN = MNPL + PKL = 105 + 15 = 120 cm². The rectangle's breadth = 120 ÷ 24 = 5 cm = KL. PKL is a triangle with base KL = 5 cm... using the key: 15 × 2 = 30, PK = 30 ÷ 5 = 6 cm.",
      "hints": ["Area PKL = 105 ÷ 7. Then KLMN area = MNPL + PKL.", "Find KL from the rectangle, then use triangle PKL's area to get PK."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q13b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q13b: 6 cm (key: KLMN = 105 + 15 = 120, 120 ÷ 24 = 5 = KL, 15 × 2 = 30, PK = 30 ÷ 5 = 6). Answer in cm."
    },
    {
      "n": 52,
      "type_id": 2,
      "question": "QRUT and QRSU are parallelograms. TUS and QUV are straight lines and QT = QU. ∠RUS = 64° and ∠UVS = 50° (∠ at V). Find ∠TQU.<br>[?]",
      "answer0": "52",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "∠TQR = 180° − 64° = 116° (co-interior in parallelogram QRUT, since ∠QUR relates to 64°). Triangle TQU is isosceles (QT = QU), so its base angles are equal: ∠TQU = 180° − (64° × 2) = 180° − 128° = 52°.",
      "hints": ["QT = QU makes triangle TQU isosceles.", "The base angles relate to 64°; angle sum gives ∠TQU = 180° − 2 × 64°."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q14a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q52.png",
      "image_page": 27,
      "image_bbox": [0.35, 0.13, 0.62, 0.36],
      "image_loc": "upper-middle of page; parallelograms QRUT and QRSU with straight lines TUS and QUV, 64° at U and 50° at V",
      "notes": "Key Paper 2 Q14a: 52° (key working 180 − 64×2). Answer in degrees."
    },
    {
      "n": 53,
      "type_id": 2,
      "question": "Using the same figure (∠RUS = 64°, ∠UVS = 50°), find ∠USV.<br>[?]",
      "answer0": "66",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "∠VUS = 180° − 64° − 52° = 64° (angles on straight line TUS / triangle relations). In triangle USV, ∠USV = 180° − ∠VUS − ∠UVS = 180° − 64° − 50° = 66°.",
      "hints": ["Find ∠VUS first using the straight line through U.", "Use the angle sum of triangle USV with ∠UVS = 50°."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q14b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q14b: 66° (key working ∠VUS = 180 − 64 − 52 = 64, then 180 − 64 − 50 = 66). Answer in degrees."
    },
    {
      "n": 54,
      "type_id": 2,
      "question": "At first, Siti had 36 more apples than oranges. After selling \\(\\dfrac{1}{3}\\) of the apples and \\(\\dfrac{1}{4}\\) of the oranges, she had 92 fruits left. How many fruits did Siti have at first?<br>[?]",
      "answer0": "132",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Apples left = \\(\\dfrac{2}{3}\\) of apples; oranges left = \\(\\dfrac{3}{4}\\) of oranges. Let oranges = O, apples = O + 36. From the key: \\(\\dfrac{2}{3}(O+36) + \\dfrac{3}{4}O = 92\\). Solving (key uses units): oranges = 48, apples = 84, total at first = 84 + 48 = 132.",
      "hints": ["Set oranges as one quantity and apples as that plus 36.", "Fruits left = 2/3 of apples + 3/4 of oranges = 92."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q15a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Paper 2 Q15a: 132 fruits (key: apples = 84, oranges = 48)."
    },
    {
      "n": 55,
      "type_id": 1,
      "question": "Siti sold \\(\\dfrac{1}{3}\\) of her 84 apples and \\(\\dfrac{1}{4}\\) of her 48 oranges. She earned 40¢ for each apple sold and $1 for each orange sold. Did she earn more from the sale of apples or oranges, and how much more?",
      "answer0": "$0.80 more from oranges",
      "answer1": "$0.80 more from apples",
      "answer2": "$1.20 more from oranges",
      "answer3": "$11.20 more from apples",
      "correct_answer": 0,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "Apples sold = \\(\\dfrac{1}{3} \\times 84 = 28\\); earnings = 28 × $0.40 = $11.20. Oranges sold = \\(\\dfrac{1}{4} \\times 48 = 12\\); earnings = 12 × $1 = $12. Oranges earned $12 − $11.20 = $0.80 more.",
      "hints": ["Find how many apples and oranges were sold, then their earnings.", "Compare $11.20 (apples) with $12 (oranges)."],
      "source": "Tao Nan 2025 P5 End-of-Year Examination Paper 2 Q15b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Fill-blank + circle-the-word format converted to MCQ. Key Paper 2 Q15b: $0.80 more from oranges."
    }
  ]
}