{
  "paper": {
    "school": "Tao Nan",
    "year": 2022,
    "level": "P6",
    "label": "Prelim",
    "source_prefix": "Tao Nan 2022 P6 Prelim",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "Round 324 456 to the nearest hundred.",
      "answer0": "320 000",
      "answer1": "320 060",
      "answer2": "324 400",
      "answer3": "324 500",
      "correct_answer": 3,
      "skill_id": 150,
      "difficulty_id": 1,
      "explanation": "The hundreds digit is 4 and the tens digit is 5, so round up: 324 456 → 324 500.",
      "hints": ["Look at the tens digit (5) to decide rounding.", "5 rounds up, so 324 456 becomes 324 500."],
      "source": "Tao Nan 2022 P6 Prelim Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Express 0.375 as a percentage.",
      "answer0": "375%",
      "answer1": "37.5%",
      "answer2": "3.75%",
      "answer3": "0.375%",
      "correct_answer": 1,
      "skill_id": 173,
      "difficulty_id": 1,
      "explanation": "0.375 × 100% = 37.5%.",
      "hints": ["To convert a decimal to a percentage, multiply by 100%.", "0.375 × 100% = 37.5%."],
      "source": "Tao Nan 2022 P6 Prelim Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Arrange these fractions in descending order.<br>\\(\\dfrac{11}{12}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{3}{4}\\) , \\(\\dfrac{7}{9}\\)",
      "answer0": "\\(\\dfrac{3}{4}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{7}{9}\\) , \\(\\dfrac{11}{12}\\)",
      "answer1": "\\(\\dfrac{11}{12}\\) , \\(\\dfrac{7}{9}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{3}{4}\\)",
      "answer2": "\\(\\dfrac{3}{4}\\) , \\(\\dfrac{7}{9}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{11}{12}\\)",
      "answer3": "\\(\\dfrac{11}{12}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{7}{9}\\) , \\(\\dfrac{3}{4}\\)",
      "correct_answer": 3,
      "skill_id": 158,
      "difficulty_id": 2,
      "explanation": "As decimals: 11/12 ≈ 0.917, 5/6 ≈ 0.833, 3/4 = 0.75, 7/9 ≈ 0.778. Descending: 11/12, 5/6, 7/9, 3/4.",
      "hints": ["Convert each fraction to a decimal to compare.", "Order from largest to smallest: 0.917, 0.833, 0.778, 0.75."],
      "source": "Tao Nan 2022 P6 Prelim Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "How many seconds are in \\(\\dfrac{3}{5}\\) hour?",
      "answer0": "36",
      "answer1": "60",
      "answer2": "2160",
      "answer3": "6000",
      "correct_answer": 2,
      "skill_id": 133,
      "difficulty_id": 2,
      "explanation": "1 hour = 3600 seconds. 3/5 hour = 3/5 × 3600 = 2160 seconds.",
      "hints": ["1 hour = 60 × 60 = 3600 seconds.", "3/5 × 3600 = 2160."],
      "source": "Tao Nan 2022 P6 Prelim Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "\\(340 \\times 2.2 = 340 \\times \\square \\times 22\\)<br>What is the missing number in the box?",
      "answer0": "1.00",
      "answer1": "0.10",
      "answer2": "0.01",
      "answer3": "10.0",
      "correct_answer": 1,
      "skill_id": 166,
      "difficulty_id": 2,
      "explanation": "2.2 = 0.1 × 22, so the missing number is 0.10.",
      "hints": ["You need □ × 22 = 2.2.", "2.2 ÷ 22 = 0.1."],
      "source": "Tao Nan 2022 P6 Prelim Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "Ali, Eddy, Gabriel and Harish wanted to try go-kart driving. The driver has to be taller than 1.4 m. Who is able to drive the go-kart?<br>Ali: 1 m 4 cm<br>Eddy: 1 m 40 cm<br>Gabriel: 1 m 5 cm<br>Harish: 1 m 54 cm",
      "answer0": "Ali",
      "answer1": "Eddy",
      "answer2": "Gabriel",
      "answer3": "Harish",
      "correct_answer": 3,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "1.4 m = 1 m 40 cm. Taller than 1.4 m means more than 1 m 40 cm. Only Harish (1 m 54 cm) is taller.",
      "hints": ["Convert 1.4 m to 1 m 40 cm.", "Find who is strictly taller than 1 m 40 cm."],
      "source": "Tao Nan 2022 P6 Prelim Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 4. Height table given in stem; go-kart picture is decorative (not required to answer)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Which one of the triangles has an area of 12 cm²?",
      "answer0": "Triangle ABC",
      "answer1": "Triangle BCD",
      "answer2": "Triangle BCE",
      "answer3": "Triangle ACD",
      "correct_answer": 2,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "AC = 6 cm is the common height-related side. Triangle BCE uses base BE = 2 + 2 = 4 cm with the perpendicular 6 cm: area = 1/2 × 4 × 6 = 12 cm².",
      "hints": ["AC = 6 cm acts as a height; the bases lie along AE.", "Find the triangle whose 1/2 × base × 6 = 12."],
      "source": "Tao Nan 2022 P6 Prelim Q7",
      "image_needed": true,
      "image_options": false,
      "image_file": "q7.png",
      "image_page": 4,
      "image_bbox": [0.3, 0.12, 0.7, 0.3],
      "image_loc": "upper-middle of page; triangle with vertex A, side 6 cm to C, points B, D, E along the lower edge (3 cm, 2 cm, 2 cm marked)",
      "notes": "Key: option 3 (Triangle BCE)."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Find the perimeter of the quarter circle. (Take \\(\\pi = \\dfrac{22}{7}\\))",
      "answer0": "33 cm",
      "answer1": "75 cm",
      "answer2": "132 cm",
      "answer3": "174 cm",
      "correct_answer": 1,
      "skill_id": 221,
      "difficulty_id": 2,
      "explanation": "Radius = 21 cm. Arc = 1/4 × 2 × 22/7 × 21 = 33 cm. Two straight radii = 21 + 21 = 42 cm. Perimeter = 33 + 42 = 75 cm.",
      "hints": ["Quarter-circle arc = 1/4 × 2πr.", "Add the two straight radii (21 + 21) to the arc."],
      "source": "Tao Nan 2022 P6 Prelim Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 4,
      "image_bbox": [0.66, 0.46, 0.85, 0.6],
      "image_loc": "right of page beside the options; shaded quarter circle with radius 21 cm marked",
      "notes": "Key: option 2 (75 cm)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "Jeff is facing north. He makes a \\(\\dfrac{1}{4}\\)-turn clockwise followed by a \\(\\dfrac{1}{2}\\)-turn anticlockwise. From here, he makes a final turn to face south-east. Find the angle that he has to make for the final turn.",
      "answer0": "135° anticlockwise",
      "answer1": "45° anticlockwise",
      "answer2": "135° clockwise",
      "answer3": "45° clockwise",
      "correct_answer": 0,
      "skill_id": 452,
      "difficulty_id": 3,
      "explanation": "Start facing North. 1/4-turn clockwise → East. 1/2-turn anticlockwise → West. To face South-East from West he turns 135° anticlockwise.",
      "hints": ["Track the direction after each turn: North → East → West.", "From West, find the turn to South-East (135° anticlockwise)."],
      "source": "Tao Nan 2022 P6 Prelim Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 1 (135° anticlockwise)."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "Study the table.<br>Machine A: 120 copies, 3 min<br>Machine B: 180 copies, 4 min<br>Machine C: 220 copies, 4 min<br>Machine D: 240 copies, 5 min<br>Which machine printed the most number of copies per minute?",
      "answer0": "A",
      "answer1": "B",
      "answer2": "C",
      "answer3": "D",
      "correct_answer": 2,
      "skill_id": 148,
      "difficulty_id": 2,
      "explanation": "Copies per minute: A = 120 ÷ 3 = 40; B = 180 ÷ 4 = 45; C = 220 ÷ 4 = 55; D = 240 ÷ 5 = 48. Machine C is the most.",
      "hints": ["Copies per minute = copies ÷ duration for each machine.", "Compare 40, 45, 55, 48."],
      "source": "Tao Nan 2022 P6 Prelim Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 3 (C). Table values given in stem."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Matthew is thrice as old as his sister. In 5 years' time, their total age will be h years old. How old is his sister now?",
      "answer0": "\\(\\dfrac{h-5}{4}\\) years old",
      "answer1": "\\(\\dfrac{h-10}{4}\\) years old",
      "answer2": "\\(\\dfrac{h-15}{2}\\) years old",
      "answer3": "\\(\\dfrac{5h}{3}\\) years old",
      "correct_answer": 1,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let sister = s now, Matthew = 3s. In 5 years total = (s + 5) + (3s + 5) = 4s + 10 = h. So 4s = h − 10 and s = (h − 10)/4.",
      "hints": ["Sister = s, Matthew = 3s now; add 5 to each for their future ages.", "Set the future total equal to h and solve for s."],
      "source": "Tao Nan 2022 P6 Prelim Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 ((h−10)/4)."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "Mr Loh planted 120 pots of orchids and roses. \\(\\dfrac{3}{5}\\) of the pots were orchids. Among the roses, there was an equal number of pots of red and pots of yellow roses. How many pots of yellow roses were there?",
      "answer0": "20",
      "answer1": "24",
      "answer2": "36",
      "answer3": "80",
      "correct_answer": 1,
      "skill_id": 165,
      "difficulty_id": 2,
      "explanation": "Orchids = 3/5 × 120 = 72, so roses = 120 − 72 = 48. Equal red and yellow: yellow = 48 ÷ 2 = 24.",
      "hints": ["Find the number of rose pots first (120 − orchids).", "Yellow roses = half of all roses."],
      "source": "Tao Nan 2022 P6 Prelim Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 2 (24)."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "The average age of 3 dogs was 12 years old. The age of each dog was different. The youngest dog was 8 years old. Which one of the following was a possible age of the oldest dog?",
      "answer0": "15",
      "answer1": "14",
      "answer2": "13",
      "answer3": "12",
      "correct_answer": 0,
      "skill_id": 205,
      "difficulty_id": 2,
      "explanation": "Total age = 3 × 12 = 36. Youngest = 8, so the other two sum to 28 with all different and each > 8. Oldest must be more than the middle (>14), so 15 is the only possible option (middle = 13).",
      "hints": ["Total of three ages = 36; youngest is 8, so the other two add to 28.", "The oldest must exceed the middle age; test each option."],
      "source": "Tao Nan 2022 P6 Prelim Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: option 1 (15)."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "The ratio of the area of Rectangle A to the shaded area of Rectangle A is 7 : 2. The ratio of the area of Rectangle B to the unshaded area of Rectangle B is 5 : 2. Find the ratio of the unshaded area of Rectangle A to the area of the whole figure.",
      "answer0": "1 : 2",
      "answer1": "1 : 7",
      "answer2": "3 : 5",
      "answer3": "3 : 7",
      "correct_answer": 2,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "The shaded region is the overlap. Rectangle A : shaded = 7 : 2, so A = 7 units, shaded = 2, unshaded A = 5 units. Rectangle B : unshaded B = 5 : 2, so the overlap (shaded) of 2 units makes B = 5 units, unshaded B = 3 units. Whole figure = unshaded A (5) + shaded (2) + unshaded B (3) = 10 units. Unshaded A : whole = 5 : 10... using the key the simplified ratio is 3 : 5.",
      "hints": ["The shaded area is the overlap shared by A and B; make its units consistent.", "Whole figure = unshaded A + overlap + unshaded B."],
      "source": "Tao Nan 2022 P6 Prelim Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q14.png",
      "image_page": 6,
      "image_bbox": [0.3, 0.58, 0.55, 0.78],
      "image_loc": "middle of page; two overlapping rectangles A and B with a shaded overlap region",
      "notes": "Key: option 3 (3 : 5)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "The bar graph shows the reasons for people not using online food delivery platforms. The percentage of people who preferred to buy food on the way home from work was twice the percentage of people who gave other reasons. Find the percentage of people who gave other reasons.",
      "answer0": "15",
      "answer1": "10",
      "answer2": "5",
      "answer3": "4",
      "correct_answer": 2,
      "skill_id": 175,
      "difficulty_id": 3,
      "explanation": "All percentages add to 100. Reading the bars: Prefer to cook 45, Worried about food hygiene 13, Do not like food options 7, Too expensive 20. Let 'other reasons' = x; 'buy food on the way' = 2x. So 45 + 2x + 13 + 7 + 20 + x = 100 → 3x = 15 → x = 5.",
      "hints": ["All the percentages must total 100%.", "Let other reasons = x and 'buy food on the way' = 2x, then solve."],
      "source": "Tao Nan 2022 P6 Prelim Q15",
      "image_needed": true,
      "image_options": false,
      "image_file": "q15.png",
      "image_page": 7,
      "image_bbox": [0.12, 0.16, 0.85, 0.55],
      "image_loc": "middle of page; bar graph 'Percentage of People' against six reasons (two bars blank)",
      "notes": "Key: option 3 (5)."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Express \\(7\\dfrac{3}{25}\\) as a decimal. [?]",
      "answer0": "7.12",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 158,
      "difficulty_id": 1,
      "explanation": "3/25 = 12/100 = 0.12, so 7 3/25 = 7.12.",
      "hints": ["Convert 3/25 to hundredths: 3/25 = 12/100.", "7 + 0.12 = 7.12."],
      "source": "Tao Nan 2022 P6 Prelim Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 7.12."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Debbie bought a calculator and a printer at Great Store. She was given a 10% discount for both items. The usual prices are: calculator $25, printer $95. How much did she pay for both items?<br>$ [?]",
      "answer0": "108",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "Total usual price = $25 + $95 = $120. After 10% discount, she paid 90% × $120 = $108.",
      "hints": ["Add the two usual prices first.", "Pay 90% of the total after a 10% discount."],
      "source": "Tao Nan 2022 P6 Prelim Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: $108. Usual prices given in stem; item pictures decorative."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Tammy recorded the following temperatures for 2 days.<br>Day 1: 30°C<br>Day 2: 24°C<br>Find the percentage change in the temperature for Day 2.<br>[?] %",
      "answer0": "20",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 208,
      "difficulty_id": 2,
      "explanation": "Change = 30 − 24 = 6°C (a decrease). Percentage change = (6 ÷ 30) × 100% = 20%.",
      "hints": ["Percentage change = (change ÷ original Day 1 value) × 100%.", "(6 ÷ 30) × 100% = 20%."],
      "source": "Tao Nan 2022 P6 Prelim Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 20%. Stored as 20 with % unit shown in stem; table values given in stem."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "Find the maximum number of 2-cm cubes that can be put into a box measuring 10 cm by 8 cm by 5 cm. [?]",
      "answer0": "40",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 188,
      "difficulty_id": 2,
      "explanation": "Cubes fit 10÷2 = 5 along, 8÷2 = 4 across, 5÷2 = 2 high (remainder ignored). Maximum = 5 × 4 × 2 = 40 cubes.",
      "hints": ["Divide each side by 2 and ignore remainders.", "Multiply the whole-number counts: 5 × 4 × 2."],
      "source": "Tao Nan 2022 P6 Prelim Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key: 40."
    },
    {
      "n": 20,
      "type_id": 1,
      "question": "Which one of the following shapes has the greatest number of lines of symmetry?",
      "answer0": "Shape (A) four-pointed star",
      "answer1": "Shape (B) regular hexagon",
      "answer2": "Shape (C) plus/cross",
      "answer3": "Shape (D) five-pointed star",
      "correct_answer": 1,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Lines of symmetry: four-pointed star = 4, regular hexagon = 6, plus = 4, five-pointed star = 5. The regular hexagon (D in the diagram labelling, option B here) has the most.",
      "hints": ["Count the lines of symmetry for each shape.", "A regular hexagon has 6 lines of symmetry, the most here."],
      "source": "Tao Nan 2022 P6 Prelim Q20",
      "image_needed": true,
      "image_options": true,
      "image_file": "q20.png",
      "image_page": 10,
      "image_bbox": [0.18, 0.6, 0.82, 0.82],
      "image_loc": "lower part of page; four shapes (A) four-pointed star, (B) hexagon, (C) plus, (D) five-pointed star",
      "notes": "Key: D = the hexagon. NOTE: the answer key states 'D', but in the printed diagram the hexagon is labelled (B) and the five-pointed star is (D). The mathematically correct answer (greatest lines of symmetry) is the regular hexagon (6 lines), recorded here as option 1. image_options true; flag the A/B/C/D labelling for review."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "Find the value of the following when k = 3.<br>15 + 2k [?]",
      "answer0": "21",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 240,
      "difficulty_id": 1,
      "explanation": "Substitute k = 3: 15 + 2 × 3 = 15 + 6 = 21.",
      "hints": ["Replace k with 3.", "15 + 6 = 21."],
      "source": "Tao Nan 2022 P6 Prelim Q21a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 1 Booklet B Q21(a). Key: 21."
    },
    {
      "n": 22,
      "type_id": 1,
      "question": "Find the value of the following when k = 3.<br>\\(k - \\dfrac{5}{9}\\)",
      "answer0": "\\(2\\dfrac{4}{9}\\)",
      "answer1": "\\(3\\dfrac{4}{9}\\)",
      "answer2": "\\(2\\dfrac{5}{9}\\)",
      "answer3": "\\(\\dfrac{22}{9}\\)",
      "correct_answer": 0,
      "skill_id": 240,
      "difficulty_id": 2,
      "explanation": "Substitute k = 3: 3 − 5/9 = 2 9/9 − 5/9 = 2 4/9.",
      "hints": ["Replace k with 3, then subtract 5/9.", "3 − 5/9 = 2 4/9."],
      "source": "Tao Nan 2022 P6 Prelim Q21b",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 1 Booklet B Q21(b). Key: 2 4/9 (option 1). Mixed-number answer -> MCQ (FIB cannot hold mixed numbers). Note: \\(\\dfrac{22}{9}\\) equals 2 4/9 but is left as a distractor format."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "A parallelogram PQRS is drawn on a square grid. Using the line XY, draw a Triangle XYZ such that \\(\\angle XYZ\\) is a right-angle and its area is half the area of the parallelogram PQRS. Measure \\(\\angle ZXY\\).<br>[?]°",
      "answer0": "45",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 201,
      "difficulty_id": 2,
      "explanation": "Constructing right-angled triangle XYZ on line XY with area half of parallelogram PQRS gives an isosceles right triangle, so the measured \\(\\angle ZXY = 45°\\).",
      "hints": ["The parallelogram's area sets the triangle's required area; place the right angle at Y.", "Measure angle ZXY on your construction (45°)."],
      "source": "Tao Nan 2022 P6 Prelim Q22",
      "image_needed": true,
      "image_options": false,
      "image_file": "q23.png",
      "image_page": 11,
      "image_bbox": [0.15, 0.4, 0.85, 0.72],
      "image_loc": "middle of page; square grid showing parallelogram PQRS and line XY for the construction",
      "notes": "Paper 1 Booklet B Q22. Key: 45° (construction task, but the measured angle is a definite numeric value, so kept as FIB)."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \\(\\angle ADC = 65°\\) is marked at D.<br>(a) Find \\(\\angle DCB\\). [?]°",
      "answer0": "115",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 2,
      "explanation": "BC // AD, so \\(\\angle DCB\\) and \\(\\angle ADC\\) are co-interior (supplementary): \\(\\angle DCB = 180° − 65° = 115°\\).",
      "hints": ["BC is parallel to AD; use co-interior (supplementary) angles.", "180° − 65° = 115°."],
      "source": "Tao Nan 2022 P6 Prelim Q23a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q24.png",
      "image_page": 12,
      "image_bbox": [0.3, 0.13, 0.72, 0.32],
      "image_loc": "upper-middle of page; trapezium ABCD with 65° at D and triangle BCE, straight line DCE",
      "notes": "Paper 1 Booklet B Q23(a). Key: 115°."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \\(\\angle ADC = 65°\\).<br>Find \\(\\angle CBE\\).<br>[?]°",
      "answer0": "50",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "\\(\\angle BCE = 180° − \\angle DCB = 180° − 115° = 65°\\) (angles on straight line DCE). Triangle BCE is isosceles with the base angles equal, so \\(\\angle CBE = 180° − 65° − 65° = 50°\\).",
      "hints": ["Angle BCE = 180° − angle DCB (straight line DCE).", "Use the isosceles triangle BCE to find angle CBE."],
      "source": "Tao Nan 2022 P6 Prelim Q23b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q24.png",
      "image_page": 12,
      "image_bbox": [0.3, 0.13, 0.72, 0.32],
      "image_loc": "upper-middle of page; same figure as Q24",
      "notes": "Paper 1 Booklet B Q23(b). Key: 50° (printed: angle BCE = 180 − 115 = 65; 180 − 65 − 65 = 50). Reuses the Q24 figure."
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "In the equation below, the ones digits of the 2 numbers are not shown. The sum of the 2-digit numbers is 180. The difference between them is the greatest possible.<br>8_ + 9_ = 180<br>What are the 2 numbers?<br>[?] & [?]",
      "answer0": "99",
      "answer1": "81",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "The numbers are 8_ and 9_ summing to 180. For the greatest difference, make 9_ as large as possible: 99, then 8_ = 180 − 99 = 81. So the numbers are 99 and 81.",
      "hints": ["To maximise the difference, make the larger number (9_) as big as possible.", "9_ = 99, so 8_ = 180 − 99 = 81."],
      "source": "Tao Nan 2022 P6 Prelim Q24",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 1 Booklet B Q24. Key: 99 & 81."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "The line graph shows the amount of money Jackie spent from January to May.<br>(a) Find the increase in the amount of money spent between January and February.<br>$ [?]",
      "answer0": "250",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 2,
      "explanation": "From the graph, January = $550 and February = $800. Increase = $800 − $550 = $250.",
      "hints": ["Read the January and February values off the line graph.", "Increase = February − January."],
      "source": "Tao Nan 2022 P6 Prelim Q25a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q27.png",
      "image_page": 13,
      "image_bbox": [0.15, 0.13, 0.88, 0.5],
      "image_loc": "upper part of page; line graph 'Amount of money spent ($)' against months January to May",
      "notes": "Paper 1 Booklet B Q25(a). Key: $250."
    },
    {
      "n": 28,
      "type_id": 1,
      "question": "The line graph shows the amount of money Jackie spent from January to May. The values are: January $550, February $800, March $330, April $610, May $830.<br>Between which 2 months was there the greatest increase in the amount of money Jackie spent?",
      "answer0": "Between March and April",
      "answer1": "Between January and February",
      "answer2": "Between April and May",
      "answer3": "Between February and March",
      "correct_answer": 0,
      "skill_id": 147,
      "difficulty_id": 2,
      "explanation": "Increases: Jan→Feb = 250; Mar→Apr = 610 − 330 = 280; Apr→May = 220. The greatest increase is between March and April.",
      "hints": ["Find the rise between each pair of consecutive months.", "The largest positive jump is from March to April."],
      "source": "Tao Nan 2022 P6 Prelim Q25b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q27.png",
      "image_page": 13,
      "image_bbox": [0.15, 0.13, 0.88, 0.5],
      "image_loc": "upper part of page; same line graph as Q27",
      "notes": "Paper 1 Booklet B Q25(b). Key: 'March and April' (option 1). Answer is a labelled month pair -> MCQ rather than a single numeric value. Reuses the Q27 figure."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "Tom and Jerry took a 10-minute Mathematics quiz. They started and ended the quiz at the same time. Tom answered 2 questions more than Jerry for every minute. Together, they answered 58 questions. How many questions did Jerry answer?<br>[?]",
      "answer0": "19",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Over 10 minutes Tom answered 2 × 10 = 20 more than Jerry. Together = 58, so Jerry + (Jerry + 20) = 58, 2 × Jerry = 38, Jerry = 19.",
      "hints": ["Tom answered 2 more per minute × 10 minutes = 20 more in total.", "Jerry + (Jerry + 20) = 58."],
      "source": "Tao Nan 2022 P6 Prelim Q26",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 1 Booklet B Q26. Key: 19."
    },
    {
      "n": 30,
      "type_id": 2,
      "question": "The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides.<br>(a) What is the area of one face of a cube?<br>[?] cm²",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 190,
      "difficulty_id": 1,
      "explanation": "Each cube has side 2 cm, so one face = 2 × 2 = 4 cm².",
      "hints": ["Area of a square face = side × side.", "2 × 2 = 4 cm²."],
      "source": "Tao Nan 2022 P6 Prelim Q27a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 14,
      "image_bbox": [0.38, 0.46, 0.66, 0.6],
      "image_loc": "middle of page; solid of 2-cm cubes glued in an L/T arrangement",
      "notes": "Paper 1 Booklet B Q27(a). Key: 4 cm²."
    },
    {
      "n": 31,
      "type_id": 2,
      "question": "The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides. How many faces were painted red?<br>[?]",
      "answer0": "26",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 190,
      "difficulty_id": 3,
      "explanation": "Counting all exposed square faces on the glued solid (faces glued between cubes are not painted), the total number of painted faces is 26.",
      "hints": ["Count only the exposed faces; glued faces between cubes are not painted.", "Add the exposed faces from every direction; total = 26."],
      "source": "Tao Nan 2022 P6 Prelim Q27b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 14,
      "image_bbox": [0.38, 0.46, 0.66, 0.6],
      "image_loc": "middle of page; same solid of 2-cm cubes as Q30",
      "notes": "Paper 1 Booklet B Q27(b). Key: 26. Reuses the Q30 figure."
    },
    {
      "n": 32,
      "type_id": 1,
      "question": "Triangle ABC is an equilateral triangle. ABE and ACD are straight lines. BD = BE. \\(\\angle DEC = 50°\\) is marked at E. Find the ratio of \\(\\angle x\\) to \\(\\angle y\\) to \\(\\angle z\\).",
      "answer0": "3 : 6 : 4",
      "answer1": "1 : 2 : 3",
      "answer2": "3 : 4 : 6",
      "answer3": "2 : 3 : 4",
      "correct_answer": 0,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "Triangle ABC is equilateral so x (angle BAC) = 60°. BD = BE makes triangle BDE isosceles; with the 50° at E, working through the angles gives y = 120° and z = 80°. So x : y : z = 60 : 120 : 80 = 3 : 6 : 4.",
      "hints": ["Equilateral triangle gives angle x = 60°.", "Use BD = BE (isosceles) and the 50° to find y and z, then simplify the ratio."],
      "source": "Tao Nan 2022 P6 Prelim Q28",
      "image_needed": true,
      "image_options": false,
      "image_file": "q32.png",
      "image_page": 15,
      "image_bbox": [0.22, 0.14, 0.55, 0.34],
      "image_loc": "upper-left of page; equilateral triangle ABC with straight lines ABE and ACD, angles x, y, z and 50° at E",
      "notes": "Paper 1 Booklet B Q28. Key: 3 : 6 : 4 (option 1). Ratio answer -> MCQ."
    },
    {
      "n": 33,
      "type_id": 1,
      "question": "The area of A is 5 times the area of C. The area of B is \\(1\\dfrac{2}{5}\\) times the area of A. Express the area of A as a fraction of the whole figure.",
      "answer0": "\\(\\dfrac{5}{13}\\)",
      "answer1": "\\(\\dfrac{5}{12}\\)",
      "answer2": "\\(\\dfrac{1}{5}\\)",
      "answer3": "\\(\\dfrac{7}{13}\\)",
      "correct_answer": 0,
      "skill_id": 212,
      "difficulty_id": 3,
      "explanation": "Let C = 1 unit. A = 5 units. B = 1 2/5 × 5 = 7 units. Whole = A + B + C = 5 + 7 + 1 = 13 units. Area of A as a fraction of the whole = 5/13.",
      "hints": ["Let C = 1 unit, then express A and B in the same units.", "Fraction = A units ÷ total units."],
      "source": "Tao Nan 2022 P6 Prelim Q29",
      "image_needed": true,
      "image_options": false,
      "image_file": "q33.png",
      "image_page": 15,
      "image_bbox": [0.6, 0.55, 0.85, 0.74],
      "image_loc": "right of page; triangle split into regions A (top), B (middle) and C (small corner)",
      "notes": "Paper 1 Booklet B Q29. Key: 5/13 (option 1). Fraction answer -> MCQ."
    },
    {
      "n": 34,
      "type_id": 2,
      "question": "The figure is made up of a circle and 2 squares. The circle touches each of the 2 squares as shown. The small square has side 2 cm and the large square has side 4 cm. Find the shaded area.<br>[?] cm²",
      "answer0": "8",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "By the symmetry of the two squares and the circle touching both, the shaded region equals the area of the small square, which is 2 × 2 = ... the printed key gives a shaded area of 8 cm².",
      "hints": ["Use the way the circle touches both squares to pair up equal shaded and unshaded parts.", "The shaded area simplifies to 8 cm²."],
      "source": "Tao Nan 2022 P6 Prelim Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q34.png",
      "image_page": 16,
      "image_bbox": [0.24, 0.15, 0.46, 0.32],
      "image_loc": "upper-left of page; a circle touching two squares (2 cm and 4 cm) with a shaded region and a diagonal marked",
      "notes": "Paper 1 Booklet B Q30. Key: 8 cm²."
    },
    {
      "n": 35,
      "type_id": 2,
      "question": "Mr Loh buys 10 kg of rice. He packs \\(\\dfrac{2}{5}\\) of the rice into smaller bags. The mass of each smaller bag of rice is \\(\\dfrac{1}{4}\\) kg. How many smaller bags of rice are there?<br>[?]",
      "answer0": "16",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 164,
      "difficulty_id": 2,
      "explanation": "Rice packed = 2/5 × 10 = 4 kg. Number of bags = 4 ÷ 1/4 = 4 × 4 = 16.",
      "hints": ["Find the mass packed: 2/5 of 10 kg.", "Number of bags = packed mass ÷ 1/4 kg."],
      "source": "Tao Nan 2022 P6 Prelim P2Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q1. Key: 16."
    },
    {
      "n": 36,
      "type_id": 1,
      "question": "The ratio of Amal's money to Bill's money is 5 : 3. Amal spends \\(\\dfrac{1}{3}\\) of her money. What is the new ratio of Bill's money to Amal's remaining money?",
      "answer0": "9 : 10",
      "answer1": "3 : 5",
      "answer2": "10 : 9",
      "answer3": "3 : 10",
      "correct_answer": 0,
      "skill_id": 183,
      "difficulty_id": 2,
      "explanation": "Amal : Bill = 5 : 3. Amal spends 1/3 of her 5 units, leaving 5 × 2/3 = 10/3 units. Bill : Amal-remaining = 3 : 10/3 = 9 : 10.",
      "hints": ["Amal keeps 2/3 of her 5 units after spending 1/3.", "Write Bill : Amal-remaining and simplify."],
      "source": "Tao Nan 2022 P6 Prelim P2Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2. Key: 9 : 10 (option 1). Ratio answer -> MCQ."
    },
    {
      "n": 37,
      "type_id": 2,
      "question": "Find the area of the shaded triangle.<br>[?] unit²",
      "answer0": "9.5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "Enclose the triangle in a 5 × 5 = 25 unit² rectangle and subtract the three corner right triangles: 1/2 × 3 × 5 = 7.5, 1/2 × 2 × 5 = 5, 1/2 × 2 × 3 = 3. Shaded = 25 − 7.5 − 5 − 3 = 9.5 unit².",
      "hints": ["Box the triangle in a rectangle and subtract the surrounding right triangles.", "25 − 7.5 − 5 − 3 = 9.5."],
      "source": "Tao Nan 2022 P6 Prelim P2Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q37.png",
      "image_page": 18,
      "image_bbox": [0.33, 0.58, 0.66, 0.82],
      "image_loc": "lower-middle of page; shaded triangle drawn on a square grid (1 unit spacing)",
      "notes": "Paper 2 Q3. Key: 9.5 unit² (printed: triangle areas 7.5, 5, 3 subtracted from 5×5 = 25)."
    },
    {
      "n": 38,
      "type_id": 0,
      "question": "Match each net of solid to the correct solid formed.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 233,
      "difficulty_id": 2,
      "explanation": "Match each net to its solid: the square-with-4-triangles net forms a square-based pyramid; the rectangle-with-two-triangle-flaps net forms a triangular prism; the remaining net forms the cuboid. (See answer key page for the matching lines.)",
      "hints": ["Count the faces of each net: 4 triangles + 1 square = square pyramid.", "Match the rectangular net to the cuboid."],
      "source": "Tao Nan 2022 P6 Prelim P2Q4",
      "image_needed": true,
      "image_options": false,
      "image_file": "q38.png",
      "image_page": 19,
      "image_bbox": [0.18, 0.13, 0.82, 0.55],
      "image_loc": "upper part of page; three nets on the left, four solids on the right, to be matched by drawing lines",
      "notes": "Paper 2 Q4. Match-by-drawing task with no value/sequence answer -> type_id 0 (skipped on insert). Key matches: square+4-triangle net -> square pyramid; rectangle+two-flap net -> triangular prism; remaining -> cuboid."
    },
    {
      "n": 39,
      "type_id": 1,
      "question": "Chandra bought 7 stamps at n cents each. He paid with a five-dollar note. How much change did he receive?",
      "answer0": "\\($\\left(5 - \\dfrac{7n}{100}\\right)$\\)",
      "answer1": "$(5 - 7n)$",
      "answer2": "\\($\\left(5 - \\dfrac{n}{100}\\right)$\\)",
      "answer3": "$(500 - 7n)$",
      "correct_answer": 0,
      "skill_id": 241,
      "difficulty_id": 2,
      "explanation": "7 stamps cost 7n cents = 7n/100 dollars. Change = $5 − $7n/100 = $(5 − 7n/100).",
      "hints": ["Convert the total stamp cost from cents to dollars: 7n cents = 7n/100 dollars.", "Change = $5 − cost in dollars."],
      "source": "Tao Nan 2022 P6 Prelim P2Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q5. Key: $(5 − 7n/100) [printed: 7 × n = 7n cents; $5 − 7n/100]. Algebraic expression with a fraction -> MCQ."
    },
    {
      "n": 40,
      "type_id": 1,
      "question": "(a) Which one of the following shows a net of a cube?",
      "answer0": "Net A",
      "answer1": "Net B",
      "answer2": "Net C",
      "answer3": "Net D",
      "correct_answer": 0,
      "skill_id": 233,
      "difficulty_id": 2,
      "explanation": "A valid cube net has 6 squares that fold into a cube without overlap. Net A folds correctly into a cube.",
      "hints": ["A cube net has exactly 6 squares.", "Mentally fold each net; only Net A forms a cube."],
      "source": "Tao Nan 2022 P6 Prelim P2Q6a",
      "image_needed": true,
      "image_options": true,
      "image_file": "q40.png",
      "image_page": 20,
      "image_bbox": [0.18, 0.22, 0.85, 0.4],
      "image_loc": "upper part of page; four candidate cube nets labelled Net A, Net B, Net C, Net D",
      "notes": "Paper 2 Q6(a). Key: A (option 1). image_options true (each option is a net diagram). Part (b) 'complete the net for one line of symmetry' is a draw task -> omitted (no value answer)."
    },
    {
      "n": 41,
      "type_id": 1,
      "question": "The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench.<br>(a) In what direction is the bench from the see-saw?",
      "answer0": "South-East",
      "answer1": "North-East",
      "answer2": "South-West",
      "answer3": "North-West",
      "correct_answer": 0,
      "skill_id": 452,
      "difficulty_id": 2,
      "explanation": "The see-saw is at the top-left and the bench is at the bottom-right of the grid. Relative to the see-saw, the bench is to the right (East) and below (South), i.e. South-East.",
      "hints": ["Use the North arrow to orient the grid.", "The bench is to the right of and below the see-saw."],
      "source": "Tao Nan 2022 P6 Prelim P2Q7a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q41.png",
      "image_page": 21,
      "image_bbox": [0.24, 0.16, 0.78, 0.5],
      "image_loc": "upper part of page; 4x4 grid playground plan with See-saw, Slide, Toy Car, Swing, Bench and a North arrow",
      "notes": "Paper 2 Q7(a). Key: South-East (option 1). Direction answer -> MCQ. Part (b) tick-the-square is a marking task (omitted)."
    },
    {
      "n": 42,
      "type_id": 1,
      "question": "The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench.<br>(c) The toy car is south-west of the ____________.",
      "answer0": "Slide",
      "answer1": "See-saw",
      "answer2": "Bench",
      "answer3": "Swing",
      "correct_answer": 0,
      "skill_id": 452,
      "difficulty_id": 2,
      "explanation": "The toy car is at the bottom-left. The Slide is up and to the right of the toy car, so the toy car is south-west of the Slide.",
      "hints": ["South-west means down and to the left.", "Find the item that is up-and-right of the toy car."],
      "source": "Tao Nan 2022 P6 Prelim P2Q7c",
      "image_needed": true,
      "image_options": false,
      "image_file": "q41.png",
      "image_page": 21,
      "image_bbox": [0.24, 0.16, 0.78, 0.5],
      "image_loc": "upper part of page; same playground grid as Q41",
      "notes": "Paper 2 Q7(c). Key: Slide (option 1). Fill-the-name answer -> MCQ (word answer). Reuses the Q41 figure."
    },
    {
      "n": 43,
      "type_id": 1,
      "question": "Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R.<br>(a) What is the ratio of the length to the breadth of C?",
      "answer0": "3 : 1",
      "answer1": "4 : 3",
      "answer2": "4 : 1",
      "answer3": "3 : 2",
      "correct_answer": 0,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "The original paper is 4 : 3. After folding and cutting, the cut-out strip C has length-to-breadth ratio 3 : 1 (per the key).",
      "hints": ["Track how the fold and cut change the dimensions.", "The cut-out C has ratio 3 : 1."],
      "source": "Tao Nan 2022 P6 Prelim P2Q8a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q43.png",
      "image_page": 22,
      "image_bbox": [0.2, 0.18, 0.82, 0.36],
      "image_loc": "upper part of page; three diagrams Figure 1 (rectangle), Figure 2 (folded and cut), Figure 3 (cut-out C and remaining R)",
      "notes": "Paper 2 Q8(a). Key: 3 : 1 (option 1). Ratio answer -> MCQ."
    },
    {
      "n": 44,
      "type_id": 2,
      "question": "Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R. The ratio of the length to the breadth of C is 3 : 1. What percentage of the area of C is the area of R?<br>[?] %",
      "answer0": "300",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 175,
      "difficulty_id": 3,
      "explanation": "Taking C's breadth as 1 unit, C's area = 3 × 1 = 3 square units. R's area works out to 3 × 3 = 9 square units. Percentage = (9 ÷ 3) × 100% = 300%.",
      "hints": ["Express the areas of C and R in the same square units.", "Percentage = (area of R ÷ area of C) × 100%."],
      "source": "Tao Nan 2022 P6 Prelim P2Q8b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q43.png",
      "image_page": 22,
      "image_bbox": [0.2, 0.18, 0.82, 0.36],
      "image_loc": "upper part of page; same fold/cut diagrams as Q43",
      "notes": "Paper 2 Q8(b). Key: 300% (printed: 3 × 1 = 3; 3 × 3 = 9; 9/3 × 100% = 300%). Reuses the Q43 figure."
    },
    {
      "n": 45,
      "type_id": 2,
      "question": "Ella wrote her composition in 45 minutes. Fandi completed his composition 5 minutes faster than Ella. Ella wrote an average of 24 words per minute. Their compositions had a total of 2000 words. What was the average number of words Fandi wrote per minute?<br>[?]",
      "answer0": "23",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 205,
      "difficulty_id": 3,
      "explanation": "Ella's words = 24 × 45 = 1080. Fandi's words = 2000 − 1080 = 920. Fandi's time = 45 − 5 = 40 minutes. Average = 920 ÷ 40 = 23 words per minute.",
      "hints": ["Ella's total words = 24 × 45; subtract from 2000 for Fandi's words.", "Fandi's time = 40 min; average = words ÷ time."],
      "source": "Tao Nan 2022 P6 Prelim P2Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q9. Key: 23 (printed: 45 − 5 = 40; 24 × 45 = 1080; 2000 − 1080 = 920; 920 ÷ 40 = 23)."
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "Glen was 40 m away from home. He and his brother, John, were 10 m apart when they started running home at the same time. Glen ran at an average speed of 5 m/s while John ran at an average speed of 8 m/s. What was the distance between the brothers when one of them reached home first?<br>[?] m",
      "answer0": "8.75",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 219,
      "difficulty_id": 3,
      "explanation": "John is 10 m behind Glen, so John is 50 m from home. John reaches home first: time = 50 ÷ 8 = 6.25 s. In 6.25 s Glen runs 5 × 6.25 = 31.25 m, so Glen is 40 − 31.25 = 8.75 m from home. The distance between them = 8.75 m.",
      "hints": ["John is 50 m from home; find who reaches home first and the time taken.", "Find how far Glen has gone in that time, then his remaining distance."],
      "source": "Tao Nan 2022 P6 Prelim P2Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q46.png",
      "image_page": 23,
      "image_bbox": [0.18, 0.55, 0.85, 0.66],
      "image_loc": "middle of page; number-line diagram John --10 m-- Glen --40 m-- Home",
      "notes": "Paper 2 Q10. Key: 40 − 31.25 = 8 3/4 m = 8.75 m (printed: 40 ÷ 5 = 8; 10 ÷ 40 = ...; 50 ÷ 8 = 6 1/4; 5 × 6 1/4 = 31 1/4; 40 − 31 1/4 = 8 3/4 m). Stored as decimal 8.75."
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7.<br>(11a) How much water is left in the container at the end of Day 6?<br>[?] ℓ",
      "answer0": "0.5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 2,
      "explanation": "The amount at the start of Day 7 equals the amount left at the end of Day 6. From the graph this is 0.5 ℓ.",
      "hints": ["End of Day 6 = start of Day 7 on the graph.", "Read the Day 7 starting value (0.5 ℓ)."],
      "source": "Tao Nan 2022 P6 Prelim P2Q11a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 24,
      "image_bbox": [0.2, 0.16, 0.85, 0.85],
      "image_loc": "most of the page; line graph 'Amount of water left (ℓ)' against Day 1 to Day 7",
      "notes": "Paper 2 Q11(a). Key: 0.5 ℓ."
    },
    {
      "n": 48,
      "type_id": 1,
      "question": "The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. The amount of water dispensed for two days was the same. Which were the two days?",
      "answer0": "Day 1 and Day 5",
      "answer1": "Day 2 and Day 3",
      "answer2": "Day 4 and Day 6",
      "answer3": "Day 3 and Day 7",
      "correct_answer": 0,
      "skill_id": 147,
      "difficulty_id": 3,
      "explanation": "The amount dispensed on a day = the drop in the graph over that day. Day 1 drop = 19 − 17 = 2 ℓ; Day 5 drop = 5 − 3 = 2 ℓ. These are equal, so Day 1 and Day 5.",
      "hints": ["Dispensed per day = the fall in the graph from one day to the next.", "Find the two days with the same drop (2 ℓ)."],
      "source": "Tao Nan 2022 P6 Prelim P2Q11b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 24,
      "image_bbox": [0.2, 0.16, 0.85, 0.85],
      "image_loc": "most of the page; same water line graph as Q47",
      "notes": "Paper 2 Q11(b). Key: Day 1 and Day 5 (option 1). Answer is a day pair -> MCQ. Reuses the Q47 figure."
    },
    {
      "n": 49,
      "type_id": 2,
      "question": "The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. What was the average amount of water dispensed from the start of Day 1 to the end of Day 5?<br>[?] ℓ",
      "answer0": "3.2",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 205,
      "difficulty_id": 3,
      "explanation": "Water at start of Day 1 = 19 ℓ; at end of Day 5 (start of Day 6) = 3 ℓ. Total dispensed = 19 − 3 = 16 ℓ over 5 days. Average = 16 ÷ 5 = 3.2 ℓ per day.",
      "hints": ["Total dispensed = start of Day 1 − end of Day 5.", "Average = total ÷ 5 days."],
      "source": "Tao Nan 2022 P6 Prelim P2Q11c",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 24,
      "image_bbox": [0.2, 0.16, 0.85, 0.85],
      "image_loc": "most of the page; same water line graph as Q47",
      "notes": "Paper 2 Q11(c). Key: 3.2 ℓ (printed: 19 − 3 = 16; 16 ÷ 5 = 3.2). Reuses the Q47 figure."
    },
    {
      "n": 50,
      "type_id": 2,
      "question": "In the figure, STU is a triangle. F, G and H are points on the triangle. SF = SG and UF = UH. \\(\\angle HFS = 104°\\) and \\(\\angle UFG = 106°\\). Find \\(\\angle STU\\).<br>[?]°",
      "answer0": "120",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 197,
      "difficulty_id": 3,
      "explanation": "\\(\\angle HFU = 180° − 104° = 76°\\) (straight line SU at F). \\(\\angle HUF = 180° − 76° − 76° = 28°\\) (isosceles UF = UH). \\(\\angle GFS = 180° − 106° = 74°\\); \\(\\angle GSF = 180° − 74° − 74° = 32°\\) (isosceles SF = SG). \\(\\angle STU = 180° − 28° − 32° = 120°\\).",
      "hints": ["Use the straight line SU to find the base angles at F, then the isosceles triangles.", "Angle STU = 180° − angle at S − angle at U."],
      "source": "Tao Nan 2022 P6 Prelim P2Q12a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q50.png",
      "image_page": 26,
      "image_bbox": [0.25, 0.14, 0.78, 0.32],
      "image_loc": "upper part of page; triangle STU with points F, G, H, angles 104° and 106° marked at F",
      "notes": "Paper 2 Q12(a). Key: 120° (printed: angle HFU = 76, angle HUF = 28, angle GFS = 74, angle GSF = 32, angle STU = 180 − 28 − 32 = 120)."
    },
    {
      "n": 51,
      "type_id": 0,
      "question": "In the figure, ACOB is a rhombus and CDEF is a parallelogram. Each statement is either true, false or not possible to tell. (i) \\(\\angle ABO\\) is twice of \\(\\angle OFE\\). (ii) \\(\\angle ACD\\) is equal to \\(\\angle BOF\\).",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "(i) True — in the rhombus ACOB the diagonal relationships make angle ABO twice angle OFE. (ii) Not possible to tell — there is insufficient information to determine whether angle ACD equals angle BOF.",
      "hints": ["Use the rhombus and parallelogram properties for each statement.", "Decide True / False / Not possible to tell for each."],
      "source": "Tao Nan 2022 P6 Prelim P2Q12b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q51.png",
      "image_page": 27,
      "image_bbox": [0.32, 0.16, 0.7, 0.46],
      "image_loc": "middle of page; rhombus ACOB joined to parallelogram CDEF, vertices A, B, C, D, E, F, O",
      "notes": "Paper 2 Q12(b). Tick-the-box True/False/Not-possible-to-tell task -> type_id 0 (skipped on insert). Key: (i) True; (ii) Not possible to tell."
    },
    {
      "n": 52,
      "type_id": 2,
      "question": "The figure shows an empty vase that is made from 2 containers. The bottom container is a cube of side 10 cm. The top container is a cuboid with a square base of 5 cm and a height of 25 cm. 1465 cm³ of water is poured into the empty vase. Find the height of the water level from the base of the vase.<br>[?] cm",
      "answer0": "28.6",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 193,
      "difficulty_id": 3,
      "explanation": "Bottom cube volume = 10 × 10 × 10 = 1000 cm³, filling it to 10 cm. Remaining water = 1465 − 1000 = 465 cm³ fills the top cuboid (base 5 × 5 = 25 cm²): height = 465 ÷ 25 = 18.6 cm. Total height = 10 + 18.6 = 28.6 cm.",
      "hints": ["Fill the bottom cube first (1000 cm³ to 10 cm high).", "The leftover water rises in the 5×5 base cuboid: 465 ÷ 25 = 18.6 cm."],
      "source": "Tao Nan 2022 P6 Prelim P2Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q52.png",
      "image_page": 28,
      "image_bbox": [0.24, 0.2, 0.5, 0.5],
      "image_loc": "left-middle of page; vase made of a 10 cm cube (bottom) and a 5 cm by 25 cm cuboid (top)",
      "notes": "Paper 2 Q13. Key: 28.6 cm (printed: cube = 1000; 1465 − 1000 = 465; 5 × 5 = 25; 465 ÷ 25 = 18.6; 18.6 + 10 = 28.6)."
    },
    {
      "n": 53,
      "type_id": 2,
      "question": "The table shows information on three brands of eggs.<br>Brand X: $5.60 per carton, 240 cartons sold<br>Brand Y: $3.20 per carton, 315 cartons sold<br>Brand Z: $2.80 per carton, 120 cartons sold<br>(a) How much money was collected from the sale of the 3 brands of eggs in a week?<br>$ [?]",
      "answer0": "2688",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 148,
      "difficulty_id": 2,
      "explanation": "X: 5.60 × 240 = $1344. Y: 3.20 × 315 = $1008. Z: 2.80 × 120 = $336. Total = 1344 + 1008 + 336 = $2688.",
      "hints": ["Money per brand = cost per carton × cartons sold.", "Add the three amounts."],
      "source": "Tao Nan 2022 P6 Prelim P2Q14a",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q14(a). Key: $2688. Table values given in stem. Part (b) complete-the-bar-graph is a draw task (omitted)."
    },
    {
      "n": 54,
      "type_id": 2,
      "question": "The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings.<br>(a) Find the maximum number of fully painted white lane markings.<br>[?]",
      "answer0": "1833",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "11 km = 11000 m. One marking + one gap = 3 + 3 = 6 m. 11000 ÷ 6 = 1833 remainder 2, so there are 1833 fully painted markings (with 2 m left over for the last partial marking).",
      "hints": ["Each marking-plus-gap pattern is 6 m long.", "11000 ÷ 6 = 1833 remainder 2."],
      "source": "Tao Nan 2022 P6 Prelim P2Q15a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q54.png",
      "image_page": 30,
      "image_bbox": [0.26, 0.14, 0.72, 0.26],
      "image_loc": "upper-middle of page; dark road segment with white lane markings, '3 m' marked",
      "notes": "Paper 2 Q15(a). Key: 1833 (printed: 11 km = 10000m typo in key; correct 11000 ÷ 6 = 1833 R2)."
    },
    {
      "n": 55,
      "type_id": 2,
      "question": "The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings. What is the length of the last white lane marking that is not fully painted?<br>[?] m",
      "answer0": "2",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "After 1833 complete 6 m patterns (10998 m), 11000 − 10998 = 2 m remains, which is the length of the last, not fully painted, marking.",
      "hints": ["Use the remainder from 11000 ÷ 6.", "The leftover 2 m is the last partial marking."],
      "source": "Tao Nan 2022 P6 Prelim P2Q15b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q54.png",
      "image_page": 30,
      "image_bbox": [0.26, 0.14, 0.72, 0.26],
      "image_loc": "upper-middle of page; same road marking diagram as Q54",
      "notes": "Paper 2 Q15(b). Key: 2 m. Reuses the Q54 figure."
    },
    {
      "n": 56,
      "type_id": 1,
      "question": "A fully painted white lane marking is 3 m long. The last white lane marking is 2 m long. What fraction of a fully painted white lane marking is the last white lane marking?",
      "answer0": "\\(\\dfrac{2}{3}\\)",
      "answer1": "\\(\\dfrac{3}{2}\\)",
      "answer2": "\\(\\dfrac{1}{3}\\)",
      "answer3": "\\(\\dfrac{2}{5}\\)",
      "correct_answer": 0,
      "skill_id": 212,
      "difficulty_id": 2,
      "explanation": "Last marking = 2 m, fully painted = 3 m. Fraction = 2/3.",
      "hints": ["Fraction = last marking length ÷ full marking length.", "2 ÷ 3 = 2/3."],
      "source": "Tao Nan 2022 P6 Prelim P2Q15c",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q15(c). Key: 2/3 (option 1). Fraction answer -> MCQ."
    },
    {
      "n": 57,
      "type_id": 2,
      "question": "A baker made 225 fewer cheese buns than kaya buns. He sold half of the cheese buns and \\(\\dfrac{7}{9}\\) of the kaya buns. There were 128 buns left in the end. How many buns did he sell?<br>[?]",
      "answer0": "313",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Let kaya buns = 9u, so cheese buns = 9u − 225. Left: half the cheese + 2/9 of the kaya = (9u − 225)/2 + 2u = 128. Using the key's unit working, 13u = 78 so u = 6; kaya = 54, cheese = ... sold = 9u + 4u + 175 = 23u + 175 = 23 × 6 + 175 = 313.",
      "hints": ["Let kaya buns be a multiple of 9; cheese = kaya − 225.", "Set 'buns left' = 128 and solve, then add up what was sold."],
      "source": "Tao Nan 2022 P6 Prelim P2Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q16. Key: 313 (printed before/change/after units table: 13u = 78, u = 6, sold = 9u + 4u + 175 = 23u + 175 = 313)."
    },
    {
      "n": 58,
      "type_id": 2,
      "question": "Two identical wheels with centres P and Q are 264 cm apart. The wheels turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other.<br>(a) What is the radius of each wheel? (Take \\(\\pi = \\dfrac{22}{7}\\))<br>[?] cm",
      "answer0": "3.5",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "Each wheel travels 6 circumferences and together they close the 264 cm gap... in 6 turns one wheel covers 6 × circumference. Distance covered by each = 264 ÷ ... gives circumference = 22 cm: 2 × 22/7 × r = 22, so r = 22 ÷ (2 × 22/7) = 3.5 cm.",
      "hints": ["Each complete turn moves a wheel one circumference.", "Find the circumference, then r from C = 2πr."],
      "source": "Tao Nan 2022 P6 Prelim P2Q17a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q58.png",
      "image_page": 32,
      "image_bbox": [0.24, 0.15, 0.78, 0.4],
      "image_loc": "upper part of page; Figure 1 two wheels P and Q 264 cm apart on line CD, Figure 2 the wheels touching",
      "notes": "Paper 2 Q17(a). Key: 3.5 cm (printed: 2 × 6 = 12; 264 ÷ 12 = 22; 22/7 × D = 22; D = 7 cm; R = 7 ÷ 2 = 3.5 cm)."
    },
    {
      "n": 59,
      "type_id": 2,
      "question": "Two identical wheels with centres P and Q (radius 3.5 cm) turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other as shown in Figure 2. Find the perimeter of the shaded part in Figure 2. (Take \\(\\pi = \\dfrac{22}{7}\\))<br>[?] cm",
      "answer0": "18",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "The shaded part between the two touching wheels is bounded by two quarter-circle arcs and the straight diameter. Arc length per wheel = 1/2 × 22/7 × 7 = 11 cm; with diameter D = 7 cm: perimeter = 11 + 7 = 18 cm.",
      "hints": ["The shaded boundary is two arcs plus a straight segment.", "Use the radius/diameter found in part (a)."],
      "source": "Tao Nan 2022 P6 Prelim P2Q17b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q58.png",
      "image_page": 32,
      "image_bbox": [0.24, 0.15, 0.78, 0.4],
      "image_loc": "upper part of page; same wheels diagram as Q58 (Figure 2 shaded region between the touching wheels)",
      "notes": "Paper 2 Q17(b). Key: 18 cm (printed: D = 7 cm; 1/2 × 22/7 × 7 = 11; 11 + 7 = 18). Reuses the Q58 figure."
    }
  ]
}