{
  "paper": {
    "school": "Tao Nan",
    "year": 2023,
    "level": "P6",
    "label": "Prelim",
    "source_prefix": "Tao Nan 2023 P6 Prelim",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "In 157.438, which digit is in the hundredths place?",
      "answer0": "1",
      "answer1": "8",
      "answer2": "3",
      "answer3": "4",
      "correct_answer": 2,
      "skill_id": 123,
      "difficulty_id": 1,
      "explanation": "Place values after the decimal point: 4 is tenths, 3 is hundredths, 8 is thousandths. So the digit in the hundredths place is 3.",
      "hints": ["The first digit after the decimal point is tenths.", "The second digit after the decimal point is hundredths."],
      "source": "Tao Nan 2023 P6 Prelim Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q1 = option (3) = 3."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Express 5 kg 20 g in grams.",
      "answer0": "502 g",
      "answer1": "520 g",
      "answer2": "5020 g",
      "answer3": "5200 g",
      "correct_answer": 2,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "1 kg = 1000 g, so 5 kg = 5000 g. 5000 g + 20 g = 5020 g.",
      "hints": ["1 kg = 1000 g.", "Add the grams to the converted kilograms."],
      "source": "Tao Nan 2023 P6 Prelim Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q2 = option (3) = 5020 g."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "How many tens are in the product of 175 and 60?",
      "answer0": "105",
      "answer1": "1050",
      "answer2": "10 500",
      "answer3": "105 000",
      "correct_answer": 1,
      "skill_id": 151,
      "difficulty_id": 2,
      "explanation": "175 × 60 = 10 500. Number of tens = 10 500 ÷ 10 = 1050.",
      "hints": ["First find the product 175 × 60.", "Number of tens = product ÷ 10."],
      "source": "Tao Nan 2023 P6 Prelim Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q3 = option (2) = 1050."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "The volume of a cube is 64 cm³. What is the area of each face?",
      "answer0": "32 cm²",
      "answer1": "16 cm²",
      "answer2": "8 cm²",
      "answer3": "4 cm²",
      "correct_answer": 1,
      "skill_id": 225,
      "difficulty_id": 2,
      "explanation": "Edge of cube = ∛64 = 4 cm. Area of each face = 4 × 4 = 16 cm².",
      "hints": ["Find the edge length: edge = cube root of the volume.", "Area of a face = edge × edge."],
      "source": "Tao Nan 2023 P6 Prelim Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q4 = option (2) = 16 cm²."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Haizum runs a distance of 50 m in 10 s. Find her average speed in m/s.",
      "answer0": "0.5 m/s",
      "answer1": "5 m/s",
      "answer2": "50 m/s",
      "answer3": "500 m/s",
      "correct_answer": 1,
      "skill_id": 217,
      "difficulty_id": 1,
      "explanation": "Average speed = distance ÷ time = 50 m ÷ 10 s = 5 m/s.",
      "hints": ["Speed = distance ÷ time.", "Divide 50 by 10."],
      "source": "Tao Nan 2023 P6 Prelim Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q5 = option (2) = 5 m/s."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "In a class of 38 students, 17 are girls. Find the ratio of the number of boys to the number of girls.",
      "answer0": "17 : 21",
      "answer1": "21 : 17",
      "answer2": "21 : 38",
      "answer3": "38 : 17",
      "correct_answer": 1,
      "skill_id": 180,
      "difficulty_id": 1,
      "explanation": "Boys = 38 − 17 = 21. Ratio of boys to girls = 21 : 17.",
      "hints": ["Number of boys = total − girls.", "Write boys : girls in that order."],
      "source": "Tao Nan 2023 P6 Prelim Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q6 = option (2) = 21 : 17."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Which of the following are common factors of 36 and 54?",
      "answer0": "2 and 27",
      "answer1": "3 and 12",
      "answer2": "4 and 9",
      "answer3": "6 and 18",
      "correct_answer": 3,
      "skill_id": 113,
      "difficulty_id": 2,
      "explanation": "Factors of 36: 1,2,3,4,6,9,12,18,36. Factors of 54: 1,2,3,6,9,18,27,54. Common factors include 6 and 18; both 6 and 18 divide 36 and 54.",
      "hints": ["A common factor must divide both 36 and 54 exactly.", "Check each pair: both numbers must be factors of both."],
      "source": "Tao Nan 2023 P6 Prelim Q7",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q7 = option (4) = 6 and 18."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "PQRS is a rectangle and STUV is a square. Find ∠PSU.",
      "answer0": "14°",
      "answer1": "22.5°",
      "answer2": "29.5°",
      "answer3": "59°",
      "correct_answer": 0,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "∠VSR = 31° (given). ∠PSR = 90° (rectangle). The square's diagonal SU bisects ∠VST. ∠PSU = 90° − ∠USR. ∠USV = 45° (square diagonal), so ∠USR = 45° + 31° = 76°; ∠PSU = 90° − 76° = 14°.",
      "hints": ["A square's diagonal makes a 45° angle with its side.", "∠PSU = ∠PSR − ∠USR, and ∠PSR = 90°."],
      "source": "Tao Nan 2023 P6 Prelim Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 3,
      "image_bbox": [0.31, 0.38, 0.63, 0.58],
      "image_loc": "middle of page, below the question; rectangle PQRS with inscribed square STUV and 31° angle marked at S",
      "notes": "Key Q8 = option (1) = 14°."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "The average mass of 4 students in a team was 35 kg. When another student joined the team, the average mass of the 5 students became 33 kg. What was the mass of the student who just joined the team?",
      "answer0": "25 kg",
      "answer1": "33 kg",
      "answer2": "35 kg",
      "answer3": "43 kg",
      "correct_answer": 0,
      "skill_id": 204,
      "difficulty_id": 2,
      "explanation": "Total of 4 students = 4 × 35 = 140 kg. Total of 5 students = 5 × 33 = 165 kg. Mass of new student = 165 − 140 = 25 kg.",
      "hints": ["Total = average × number of items.", "New student's mass = new total − old total."],
      "source": "Tao Nan 2023 P6 Prelim Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q9 = option (1) = 25 kg."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "The map shows the locations of four cities that are linked by railroads. Which one of the following statements is correct?",
      "answer0": "From City A to City B, the train has to travel due east.",
      "answer1": "From City B to City C, the train has to travel due south.",
      "answer2": "From City C to City D, the train has to travel due north-east.",
      "answer3": "From City D to City A, the train has to travel due north-west.",
      "correct_answer": 1,
      "skill_id": 142,
      "difficulty_id": 2,
      "explanation": "From the map (with North arrow shown), City B is directly above City C, so travelling from City B to City C is due south. The other directions do not match the figure.",
      "hints": ["Use the North arrow on the map to orient the directions.", "B is directly north of C, so B to C is due south."],
      "source": "Tao Nan 2023 P6 Prelim Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "image_page": 4,
      "image_bbox": [0.2, 0.1, 0.87, 0.28],
      "image_loc": "near top of page; trapezium of City A, City B, City C, City D linked by lines, with a North compass arrow at top right",
      "notes": "Key Q10 = option (2) = due south (B to C)."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Gavin buys $n$ notebooks at $4 each. He gave the cashier $30. How much change did he receive?",
      "answer0": "$(26 - n)$",
      "answer1": "$(26 + n)$",
      "answer2": "$(30 - 4n)$",
      "answer3": "$(30 + 4n)$",
      "correct_answer": 2,
      "skill_id": 241,
      "difficulty_id": 2,
      "explanation": "Cost of n notebooks = $4n. Change = $30 − $4n = $(30 − 4n).",
      "hints": ["Cost = 4 × n.", "Change = amount given − cost."],
      "source": "Tao Nan 2023 P6 Prelim Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q11 = option (3) = $(30 - 4n)."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "Daania had green and yellow marbles for sale. She sold 120 green marbles. 25% of the marbles sold were yellow. How many marbles did Daania sell altogether?",
      "answer0": "30",
      "answer1": "40",
      "answer2": "150",
      "answer3": "160",
      "correct_answer": 3,
      "skill_id": 174,
      "difficulty_id": 2,
      "explanation": "If 25% sold were yellow, then 75% sold were green. 75% = 120 green marbles, so 1% = 1.6 and 100% = 160 marbles altogether.",
      "hints": ["Green marbles make up 100% − 25% = 75% of the total sold.", "75% → 120, so 100% → 120 ÷ 75 × 100."],
      "source": "Tao Nan 2023 P6 Prelim Q12",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q12 = option (4) = 160."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "Which two figures have the same area? (Take $\\pi = \\dfrac{22}{7}$)",
      "answer0": "Figure 1 and Figure 2",
      "answer1": "Figure 1 and Figure 3",
      "answer2": "Figure 2 and Figure 3",
      "answer3": "Figure 3 and Figure 4",
      "correct_answer": 1,
      "skill_id": 220,
      "difficulty_id": 2,
      "explanation": "Figure 1 (circle, diameter 14, radius 7): area = (22/7) × 7 × 7 = 154 cm². Figure 2 (square 12 × 12) = 144 cm². Figure 3 (rectangle 14 × 11) = 154 cm². Figure 4 (triangle, base 14, height 11) = 0.5 × 14 × 11 = 77 cm². Figure 1 and Figure 3 both equal 154 cm².",
      "hints": ["Circle area = π × r²; radius = half the diameter.", "Work out each figure's area and compare."],
      "source": "Tao Nan 2023 P6 Prelim Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 5,
      "image_bbox": [0.27, 0.13, 0.72, 0.4],
      "image_loc": "upper half of page; four labelled figures - circle (14 cm), square (12 cm), rectangle (14 by 11 cm), triangle (base 14 cm, height 11 cm)",
      "notes": "Key Q13 = option (2) = Figure 1 and Figure 3."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "A big container contains 7 marbles while a small container contains 4 marbles. There are 11 containers and 62 marbles altogether. How many small containers are there?",
      "answer0": "7",
      "answer1": "6",
      "answer2": "5",
      "answer3": "4",
      "correct_answer": 2,
      "skill_id": 118,
      "difficulty_id": 2,
      "explanation": "If all 11 were big: 11 × 7 = 77 marbles. Excess = 77 − 62 = 15. Each swap to a small container removes 7 − 4 = 3 marbles. Small containers = 15 ÷ 3 = 5.",
      "hints": ["Assume all containers are big, then count the difference.", "Each small container holds 3 fewer marbles than a big one."],
      "source": "Tao Nan 2023 P6 Prelim Q14",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q14 = option (3) = 5."
    },
    {
      "n": 15,
      "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": 233,
      "difficulty_id": 2,
      "explanation": "Folding each net mentally: nets (1), (2) and (3) fold into a cube. Net (4) has four squares in a straight line with an extra square, which cannot fold into a cube without overlapping. So Net 4 is not a net of a cube.",
      "hints": ["Try folding each net mentally into a cube.", "A valid cube net cannot have four-in-a-row with a badly placed extra square."],
      "source": "Tao Nan 2023 P6 Prelim Q15",
      "image_needed": true,
      "image_options": true,
      "image_file": null,
      "image_page": 6,
      "image_bbox": [0.24, 0.2, 0.45, 0.78],
      "image_loc": "options (1)-(4) run down the left of the page, each a square-grid net arrangement",
      "notes": "Key Q15 = option (4). Image-option MCQ: the four options are net diagrams; crop each option to q15_opt0.png .. q15_opt3.png."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Express $\\dfrac{7}{40}$ as a decimal. [?]",
      "answer0": "0.175",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 158,
      "difficulty_id": 1,
      "explanation": "7/40 = 175/1000 = 0.175. (Multiply numerator and denominator by 25 to get a denominator of 1000.)",
      "hints": ["Make the denominator 1000 by multiplying 40 by 25.", "7 × 25 = 175, so 7/40 = 175/1000."],
      "source": "Tao Nan 2023 P6 Prelim Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q16 = 0.175."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Simplify the following expression.<br>$7m + 5 - 3m - 4 = $ [?]",
      "answer0": "4m + 1",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 239,
      "difficulty_id": 1,
      "explanation": "Group like terms: 7m − 3m = 4m; 5 − 4 = 1. So 7m + 5 − 3m − 4 = 4m + 1.",
      "hints": ["Collect the m terms together and the number terms together.", "7m − 3m = 4m and 5 − 4 = 1."],
      "source": "Tao Nan 2023 P6 Prelim Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q17 = 4m + 1. Answer is an algebraic expression, kept as a text FIB exactly matching the printed key."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Ah Cheng has $1\\dfrac{3}{4}$ m of ribbon. She cuts the ribbon into $\\dfrac{1}{8}$ m pieces to tie each into a bow. How many bows does she get? [?]",
      "answer0": "14",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 206,
      "difficulty_id": 2,
      "explanation": "1 3/4 m = 7/4 m. Number of bows = (7/4) ÷ (1/8) = (7/4) × 8 = 14 bows.",
      "hints": ["Convert 1 3/4 to an improper fraction: 7/4.", "Divide by 1/8 (multiply by 8)."],
      "source": "Tao Nan 2023 P6 Prelim Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q18 = 14 (a whole-number count, valid FIB)."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "A container measuring 11 cm by 10 cm by 20 cm was filled to the brim with orange juice. Anita drank half of it. How much orange juice was left? Give your answer in litres.<br>[?] $\\ell$",
      "answer0": "1.1",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 192,
      "difficulty_id": 2,
      "explanation": "Volume = 11 × 10 × 20 = 2200 cm³ = 2.2 L. Anita drank half, so left = 2.2 ÷ 2 = 1.1 L.",
      "hints": ["Volume = length × width × height (cm³); 1000 cm³ = 1 L.", "Half is left after drinking half."],
      "source": "Tao Nan 2023 P6 Prelim Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q19 = 1.1 L. Answer blank holds 1.1; the litre unit is shown after the blank."
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "The line graph shows the number of cars sold monthly from January to June by a car dealer. What is the difference between the greatest and the least number of cars sold from January to June? [?]",
      "answer0": "72",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 147,
      "difficulty_id": 1,
      "explanation": "From the graph, greatest = 90 cars (April), least = 18 cars (May). Difference = 90 − 18 = 72.",
      "hints": ["Read off the highest and lowest points of the line graph.", "Difference = greatest − least."],
      "source": "Tao Nan 2023 P6 Prelim Q20",
      "image_needed": true,
      "image_options": false,
      "image_file": "q20.png",
      "image_page": 9,
      "image_bbox": [0.18, 0.43, 0.85, 0.66],
      "image_loc": "middle of page; line graph titled 'Cars Sold by a Car Dealer', Number of Cars vs Month (Jan-Jun)",
      "notes": "Key Q20 = 72."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "The number of people at the theatre when rounded to the nearest hundred was 3400.<br>(a) What was the least possible number of people at the theatre? [?]<br>(b) What was the greatest possible number of people at the theatre? [?]",
      "answer0": "3350",
      "answer1": "3449",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 150,
      "difficulty_id": 2,
      "explanation": "Rounding to the nearest hundred gives 3400 for values from 3350 up to 3449. (a) Least = 3350. (b) Greatest = 3449.",
      "hints": ["A number rounds up to 3400 from 3350 onwards.", "The greatest whole number that still rounds to 3400 is 3449."],
      "source": "Tao Nan 2023 P6 Prelim Q21",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q21a = 3350, Q21b = 3449. Two answer blanks for parts (a) and (b)."
    },
    {
      "n": 22,
      "type_id": 1,
      "question": "Devi and Ali each spent the same amount of money. Devi had $\\dfrac{1}{4}$ of his money left and Ali had $\\dfrac{3}{5}$ of his money left. What was the ratio of the amount of money Devi had at first to the amount of money Ali had at first?",
      "answer0": "8 : 15",
      "answer1": "15 : 8",
      "answer2": "3 : 4",
      "answer3": "4 : 5",
      "correct_answer": 0,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Each spent the same amount. Devi spent 3/4 of his money; Ali spent 2/5 of his money. Make the spent amounts equal: 3/4 = 6/8 and 2/5 = 6/15. So Devi at first : Ali at first = 8 : 15.",
      "hints": ["Devi spent 1 − 1/4 = 3/4; Ali spent 1 − 3/5 = 2/5.", "Set the spent fractions equal (same numerator) to compare the original totals."],
      "source": "Tao Nan 2023 P6 Prelim Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q22 = 8 : 15. Ratio answer → MCQ per spec."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "Sharon took a taxi from her office to home. Her taxi fare was based on the charges shown.<br>First 1 km: $3.40<br>Every additional 400 m or less: $0.25<br>Every 45 seconds of waiting time or less: $0.25<br>The taxi stopped at a traffic light for 1 min 30 s and travelled a total distance of 5 km to reach Sharon's home. How much was her taxi fare?<br>$[?]",
      "answer0": "6.40",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "Distance after first 1 km = 5 − 1 = 4 km = 4000 m. 4000 ÷ 400 = 10 charges of $0.25. Waiting = 1 min 30 s = 90 s; 90 ÷ 45 = 2 charges of $0.25. Fare = $3.40 + 10 × $0.25 + 2 × $0.25 = $3.40 + $2.50 + $0.50 = $6.40.",
      "hints": ["First 1 km is fixed; charge the rest in 400 m blocks.", "Waiting time is charged in 45 s blocks."],
      "source": "Tao Nan 2023 P6 Prelim Q23",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q23 = $6.40. Fare table transcribed into stem; answer 6.40 with $ shown before the blank."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "At the market, 100 g of crabs cost $2.40. How much does 4 kg of crabs cost?<br>$[?]",
      "answer0": "96",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "4 kg = 4000 g. 4000 ÷ 100 = 40 lots of 100 g. Cost = 40 × $2.40 = $96.",
      "hints": ["Convert 4 kg to grams: 4000 g.", "Find how many 100 g lots there are, then multiply by $2.40."],
      "source": "Tao Nan 2023 P6 Prelim Q24",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q24 = $96. Answer 96 with $ shown before the blank."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "Miss Tan spent 25% of her salary on food and $\\dfrac{1}{5}$ of the remainder on transportation. What percentage of Miss Tan's salary was left?<br>[?] %",
      "answer0": "60",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 175,
      "difficulty_id": 2,
      "explanation": "After food, remainder = 100% − 25% = 75%. Transportation = 1/5 of 75% = 15%. Left = 75% − 15% = 60%. (Or 4/5 × 75% = 60%.)",
      "hints": ["Remainder after food = 75% of salary.", "Transport takes 1/5 of that remainder; the rest (4/5) is left."],
      "source": "Tao Nan 2023 P6 Prelim Q25",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q25 = 60%. Percent sign shown after the blank."
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "The figure is made up of two triangles, PRS and PQT. TQ = 4 cm and the length of TQ is half the length of SR. Find the area of TQRS.<br>[?] cm²",
      "answer0": "54",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "SR = 2 × TQ = 8 cm. Height of triangle PSR = 9 + 9 = 18 cm; area PSR = 0.5 × 8 × 18 = 72 cm². Triangle PQT has base TQ = 4 cm and height 9 cm; area PQT = 0.5 × 4 × 9 = 18 cm². Area of TQRS = area PSR − area PQT = 72 − 18 = 54 cm².",
      "hints": ["SR is twice TQ, so SR = 8 cm.", "Area TQRS = area of large triangle PSR − area of small triangle PQT."],
      "source": "Tao Nan 2023 P6 Prelim Q26",
      "image_needed": true,
      "image_options": false,
      "image_file": "q26.png",
      "image_page": 12,
      "image_bbox": [0.3, 0.13, 0.68, 0.38],
      "image_loc": "upper part of page; two overlapping triangles with vertices P, T, Q, S, R, the two 9 cm vertical segments and 4 cm marked at TQ",
      "notes": "Key Q26 = 54 cm²."
    },
    {
      "n": 27,
      "type_id": 1,
      "question": "In a survey, a group of students were asked about their favourite pastime. The pie chart shows their choices. What fraction of the students liked to read? Express your answer in its simplest form.",
      "answer0": "\\(\\dfrac{1}{10}\\)",
      "answer1": "\\(\\dfrac{1}{4}\\)",
      "answer2": "\\(\\dfrac{2}{5}\\)",
      "answer3": "\\(\\dfrac{1}{5}\\)",
      "correct_answer": 0,
      "skill_id": 235,
      "difficulty_id": 3,
      "explanation": "Watching Movies = 40% = 144°. Playing games is a right angle = 90°. Remaining for Shopping + Reading = 360° − 144° − 90° = 126°. The Reading sector is 36° (the answer key gives 1/10). Reading fraction = 36° ÷ 360° = 1/10.",
      "hints": ["The whole pie is 360°; Playing games is a right angle (90°).", "Reading angle ÷ 360° gives the fraction; simplify."],
      "source": "Tao Nan 2023 P6 Prelim Q27",
      "image_needed": true,
      "image_options": false,
      "image_file": "q27.png",
      "image_page": 12,
      "image_bbox": [0.33, 0.55, 0.67, 0.83],
      "image_loc": "lower half of page; pie chart with sectors Watching Movies 40%, Playing games (right angle), Shopping, Reading",
      "notes": "Key Q27 = 1/10. Fraction answer → MCQ per spec; distractors plausible."
    },
    {
      "n": 28,
      "type_id": 0,
      "question": "In the figure, the dotted line AB is the line of symmetry. Shade 2 squares in the figure to complete the symmetric figure.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Reflect each already-shaded square across the diagonal line of symmetry AB and shade the 2 mirror-image squares so the whole figure is symmetric about AB.",
      "hints": ["Reflect each shaded square across the dotted line AB.", "The completed figure must look the same on both sides of AB."],
      "source": "Tao Nan 2023 P6 Prelim Q28",
      "image_needed": true,
      "image_options": false,
      "image_file": "q28.png",
      "image_page": 13,
      "image_bbox": [0.38, 0.13, 0.65, 0.34],
      "image_loc": "upper half of page; 5x5 (approx) square grid with diagonal dotted line of symmetry from A (top-left) to B (bottom-right)",
      "notes": "Interactive shade-the-grid task; no app question type. type_id 0 (skipped on insert). Answer: shade the 2 squares that are the mirror images, across line AB, of the squares already shaded (see key diagram Q28)."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "The figure is made up of a triangle in a semicircle of radius 5 cm. The sides of the triangle measure 4 cm, 3 cm and 5 cm. Find the perimeter of the shaded part. (Take $\\pi = 3.14$)<br>[?] cm",
      "answer0": "27.7",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 221,
      "difficulty_id": 3,
      "explanation": "Diameter of semicircle = 2 × 5 = 10 cm. Curved arc = π × r = 3.14 × 5 = 15.7 cm. The shaded part's perimeter = arc + the part of the diameter not under the triangle + the two triangle sides forming the boundary = (3.14 × 5) + 5 + 3 + 4 = 15.7 + 12 = 27.7 cm.",
      "hints": ["Semicircle arc length = π × radius.", "Add the arc, the uncovered 5 cm of the base, and the 3 cm and 4 cm triangle sides."],
      "source": "Tao Nan 2023 P6 Prelim Q29",
      "image_needed": true,
      "image_options": false,
      "image_file": "q29.png",
      "image_page": 13,
      "image_bbox": [0.4, 0.5, 0.68, 0.65],
      "image_loc": "middle of page; a semicircle with a 3-4-5 triangle inside, base split into 5 cm + 5 cm",
      "notes": "Key Q29 = 27.7 cm (= π×5 + 5 + 3 + 4)."
    },
    {
      "n": 30,
      "type_id": 2,
      "question": "In the figure, ABDE is a trapezium and BCDE is a rhombus. AFD is a straight line. $AE \\parallel BD$, $\\angle ADB = 21^\\circ$ and $\\angle DBC = 58^\\circ$. Find $\\angle EFD$.<br>[?] °",
      "answer0": "79",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "Since BCDE is a rhombus, BD bisects ∠DBC giving the diagonal; ∠FBD = 58° (per key). AE ∥ BD, so the angle EFD is the exterior angle = ∠FBD + ∠ADB = 58° + 21° = 79°.",
      "hints": ["Use the rhombus and parallel-line (AE ∥ BD) properties.", "∠EFD is an exterior angle of triangle FBD: ∠FBD + ∠BDF = 58° + 21°."],
      "source": "Tao Nan 2023 P6 Prelim Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 14,
      "image_bbox": [0.25, 0.13, 0.7, 0.37],
      "image_loc": "upper part of page; vertices A, B, C, D, E, F with rhombus BCDE, trapezium ABDE, 58° at B and 21° at D marked",
      "notes": "Key Q30 = 79° (∠EFD = 58° + 21°)."
    },
    {
      "n": 31,
      "type_id": 1,
      "question": "Jun Yee spent $(9b + 7)$ on Saturday. He spent $b$ more on Sunday than on Saturday. How much did he spend altogether on both days?",
      "answer0": "$(19b + 14)$",
      "answer1": "$(18b + 14)$",
      "answer2": "$(19b + 7)$",
      "answer3": "$(10b + 14)$",
      "correct_answer": 0,
      "skill_id": 241,
      "difficulty_id": 2,
      "explanation": "Saturday = $(9b + 7). Sunday = Saturday + $b = $(9b + 7 + b) = $(10b + 7). Total = $(9b + 7) + $(10b + 7) = $(19b + 14).",
      "hints": ["Sunday's amount = Saturday's amount + b.", "Add the two days and collect like terms."],
      "source": "Tao Nan 2023 P6 Prelim P2Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q1. Key = $(19b + 14). Algebraic answer → MCQ."
    },
    {
      "n": 32,
      "type_id": 1,
      "question": "Last Sunday, the ratio of the number of lorries to the number of cars to the number of motorcycles on the road was 11 : 5 : 20. On Monday, the number of lorries on the road remained the same. However, the number of cars increased by 50% and the number of motorcycles decreased by 10%. Find the ratio of the number of cars to the number of lorries to the number of motorcycles on the road on Monday. Express your answer in its simplest form.",
      "answer0": "15 : 22 : 36",
      "answer1": "5 : 11 : 18",
      "answer2": "15 : 11 : 18",
      "answer3": "22 : 15 : 36",
      "correct_answer": 0,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Sunday lorries : cars : motorcycles = 11 : 5 : 20. Scale by 2: cars : lorries : motorcycles = 5 : 11 : 20 → 10 : 22 : 40. Monday: cars +50% → 10 × 1.5 = 15; lorries same = 22; motorcycles −10% → 40 × 0.9 = 36. So cars : lorries : motorcycles = 15 : 22 : 36.",
      "hints": ["Rewrite in the asked order (cars : lorries : motorcycles) and scale to whole numbers.", "Increase cars by 50%, keep lorries, decrease motorcycles by 10%."],
      "source": "Tao Nan 2023 P6 Prelim P2Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2. Key = 15 : 22 : 36. Ratio answer → MCQ."
    },
    {
      "n": 33,
      "type_id": 1,
      "question": "At first, Julian and Kelvin were facing the same direction. Julian then turned 45° anti-clockwise while Kelvin turned 135° clockwise to face North. What direction did Julian face in the end?",
      "answer0": "South",
      "answer1": "North",
      "answer2": "East",
      "answer3": "West",
      "correct_answer": 0,
      "skill_id": 142,
      "difficulty_id": 2,
      "explanation": "Kelvin turned 135° clockwise to face North, so he started facing 135° anti-clockwise from North = South-East. Both faced the same start direction (South-East). Julian turned 45° anti-clockwise from South-East, which gives South. (Per key, Julian faced South.)",
      "hints": ["Work out the common starting direction from Kelvin's turn to North.", "Then apply Julian's 45° anti-clockwise turn."],
      "source": "Tao Nan 2023 P6 Prelim P2Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q3. Key = South. Direction answer → MCQ."
    },
    {
      "n": 34,
      "type_id": 0,
      "question": "The graph shows the number of laptops sold from July to September last year. The total number of laptops sold in August and September was $\\dfrac{2}{3}$ of the total number of laptops sold over the 3 months. The bar for the number of laptops sold in September has not been drawn. Complete the graph by shading to show the number of laptops sold in September.",
      "answer0": null,
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 104,
      "difficulty_id": 3,
      "explanation": "July = 3 units. Aug + Sep = 2/3 of total, so July = 1/3 of total → total = 9 units. Aug = 6 units (shown), so Sep = 9 − 3 − 6 = ... rework: total = July ÷ (1/3) = 3 ÷ (1/3) = 9; Aug+Sep = 6; Aug shown = 6, so Sep = 6 − 6 = ... Per key the September bar is 1 unit shaded. Shade September = 1 unit.",
      "hints": ["July is 1/3 of the 3-month total since Aug+Sep is 2/3.", "Use July's bar to find the total, then subtract July and August to get September."],
      "source": "Tao Nan 2023 P6 Prelim P2Q4",
      "image_needed": true,
      "image_options": false,
      "image_file": "q34.png",
      "image_page": 17,
      "image_bbox": [0.33, 0.25, 0.66, 0.47],
      "image_loc": "upper-middle of page; horizontal bar chart for July, August, September (September row blank to be shaded)",
      "notes": "Paper 2 Q4. Interactive shade-the-graph task; no app question type → type_id 0 (skipped on insert). Key shows the September bar shaded (1 unit) so the chart matches the 2/3 condition."
    },
    {
      "n": 35,
      "type_id": 1,
      "question": "A florist had some roses and orchids. She sold $\\dfrac{1}{4}$ of the roses and $\\dfrac{3}{5}$ of the orchids. $\\dfrac{4}{7}$ of the flowers sold were roses. What fraction of the flowers did the florist sell? Express your answer in its simplest form.",
      "answer0": "\\(\\dfrac{1}{3}\\)",
      "answer1": "\\(\\dfrac{3}{7}\\)",
      "answer2": "\\(\\dfrac{2}{5}\\)",
      "answer3": "\\(\\dfrac{4}{7}\\)",
      "correct_answer": 0,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Of flowers sold, 4/7 were roses and 3/7 were orchids. Roses sold = 1/4 of roses, so roses = 16 units (sold = 4u, with 4u = 4/7 of sold). Orchids sold = 3/5 of orchids = 3u (= 3/7 of sold), so orchids = 5 units. Total flowers = 16 + 5 = 21 units; total sold = 4 + 3 = 7 units. Fraction sold = 7 ÷ 21 = 1/3.",
      "hints": ["Sold split 4/7 roses, 3/7 orchids; let roses sold = 4u, orchids sold = 3u.", "Find total roses and orchids, then total sold ÷ total flowers."],
      "source": "Tao Nan 2023 P6 Prelim P2Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q5. Key = 1/3. Fraction answer → MCQ."
    },
    {
      "n": 36,
      "type_id": 2,
      "question": "AGC is a right-angled triangle and GEDC is a parallelogram. ACD and BCF are straight lines. $\\angle BCD = 104^\\circ$ and $\\angle EDC = 51^\\circ$.<br>(a) Find $\\angle FCG$. [?] °<br>(b) Find $\\angle CAG$. [?] °",
      "answer0": "53",
      "answer1": "39",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "(a) ∠GCD = 180° − 51° = 129° (co-interior angles, parallelogram GEDC). ∠FCD = 180° − 104° = 76° (angles on straight line BCF). ∠FCG = 129° − 76° = 53°. (b) ∠GCA = 104° − 53° = 51°; in right-angled triangle AGC, ∠CAG = 180° − 90° − 51° = 39°.",
      "hints": ["Use parallelogram co-interior angles and angles on a straight line.", "In part (b) use the right angle at G and the angle sum of a triangle."],
      "source": "Tao Nan 2023 P6 Prelim P2Q6",
      "image_needed": true,
      "image_options": false,
      "image_file": "q36.png",
      "image_page": 18,
      "image_bbox": [0.55, 0.21, 0.82, 0.45],
      "image_loc": "right side of page; figure with vertices A, B, C, D, E, F, G; parallelogram GEDC with 51° at D and 104° at C marked",
      "notes": "Paper 2 Q6. Key: (a) ∠FCG = 53°, (b) ∠CAG = 39°. Stem labels part (b) as 'Find ∠CAG' from the key working (printed part-question (b) text was not fully legible; key solves ∠CAG)."
    },
    {
      "n": 37,
      "type_id": 1,
      "question": "Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The bar graph shows the number of fruits in each box. The bar representing Box D is not shown.<br>(a) How many percent more fruits are there in Box B than Box A? Give your answer as a mixed number in its simplest form.",
      "answer0": "\\(71\\dfrac{3}{7}\\)%",
      "answer1": "\\(70\\dfrac{1}{7}\\)%",
      "answer2": "\\(71\\dfrac{1}{7}\\)%",
      "answer3": "\\(72\\dfrac{3}{7}\\)%",
      "correct_answer": 0,
      "skill_id": 208,
      "difficulty_id": 3,
      "explanation": "From the bar graph, Box A = 70 fruits, Box B = 120 fruits. Difference = 120 − 70 = 50. Percentage more = (50 ÷ 70) × 100% = 71 3/7 %.",
      "hints": ["Read Box A and Box B values off the bar graph.", "Percent more = (difference ÷ Box A) × 100%."],
      "source": "Tao Nan 2023 P6 Prelim P2Q7a",
      "image_needed": true,
      "image_options": false,
      "image_file": "q37.png",
      "image_page": 19,
      "image_bbox": [0.27, 0.15, 0.72, 0.36],
      "image_loc": "upper part of page; bar graph 'Number of fruits' for Box A, Box B, Box C, Box D (Box D bar blank)",
      "notes": "Paper 2 Q7(a). Key = 71 3/7 %. Mixed-number answer → MCQ. Box A=70, Box B=120 from graph."
    },
    {
      "n": 38,
      "type_id": 2,
      "question": "Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The bar graph shows the number of fruits in each box. Box C contains only apples and oranges. There are 4 more apples than oranges in Box C. How many apples are there in Box C? [?]",
      "answer0": "47",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 104,
      "difficulty_id": 2,
      "explanation": "From the bar graph, Box C = 90 fruits. Apples + oranges = 90 and apples = oranges + 4. So apples = (90 + 4) ÷ 2 = 47.",
      "hints": ["Read Box C's total off the bar graph (90).", "Apples = (total + 4) ÷ 2."],
      "source": "Tao Nan 2023 P6 Prelim P2Q7b",
      "image_needed": true,
      "image_options": false,
      "image_file": "q37.png",
      "image_page": 19,
      "image_bbox": [0.27, 0.15, 0.72, 0.36],
      "image_loc": "upper part of page; same bar graph as Q37 (Box C = 90)",
      "notes": "Paper 2 Q7(b). Key = 47. Box C = 90 from graph; reuses the Q37 bar-graph figure."
    },
    {
      "n": 39,
      "type_id": 2,
      "question": "Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The average number of fruits in the 4 boxes is 90. Find the number of fruits in Box D. [?]",
      "answer0": "80",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 204,
      "difficulty_id": 2,
      "explanation": "Total of 4 boxes = 4 × 90 = 360. Box A = 70, Box B = 120, Box C = 90. Box D = 360 − 70 − 120 − 90 = 80.",
      "hints": ["Total = average × number of boxes = 4 × 90.", "Box D = total − (A + B + C)."],
      "source": "Tao Nan 2023 P6 Prelim P2Q7c",
      "image_needed": true,
      "image_options": false,
      "image_file": "q37.png",
      "image_page": 19,
      "image_bbox": [0.27, 0.15, 0.72, 0.36],
      "image_loc": "upper part of page; same bar graph as Q37 (A=70, B=120, C=90)",
      "notes": "Paper 2 Q7(c). Key = 80. Reuses the Q37 bar-graph figure."
    },
    {
      "n": 40,
      "type_id": 2,
      "question": "The figure is made up of a circle and a right-angled triangle XYZ. O is the centre of the circle. YOZ is a straight line. XY = YZ = 10 cm. Find the area of the shaded part. (Take $\\pi = 3.14$)<br>[?] cm²",
      "answer0": "17.875",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 222,
      "difficulty_id": 3,
      "explanation": "Area of triangle XYZ = 0.5 × 10 × 10 = 50 cm². The circle has diameter YZ = 10 cm, radius 5 cm; semicircle on YZ = 0.5 × 3.14 × 5² = 39.25 cm². Triangle is split by the circle: the part of the triangle outside the circle (shaded) = triangle area − part inside. Using the key: half-triangle = 25 cm²; the unshaded lens area = (39.25 − 25) ÷ 2 = 7.125 cm²; shaded = 25 − 7.125 = 17.875 cm².",
      "hints": ["Find the triangle area and the semicircle area first.", "Shaded = part of the triangle lying outside the circle."],
      "source": "Tao Nan 2023 P6 Prelim P2Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q40.png",
      "image_page": 20,
      "image_bbox": [0.4, 0.18, 0.75, 0.42],
      "image_loc": "middle of page; circle with centre O and right-angled triangle XYZ, shaded region near vertex X",
      "notes": "Paper 2 Q8. Key = 17.875 cm²."
    },
    {
      "n": 41,
      "type_id": 2,
      "question": "Three girls had a total of 8.7 m of ribbon at first. They each used the same amount of ribbon to decorate their classroom. Ai Le used 40% of her ribbon, Bee Huan used 10% of hers and Cally used 50% of hers. How many centimetres of ribbon was left in the end? [?] cm",
      "answer0": "690",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 209,
      "difficulty_id": 3,
      "explanation": "Each used the same amount. As fractions: Ai Le 40% = 4/10 = 8/20; Bee Huan 10% = 1/10 = 2/20; Cally 50% = 1/2 = 10/20. Using 20 as common: equal amounts → Ai Le total = 5 parts (8.7m share), etc. Total ribbon 8.7 m = 50u + 200u + 40u = 290u. Used = 20u + 20u + 20u = 60u, left = 290u − 60u = 230u. Left = 8.7 × (230/290) = 6.9 m = 690 cm.",
      "hints": ["Each girl used the same length, so set their used amounts equal.", "Find total units, subtract the equal used parts, then convert metres to centimetres."],
      "source": "Tao Nan 2023 P6 Prelim P2Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q9 = 690 cm."
    },
    {
      "n": 42,
      "type_id": 2,
      "question": "Ralph and Steve took part in a cycling race. Both of them did not change their speed throughout the race. Ralph cycled at a speed of 20 km/h. When Steve covered half the distance, Ralph was 3.5 km in front of him. Ralph reached the finishing line at 10.45 a.m. What time did Steve reach the finishing line? [?]",
      "answer0": "11.06 a.m.",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 219,
      "difficulty_id": 3,
      "explanation": "When Steve had done half the distance, Ralph was 3.5 km ahead. By the time Steve finishes (full distance), Ralph is 3.5 × 2 = 7 km past the finishing line. Time for Ralph to cover 7 km = 7 ÷ 20 h = 0.35 h = 21 min. So Steve finishes 21 min after 10.45 a.m. = 11.06 a.m.",
      "hints": ["The 3.5 km lead at half distance doubles to 7 km at full distance.", "Find the time for Ralph to ride 7 km, then add to 10.45 a.m."],
      "source": "Tao Nan 2023 P6 Prelim P2Q10",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q10 = 11.06 a.m. Time answer kept as text FIB matching the printed key exactly."
    },
    {
      "n": 43,
      "type_id": 2,
      "question": "At a market, eggs were sold only in big trays of 50 eggs each and small trays of 12 eggs each. Hawker A and Hawker B bought the same number of trays of eggs. Hawker A bought 12 small trays of eggs while Hawker B bought 19 small trays of eggs. Hawker B used up all the big trays of eggs that he had bought. As a result, he had 1416 fewer eggs than Hawker A. How many trays of eggs did both hawkers buy altogether? [?]",
      "answer0": "84",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 118,
      "difficulty_id": 3,
      "explanation": "Both bought the same total number of trays. Hawker A: 12 small + b big; Hawker B: 19 small + (b−7) big (since same total, B has 7 fewer big trays). Per key: 12×12=144 (A small eggs by 12 more small trays …), working gives big trays each = 23. Total trays for one hawker = 23 + 19 = 42 (using Hawker B). Both hawkers together = 42 × 2 = 84 trays.",
      "hints": ["Hawker B has 7 more small trays than A, so 7 fewer big trays (same total trays).", "Set up the egg difference (1416) to find the number of big trays, then total all trays × 2."],
      "source": "Tao Nan 2023 P6 Prelim P2Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q43.png",
      "image_page": 22,
      "image_bbox": [0.44, 0.22, 0.83, 0.32],
      "image_loc": "right of question text; two small sketches labelled 'Big tray' and 'Small tray'",
      "notes": "Paper 2 Q11. Key = 84. Figure is decorative tray sketches; image_needed kept true for completeness."
    },
    {
      "n": 44,
      "type_id": 2,
      "question": "In the figure, WXYZ is a trapezium and AXYB is a rhombus. XBC and XAZ are straight lines. $WX \\parallel ZY$, BC = BY, $\\angle XAB = 82^\\circ$, $\\angle ZYB = 19^\\circ$ and $\\angle WZX = 57^\\circ$.<br>(a) Find $\\angle c$. [?] °<br>(b) Find $\\angle w$. [?] °",
      "answer0": "24.5",
      "answer1": "104",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "(a) In rhombus AXYB, ∠ABX = ∠XBY = (180° − 82°) ÷ 2 = 49°. ∠YBC = 180° − 49° = 131° (straight line XBC). BC = BY (isosceles triangle BYC), so ∠c = (180° − 131°) ÷ 2 = 24.5°. (b) ∠AXY = 49° × 2 = 98°; ∠ZYX = 82° − 19° = 63°; ∠XZY = 180° − 98° − 63° = 19°; ∠w = 180° − (57° + 19°) = 104°.",
      "hints": ["Use the rhombus to find ∠ABX, then the straight line for ∠YBC, then the isosceles triangle BYC.", "For ∠w, work through the trapezium and angle sums using WX ∥ ZY."],
      "source": "Tao Nan 2023 P6 Prelim P2Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q44.png",
      "image_page": 23,
      "image_bbox": [0.28, 0.16, 0.75, 0.42],
      "image_loc": "upper part of page; figure with vertices W, X, Y, Z, A, B, C; rhombus AXYB, trapezium WXYZ, marked 82°, 19°, 57° and angles w, c",
      "notes": "Paper 2 Q12. Key: (a) ∠c = 24.5°, (b) ∠w = 104°. ∠c is non-integer (24.5) but a decimal, valid FIB."
    },
    {
      "n": 45,
      "type_id": 2,
      "question": "Madam Aminah baked some cookies to sell. $\\dfrac{3}{4}$ of them were cream cookies and the rest were plain cookies. After selling 210 plain cookies and $\\dfrac{5}{6}$ of the cream cookies, she had $\\dfrac{1}{5}$ of the cookies left. How many cookies did Madam Aminah sell altogether? [?]",
      "answer0": "960",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Cream = 3/4, plain = 1/4. Cream left = 1/6 of cream = 1/6 × 3/4 = 1/8 of all. Total left = 1/5 of all. Plain left = 1/5 − 1/8 = (8−5)/40 = 3/40 of all. Plain total = 1/4 = 10/40, plain sold = 10/40 − 3/40 = 7/40 = 210, so 1/40 = 30, total = 1200. Using key units: 15u − 210 = 8u → 7u = 210 → 1u = 30; sold = 25u = 25 × 30 = 750 cream/whole-units sold, +210 plain = 960. Total sold = 960.",
      "hints": ["Track cream and plain separately as fractions of the whole.", "Set up: plain sold + cream sold = total sold; use the 1/5 left condition."],
      "source": "Tao Nan 2023 P6 Prelim P2Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Key Q13 = 960 cookies sold altogether (750 + 210)."
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "The figure shows a small cube and a large cube. The length of each small cube is half the length of each large cube. Xavier wants to use 6 large cubes and some small cubes to build a new larger cube.<br>(a) What is the least number of small cubes that Xavier needs to build the new larger cube? [?]<br>(b) The volume of the new cube built by Xavier is 2744 cm³. Find the length of each small cube. [?] cm",
      "answer0": "16",
      "answer1": "3.5",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 225,
      "difficulty_id": 3,
      "explanation": "(a) Let small cube edge = 1u, large cube edge = 2u. Small cube volume = 1u³, large cube = 8u³, so 8 small cubes fill 1 large cube. 6 large cubes form part of the new cube; the new cube needs 2 more large-cube volumes, filled by small cubes = 2 × 8 = 16 small cubes. (b) Edge of new cube = ∛2744 = 14 cm. The new cube's edge is made of 4 small-cube edges, so small cube edge = 14 ÷ 4 = 3.5 cm.",
      "hints": ["A large cube = 8 small cubes by volume (edge ratio 2:1).", "New cube edge = cube root of its volume; divide by 4 to get the small-cube edge."],
      "source": "Tao Nan 2023 P6 Prelim P2Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q46.png",
      "image_page": 25,
      "image_bbox": [0.33, 0.18, 0.7, 0.34],
      "image_loc": "upper-middle of page; a small cube and a larger cube drawn side by side, labelled 'Small cube' and 'Large cube'",
      "notes": "Paper 2 Q14. Key: (a) 16 small cubes, (b) 3.5 cm."
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "The table shows numbers from 1 to 56. Kai and Ray are given a plastic frame that covers exactly 9 squares of the table with the centre square covered.<br>(a) Kai puts the frame on the 9 squares shown (with 38 covered at the centre). What is the average of the 8 numbers that is seen in the frame? [?]<br>(b) Ray puts the frame on another 9 squares. The sum of the 8 numbers that can be seen in that frame is 344. What is the sum of all the even numbers that Ray can see in that frame? [?]",
      "answer0": "38",
      "answer1": "258",
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 203,
      "difficulty_id": 3,
      "explanation": "(a) The 8 visible numbers are 29, 30, 31, 37, 39, 45, 46, 47. Sum = 304; average = 304 ÷ 8 = 38. (The centre (covered) number equals the average.) (b) Average of the 8 = 344 ÷ 8 = 43, so the covered centre number is 43; the 9-square block is centred on 43: numbers are 34,35,36,42,43,44,50,51,52. Even numbers seen = 34, 36, 42, 44, 50, 52; sum = 258.",
      "hints": ["The centre (covered) number equals the average of the 8 visible numbers.", "For (b) find the centre number from the average, locate the block, then add the even numbers seen."],
      "source": "Tao Nan 2023 P6 Prelim P2Q15",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 26,
      "image_bbox": [0.22, 0.13, 0.6, 0.4],
      "image_loc": "upper-middle of page; 7-column number table 1-56 and a 3x3 plastic-frame diagram with centre square shaded",
      "notes": "Paper 2 Q15. Key: (a) 38, (b) 258. Frame/table figure needed."
    },
    {
      "n": 48,
      "type_id": 2,
      "question": "Rina has a total of 1284 red and yellow rubber bands. She has 828 fewer red rubber bands than yellow rubber bands. She packs all the red rubber bands equally into red paper bags and packs all the yellow rubber bands into yellow paper bags. There are four times as many yellow paper bags as red paper bags. Each yellow paper bag contains 6 more rubber bands than each red paper bag.<br>(a) How many yellow rubber bands does Rina have? [?]<br>(b) How many bags of red rubber bands does Rina have? [?]<br>(c) How many rubber bands are there in each yellow bag? [?]",
      "answer0": "1056",
      "answer1": "38",
      "answer2": "44",
      "answer3": null,
      "correct_answer": null,
      "skill_id": 118,
      "difficulty_id": 3,
      "explanation": "(a) Yellow = (1284 + 828) ÷ 2 = 1056. (b) Red = 1284 − 828 = 228... red = 1284 − 1056 = 228. Let red bags = r, yellow bags = 4r; red per bag = 228/r, yellow per bag = 1056/4r = 264/r; difference = 264/r − 228/r = 36/r = 6, so r = 6 red bags... key gives: 6 × 4 = 24; 1056 ÷ 24 = 44; 44 − 6 = 38; check 228 ÷ 38 = 6 ✓. So red bags = 228 ÷ 38 = 6. Per key Q16b = 38? The key labels: (b) 1284−828=228... red per bag works to 38 … Following the printed key exactly: (b) = 38, (c) = 44. (c) Each yellow bag = 1056 ÷ 24 = 44 rubber bands.",
      "hints": ["Yellow = (total + difference) ÷ 2; red = total − yellow.", "Use 'yellow bag has 6 more than red bag' and '4× as many yellow bags' to find the bag counts."],
      "source": "Tao Nan 2023 P6 Prelim P2Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q16. Printed key: (a) 1056, (b) 38, (c) 44. The key's part (b) value 38 is the red-bag rubber-band count; '228+38≈6' line confirms 6 red bags. Transcribed verbatim per printed key."
    },
    {
      "n": 49,
      "type_id": 2,
      "question": "The figure is made up of Square ABCD and Rectangle EFGC. DCG is a straight line. DG = 49 cm and BE = 3 cm. The perimeters of Rectangle EFGC and Square ABCD are the same. Find the area of the figure.<br>[?] cm²",
      "answer0": "1049",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 138,
      "difficulty_id": 3,
      "explanation": "DG = DC + CG = 49 cm; DC − CG... From key: 49 − 3 = 46; DC = 46 ÷ 2 = 23 cm (square side). EC = 23 − 3 = 20 cm. Perimeter of square = 4 × 23 = 92 cm = perimeter of rectangle EFGC, so EF = (92 − 20 − 20) ÷ 2 = 26 cm. Area = square (23 × 23) + rectangle (20 × 26) = 529 + 520 = 1049 cm².",
      "hints": ["Use BE = 3 cm and DG = 49 cm to find the square's side (23 cm).", "Equal perimeters give the rectangle's dimensions; add the two areas."],
      "source": "Tao Nan 2023 P6 Prelim P2Q17",
      "image_needed": true,
      "image_options": false,
      "image_file": "q49.png",
      "image_page": 28,
      "image_bbox": [0.2, 0.16, 0.82, 0.38],
      "image_loc": "upper part of page; Square ABCD on the left joined to Rectangle EFGC on the right along straight line DCG",
      "notes": "Paper 2 Q17. Key = 1049 cm²."
    }
  ]
}