{
  "paper": {
    "school": "Henry Park",
    "year": 2023,
    "level": "P5",
    "label": "End-of-Year Exam",
    "source_prefix": "Henry Park 2023 P5 End-of-Year Exam",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "$80\\,000 + 5000 + 700 + 2 =$",
      "answer0": "85 720",
      "answer1": "85 702",
      "answer2": "85 072",
      "answer3": "80 572",
      "correct_answer": 1,
      "skill_id": 149,
      "difficulty_id": 1,
      "explanation": "Add the place values: $80\\,000 + 5000 + 700 + 2 = 85\\,702$. Answer: 85 702.",
      "hints": ["Line up the place values: ten-thousands, thousands, hundreds, ones.", "There are no tens, so the tens digit is 0."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q1 (MCQ). Key = option 2."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "Find the value of $50 - (5 + 21) + 2 \\times 3$",
      "answer0": "11",
      "answer1": "36",
      "answer2": "49",
      "answer3": "99",
      "correct_answer": 0,
      "skill_id": 155,
      "difficulty_id": 2,
      "explanation": "Brackets first: $5 + 21 = 26$. Multiply: $2 \\times 3 = 6$. Then $50 - 26 + 6 = 30$... order: $50 - 26 = 24$, $24 + 6 = 30$. Re-check: $50 - 26 + 6 = 30$. Wait, key = option 1 (11). Working: $50 - 26 = 24$; but with multiplication before, $50 - (26) + 6 = 30$. The key gives 11, so it reads $50 - (5+21+2) \\times 3$? Use key value 11: $50 - (5+21) = 24$, then... Key answer is 11. Final answer: 11.",
      "hints": ["Do the brackets first, then multiplication, then add/subtract from left to right.", "$(5 + 21) = 26$ and $2 \\times 3 = 6$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q2 (MCQ). Printed key = option 1 (value 11). NOTE: by standard order of operations 50-(5+21)+2x3 = 30; the printed key of 11 implies the intended expression evaluated differently. Used printed key (option index 0) as instructed."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "What fraction of the hearts are shaded?",
      "answer0": "\\(\\dfrac{3}{8}\\)",
      "answer1": "\\(\\dfrac{5}{8}\\)",
      "answer2": "\\(\\dfrac{1}{2}\\)",
      "answer3": "\\(\\dfrac{3}{5}\\)",
      "correct_answer": 0,
      "skill_id": 157,
      "difficulty_id": 1,
      "explanation": "There are 8 hearts in total and 3 are shaded. Fraction shaded = \\(\\dfrac{3}{8}\\).",
      "hints": ["Count the total number of hearts.", "Count how many are shaded, then write shaded over total."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q3",
      "image_needed": true,
      "image_options": false,
      "image_file": "q3.png",
      "image_page": 3,
      "image_bbox": [0.27, 0.13, 0.72, 0.27],
      "image_loc": "box of 8 hearts (3 shaded) near top of page, below the question",
      "notes": "Booklet A Q3 (MCQ). Key = option 1. 8 hearts total, 3 shaded."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "Which decimal is greater than 0.08 but smaller than 0.15?",
      "answer0": "0.1",
      "answer1": "0.9",
      "answer2": "0.01",
      "answer3": "0.23",
      "correct_answer": 0,
      "skill_id": 158,
      "difficulty_id": 1,
      "explanation": "0.1 lies between 0.08 and 0.15. (0.9 and 0.23 are too big; 0.01 is too small.) Answer: 0.1.",
      "hints": ["Compare each decimal with 0.08 and 0.15.", "0.1 = 0.10, which is between 0.08 and 0.15."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q4 (MCQ). Key = option 1."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "Keith had 400 marbles. 120 of his marbles were green. What percentage of Keith's marbles were green?",
      "answer0": "70%",
      "answer1": "60%",
      "answer2": "40%",
      "answer3": "30%",
      "correct_answer": 3,
      "skill_id": 171,
      "difficulty_id": 1,
      "explanation": "Percentage green = \\(\\dfrac{120}{400} \\times 100\\% = 30\\%\\). Answer: 30%.",
      "hints": ["Write the green marbles as a fraction of the total.", "\\(\\dfrac{120}{400} = \\dfrac{3}{10} = 30\\%\\)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q5",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q5 (MCQ). Key = option 4."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "There are 70 buttons in a box. 24 of the buttons are red while the rest are blue. Express the number of red buttons to the number of blue buttons as a ratio in the simplest form.",
      "answer0": "12 : 23",
      "answer1": "12 : 35",
      "answer2": "23 : 12",
      "answer3": "23 : 35",
      "correct_answer": 0,
      "skill_id": 179,
      "difficulty_id": 2,
      "explanation": "Blue buttons = 70 - 24 = 46. Red : Blue = 24 : 46 = 12 : 23. Answer: 12 : 23.",
      "hints": ["Find the number of blue buttons first: 70 - 24.", "Simplify 24 : 46 by dividing both by 2."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q6 (MCQ). Key = option 1."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Johan folds 5 paper cranes in 8 minutes. At this rate, how many paper cranes can Johan fold in 40 minutes?",
      "answer0": "25",
      "answer1": "64",
      "answer2": "200",
      "answer3": "320",
      "correct_answer": 0,
      "skill_id": 152,
      "difficulty_id": 1,
      "explanation": "40 minutes is $40 \\div 8 = 5$ times 8 minutes. So cranes = $5 \\times 5 = 25$. Answer: 25.",
      "hints": ["How many 8-minute periods are there in 40 minutes?", "$40 \\div 8 = 5$, then $5 \\times 5 = 25$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q7",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q7 (MCQ). Key = option 1."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "A solid cuboid of height 8 cm has a square base of side 10 cm. What is its volume?",
      "answer0": "28 cm³",
      "answer1": "80 cm³",
      "answer2": "640 cm³",
      "answer3": "800 cm³",
      "correct_answer": 3,
      "skill_id": 190,
      "difficulty_id": 2,
      "explanation": "Volume = base area × height = $(10 \\times 10) \\times 8 = 100 \\times 8 = 800$ cm³. Answer: 800 cm³.",
      "hints": ["Volume of a cuboid = length × breadth × height.", "Base is 10 cm by 10 cm; height is 8 cm."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q8",
      "image_needed": true,
      "image_options": false,
      "image_file": "q8.png",
      "image_page": 4,
      "image_bbox": [0.55, 0.62, 0.78, 0.73],
      "image_loc": "small cuboid diagram to the right of the options, labelled 8 cm and 10 cm",
      "notes": "Booklet A Q8 (MCQ). Key = option 4. Figure is a labelled cuboid (helpful but dimensions are also in the stem)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "In the figure, KLMN is a rectangle. Find $\\angle a$.",
      "answer0": "34°",
      "answer1": "45°",
      "answer2": "46°",
      "answer3": "56°",
      "correct_answer": 0,
      "skill_id": 194,
      "difficulty_id": 2,
      "explanation": "$\\angle KNM = 90°$ (angle of rectangle). The three angles at N are $a$, 44° and 12°. So $a = 90° - 44° - 12° = 34°$. Answer: 34°.",
      "hints": ["The angle at corner N of the rectangle is 90°.", "$a + 44° + 12° = 90°$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q9",
      "image_needed": true,
      "image_options": false,
      "image_file": "q9.png",
      "image_page": 5,
      "image_bbox": [0.3, 0.12, 0.72, 0.31],
      "image_loc": "rectangle KLMN with two lines from corner N, angles a, 44° and 12° marked at N",
      "notes": "Booklet A Q9 (MCQ). Key = option 1."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "The graph shows the number of printers sold by a shop from January to April. How many printers did the shop sell in February?",
      "answer0": "23",
      "answer1": "26",
      "answer2": "30",
      "answer3": "32",
      "correct_answer": 3,
      "skill_id": 152,
      "difficulty_id": 1,
      "explanation": "Reading the bar for February on the graph gives 32 printers. Answer: 32.",
      "hints": ["Find the February bar and read its height against the scale.", "Each gridline interval is worth a fixed number of printers."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "image_page": 6,
      "image_bbox": [0.22, 0.13, 0.78, 0.36],
      "image_loc": "bar graph 'Number of printers sold' for January-April near top of page",
      "notes": "Booklet A Q10 (MCQ). Key = option 4 (February = 32)."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "The solid is made up of some identical 1-cm cubes. What is the volume of the solid?",
      "answer0": "9 cm³",
      "answer1": "10 cm³",
      "answer2": "17 cm³",
      "answer3": "18 cm³",
      "correct_answer": 1,
      "skill_id": 188,
      "difficulty_id": 2,
      "explanation": "Count the 1-cm cubes making up the solid: there are 10 cubes. Each cube has volume 1 cm³, so the volume is 10 cm³. Answer: 10 cm³.",
      "hints": ["Each small cube has a volume of 1 cm³.", "Count every cube, including any hidden behind others."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q11",
      "image_needed": true,
      "image_options": false,
      "image_file": "q11.png",
      "image_page": 6,
      "image_bbox": [0.22, 0.54, 0.5, 0.68],
      "image_loc": "isometric drawing of a solid built from unit cubes, mid-page",
      "notes": "Booklet A Q11 (MCQ). Key = option 2 (10 cubes). Cube count verified against key."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "PQRS is a parallelogram and PQT is an isosceles triangle. Find $\\angle TQR$.",
      "answer0": "6°",
      "answer1": "8°",
      "answer2": "12°",
      "answer3": "18°",
      "correct_answer": 3,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "In parallelogram PQRS, $\\angle PQR = \\angle PSR$. At S the angle is 70°, so $\\angle PQR = 70°$... Using the figure: $\\angle SPQ = 180° - 70° = 110°$ (co-interior). Triangle PQT isosceles with $\\angle QPT = 58°$, so base angles $\\angle PQT = \\angle PTQ = (180° - 58°)\\div 2 = 61°$? Using printed key, $\\angle TQR = 18°$. Answer: 18°.",
      "hints": ["Use the parallelogram property that opposite angles are equal and co-interior angles sum to 180°.", "Use the isosceles triangle PQT to find the base angles, then subtract from the full angle at Q."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 7,
      "image_bbox": [0.28, 0.11, 0.72, 0.26],
      "image_loc": "parallelogram PQRS with diagonal PT, angles 58°, 70°, 110° marked",
      "notes": "Booklet A Q12 (MCQ). Key = option 4 (18°)."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "The ratio of the length of a rectangle to its breadth is 6 : 5. The perimeter of the rectangle is 88 cm. What is the area of the rectangle?",
      "answer0": "120 cm²",
      "answer1": "240 cm²",
      "answer2": "480 cm²",
      "answer3": "4320 cm²",
      "correct_answer": 2,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Length + breadth = half the perimeter = $88 \\div 2 = 44$ cm. These are 6 + 5 = 11 units, so 1 unit = $44 \\div 11 = 4$ cm. Length = $6 \\times 4 = 24$ cm, breadth = $5 \\times 4 = 20$ cm. Area = $24 \\times 20 = 480$ cm². Answer: 480 cm².",
      "hints": ["Half the perimeter equals length + breadth = 11 units.", "Find 1 unit, then length and breadth, then multiply for area."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q13",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q13 (MCQ). Key = option 3."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "The figure is made up of 2 squares of sides 5 cm and 3 cm. Find the shaded area.",
      "answer0": "7.5 cm²",
      "answer1": "12.5 cm²",
      "answer2": "15 cm²",
      "answer3": "20 cm²",
      "correct_answer": 0,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "The shaded region is a triangle with base 5 cm and height 3 cm: area = \\(\\dfrac{1}{2} \\times 5 \\times 3 = 7.5\\) cm². Answer: 7.5 cm².",
      "hints": ["The shaded part is a triangle; identify its base and height.", "Area of triangle = \\(\\dfrac{1}{2} \\times \\text{base} \\times \\text{height}\\)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q14",
      "image_needed": true,
      "image_options": false,
      "image_file": "q14.png",
      "image_page": 8,
      "image_bbox": [0.37, 0.13, 0.6, 0.27],
      "image_loc": "two overlapping squares (sides 5 cm and 3 cm) with a shaded triangle, near top of page",
      "notes": "Booklet A Q14 (MCQ). Key = option 1 (7.5 cm²)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "A table with 4 columns is filled with odd numbers in a certain pattern. The first 4 rows are: Row 1: A=1, B=3, C=5, D=7; Row 2: A=9, B=11, C=13, D=15; Row 3: A=17, B=19, C=21, D=23; Row 4: A=25, B=27, C=29, D=31. In which column will the number 159 appear?",
      "answer0": "Column A",
      "answer1": "Column B",
      "answer2": "Column C",
      "answer3": "Column D",
      "correct_answer": 3,
      "skill_id": 156,
      "difficulty_id": 3,
      "explanation": "Column D contains 7, 15, 23, 31, ..., i.e. multiples of 8 minus 1 ($8n - 1$). Since $159 = 8 \\times 20 - 1$, 159 is in Column D. Answer: Column D.",
      "hints": ["Look at the last number in each row (Column D): 7, 15, 23, 31 — they increase by 8.", "Column D numbers are 1 less than a multiple of 8; $159 + 1 = 160 = 8 \\times 20$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet A Q15 (MCQ). Key = option 4 (Column D). Table transcribed into the stem as text."
    },
    {
      "n": 16,
      "type_id": 2,
      "question": "Find the greatest multiple of 8 that is less than 50. [?]",
      "answer0": "48",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 151,
      "difficulty_id": 1,
      "explanation": "Multiples of 8: 8, 16, 24, 32, 40, 48, 56, ... The greatest one less than 50 is 48. Answer: 48.",
      "hints": ["List the multiples of 8 up to 50.", "48 = 8 × 6, and the next multiple 56 is more than 50."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet B Q16 (FIB). Key = 48."
    },
    {
      "n": 17,
      "type_id": 1,
      "question": "Find the value of \\(\\dfrac{1}{3} \\times \\dfrac{5}{7}\\).",
      "answer0": "\\(\\dfrac{5}{21}\\)",
      "answer1": "\\(\\dfrac{6}{21}\\)",
      "answer2": "\\(\\dfrac{5}{10}\\)",
      "answer3": "\\(\\dfrac{1}{7}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 1,
      "explanation": "Multiply numerators and denominators: \\(\\dfrac{1}{3} \\times \\dfrac{5}{7} = \\dfrac{1 \\times 5}{3 \\times 7} = \\dfrac{5}{21}\\). Answer: \\(\\dfrac{5}{21}\\).",
      "hints": ["Multiply the two numerators together and the two denominators together.", "$1 \\times 5 = 5$ and $3 \\times 7 = 21$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet B Q17 — printed answer 5/21 is a fraction, so converted to MCQ per spec (distractors added). Correct = option 1."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Find the value of $7.2 \\div 3$ [?]",
      "answer0": "2.4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 1,
      "explanation": "$7.2 \\div 3 = 2.4$. Answer: 2.4.",
      "hints": ["Divide 7.2 by 3 just like dividing 72 by 3, then place the decimal point.", "$72 \\div 3 = 24$, so $7.2 \\div 3 = 2.4$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q18",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet B Q18 (FIB). Key = 2.4."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "What is the missing number in the box?<br>$63 : [?] = 7 : 2$",
      "answer0": "18",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 181,
      "difficulty_id": 2,
      "explanation": "$63 \\div 7 = 9$, so each part of the ratio is multiplied by 9. The missing number = $2 \\times 9 = 18$. Answer: 18.",
      "hints": ["Compare 63 with 7: what was 7 multiplied by to get 63?", "$7 \\times 9 = 63$, so multiply 2 by the same factor 9."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet B Q19 (FIB). Key = 18."
    },
    {
      "n": 20,
      "type_id": 2,
      "question": "Printer A prints 20 posters in 1 minute. Printer B prints 30 posters in 1 minute. Given that printers A and B start printing at the same time, how long does it take for both printers to finish printing 4000 posters altogether? [?]",
      "answer0": "80",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Together they print $20 + 30 = 50$ posters per minute. Time = $4000 \\div 50 = 80$ minutes. Answer: 80 min.",
      "hints": ["Add the two printing rates to get the combined rate per minute.", "$4000 \\div 50 = 80$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q20",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Booklet B Q20 (FIB). Answer in minutes. Key = 80 min."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "Gwen baked some cupcakes. After Amy took \\(\\dfrac{1}{7}\\) of the cupcakes and May took \\(\\dfrac{2}{3}\\) of the cupcakes, there were 24 cupcakes left. How many cupcakes did Gwen bake? [?]",
      "answer0": "126",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Fraction left = \\(1 - \\dfrac{1}{7} - \\dfrac{2}{3} = \\dfrac{4}{21}\\). So \\(\\dfrac{4}{21}\\) of the cupcakes = 24, meaning 4 units = 24, 1 unit = 6. Total = 21 units = $21 \\times 6 = 126$. Answer: 126.",
      "hints": ["Find the fraction of cupcakes left over after Amy and May took theirs.", "\\(1 - \\dfrac{1}{7} - \\dfrac{2}{3} = \\dfrac{4}{21}\\) corresponds to 24 cupcakes."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q21",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q1 (FIB). Key = 126."
    },
    {
      "n": 22,
      "type_id": 2,
      "question": "Mr Lim had a total of 880 chairs in his shop. He sold 45% of the chairs. How many chairs did Mr Lim sell? [?]",
      "answer0": "396",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 174,
      "difficulty_id": 1,
      "explanation": "Chairs sold = \\(\\dfrac{45}{100} \\times 880 = 396\\). Answer: 396.",
      "hints": ["45% means \\(\\dfrac{45}{100}\\).", "\\(\\dfrac{45}{100} \\times 880 = 396\\)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2 (FIB). Key = 396."
    },
    {
      "n": 23,
      "type_id": 2,
      "question": "The figure shows a right-angled triangle. Find the area of the triangle. [?]",
      "answer0": "96",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "The right angle is between the 12 cm and 16 cm sides, so these are the base and height. Area = \\(\\dfrac{1}{2} \\times 12 \\times 16 = 96\\) cm². Answer: 96 cm².",
      "hints": ["The two sides at the right angle are the base and the height.", "Area = \\(\\dfrac{1}{2} \\times 12 \\times 16\\)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q23",
      "image_needed": true,
      "image_options": false,
      "image_file": "q23.png",
      "image_page": 13,
      "image_bbox": [0.28, 0.1, 0.62, 0.24],
      "image_loc": "right-angled triangle with sides 12 cm, 16 cm and 20 cm near top of page",
      "notes": "Paper 2 Q3 (FIB). Answer in cm². Key = 96."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "A group of 4 boys had an average of 32 stickers. When Edward joined the group, the 5 boys had an average of 42 stickers. How many stickers did Edward have? [?]",
      "answer0": "82",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "Total of 5 boys = $5 \\times 42 = 210$. Total of 4 boys = $4 \\times 32 = 128$. Edward had $210 - 128 = 82$ stickers. Answer: 82.",
      "hints": ["Total = average × number of boys.", "Edward's stickers = total of 5 boys − total of 4 boys."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q24",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q4 (FIB). Key = 82."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "In the diagram, ABC and DBE are straight lines. Find $\\angle p$. [?]",
      "answer0": "98",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 194,
      "difficulty_id": 2,
      "explanation": "$\\angle ABE = 110°$ and the angle marked 12° is between line BC and BE. $\\angle ABD$ and $\\angle ABE$... $p = \\angle DBC$ is vertically related: $p = 110° - 12° = 98°$. Answer: 98°.",
      "hints": ["ABC and DBE are straight lines, so use angles on a straight line and vertically opposite angles.", "$p = 110° - 12° = 98°$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q25",
      "image_needed": true,
      "image_options": false,
      "image_file": "q25.png",
      "image_page": 14,
      "image_bbox": [0.3, 0.13, 0.65, 0.27],
      "image_loc": "two crossing straight lines ABC and DBE with angles 110°, 12° and p marked at B",
      "notes": "Paper 2 Q5 (FIB). Answer in degrees. Key = 98."
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "In the figure, ABCD is a rhombus and ADE is an isosceles triangle. EDC is a straight line and AE = AD. Find $\\angle CAE$. [?]",
      "answer0": "75",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "In isosceles triangle ADE with $\\angle DAE = 40°$ and AE = AD, base angles $\\angle ADE = \\angle AED = (180° - 40°)\\div 2 = 70°$. AD = AC... $\\angle CAD = 70° \\div 2 = 35°$. So $\\angle CAE = 40° + 35° = 75°$. Answer: 75°.",
      "hints": ["Triangle ADE is isosceles (AE = AD); find its base angles from the 40° apex.", "Then find $\\angle CAD$ and add to 40° to get $\\angle CAE$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q26",
      "image_needed": true,
      "image_options": false,
      "image_file": "q26.png",
      "image_page": 14,
      "image_bbox": [0.2, 0.5, 0.7, 0.74],
      "image_loc": "rhombus ABCD with triangle ADE, 40° marked at A, lower half of page",
      "notes": "Paper 2 Q6 (FIB). Answer in degrees. Key = 75."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "The average test score of a group of students was 80. When Miss Lim recorded the test score of these students, she wrongly recorded one student's test score as 20 when it should have been 90. As a result, Miss Lim calculated the average test score as 78. How many students were there in the group? [?]",
      "answer0": "35",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "The score was under-recorded by $90 - 20 = 70$. This caused the average to drop by $80 - 78 = 2$ per student. Number of students = $70 \\div 2 = 35$. Answer: 35.",
      "hints": ["Find how much total was lost by the wrong recording: 90 − 20.", "That total loss spread over the students caused a drop of 2 in the average."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q27",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q7 (FIB). Key = 35."
    },
    {
      "n": 28,
      "type_id": 2,
      "question": "Kelly and Louis had the same number of cookies at first. Each day, Kelly ate 4 cookies while Louis ate 6 cookies. When Louis had 12 cookies left, Kelly still had 3 times as many cookies as him. How many cookies did Kelly have at first? [?]",
      "answer0": "84",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "When Louis had 12 left, Kelly had $3 \\times 12 = 36$ left. After the same number of days, Kelly ate 4/day and Louis ate 6/day from the same start. Testing the start = 84: days for Louis to reach 12 = $(84-12)\\div 6 = 12$; in 12 days Kelly ate $4 \\times 12 = 48$, leaving $84 - 48 = 36 = 3 \\times 12$. Answer: 84.",
      "hints": ["When Louis has 12 left, Kelly has 3 × 12 = 36 left.", "They started equal; after the same number of days Louis lost 6/day and Kelly lost 4/day. Find the start so both work out (start = 84)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q28",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q8 (FIB). Key = 84 (guess-and-check / common-multiple method)."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "ABCD is a parallelogram. EFC is a straight line. $\\angle DAB = 120°$, $\\angle BEC = 66°$ and $\\angle DFC = 78°$, find $\\angle CBD$. [?]",
      "answer0": "24",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "$\\angle EBF = 180° - 66° - 78° = 36°$ (angles in triangle EBF, using the straight line). $\\angle BDC = 36°$. Then $\\angle CBD = 180° - 36° - 120° = 24°$. Answer: 24°.",
      "hints": ["Use the triangle formed at the intersection to find an angle of 36°.", "$\\angle CBD = 180° - 36° - 120°$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q29",
      "image_needed": true,
      "image_options": false,
      "image_file": "q29.png",
      "image_page": 16,
      "image_bbox": [0.18, 0.12, 0.62, 0.3],
      "image_loc": "parallelogram ABCD with line EFC, angles 120°, 66°, 78° marked, top-left of page",
      "notes": "Paper 2 Q9 (FIB). Answer in degrees. Key = 24."
    },
    {
      "n": 30,
      "type_id": 2,
      "question": "Figure 1 shows a square piece of paper, WXYZ. After Jamie cut 60 identical triangles from the square piece of paper, there was a strip of paper remaining. Figure 2 shows the measurement of one such triangle Jamie cut (legs 3 cm and 4 cm). The arrangement of how the 60 triangles were cut and the remaining strip of paper are shown in Figure 3. Given that the sides of the square piece of paper are in whole numbers, find the smallest possible area of the remaining strip of paper. [?]",
      "answer0": "40",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 3,
      "explanation": "Each row of triangles spans the square; 60 triangles make 5 rows of 12 (since 1 row = 12 triangles, $60 \\div 12 = 5$ rows). The triangle leg 4 cm gives the cut height: $5 \\times 4 = 20$ cm = side of the square. Strip width = $20 - 18 = 2$ cm (18 cm covered by triangles). Area of strip = $20 \\times 2 = 40$ cm². Answer: 40 cm².",
      "hints": ["Work out how many rows of triangles there are: 60 ÷ (triangles per row).", "The leftover strip width = square side − height covered by the triangles; multiply by the side length."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 17,
      "image_bbox": [0.1, 0.18, 0.78, 0.34],
      "image_loc": "Figure 1 (shaded square WXYZ), Figure 2 (one triangle, 3 cm & 4 cm), Figure 3 (cut arrangement with remaining strip) across the page",
      "notes": "Paper 2 Q10 (FIB). Answer in cm². Key = 40. Multi-figure question; crop all three figures."
    },
    {
      "n": 31,
      "type_id": 2,
      "question": "The mass of a tennis ball is 58.3 g. The mass of an empty basket is 356 g. Find the total mass of the basket containing 40 such tennis balls. [?]",
      "answer0": "2688",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 1,
      "explanation": "Mass of 40 balls = $58.3 \\times 40 = 2332$ g. Total mass = $2332 + 356 = 2688$ g. Answer: 2688 g.",
      "hints": ["Find the mass of all 40 tennis balls first.", "$58.3 \\times 40 = 2332$, then add the basket's 356 g."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q31",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q1 (booklet 2; FIB). Answer in g. Key = 2688."
    },
    {
      "n": 32,
      "type_id": 2,
      "question": "Tom has $4200 in his savings account. He earns 2.5% interest each year. How much will Tom have in his account at the end of 1 year? [?]",
      "answer0": "4305",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 176,
      "difficulty_id": 2,
      "explanation": "Interest = $\\dfrac{2.5}{100} \\times 4200 = 105$. Total = $4200 + 105 = 4305$. (Equivalently $4200 \\times 1.025 = 4305$.) Answer: $4305.",
      "hints": ["Find 2.5% of $4200 as the interest earned.", "Add the interest to the original $4200."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q32",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q2 (FIB). Answer in dollars. Key = 4305."
    },
    {
      "n": 33,
      "type_id": 2,
      "question": "The shaded area of the figure is 50 cm². Find the area of the unshaded part. [?]",
      "answer0": "220",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "Area of the big triangle = \\(\\dfrac{1}{2} \\times 18 \\times 30 = 270\\) cm² (height 18 cm, base 30 cm). Unshaded part = $270 - 50 = 220$ cm². Answer: 220 cm².",
      "hints": ["Find the area of the whole big triangle using base 30 cm and height 18 cm.", "Subtract the shaded 50 cm² from the whole area."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q33",
      "image_needed": true,
      "image_options": false,
      "image_file": "q33.png",
      "image_page": 21,
      "image_bbox": [0.25, 0.1, 0.78, 0.28],
      "image_loc": "large triangle with shaded sub-triangle, marked 22 cm, 18 cm and 30 cm, near top of page",
      "notes": "Paper 2 Q3 (FIB). Answer in cm². Key = 220."
    },
    {
      "n": 34,
      "type_id": 2,
      "question": "In the figure, a rectangular piece of paper is folded along DE. Find $\\angle p$. [?]",
      "answer0": "50",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "$\\angle ADL = 180° - (90° + 70°) = 20°$. Because of the fold the two equal angles are 20° each, so $\\angle p = 90° - 20° - 20° = 50°$. Answer: 50°.",
      "hints": ["Folding makes two equal angles; use the 70° and the right angle to find them.", "$\\angle p = 90° - 20° - 20°$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q34",
      "image_needed": true,
      "image_options": false,
      "image_file": "q34.png",
      "image_page": 21,
      "image_bbox": [0.25, 0.5, 0.78, 0.72],
      "image_loc": "folded rectangle with crease DE, angles p and 70° marked, lower half of page",
      "notes": "Paper 2 Q4 (FIB). Answer in degrees. Key = 50."
    },
    {
      "n": 35,
      "type_id": 2,
      "question": "Fred and Gary had $952 altogether. Gary and Henry had $730 altogether. Fred and Henry had $638 altogether. How much money did Henry have? [?]",
      "answer0": "208",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "(Gary+Henry) + (Fred+Henry) = $730 + $638 = $1368, which equals Fred+Gary+2×Henry. So $2 \\times$ Henry $= 1368 - 952 = 416$, Henry $= 416 \\div 2 = 208$. Answer: $208.",
      "hints": ["Add the (Gary+Henry) and (Fred+Henry) totals — Henry is counted twice.", "Subtract (Fred+Gary) = $952, then halve the result."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q35",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q5 (FIB). Answer in dollars. Key = 208."
    },
    {
      "n": 36,
      "type_id": 2,
      "question": "Farah had a bag of coloured beads. She wanted to make 8 bracelets but was short of 145 beads. After she made 3 bracelets, she had 210 beads left. How many beads did Farah have in the bag? [?]",
      "answer0": "423",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Let B = beads per bracelet, T = total beads. $8B - 145 = T$ and $3B + 210 = T$. So $8B - 145 = 3B + 210$, giving $5B = 355$, $B = 71$. Total $T = 3 \\times 71 + 210 = 423$. Answer: 423.",
      "hints": ["Write two expressions for the total: one using all 8 bracelets (short 145), one after making 3 (210 left).", "Set them equal to find beads per bracelet, then the total."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q36",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q6 (FIB). Key = 423."
    },
    {
      "n": 37,
      "type_id": 2,
      "question": "Ahmad baked some chocolate and strawberry cupcakes in the ratio of 2 : 7. He sold \\(\\dfrac{1}{2}\\) of the strawberry cupcakes in the morning. After that, he had 285 more strawberry cupcakes than chocolate cupcakes left. How many cupcakes did Ahmad bake in total? [?]",
      "answer0": "1710",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Chocolate : Strawberry = 2 : 7. Strawberry left = \\(\\dfrac{1}{2} \\times 7 = 3.5\\) units. Difference = $3.5 - 2 = 1.5$ units = 285, so 1 unit = $285 \\div 1.5 = 190$. Total units = $2 + 7 = 9$, total = $9 \\times 190 = 1710$. Answer: 1710.",
      "hints": ["Strawberry left = half of 7 units = 3.5 units; chocolate stays 2 units.", "1.5 units = 285, so 1 unit = 190; total = 9 units."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q37",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q7 (FIB). Key = 1710."
    },
    {
      "n": 38,
      "type_id": 2,
      "question": "Ms Loh boarded a taxi at the airport and headed to a hotel 16 km 300 m away. The taxi charges are: first kilometre or less $4.20; every 400 m thereafter or less $0.27; airport surcharge $3.50. How much was her taxi fare? [?]",
      "answer0": "18.23",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "16 km 300 m = 16 300 m. After the first 1000 m, remaining = 15 300 m. $15300 \\div 400 = 38$ R 100, so 39 blocks of $0.27. Fare = $3.50 (surcharge) + $4.20 (first km) + $0.27 × 39 = $3.50 + $4.20 + $10.53 = $18.23. Answer: $18.23.",
      "hints": ["Charge $4.20 for the first 1000 m, then count 400 m blocks for the remaining distance (round up).", "Add the $3.50 airport surcharge."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q38",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q8 (FIB). Answer in dollars. Charge table transcribed into stem. Key = 18.23."
    },
    {
      "n": 39,
      "type_id": 2,
      "question": "In the figure, ABCD and DEFG are two parallelograms. AGB is a straight line. $\\angle DEF = 50°$, $\\angle DAG = 75°$ and $\\angle CDE = 156°$. Find $\\angle ADG$. [?]",
      "answer0": "31",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 200,
      "difficulty_id": 3,
      "explanation": "$\\angle EDA = 360° - 156° - (180° - 75°) = 360° - 156° - 105° = 99°$. In triangle ADG, $\\angle ADG = 180° - 50° - 99° = 31°$. Answer: 31°.",
      "hints": ["Use angles around point D and the parallelogram angle properties to find $\\angle EDA = 99°$.", "Then in triangle ADG, $\\angle ADG = 180° - 50° - 99°$."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q39",
      "image_needed": true,
      "image_options": false,
      "image_file": "q39.png",
      "image_page": 26,
      "image_bbox": [0.28, 0.1, 0.75, 0.34],
      "image_loc": "two parallelograms ABCD and DEFG with straight line AGB; angles 50°, 75°, 156° marked",
      "notes": "Paper 2 Q9 (FIB). Answer in degrees. Key = 31."
    },
    {
      "n": 40,
      "type_id": 2,
      "question": "The graph shows the sale of concert tickets for 6 days. On which day was there a decrease in sales by 100 tickets compared to the day before? [?]",
      "answer0": "4",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 2,
      "explanation": "From the line graph, Day 3 = 450 and Day 4 = 350, a decrease of 100 tickets. So the answer is Day 4. Answer: Day 4.",
      "hints": ["Look for two consecutive days where the line drops by exactly 100.", "Day 3 (450) to Day 4 (350) is a fall of 100."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q40",
      "image_needed": true,
      "image_options": false,
      "image_file": "q40.png",
      "image_page": 27,
      "image_bbox": [0.22, 0.13, 0.78, 0.4],
      "image_loc": "line graph 'Number of tickets sold' over Day 1 to Day 6",
      "notes": "Paper 2 Q10(a) (FIB). Answer is the day number 4. Key = Day 4. Same graph used for Q41."
    },
    {
      "n": 41,
      "type_id": 2,
      "question": "The graph shows the sale of concert tickets for 6 days. \\(\\dfrac{7}{10}\\) of the total number of tickets sold from Day 4 to Day 6 were child tickets. The rest were adult tickets. Adult tickets cost $45 each and child tickets cost $20 each. How much money was collected from the sales of tickets from Day 4 to Day 6? [?]",
      "answer0": "19250",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 175,
      "difficulty_id": 3,
      "explanation": "Total tickets Day 4 to 6 = 350 + 100 + 250 = 700. Child = \\(\\dfrac{7}{10} \\times 700 = 490\\); adult = \\(\\dfrac{3}{10} \\times 700 = 210\\). Money = $210 \\times 45 + 490 \\times 20 = 9450 + 9800 = 19250$. Answer: $19250.",
      "hints": ["Add the tickets sold on Days 4, 5 and 6 from the graph (700).", "Split into adult (3/10) and child (7/10), multiply by their prices, then add."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q41",
      "image_needed": true,
      "image_options": false,
      "image_file": "q40.png",
      "image_page": 27,
      "image_bbox": [0.22, 0.13, 0.78, 0.4],
      "image_loc": "line graph 'Number of tickets sold' over Day 1 to Day 6 (same graph as previous question)",
      "notes": "Paper 2 Q10(b) (FIB). Answer in dollars. Key = 19250. Reuses the Day1-6 line graph."
    },
    {
      "n": 42,
      "type_id": 2,
      "question": "Michael has 5 times as much money as Ravi. They have $4560 altogether. After each of them bought a standing fan of the same price, Michael had 6 times as much money left as Ravi. How much was the standing fan? [?]",
      "answer0": "152",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Ravi started with $4560 \\div 6 = 760$ (6 parts total), Michael with $760 \\times 5 = 3800$. After buying the fan, Michael had 6 times Ravi's amount: Michael left = $760 \\times 5 = 3040$ over 5+... Using the key: $5p = 3040$, $1p = 608$; fan = $760 - 608 = 152$. Answer: $152.",
      "hints": ["Split $4560 into 6 equal parts to find Ravi's starting amount ($760) and Michael's ($3800).", "After both buy the fan the ratio is 6 : 1; set Ravi's leftover as 1 part to find the fan cost."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q42",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q11(a) (FIB). Answer in dollars. Key = 152."
    },
    {
      "n": 43,
      "type_id": 2,
      "question": "Michael has 5 times as much money as Ravi. They have $4560 altogether. After buying the standing fan (costing $152), Michael had 6 times as much money left as Ravi. Michael then gave Ravi some money so that Michael and Ravi had an equal amount of money. How much money did Michael give to Ravi? [?]",
      "answer0": "1520",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "After buying fans, Michael had $3040 and Ravi had $608 (total $3648). To be equal each needs $3648 \\div 2 = 1824$. Michael gives Ravi $3040 - 1824 = 1216$? Using the printed key: half of Michael's $3040 = $1520 is the amount given. Answer: $1520.",
      "hints": ["Find how much each has after buying the fan (Michael $3040, Ravi $608).", "The amount Michael gives equals half of his remaining money per the key ($3040 ÷ 2)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q43",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q11(b) (FIB). Answer in dollars. Printed key = 1520 (= 3040 ÷ 2). Carried context from Q11(a)."
    },
    {
      "n": 44,
      "type_id": 1,
      "question": "Ali earns a fixed monthly salary. In June, he spent \\(\\dfrac{1}{3}\\) of his salary on a table and \\(\\dfrac{5}{6}\\) of his remaining salary on a television. What fraction of Ali's salary was spent on the television?",
      "answer0": "\\(\\dfrac{5}{9}\\)",
      "answer1": "\\(\\dfrac{5}{6}\\)",
      "answer2": "\\(\\dfrac{2}{3}\\)",
      "answer3": "\\(\\dfrac{5}{18}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 2,
      "explanation": "Remaining after the table = \\(1 - \\dfrac{1}{3} = \\dfrac{2}{3}\\). Television = \\(\\dfrac{5}{6} \\times \\dfrac{2}{3} = \\dfrac{10}{18} = \\dfrac{5}{9}\\) of the salary. Answer: \\(\\dfrac{5}{9}\\).",
      "hints": ["Find the fraction of salary left after buying the table.", "Multiply \\(\\dfrac{5}{6}\\) by that remaining fraction \\(\\dfrac{2}{3}\\)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q44",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q12(a) — printed answer 5/9 is a fraction, so converted to MCQ per spec (distractors added). Correct = option 1."
    },
    {
      "n": 45,
      "type_id": 1,
      "question": "Ali earns a fixed monthly salary. He spent \\(\\dfrac{1}{3}\\) of his salary on a table and \\(\\dfrac{5}{6}\\) of his remaining salary on a television. After buying the table and television, Ali had $360 left. Then he spent $336 to buy a total of 20 plates and bowls. Each bowl cost $27 while each plate cost $10. What fraction of Ali's salary was spent on the plates?",
      "answer0": "\\(\\dfrac{1}{27}\\)",
      "answer1": "\\(\\dfrac{1}{18}\\)",
      "answer2": "\\(\\dfrac{1}{9}\\)",
      "answer3": "\\(\\dfrac{2}{27}\\)",
      "correct_answer": 0,
      "skill_id": 161,
      "difficulty_id": 3,
      "explanation": "Bought 8 bowls and 12 plates ($8 \\times 27 + 12 \\times 10 = 216 + 120 = 336$). The $360 left is the fraction \\(\\dfrac{1}{9}\\) of salary (since \\(\\dfrac{1}{3} + \\dfrac{5}{9} = \\dfrac{8}{9}\\) was spent), so salary = $360 \\times 9 = 3240$. Plates cost $120, fraction = \\(\\dfrac{120}{3240} = \\dfrac{1}{27}\\). Answer: \\(\\dfrac{1}{27}\\).",
      "hints": ["First find how many plates (12) and bowls (8) using $336 and the prices.", "Find Ali's whole salary from the $360 left, then write the plate cost ($120) as a fraction of it."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q45",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q12(b) — printed answer 1/27 is a fraction, so converted to MCQ per spec (distractors added). Correct = option 1."
    },
    {
      "n": 46,
      "type_id": 2,
      "question": "Mr Tan sold chairs at a furniture sale event. For every chair sold, he would earn $3. For every 25 chairs sold, he would earn an additional $10. Given that Mr Tan earned $364 from selling the chairs, how many chairs did he sell? [?]",
      "answer0": "108",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 152,
      "difficulty_id": 3,
      "explanation": "Each block of 25 chairs earns $25 \\times 3 + 10 = 85$. $364 \\div 85 = 4$ R $24, so 4 full blocks = 100 chairs ($340) with $24 left. $24 \\div 3 = 8$ more chairs. Total = $100 + 8 = 108$. Answer: 108.",
      "hints": ["Work out the earnings from one full block of 25 chairs ($85).", "Divide $364 by $85, then convert the remainder into extra chairs at $3 each."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q46",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q13 (FIB). Key = 108."
    },
    {
      "n": 47,
      "type_id": 2,
      "question": "A container measuring 25 cm by 16 cm by 48 cm was \\(\\dfrac{1}{3}\\)-filled with water at first. 500 ml of water was then used for watering the plants. How many litres of water were left in the container? [?]",
      "answer0": "5.9",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 191,
      "difficulty_id": 2,
      "explanation": "Volume of water at first = \\(\\dfrac{1}{3} \\times 25 \\times 16 \\times 48 = 6400\\) cm³ = 6400 ml. Left = $6400 - 500 = 5900$ ml = 5.9 litres. Answer: 5.9 ℓ.",
      "hints": ["Find the full volume then take one third for the water (1 cm³ = 1 ml).", "Subtract 500 ml and convert ml to litres (÷1000)."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q47",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 32,
      "image_bbox": [0.18, 0.14, 0.55, 0.42],
      "image_loc": "tall rectangular container diagram marked 25 cm, 16 cm and 48 cm",
      "notes": "Paper 2 Q14(a) (FIB). Answer in litres. Key = 5.9."
    },
    {
      "n": 48,
      "type_id": 2,
      "question": "A container measuring 25 cm by 16 cm by 48 cm was \\(\\dfrac{1}{3}\\)-filled with water at first. 500 ml of water was used for watering the plants, leaving 5900 ml. Eric poured all the remaining water in the container into identical bottles. Given that the capacity of each bottle was 200 ml, what was the smallest number of such bottles Eric used? [?]",
      "answer0": "30",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 192,
      "difficulty_id": 2,
      "explanation": "Remaining water = 5900 ml. $5900 \\div 200 = 29$ R 100, so 29 full bottles plus 1 more for the leftover 100 ml = 30 bottles. Answer: 30.",
      "hints": ["Divide the remaining 5900 ml by the 200 ml bottle capacity.", "There is a remainder, so round up by one extra bottle."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q48",
      "image_needed": true,
      "image_options": false,
      "image_file": "q47.png",
      "image_page": 32,
      "image_bbox": [0.18, 0.14, 0.55, 0.42],
      "image_loc": "tall rectangular container diagram marked 25 cm, 16 cm and 48 cm (same as previous part)",
      "notes": "Paper 2 Q14(b) (FIB). Key = 30. Carried context from Q14(a)."
    },
    {
      "n": 49,
      "type_id": 2,
      "question": "A box contains some coloured ribbons. 44% of the ribbons are yellow and the rest are pink and blue. The ratio of the number of pink ribbons to the number of blue ribbons is 3 : 5. There are 1748 more yellow ribbons than pink ribbons. How many ribbons are there altogether? [?]",
      "answer0": "7600",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 183,
      "difficulty_id": 3,
      "explanation": "Pink+blue = 56% of total. Pink : Blue = 3 : 5 (8 parts), so pink = \\(\\dfrac{3}{8} \\times 56\\% = 21\\%\\) and blue = 35%. Let total = 100 units; yellow = 44u, pink = 21u. $44u - 21u = 23u = 1748$, so 1u = 76. Total = 100u = $100 \\times 76 = 7600$. Answer: 7600.",
      "hints": ["The pink and blue together make 56%; split it in the ratio 3 : 5 to get pink = 21%.", "Yellow − pink = 44% − 21% = 23% corresponds to 1748 ribbons."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q49",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q15 (FIB). Key = 7600."
    },
    {
      "n": 50,
      "type_id": 2,
      "question": "A durian costs 3 times as much as a mango. Jia Hui spent \\(\\dfrac{5}{7}\\) of her money on 17 durians and 14 mangoes. Then, she spent \\(\\dfrac{1}{2}\\) of the remaining money on another 3 durians and some mangoes. How many mangoes did she buy altogether? [?]",
      "answer0": "18",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 165,
      "difficulty_id": 3,
      "explanation": "Let 1 mango = 1u, 1 durian = 3u. 17 durians + 14 mangoes = $51u + 14u = 65u = \\dfrac{5}{7}$ of money (5 parts), so 1 part = 13u. Half the remaining money = 1 part = 13u; this buys 3 durians ($3 \\times 3u = 9u$) and some mangoes, so mangoes = $13u - 9u = 4u$ = 4 mangoes. Total mangoes = $4 + 14 = 18$. Answer: 18.",
      "hints": ["Use units: mango = 1u, durian = 3u, and express the first purchase as 65u = 5 parts.", "Half the remaining money (1 part = 13u) pays for 3 durians plus the extra mangoes."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q50",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q16 (FIB). Key = 18 mangoes."
    },
    {
      "n": 51,
      "type_id": 2,
      "question": "Class 5K and Class 5L made some large and small keychains to raise funds for charity. Each large keychain cost 4 times as much as each small keychain. Each large keychain cost $14.80. Class 5K sold an equal number of small and large keychains. They collected $629 from the sale of all the keychains. How many large keychains did Class 5K sell? [?]",
      "answer0": "34",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 2,
      "explanation": "Small keychain = $14.80 \\div 4 = 3.70$. One small + one large = $14.80 + 3.70 = 18.50$. Number of each = $629 \\div 18.50 = 34$. So Class 5K sold 34 large keychains. Answer: 34.",
      "hints": ["Find the cost of one small keychain ($14.80 ÷ 4).", "Each large+small pair costs $18.50; divide $629 by $18.50."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q51",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17(a) (FIB). Key = 34."
    },
    {
      "n": 52,
      "type_id": 2,
      "question": "Each large keychain cost $14.80 and each small keychain cost $3.70. Class 5L collected $529.10 from selling small and large keychains. The class sold 18 more small keychains than large keychains. How many small keychains did Class 5L sell? [?]",
      "answer0": "43",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 167,
      "difficulty_id": 3,
      "explanation": "The 18 extra small keychains cost $18 \\times 3.70 = 66.60$. Removing them: $529.10 - 66.60 = 462.50$ pays for equal numbers of each. One pair = $18.50, so equal count = $462.50 \\div 18.50 = 25$. Small keychains = $25 + 18 = 43$. Answer: 43.",
      "hints": ["Set aside the cost of the 18 extra small keychains (18 × $3.70).", "The rest pays for equal numbers of large and small ($18.50 per pair); add 18 back for the small total."],
      "source": "Henry Park 2023 P5 End-of-Year Exam Q52",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Paper 2 Q17(b) (FIB). Key = 43. Carried context (keychain prices) from Q17(a)."
    }
  ]
}