{ "paper": { "school": "Singapore Chinese Girls' School", "year": 2022, "level": "P6", "label": "Prelim", "source_prefix": "SCGS 2022 P6 Prelim", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 1, "question": "3 thousands, 57 tens, and 3 ones is", "answer0": "3060", "answer1": "3573", "answer2": "8703", "answer3": "35 703", "correct_answer": 1, "skill_id": 149, "difficulty_id": 2, "explanation": "3 thousands = 3000, 57 tens = 570, 3 ones = 3. Total = 3000 + 570 + 3 = 3573.", "hints": ["57 tens = 57 × 10 = 570.", "Add the thousands, the tens and the ones."], "source": "SCGS 2022 P6 Prelim P1 Q1", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 2 (1-based) = 3573." }, { "n": 2, "type_id": 1, "question": "Which of the following is equivalent to \\(2\\dfrac{5}{6}\\)?", "answer0": "\\(\\dfrac{7}{6}\\)", "answer1": "\\(\\dfrac{13}{6}\\)", "answer2": "\\(\\dfrac{17}{6}\\)", "answer3": "\\(\\dfrac{32}{6}\\)", "correct_answer": 2, "skill_id": 119, "difficulty_id": 1, "explanation": "\\(2\\dfrac{5}{6} = \\dfrac{2 \\times 6 + 5}{6} = \\dfrac{17}{6}\\).", "hints": ["Multiply the whole number by the denominator, then add the numerator.", "Keep the same denominator 6."], "source": "SCGS 2022 P6 Prelim P1 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 3 (1-based) = 17/6." }, { "n": 3, "type_id": 1, "question": "In 52.79, what does the digit 7 stand for?", "answer0": "7 tens", "answer1": "7 ones", "answer2": "7 tenths", "answer3": "7 hundredths", "correct_answer": 2, "skill_id": 123, "difficulty_id": 1, "explanation": "In 52.79, the 7 is the first digit after the decimal point, so it is in the tenths place: 7 tenths.", "hints": ["The first place after the decimal point is tenths.", "Identify the position of the 7."], "source": "SCGS 2022 P6 Prelim P1 Q3", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 3 (1-based) = 7 tenths." }, { "n": 4, "type_id": 1, "question": "Which of the following when divided by 6 gives a quotient of 3 and a remainder of 2?", "answer0": "6", "answer1": "9", "answer2": "15", "answer3": "20", "correct_answer": 3, "skill_id": 74, "difficulty_id": 1, "explanation": "Number = 6 × quotient + remainder = 6 × 3 + 2 = 18 + 2 = 20.", "hints": ["Dividend = divisor × quotient + remainder.", "6 × 3 + 2 = 20."], "source": "SCGS 2022 P6 Prelim P1 Q4", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 4 (1-based) = 20." }, { "n": 5, "type_id": 1, "question": "Arrange the following numbers in ascending order.
2.1, 2.01, 2.21", "answer0": "2.01, 2.1, 2.21", "answer1": "2.1, 2.01, 2.21", "answer2": "2.1, 2.21, 2.01", "answer3": "2.21, 2.1, 2.01", "correct_answer": 0, "skill_id": 124, "difficulty_id": 1, "explanation": "Ascending (smallest first): 2.01 < 2.1 < 2.21.", "hints": ["Line up the decimal points and compare.", "2.01 is the smallest, 2.21 is the largest."], "source": "SCGS 2022 P6 Prelim P1 Q5", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 1 (1-based) = 2.01, 2.1, 2.21." }, { "n": 6, "type_id": 1, "question": "What is the closest estimation of the reading shown?", "answer0": "4750 g", "answer1": "5225 g", "answer2": "5500 g", "answer3": "5750 g", "correct_answer": 1, "skill_id": 86, "difficulty_id": 2, "explanation": "The weighing scale dial (0 to 10 kg) shows the needle just past 5 kg; the closest estimate is about 5225 g.", "hints": ["Read where the needle points on the kilogram dial.", "Convert the kilogram reading to grams."], "source": "SCGS 2022 P6 Prelim P1 Q6", "image_needed": true, "image_options": false, "image_file": "q6.png", "image_page": 4, "image_bbox": [0.44, 0.58, 0.74, 0.82], "image_loc": "lower-right of page, circular weighing scale dial (0 to 10 kg)", "notes": "Answer key: option 2 (1-based) = 5225 g. Read needle from figure." }, { "n": 7, "type_id": 1, "question": "Peter had 15 sweets and 9 chocolates. What fraction of the snacks Peter had are chocolates?", "answer0": "\\(\\dfrac{2}{3}\\)", "answer1": "\\(\\dfrac{3}{5}\\)", "answer2": "\\(\\dfrac{3}{8}\\)", "answer3": "\\(\\dfrac{5}{8}\\)", "correct_answer": 2, "skill_id": 120, "difficulty_id": 2, "explanation": "Total snacks = 15 + 9 = 24. Chocolates fraction = \\(\\dfrac{9}{24} = \\dfrac{3}{8}\\).", "hints": ["Total snacks = sweets + chocolates.", "Simplify \\(\\dfrac{9}{24}\\) by dividing by 3."], "source": "SCGS 2022 P6 Prelim P1 Q7", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 3 (1-based) = 3/8." }, { "n": 8, "type_id": 1, "question": "In the figure, the length of AD is thrice of AC. Find the area of triangle BCD.", "answer0": "16 cm²", "answer1": "24 cm²", "answer2": "32 cm²", "answer3": "48 cm²", "correct_answer": 0, "skill_id": 186, "difficulty_id": 3, "explanation": "AC : AD = 1 : 3, so CD = AD − AC = 2 parts. Triangle BCD has base CD and height BA = 4 cm. With the marked equal segments (AC = CD-parts), CD works out so that area BCD = \\(\\dfrac{1}{2} \\times CD \\times 4 = 16\\) cm².", "hints": ["CD = AD − AC; use AD = 3 × AC.", "Area of triangle BCD = \\(\\dfrac{1}{2} \\times CD \\times BA\\)."], "source": "SCGS 2022 P6 Prelim P1 Q8", "image_needed": true, "image_options": false, "image_file": "q8.png", "image_page": 5, "image_bbox": [0.22, 0.42, 0.74, 0.62], "image_loc": "middle of page, right-angled triangle with B at top, base A-C-D with equal tick marks, BA = 4 cm", "notes": "Answer key: option 1 (1-based) = 16 cm²." }, { "n": 9, "type_id": 1, "question": "Find the value of \\(\\dfrac{5w}{2} - w + 2\\) when \\(w = 10\\).", "answer0": "13", "answer1": "17", "answer2": "21", "answer3": "22", "correct_answer": 1, "skill_id": 240, "difficulty_id": 2, "explanation": "\\(\\dfrac{5 \\times 10}{2} - 10 + 2 = 25 - 10 + 2 = 17\\).", "hints": ["Substitute w = 10 into the expression.", "\\(\\dfrac{5 \\times 10}{2} = 25\\)."], "source": "SCGS 2022 P6 Prelim P1 Q9", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 2 (1-based) = 17." }, { "n": 10, "type_id": 1, "question": "The figure shows 2 parallelograms, ABCD and JKLM. Which of the following statements is true?", "answer0": "Line AB is parallel to line JK.", "answer1": "Line CD is perpendicular to Line JM.", "answer2": "Parallelogram JKLM is also a rectangle.", "answer3": "The angle \\(\\angle ABC\\) is equal to angle \\(\\angle KJM\\).", "correct_answer": 1, "skill_id": 200, "difficulty_id": 3, "explanation": "Comparing the gradients on the grid, line CD is perpendicular to line JM, so statement (2) is true.", "hints": ["Compare the slopes of the lines on the grid.", "Perpendicular lines have slopes that are negative reciprocals."], "source": "SCGS 2022 P6 Prelim P1 Q10", "image_needed": true, "image_options": false, "image_file": "q10.png", "image_page": 6, "image_bbox": [0.14, 0.18, 0.86, 0.46], "image_loc": "top of page, square grid with parallelograms ABCD and JKLM", "notes": "Answer key: option 2 (1-based) = Line CD is perpendicular to Line JM." }, { "n": 11, "type_id": 1, "question": "How many of the following shapes have at least a line of symmetry? (isosceles triangle, parallelogram, trapezium, rhombus)", "answer0": "1", "answer1": "2", "answer2": "3", "answer3": "4", "correct_answer": 1, "skill_id": 144, "difficulty_id": 2, "explanation": "Isosceles triangle has a line of symmetry; a general parallelogram has none; a general trapezium has none; a rhombus has lines of symmetry (along its diagonals). So 2 shapes have at least one line of symmetry.", "hints": ["Check each shape for a line that folds it onto itself.", "Parallelograms (non-special) and scalene trapeziums have no line of symmetry."], "source": "SCGS 2022 P6 Prelim P1 Q11", "image_needed": true, "image_options": false, "image_file": "q11.png", "image_page": 6, "image_bbox": [0.12, 0.66, 0.86, 0.80], "image_loc": "lower part of page, four shapes labelled isosceles triangle, parallelogram, trapezium, rhombus", "notes": "Answer key: option 2 (1-based) = 2 (isosceles triangle and rhombus)." }, { "n": 12, "type_id": 1, "question": "The figure is made up of a large semi-circle and 3 small identical semi-circles. Given that the length of AB is 12 cm, find the area of the shaded part in terms of \\(\\pi\\).", "answer0": "\\(12\\pi\\) cm²", "answer1": "\\(18\\pi\\) cm²", "answer2": "\\(24\\pi\\) cm²", "answer3": "\\(48\\pi\\) cm²", "correct_answer": 0, "skill_id": 222, "difficulty_id": 3, "explanation": "Large semi-circle: diameter AB = 12, radius 6, area = \\(\\dfrac{1}{2}\\pi(6^2) = 18\\pi\\). Three small semi-circles: each diameter = 4, radius 2, area = \\(\\dfrac{1}{2}\\pi(2^2) = 2\\pi\\), total = 6\\(\\pi\\). Shaded = large − 3 small = 18\\(\\pi\\) − 6\\(\\pi\\) = 12\\(\\pi\\) cm².", "hints": ["Find the large semicircle's area, then subtract the three small semicircles.", "Each small semicircle has diameter 12 ÷ 3 = 4 cm."], "source": "SCGS 2022 P6 Prelim P1 Q12", "image_needed": true, "image_options": false, "image_file": "q12.png", "image_page": 7, "image_bbox": [0.48, 0.16, 0.78, 0.30], "image_loc": "upper-right of page, large semicircle with 3 small semicircles cut out along diameter AB (12 cm)", "notes": "Answer key: option 1 (1-based) = 12π cm²." }, { "n": 13, "type_id": 1, "question": "The distance between Point A and B is 480 m. John started cycling from point A to B at an average speed of 3 m/s while Peter started cycling from point B to A at an average speed of 2 m/s. How far apart will they be after 40 seconds?", "answer0": "40 m", "answer1": "80 m", "answer2": "120 m", "answer3": "280 m", "correct_answer": 3, "skill_id": 217, "difficulty_id": 2, "explanation": "In 40 s: John covers 3 × 40 = 120 m; Peter covers 2 × 40 = 80 m. Together they close 120 + 80 = 200 m. Distance apart = 480 − 200 = 280 m.", "hints": ["Find how far each cyclist travels in 40 seconds.", "They move toward each other, so subtract the total covered from 480 m."], "source": "SCGS 2022 P6 Prelim P1 Q13", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: option 4 (1-based) = 280 m." }, { "n": 14, "type_id": 1, "question": "The figure shows 10 cubes glued together to form a solid. The entire solid, including the base, was then painted red. How many cubes have only 3 of the faces painted?", "answer0": "1", "answer1": "2", "answer2": "3", "answer3": "4", "correct_answer": 1, "skill_id": 190, "difficulty_id": 3, "explanation": "Counting the cubes whose exactly 3 faces are exposed (corner-type cubes given the arrangement and that the base is painted) gives 2 cubes.", "hints": ["A cube with exactly 3 painted faces sits where three faces are exposed.", "Trace each cube's exposed faces in the arrangement (base counts as painted)."], "source": "SCGS 2022 P6 Prelim P1 Q14", "image_needed": true, "image_options": false, "image_file": "q14.png", "image_page": 7, "image_bbox": [0.44, 0.60, 0.82, 0.86], "image_loc": "lower-right of page, 3D arrangement of 10 unit cubes", "notes": "Answer key: option 2 (1-based) = 2. Count-to-verify against figure." }, { "n": 15, "type_id": 1, "question": "The bar graph shows the result of 40 students voting for their favourite type of food, A to E. Which pie chart best represents the information in the bar graph?", "answer0": "Pie chart 1", "answer1": "Pie chart 2", "answer2": "Pie chart 3", "answer3": "Pie chart 4", "correct_answer": 1, "skill_id": 235, "difficulty_id": 3, "explanation": "Bar values: A = 4, B = 10, C = 8, D = 6, E = 12. The correct pie chart has the matching proportional sectors with the order A, B, C, D, E and B's right-angle (10 of 40 = 90°). That is pie chart 2.", "hints": ["Read the number of votes for each food from the bar graph.", "Match the sector sizes (and the 90° quarter for the value of 10) to the correct pie chart."], "source": "SCGS 2022 P6 Prelim P1 Q15", "image_needed": true, "image_options": true, "image_file": "q15.png", "image_page": 8, "image_bbox": [0.16, 0.18, 0.74, 0.40], "image_loc": "top of page, bar graph of 40 students' favourite food (A-E); options are four pie charts below", "notes": "Answer key: option 2 (1-based). The four options are pictured pie charts (image_options true); answer labels are placeholders 'Pie chart 1-4'." }, { "n": 16, "type_id": 2, "question": "Express \\(7\\dfrac{3}{5}\\) as a decimal. [?]", "answer0": "7.6", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 158, "difficulty_id": 1, "explanation": "\\(\\dfrac{3}{5} = \\dfrac{6}{10} = 0.6\\). So \\(7\\dfrac{3}{5} = 7.6\\).", "hints": ["Convert \\(\\dfrac{3}{5}\\) to tenths.", "\\(\\dfrac{6}{10} = 0.6\\)."], "source": "SCGS 2022 P6 Prelim P1 Q16", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 7.6." }, { "n": 17, "type_id": 2, "question": "Find the value of \\(2.6 \\times 40\\). [?]", "answer0": "104", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 166, "difficulty_id": 1, "explanation": "\\(2.6 \\times 40 = 2.6 \\times 4 \\times 10 = 10.4 \\times 10 = 104\\).", "hints": ["Multiply 2.6 by 4 first.", "Then multiply by 10."], "source": "SCGS 2022 P6 Prelim P1 Q17", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 104." }, { "n": 18, "type_id": 2, "question": "Express \\(\\dfrac{11}{20}\\) as a percentage. [?] %", "answer0": "55", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 173, "difficulty_id": 1, "explanation": "\\(\\dfrac{11}{20} = \\dfrac{55}{100} = 55\\%\\).", "hints": ["Make the denominator 100 by multiplying by \\(\\dfrac{5}{5}\\).", "\\(\\dfrac{55}{100} = 55\\%\\)."], "source": "SCGS 2022 P6 Prelim P1 Q18", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 55%." }, { "n": 19, "type_id": 2, "question": "Measure and write down the size of \\(\\angle u\\). [?]°", "answer0": "142", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 140, "difficulty_id": 1, "explanation": "Measuring the printed angle with a protractor gives \\(\\angle u = 142°\\).", "hints": ["Place the protractor centre at the vertex.", "Read the scale where the second arm crosses."], "source": "SCGS 2022 P6 Prelim P1 Q19", "image_needed": true, "image_options": false, "image_file": "q19.png", "image_page": 11, "image_bbox": [0.18, 0.22, 0.66, 0.38], "image_loc": "upper part of page, an obtuse angle u drawn with two rays", "notes": "Answer key: 142°. Requires measuring the printed angle." }, { "n": 20, "type_id": 1, "question": "3 ℓ of water was poured into 4 glasses equally. What is the volume of water in each glass?", "answer0": "\\(\\dfrac{3}{4}\\,\\ell\\)", "answer1": "\\(\\dfrac{1}{4}\\,\\ell\\)", "answer2": "\\(\\dfrac{4}{3}\\,\\ell\\)", "answer3": "\\(1\\dfrac{1}{4}\\,\\ell\\)", "correct_answer": 0, "skill_id": 164, "difficulty_id": 1, "explanation": "3 ℓ ÷ 4 = \\(\\dfrac{3}{4}\\,\\ell\\) in each glass.", "hints": ["Divide the total volume by the number of glasses.", "3 ÷ 4 = \\(\\dfrac{3}{4}\\)."], "source": "SCGS 2022 P6 Prelim P1 Q20", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer is a fraction \\(\\dfrac{3}{4}\\,\\ell\\), so rendered as MCQ per FIB-fraction rule. Answer key: 3/4 ℓ. Distractors constructed." }, { "n": 21, "type_id": 2, "question": "John received the following test results: English Language 68, Mathematics 74, Science 83, Chinese Language ?. What did John get for his Chinese Language marks if the average marks for all four subjects is 75? [?]", "answer0": "75", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 204, "difficulty_id": 2, "explanation": "Total for 4 subjects = 75 × 4 = 300. Known three = 68 + 74 + 83 = 225. Chinese = 300 − 225 = 75.", "hints": ["Total marks = average × number of subjects.", "Subtract the three known marks from the total."], "source": "SCGS 2022 P6 Prelim P1 Q21", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 75 × 4 = 300; 300 − 225 = 75." }, { "n": 22, "type_id": 2, "question": "Farhana took 8 minutes to walk home from school, which was 1.2 km away. What was her average speed? [?] m/min", "answer0": "150", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 218, "difficulty_id": 2, "explanation": "1.2 km = 1200 m. Average speed = 1200 ÷ 8 = 150 m/min.", "hints": ["Convert 1.2 km to metres.", "Speed = distance ÷ time."], "source": "SCGS 2022 P6 Prelim P1 Q22", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 1200 ÷ 8 = 150 m/min." }, { "n": 23, "type_id": 2, "question": "In the figure, not drawn to scale, AD is parallel to BC. With \\(\\angle ADB = 60°\\) and the right angle at B, find \\(\\angle x\\). [?]°", "answer0": "30", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 194, "difficulty_id": 3, "explanation": "Since AD // BC, the angle at B between BD and the perpendicular is 90°. \\(\\angle x = 90° - 60° = 30°\\).", "hints": ["Use the parallel lines AD and BC to transfer the 60° angle.", "Subtract from the right angle at B."], "source": "SCGS 2022 P6 Prelim P1 Q23", "image_needed": true, "image_options": false, "image_file": "q23.png", "image_page": 13, "image_bbox": [0.18, 0.12, 0.56, 0.30], "image_loc": "upper part of page, figure with AD parallel to BC, x at B, 60° at D, 76° at C", "notes": "Answer key: 180 − 90 = 90; 90 − 60 = 30°." }, { "n": 24, "type_id": 2, "question": "Mr Chua had 36 kg of rice. He wanted to pack them into smaller bags of \\(\\dfrac{4}{5}\\) kg each. How many packets of rice will he get? [?]", "answer0": "45", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 2, "explanation": "\\(36 \\div \\dfrac{4}{5} = 36 \\times \\dfrac{5}{4} = 45\\) packets.", "hints": ["Dividing by \\(\\dfrac{4}{5}\\) is the same as multiplying by \\(\\dfrac{5}{4}\\).", "36 × 5 ÷ 4 = 45."], "source": "SCGS 2022 P6 Prelim P1 Q24", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 36 ÷ 4/5 = 45 packets." }, { "n": 25, "type_id": 2, "question": "A vase was sold at a 40% discount for $48. What was the original price of the vase before the discount? [?]", "answer0": "80", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 176, "difficulty_id": 2, "explanation": "A 40% discount means $48 is 60% of the original price. 1% = 48 ÷ 60 = $0.80. Original = 0.80 × 100 = $80.", "hints": ["After a 40% discount, the price paid is 60% of the original.", "Find 1%, then multiply by 100."], "source": "SCGS 2022 P6 Prelim P1 Q25", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 60% = $48; 1% = 0.80; original = $80." }, { "n": 26, "type_id": 1, "question": "A container completely filled with water weighed \\(1\\dfrac{4}{5}\\) kg. After pouring out \\(\\dfrac{2}{3}\\) of the water, it weighed 1 kg. What was the mass of the container?", "answer0": "\\(\\dfrac{3}{5}\\) kg", "answer1": "\\(\\dfrac{2}{5}\\) kg", "answer2": "\\(\\dfrac{4}{5}\\) kg", "answer3": "\\(\\dfrac{1}{5}\\) kg", "correct_answer": 0, "skill_id": 165, "difficulty_id": 3, "explanation": "Water + container = \\(1\\dfrac{4}{5}\\) kg. Poured out \\(\\dfrac{2}{3}\\) of the water = the drop in mass: \\(1\\dfrac{4}{5} - 1 = \\dfrac{4}{5}\\) kg is \\(\\dfrac{2}{3}\\) of the water. So all the water = \\(\\dfrac{4}{5} \\div \\dfrac{2}{3} = \\dfrac{6}{5}\\) kg... container = \\(1\\dfrac{4}{5} - \\dfrac{6}{5} = \\dfrac{9}{5} - \\dfrac{6}{5} = \\dfrac{3}{5}\\) kg. (Key: 1/3 of water remaining = 2/5 kg; container = 1 − 2/5 = 3/5 kg.)", "hints": ["The mass poured out equals \\(\\dfrac{2}{3}\\) of the water.", "Find the remaining water mass, then subtract from the 1 kg to get the container."], "source": "SCGS 2022 P6 Prelim P1 Q26", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer is a fraction \\(\\dfrac{3}{5}\\) kg, so rendered as MCQ per FIB-fraction rule. Answer key working: 2/3 of water = 4/5; 1/3 of water = 2/5; container = 1 − 2/5 = 3/5 kg." }, { "n": 27, "type_id": 0, "question": "The net of a pyramid drawn has 2 missing faces. Shade 2 faces to complete the net of the pyramid.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 233, "difficulty_id": 2, "explanation": "Two triangular faces must be shaded/added in the correct positions to complete the pyramid net (see answer key figure).", "hints": [], "source": "SCGS 2022 P6 Prelim P1 Q27", "image_needed": true, "image_options": false, "image_file": "q27.png", "image_page": 15, "image_bbox": [0.18, 0.14, 0.70, 0.46], "image_loc": "upper part of page, incomplete net of a pyramid with shaded and dashed triangular faces", "notes": "Interactive shade-to-complete-net task — type_id 0, skipped on insert. Answer key shows the two faces to shade." }, { "n": 28, "type_id": 2, "question": "The figure shows a semi-circle overlapping with a quadrant. Find the perimeter of the shaded part. (Take \\(\\pi = \\dfrac{22}{7}\\), the diameter / radius marked is 28 cm) [?] cm", "answer0": "72", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 3, "explanation": "Semi-circle arc (radius 14): \\(2 \\times \\dfrac{22}{7} \\times 14 \\times \\dfrac{1}{2} = 44\\) cm. Adding the two straight 14 cm edges: 44 + 14 + 14 = ... the key combines the semicircle arc (44) with the quadrant straight sides: 44 + 28 = 72 cm.", "hints": ["Find the semicircle arc length using \\(\\dfrac{1}{2} \\times 2\\pi r\\).", "Add the straight edges that bound the shaded region."], "source": "SCGS 2022 P6 Prelim P1 Q28", "image_needed": true, "image_options": false, "image_file": "q28.png", "image_page": 15, "image_bbox": [0.20, 0.50, 0.52, 0.72], "image_loc": "lower part of page, a semicircle overlapping a quadrant, 28 cm marked across the base", "notes": "Answer key: semicircle = 2(22/7)×14×½ = 44; 44 + 28 = 72 cm." }, { "n": 29, "type_id": 2, "question": "Charmaine read 30 pages on Monday and \\(\\dfrac{1}{2}\\) of the remaining book on Tuesday. She was then left with 20% of the book unread. How many pages does the book have? [?]", "answer0": "50", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "After Monday, the remaining book is read half on Tuesday, leaving the other half = 20% of the whole. So \\(\\dfrac{1}{2}\\) of the Monday-remainder = 20%, meaning the Monday-remainder = 40%; thus Monday's 30 pages = 60% of the book... using the key units: 5u − 2u = 3u corresponds to 30, 1u = 10, total 5u = 50 pages.", "hints": ["The unread 20% is half of what was left after Monday.", "Set up units so Monday's 30 pages correspond to the remaining units; solve for the whole."], "source": "SCGS 2022 P6 Prelim P1 Q29", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 5u − 2u = 3u; 3u = 30; u = 10; 5u = 50 pages." }, { "n": 30, "type_id": 0, "question": "The figure, not drawn to scale, shows a semi-circle with centre O and straight lines AD, OB and CD. For each statement, tick True, False or Not possible to tell: (a) \\(\\angle OAB\\) is equal to \\(\\angle OBA\\). (b) Triangle OAB is an equilateral triangle.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 195, "difficulty_id": 3, "explanation": "(a) OA and OB are both radii, so triangle OAB is isosceles and \\(\\angle OAB = \\angle OBA\\) → True. (b) Without more information it cannot be determined whether all sides are equal → Not possible to tell.", "hints": [], "source": "SCGS 2022 P6 Prelim P1 Q30", "image_needed": true, "image_options": false, "image_file": "q30.png", "image_page": 16, "image_bbox": [0.18, 0.40, 0.62, 0.58], "image_loc": "middle of page, semicircle centre O with points A, B on arc and straight lines AD, OB, CD", "notes": "Interactive tick-True/False/Not-possible table — type_id 0, skipped on insert. Answer key: (a) True, (b) Not possible to tell." }, { "n": 31, "type_id": 1, "question": "A crate contains apples, oranges and pears. \\(\\dfrac{1}{2}\\) of the fruits are pears. The ratio of the number of apples to oranges is 3 : 4. What is the ratio of the number of pears to the number of oranges?", "answer0": "7 : 4", "answer1": "4 : 7", "answer2": "1 : 4", "answer3": "7 : 8", "correct_answer": 0, "skill_id": 183, "difficulty_id": 3, "explanation": "Pears = \\(\\dfrac{1}{2}\\) of the fruits, so apples + oranges = the other half. Apples : oranges = 3 : 4, total 7 units = the non-pear half, so pears = 7 units too. Pears : oranges = 7 : 4.", "hints": ["Apples and oranges together make half the crate (7 units in the ratio 3:4).", "Pears equal that same half (7 units); compare to oranges (4 units)."], "source": "SCGS 2022 P6 Prelim P2 Q1", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer is a ratio, so rendered as MCQ per FIB rule. Answer key: P : O = 7 : 4." }, { "n": 32, "type_id": 2, "question": "The exchange rate for Singapore dollar (SGD) to Malaysia ringgit (MYR) is 10 SGD = 32.35 MYR. How much MYR will I get if I exchange 220 SGD? [?] MYR", "answer0": "711.70", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 2, "explanation": "220 SGD = 22 × 10 SGD, so 22 × 32.35 = 711.70 MYR.", "hints": ["220 is 22 times 10.", "Multiply 32.35 by 22."], "source": "SCGS 2022 P6 Prelim P2 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 32.35 × 22 = MYR 711.70." }, { "n": 33, "type_id": 2, "question": "The figure, not drawn to scale, shows a square in a right-angle triangle (legs 3 cm and including a 4 cm side, hypotenuse-region 10 cm). Find the area of the shaded part. [?] cm²", "answer0": "20", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 186, "difficulty_id": 3, "explanation": "The shaded part is the large triangle minus the square. Using the key: shaded area = \\(\\dfrac{1}{2} \\times 10 \\times 4 = 20\\) cm² (the shaded triangle has base 10 cm and height 4 cm).", "hints": ["Identify the base and height of the shaded triangle.", "Area of a triangle = \\(\\dfrac{1}{2} \\times base \\times height\\)."], "source": "SCGS 2022 P6 Prelim P2 Q3", "image_needed": true, "image_options": false, "image_file": "q33.png", "image_page": 19, "image_bbox": [0.10, 0.60, 0.48, 0.84], "image_loc": "lower-left of page, right-angle triangle with an inscribed square, 4 cm, 3 cm and 10 cm marked", "notes": "Answer key: ∠x + ∠y = 90°; shaded area = ½ × 10 × 4 = 20 cm²." }, { "n": 34, "type_id": 2, "question": "Water was flowing out from a leaking tap at a rate of 270 ml per minute, filling up the container shown. It took 27 minutes for the container to be completely filled with water. The base is 27 cm by 12 cm. What is the height of the container? [?] cm", "answer0": "22.5", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 191, "difficulty_id": 3, "explanation": "Total volume = 270 × 27 = 7290 ml = 7290 cm³. Height = volume ÷ base area = 7290 ÷ (27 × 12) = 7290 ÷ 324 = 22.5 cm.", "hints": ["Total volume = rate × time; 1 ml = 1 cm³.", "Height = volume ÷ base area."], "source": "SCGS 2022 P6 Prelim P2 Q4", "image_needed": true, "image_options": false, "image_file": "q34.png", "image_page": 20, "image_bbox": [0.10, 0.16, 0.46, 0.36], "image_loc": "upper-left of page, cuboid container with base 27 cm by 12 cm and height '?'", "notes": "Answer key: 27 × 270 cm³ = 7290; 7290 ÷ (27×12) = 22.5 cm." }, { "n": 35, "type_id": 0, "question": "Triangle ABC is drawn on the square grid. By joining dots on the grid with straight lines: (a) draw and label a trapezium CABF such that the length of BF is half of AC. (b) draw and label Triangle ABD such that its area is half of Triangle ABC. Triangle ABD must not overlap with trapezium CABF.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 179, "difficulty_id": 3, "explanation": "Draw the trapezium and triangle on the grid as specified (see answer key figure).", "hints": [], "source": "SCGS 2022 P6 Prelim P2 Q5", "image_needed": true, "image_options": false, "image_file": "q35.png", "image_page": 20, "image_bbox": [0.16, 0.62, 0.62, 0.92], "image_loc": "lower part of page, triangle ABC on a dotted square grid", "notes": "Interactive draw-on-grid task — type_id 0, skipped on insert." }, { "n": 36, "type_id": 2, "question": "Helen is \\((y + 8)\\) years old now. She is 3 years older than Bonny.
(a) What will be their total age in 2 years' time in terms of \\(y\\)? [?]
(b) If \\(y = 5\\), find their total age in 2 years' time. [?]", "answer0": "2y + 17", "answer1": "27", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 241, "difficulty_id": 3, "explanation": "Bonny now = \\((y + 8) - 3 = y + 5\\). In 2 years: Helen = \\(y + 10\\), Bonny = \\(y + 7\\). Total = \\((y + 10) + (y + 7) = 2y + 17\\). (b) When y = 5: \\(2(5) + 17 = 27\\) years.", "hints": ["Bonny is 3 years younger than Helen; find Bonny's age now.", "Add 2 to each age, then sum; substitute y = 5 for part (b)."], "source": "SCGS 2022 P6 Prelim P2 Q6", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: (a) (2y + 17) yrs; (b) 2×5 + 17 = 27 yrs." }, { "n": 37, "type_id": 2, "question": "The bar graph shows the number of cakes a bakery sold from January to March. The number of cakes sold in March was 15% of the total number of cakes sold from January to April. What was the total number of cakes sold from January to April? [?]", "answer0": "60", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "From the graph, March = 9 cakes. March is 15% of the Jan-Apr total, so 15% = 9, 1% = 0.6, total = 0.6 × 100 = 60 cakes.", "hints": ["Read the March bar value.", "If 15% = the March value, find 100%."], "source": "SCGS 2022 P6 Prelim P2 Q7", "image_needed": true, "image_options": false, "image_file": "q37.png", "image_page": 22, "image_bbox": [0.12, 0.22, 0.78, 0.46], "image_loc": "upper part of page, bar graph of cakes sold (Jan, Feb, Mar, Apr) with April blank", "notes": "Answer key: 15% = 9; 100% = 9/15 × 100 = 60 cakes. Part (b) draws the April bar (= 60 − (22+16+9) = 13) — interactive, captured in notes. April bar = 13." }, { "n": 38, "type_id": 2, "question": "In the figure, not drawn to scale, ABCD is a square with a length of 14 cm. Given that BCF is a straight line, and the area of triangle AED is 36.75 cm², find the area of triangle EFD. [?] cm²", "answer0": "61.25", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 186, "difficulty_id": 3, "explanation": "Area of triangle AFD = \\(\\dfrac{1}{2} \\times 14 \\times 14 = 98\\) cm² (base AD = 14, height = 14). Triangle EFD = triangle AFD − triangle AED = 98 − 36.75 = 61.25 cm².", "hints": ["Triangle AFD has base AD and the full height of the square.", "Subtract the area of triangle AED."], "source": "SCGS 2022 P6 Prelim P2 Q8", "image_needed": true, "image_options": false, "image_file": "q38.png", "image_page": 23, "image_bbox": [0.40, 0.14, 0.80, 0.34], "image_loc": "upper part of page, square ABCD (14 cm) with straight line BCF and shaded triangle EFD", "notes": "Answer key: area AFD = ½×14×14 = 98; 98 − 36.75 = 61.25 cm²." }, { "n": 39, "type_id": 2, "question": "In the figure, not drawn to scale, ABCD is a rectangle. EF is parallel to BD and AD = DG. With \\(\\angle AFH = 110°\\) and \\(\\angle BFK = 42°\\),
(a) Find \\(\\angle AFE. [?]°
(b) Find \\(\\angle BKC. [?]°", "answer0": "25", "answer1": "67", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "(a) Angles on the straight line at F: \\(\\angle AFE = 180° - 110° - 45° = 25°\\). (b) \\(\\angle BKC = 42° + 25° = 67°\\) (exterior angle / using the parallel lines).", "hints": ["Use angles on a straight line at point F for part (a).", "Use the parallel lines EF // BD and the given angles for part (b)."], "source": "SCGS 2022 P6 Prelim P2 Q9", "image_needed": true, "image_options": false, "image_file": "q39.png", "image_page": 24, "image_bbox": [0.12, 0.12, 0.70, 0.34], "image_loc": "upper part of page, rectangle ABCD with EF parallel to BD, points F, H, J, K, G marked, 110° and 42° given", "notes": "Answer key: (a) 180 − 110 − 45 = 25°; (b) 42 + 25 = 67°." }, { "n": 40, "type_id": 2, "question": "The cost of an adult ticket to a concert was $68.80. The cost of a child ticket was $32.80. The total amount of money collected from ticket sales was $28 100 for a capacity of 500 people. How many adults attended the concert? [?]", "answer0": "325", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "If all 500 were children: 32.80 × 500 = $16 400. Difference = 28 100 − 16 400 = $11 700. Each adult costs 68.80 − 32.80 = $36 more than a child. Number of adults = 11 700 ÷ 36 = 325.", "hints": ["Assume all 500 are children and find the shortfall in money collected.", "Each adult adds (68.80 − 32.80) more; divide the shortfall by that."], "source": "SCGS 2022 P6 Prelim P2 Q10", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 32.80×500 = 16400; 28100 − 16400 = 11700; 11700 ÷ 36 = 325 adults." }, { "n": 41, "type_id": 1, "question": "A baker had a 50-kg sack of flour at first. The graph shows the amount of flour left at the end of each day for 5 days. Which day did the baker use the greatest amount of flour?", "answer0": "Day 1", "answer1": "Day 2", "answer2": "Day 3", "answer3": "Day 4", "correct_answer": 2, "skill_id": 147, "difficulty_id": 2, "explanation": "The greatest usage is on the day with the steepest fall in the graph. The steepest drop is between Day 3 and Day 4, i.e. the flour used on Day 4... per the key the greatest amount used corresponds to the steepest segment (between Day 3 and Day 4), pointing to Day 4. Using the printed key the answer is the steepest-line day.", "hints": ["The most flour is used where the graph drops the most.", "Compare the steepness of each day's segment."], "source": "SCGS 2022 P6 Prelim P2 Q11a", "image_needed": true, "image_options": false, "image_file": "q41.png", "image_page": 26, "image_bbox": [0.10, 0.14, 0.78, 0.40], "image_loc": "upper part of page, line graph of flour left (kg) over Day 1 to Day 5", "notes": "Answer key part (a): steepest line between Day 3 and Day 4 → the day of greatest use is Day 4. Day-name answer → MCQ. (Part (b) is a separate FIB, see Q42.)" }, { "n": 42, "type_id": 2, "question": "A baker had a 50-kg sack of flour at first. The graph shows the amount of flour left at the end of each day for 5 days. What percentage of the 50-kg sack of flour was used by day 5? [?] %", "answer0": "80", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 2, "explanation": "At the end of Day 5, 10 kg of flour is left, so flour used = 50 − 10 = 40 kg. Percentage used = \\(\\dfrac{40}{50} \\times 100 = 80\\%\\).", "hints": ["Read the flour left at the end of Day 5 from the graph.", "Percentage used = used ÷ 50 × 100."], "source": "SCGS 2022 P6 Prelim P2 Q11b", "image_needed": true, "image_options": false, "image_file": "q42.png", "image_page": 26, "image_bbox": [0.10, 0.14, 0.78, 0.40], "image_loc": "upper part of page, same line graph of flour left over Day 1 to Day 5 as Q41", "notes": "Answer key part (b): 50 − 10 = 40 used; 40/50 = 80%." }, { "n": 43, "type_id": 2, "question": "In the figure, ABCD is a square. ADF is an equilateral triangle and DECF is a trapezium. DE // FC and \\(\\angle DEC = 50°\\).
(a) Find \\(\\angle DCE. [?]°
(b) Find \\(\\angle BFC. [?]°", "answer0": "55", "answer1": "150", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "(a) \\(\\angle FDC = 90° - 60° = 30°\\) (square corner minus equilateral angle); \\(\\angle DCE = (180° - 30°) \\div 2 = 75°\\)? Per the key: \\(\\angle DCF = (180° - 30°) \\div 2 = 75°\\), and \\(\\angle DCE = 180° - 75° - 50° = 55°\\). (b) \\(\\angle BFC = 360° - 75° - 75° - 60° = 150°\\).", "hints": ["Use the square corner (90°) and equilateral triangle (60°) to find \\(\\angle FDC\\).", "Use the isosceles trapezium and angles around point F for part (b)."], "source": "SCGS 2022 P6 Prelim P2 Q12", "image_needed": true, "image_options": false, "image_file": "q43.png", "image_page": 27, "image_bbox": [0.40, 0.14, 0.80, 0.56], "image_loc": "right part of page, square ABCD with equilateral triangle ADF, trapezium DECF and E below with 50° marked", "notes": "Answer key: (a) ∠FDC = 90−60 = 30; ∠DCF = (180−30)÷2 = 75; ∠DCE = 180−75−50 = 55°; (b) ∠BFC = 360−75−75−60 = 150°." }, { "n": 44, "type_id": 2, "question": "Fred, Gerald and Harry shared $123 altogether. At a toy shop, Fred spent \\(\\dfrac{2}{5}\\) of his money, Gerald spent \\(\\dfrac{3}{4}\\) of his money and Harry spent \\(\\dfrac{2}{3}\\) of his money. Fred and Gerald spent the same amount of money and Harry spent twice of what Fred spent. Find the amount of money Gerald had at first. [?]", "answer0": "24", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "Let Fred spent = \\(\\dfrac{2}{5}F\\). Gerald spent = \\(\\dfrac{3}{4}G\\) = Fred's spending. Harry spent = \\(\\dfrac{2}{3}H\\) = twice Fred's. Using the key units: \\(\\dfrac{2}{5}F = \\dfrac{6}{15}F\\); making Fred = 15u, Gerald = 8u, Harry = 18u; total = 41u = $123, so 1u = $3. Gerald = 8u = $24.", "hints": ["Express each person's spending as a fraction and set Fred's = Gerald's, Harry's = 2 × Fred's.", "Find the money units; the total is $123."], "source": "SCGS 2022 P6 Prelim P2 Q13", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: Fred 15u, Gerald 8u, Harry 18u; 41u = $123; 1u = $3; Gerald = 8 × 3 = $24." }, { "n": 45, "type_id": 2, "question": "Mdm Pang baked some cookies. She gave \\(\\dfrac{1}{4}\\) of it to her relatives and gave 80 cookies to her friends. She was left with \\(\\dfrac{1}{3}\\) of it.
(a) How many cookies had Mdm Pang left? [?]
(b) Mdm Pang packed the leftover cookies into 10 small and large bags. The number of cookies in each large bag is twice the number of cookies in each small bag. How many large bags of cookies were there? [?]", "answer0": "64", "answer1": "6", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "(a) Gave \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\), left \\(\\dfrac{1}{3} = \\dfrac{4}{12}\\), so friends got \\(1 - \\dfrac{3}{12} - \\dfrac{4}{12} = \\dfrac{5}{12}\\) = 80 cookies. \\(\\dfrac{1}{12} = 16\\); left = \\(\\dfrac{4}{12} = 64\\) cookies. (b) Let small bags = s, large = 10 − s. With each large = 2 × small (per bag), total cookies 64 leads to: small-bag count gives 8 small + ... key: 4 large bags... actually key answer is 6 large bags (4 small + 6 large with 64 divisible appropriately).", "hints": ["For (a), find the friends' fraction, set it equal to 80, and find the left fraction.", "For (b), set up small and large bags summing to 10 with each large holding twice a small."], "source": "SCGS 2022 P6 Prelim P2 Q14", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: (a) 5/12 = 80, 1/12 = 16, left 4/12 = 64 cookies; (b) testing bag splits, 4 small + 6 large with 64 divisible by 16 → 6 large bags. Answer (b) = 6." }, { "n": 46, "type_id": 2, "question": "Celine took a square piece of paper and cut along the dotted line. As a result, she got a small square of area 49 cm² and 8 identical right-angled triangles. Triangle EFG is one such right-angled triangle with hypotenuse EF = 17 cm.
(a) Find the area of the square ABCD. [?] cm²
(b) Find the length of FG. [?] cm", "answer0": "289", "answer1": "8", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 138, "difficulty_id": 3, "explanation": "(a) Square ABCD has side 17 cm (the hypotenuse), so area = 17 × 17 = 289 cm². (b) 4 triangles = 289 − 49 = 240 cm²; 8 triangles = 480 cm². Area of the large square = 480 + 49 = 529 cm², side = \\(\\sqrt{529} = 23\\) cm; small square side = \\(\\sqrt{49} = 7\\); FG = (23 − 7) ÷ 2 = 8 cm.", "hints": ["The side of ABCD is the 17 cm hypotenuse.", "Use the total area to find the large square's side, then FG = (large side − small side) ÷ 2."], "source": "SCGS 2022 P6 Prelim P2 Q15", "image_needed": true, "image_options": false, "image_file": "q46.png", "image_page": 30, "image_bbox": [0.10, 0.18, 0.74, 0.40], "image_loc": "upper part of page, square ABCD with dotted cut lines and a separate right-angled triangle EFG (17 cm)", "notes": "Answer key: (a) 17×17 = 289 cm²; (b) 4 triangles = 240, 8 = 480, large square = 529, side 23, small side 7, FG = (23−7)/2 = 8 cm." }, { "n": 47, "type_id": 2, "question": "There were 75 more children than adults at a funfair on Saturday. On Sunday, the number of children increased by 24% while the number of adults decreased by 15%. There were 2810 people on Sunday. How many people were there at the funfair on Saturday? [?]", "answer0": "2675", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "Let Saturday adults = 100% = A. Children = A + 75 = 100%. Sunday children = 124% and adults = 85%, with the 75-difference: Sunday total = 124% (children) + 85% (adults), and the children are 75 more in base. Using the key: 209% corresponds to (2810 − 93) = 2717, so 1% = 13, 200% = 2600, Saturday total = 2600 + 75 = 2675.", "hints": ["Write Sunday's children and adults as percentages of Saturday's, accounting for the 75 difference.", "Solve for 1%, then build back Saturday's total."], "source": "SCGS 2022 P6 Prelim P2 Q16", "image_needed": false, "image_options": false, "image_file": null, "notes": "Answer key: 209% = 2810 − 93 = 2717; 1% = 13; 200% = 2600; Saturday = 2600 + 75 = 2675." }, { "n": 48, "type_id": 2, "question": "The diagram shows figures made up of dots and lines. The table records: Figure 1 → 2 dots, 1 line; Figure 2 → 6 dots, 7 lines; Figure 3 → 12 dots, 17 lines; Figure 4 → 20 dots, 31 lines.
(a) For Figure 5, complete the table: [?] dots and [?] lines.
(b) Which figure no. will it be where there are 156 dots? [?]", "answer0": "30", "answer1": "49", "answer2": "12", "answer3": null, "correct_answer": null, "skill_id": 6, "difficulty_id": 3, "explanation": "Number of dots = figure number × (figure number + 1). Figure 5 dots = 5 × 6 = 30. Number of lines = (figure no.)² + (figure no. − 1) × (figure no. + 1); Figure 5 lines = 5×5 + 4×6 = 25 + 24 = 49. (b) Dots = n(n+1) = 156 → 12 × 13 = 156, so Figure 12.", "hints": ["Dots follow n × (n + 1); lines follow n² + (n − 1)(n + 1).", "For 156 dots, find n with n(n+1) = 156."], "source": "SCGS 2022 P6 Prelim P2 Q17", "image_needed": true, "image_options": false, "image_file": "q48.png", "image_page": 32, "image_bbox": [0.16, 0.12, 0.80, 0.24], "image_loc": "upper part of page, Figures 1-4 made of dots and connecting lines", "notes": "Answer key: (a) Figure 5 = 30 dots, 49 lines; (b) 12 × 13 = 156 → Figure 12. The three blanks are Figure-5 dots (30), Figure-5 lines (49), and the figure number for 156 dots (12)." } ] }