{ "paper": { "school": "Nanyang", "year": 2025, "level": "P6", "label": "Weighted Assessment 1", "source_prefix": "Nanyang 2025 P6 Weighted Assessment 1", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 1, "question": "In the number line, what is the value of Y?", "answer0": "2.85", "answer1": "2.8", "answer2": "2.75", "answer3": "2.7", "correct_answer": 0, "skill_id": 158, "difficulty_id": 2, "explanation": "The interval from 2.6 to 3.0 is divided into equal parts of 0.05. Y sits at the mark reading 2.85.", "hints": ["Find the value of each small division between 2.6 and 3.0.", "Count the divisions up to Y's arrow."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q1", "image_needed": true, "image_options": false, "image_file": "q1.png", "image_page": 2, "image_bbox": [0.20, 0.27, 0.82, 0.37], "image_loc": "number line from 2.6 to 3.0 with arrow labelled Y, upper-middle of page", "notes": "Key answer (1) = 2.85." }, { "n": 2, "type_id": 1, "question": "Find the value of \\(\\dfrac{5}{6}\\div\\dfrac{1}{4}\\).", "answer0": "\\(\\dfrac{10}{3}\\)", "answer1": "\\(\\dfrac{5}{24}\\)", "answer2": "\\(\\dfrac{3}{10}\\)", "answer3": "\\(\\dfrac{24}{5}\\)", "correct_answer": 0, "skill_id": 206, "difficulty_id": 1, "explanation": "\\(\\dfrac{5}{6}\\div\\dfrac{1}{4}=\\dfrac{5}{6}\\times4=\\dfrac{20}{6}=\\dfrac{10}{3}\\).", "hints": ["Dividing by 1/4 is the same as multiplying by 4.", "Simplify 20/6."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (1) = 10/3. Fraction options so MCQ." }, { "n": 3, "type_id": 1, "question": "Which of the following is the same as 25% of 20%?", "answer0": "\\(\\dfrac{1}{4}\\times\\dfrac{1}{5}\\)", "answer1": "\\(\\dfrac{3}{4}\\times\\dfrac{1}{5}\\)", "answer2": "\\(\\dfrac{1}{4}\\times\\dfrac{4}{5}\\)", "answer3": "\\(\\dfrac{3}{4}\\times\\dfrac{4}{5}\\)", "correct_answer": 0, "skill_id": 173, "difficulty_id": 1, "explanation": "25% = \\(\\dfrac{1}{4}\\) and 20% = \\(\\dfrac{1}{5}\\). 25% of 20% = \\(\\dfrac{1}{4}\\times\\dfrac{1}{5}\\).", "hints": ["Convert each percentage to a fraction.", "25% = 1/4 and 20% = 1/5."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q3", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (1) = 1/4 × 1/5." }, { "n": 4, "type_id": 1, "question": "The square grid shows Triangle P. What type of triangle is Triangle P?", "answer0": "Obtuse-angled triangle", "answer1": "Right-angled triangle", "answer2": "Equilateral triangle", "answer3": "Isosceles triangle", "correct_answer": 3, "skill_id": 195, "difficulty_id": 1, "explanation": "On the grid the triangle has two equal sloping sides, making it an isosceles triangle.", "hints": ["Count grid units to compare the side lengths.", "Two equal sides means isosceles."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q4", "image_needed": true, "image_options": false, "image_file": "q4.png", "image_page": 3, "image_bbox": [0.38, 0.43, 0.66, 0.62], "image_loc": "dashed square grid with Triangle P (tall narrow triangle) labelled P, middle of page", "notes": "Key answer (4) = Isosceles triangle." }, { "n": 5, "type_id": 1, "question": "In the square grid, which shape is a rhombus?", "answer0": "Shape (1)", "answer1": "Shape (2)", "answer2": "Shape (3)", "answer3": "Shape (4)", "correct_answer": 3, "skill_id": 199, "difficulty_id": 2, "explanation": "A rhombus has 4 equal sides. Only Shape (4) has all four sides equal in length.", "hints": ["A rhombus has all four sides equal.", "Check the side lengths of each shape on the grid."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q5", "image_needed": true, "image_options": true, "image_file": null, "image_page": 4, "image_bbox": [0.24, 0.16, 0.82, 0.34], "image_loc": "four shapes (1)-(4) on square grids in a row, upper part of page", "notes": "Key answer (4). Options are shapes drawn on grids; crop each option q5_opt0..3." }, { "n": 6, "type_id": 1, "question": "What is the area of triangle ABC as shown in the figure?", "answer0": "18 cm²", "answer1": "20 cm²", "answer2": "30 cm²", "answer3": "36 cm²", "correct_answer": 0, "skill_id": 186, "difficulty_id": 1, "explanation": "Base CB = 4 + 8 = 12 cm, height = 3 cm. Area = \\(\\dfrac{1}{2}\\times12\\times3 = 18\\) cm².", "hints": ["The base is the whole bottom CB = 4 + 8.", "Area of a triangle = 1/2 × base × height."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q6", "image_needed": true, "image_options": false, "image_file": "q6.png", "image_page": 4, "image_bbox": [0.26, 0.45, 0.72, 0.60], "image_loc": "triangle ABC with 5 cm slant, 3 cm height, base split 4 cm and 8 cm, lower-middle of page", "notes": "Key answer (1) = 18 cm². The 5 cm side is a distractor." }, { "n": 7, "type_id": 1, "question": "What is the area of a circle with diameter 60 cm? (Take \\(\\pi=3.14\\))", "answer0": "94.2 cm²", "answer1": "188.4 cm²", "answer2": "2826 cm²", "answer3": "11 304 cm²", "correct_answer": 2, "skill_id": 220, "difficulty_id": 1, "explanation": "Radius = 60 ÷ 2 = 30 cm. Area = \\(3.14\\times30\\times30 = 3.14\\times900 = 2826\\) cm².", "hints": ["The radius is half the diameter (30 cm).", "Area = π × r × r."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q7", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (3) = 2826 cm²." }, { "n": 8, "type_id": 1, "question": "Which of the following is likely to be the length of an approved scientific calculator for PSLE?", "answer0": "0.018 m", "answer1": "0.18 m", "answer2": "1.8 m", "answer3": "18 m", "correct_answer": 1, "skill_id": 184, "difficulty_id": 1, "explanation": "A scientific calculator is about 18 cm long = 0.18 m.", "hints": ["A calculator is roughly the length of your hand.", "0.18 m = 18 cm."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q8", "image_needed": true, "image_options": false, "image_file": "q8.png", "image_page": 6, "image_bbox": [0.44, 0.20, 0.64, 0.48], "image_loc": "scientific calculator with a '?' length arrow on the left, upper-middle of page", "notes": "Key answer (2) = 0.18 m." }, { "n": 9, "type_id": 1, "question": "The bar graph shows the number of cars sold from June to September. In which month was the number of cars sold half as many as the number of cars sold in September?", "answer0": "June", "answer1": "July", "answer2": "August", "answer3": "September", "correct_answer": 2, "skill_id": 147, "difficulty_id": 2, "explanation": "September sold about 6000 cars; half of that is about 3000, which matches August's bar.", "hints": ["Read off the height of the September bar first.", "Find the month whose bar is half that height."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q9", "image_needed": true, "image_options": false, "image_file": "q9.png", "image_page": 7, "image_bbox": [0.16, 0.20, 0.82, 0.44], "image_loc": "bar graph of cars sold for June, July, August, September, upper part of page", "notes": "Key answer (3) = August. Graph shared with Q10." }, { "n": 10, "type_id": 1, "question": "Which one of the following statements is true?", "answer0": "The number of cars sold in June was 8500.", "answer1": "The number of cars sold in July is \\(\\dfrac{3}{4}\\) the number of cars sold in June.", "answer2": "The increase in the number of cars sold from August to September was 9000.", "answer3": "The total number of cars sold in June and August is the same as the number of cars sold in July.", "correct_answer": 3, "skill_id": 147, "difficulty_id": 2, "explanation": "From the graph: June ≈ 9000, July ≈ 12000, August ≈ 3000, September ≈ 6000. June + August = 9000 + 3000 = 12000 = July. Statement (4) is true.", "hints": ["Read each bar's value carefully.", "Check each statement against the readings."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q10", "image_needed": true, "image_options": false, "image_file": "q9.png", "image_page": 7, "image_bbox": [0.16, 0.20, 0.82, 0.44], "image_loc": "same bar graph as Q9 (June-September cars sold), on the previous page", "notes": "Key answer (4). Uses the same bar graph as Q9 (page 7); readings June 9000, July 12000, August 3000, September 6000." }, { "n": 11, "type_id": 1, "question": "Last month, a florist sold 800 roses. This month, she sold 1000 roses. What was the percentage increase in the number of roses sold?", "answer0": "20%", "answer1": "25%", "answer2": "80%", "answer3": "200%", "correct_answer": 1, "skill_id": 208, "difficulty_id": 2, "explanation": "Increase = 1000 − 800 = 200. Percentage increase = \\(\\dfrac{200}{800}\\times100\\% = 25\\%\\).", "hints": ["Find the increase in roses first (200).", "Percentage increase = increase ÷ original × 100%."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q11", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (2) = 25%." }, { "n": 12, "type_id": 1, "question": "The figure is made up of 3 identical quarter circles of radius 28 cm. Find its perimeter. (Take \\(\\pi=\\dfrac{22}{7}\\))", "answer0": "132 cm", "answer1": "176 cm", "answer2": "188 cm", "answer3": "232 cm", "correct_answer": 2, "skill_id": 221, "difficulty_id": 3, "explanation": "Each quarter-circle arc = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times2\\times28 = 44\\) cm. Three arcs = 3 × 44 = 132 cm. The straight edges add 28 + 28 = 56 cm; per the key the total perimeter = 188 cm.", "hints": ["Find one quarter-circle arc length (radius 28 cm).", "Add the three arcs and the straight radii that form the outline."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q12", "image_needed": true, "image_options": false, "image_file": "q12.png", "image_page": 9, "image_bbox": [0.42, 0.16, 0.66, 0.34], "image_loc": "figure of 3 identical quarter circles (radius 28 cm) joined, upper-middle of page", "notes": "Key answer (3) = 188 cm." }, { "n": 13, "type_id": 1, "question": "A lollipop cost $0.70. There were 80 lollipops in a box. Janie bought 8 such boxes of lollipops for her class party. How much did she spend on the lollipops?", "answer0": "$408", "answer1": "$428", "answer2": "$448", "answer3": "$560", "correct_answer": 2, "skill_id": 167, "difficulty_id": 1, "explanation": "Total lollipops = 80 × 8 = 640. Cost = 640 × $0.70 = $448.", "hints": ["Find the total number of lollipops first.", "Multiply by $0.70."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q13", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (3) = $448." }, { "n": 14, "type_id": 1, "question": "An empty rectangular tank was 40 cm long, 20 cm wide and 80 cm high. Mary poured some water into it and the water level reached a height of 30 cm. How many litres of water were there in the tank?", "answer0": "64 000", "answer1": "24 000", "answer2": "64", "answer3": "24", "correct_answer": 3, "skill_id": 191, "difficulty_id": 2, "explanation": "Volume of water = 40 × 20 × 30 = 24 000 cm³. 24 000 cm³ ÷ 1000 = 24 litres.", "hints": ["Use the water height (30 cm), not the tank height.", "1000 cm³ = 1 litre."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q14", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (4) = 24 litres." }, { "n": 15, "type_id": 1, "question": "In a basket, \\(\\dfrac{5}{9}\\) of the fruits are apples and the rest are oranges. \\(\\dfrac{3}{10}\\) of the apples are green in colour. There are 15 green apples. How many fruits are there in the basket?", "answer0": "45", "answer1": "50", "answer2": "90", "answer3": "135", "correct_answer": 2, "skill_id": 165, "difficulty_id": 2, "explanation": "Green apples = \\(\\dfrac{3}{10}\\) of apples = 15, so apples = 15 × \\(\\dfrac{10}{3}\\) = 50. Apples are \\(\\dfrac{5}{9}\\) of the fruits, so fruits = 50 × \\(\\dfrac{9}{5}\\) = 90.", "hints": ["Find the number of apples from the 15 green apples.", "Apples are 5/9 of all the fruits."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q15", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer (3) = 90." }, { "n": 16, "type_id": 2, "question": "Express \\(3\\dfrac{1}{4}\\) as a decimal.
[?]", "answer0": "3.25", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 158, "difficulty_id": 1, "explanation": "\\(\\dfrac{1}{4}=0.25\\), so \\(3\\dfrac{1}{4}=3.25\\).", "hints": ["Convert 1/4 to a decimal (0.25).", "Add it to the whole number 3."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q16", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer 3.25. Decimal answer so FIB." }, { "n": 17, "type_id": 2, "question": "The volume of a cube is 1000 cm³. Find the length of one edge of the cube.
[?] cm", "answer0": "10", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 225, "difficulty_id": 1, "explanation": "Edge = \\(\\sqrt[3]{1000}=10\\) cm (since 10 × 10 × 10 = 1000).", "hints": ["The edge is the cube root of the volume.", "Find the number that, cubed, gives 1000."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q17", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer 10 cm." }, { "n": 18, "type_id": 0, "question": "6 unit cubes were stacked and glued together to form the solid shown (with Top view, Front view and Side view arrows marked). Draw the side view of the solid on the grid.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 187, "difficulty_id": 2, "explanation": "Looking from the side, the solid projects to an L-shaped outline of unit squares (see the answer key figure).", "hints": ["Imagine looking at the solid from the side-view arrow.", "Shade the unit squares that you would see."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q18", "image_needed": true, "image_options": false, "image_file": "q18.png", "image_page": 13, "image_bbox": [0.36, 0.16, 0.66, 0.38], "image_loc": "isometric drawing of 6 unit cubes with Top/Front/Side view arrows, upper-middle of page", "notes": "Interactive draw-the-side-view task with no app question type — type_id 0, skipped on insert. Key: see figure (L-shaped side view)." }, { "n": 19, "type_id": 2, "question": "ACD and BCE are straight lines. \\(\\angle ABE = 55\\degree\\), \\(\\angle DEB = 114\\degree\\) and \\(\\angle DAB = 90\\degree\\). Find \\(\\angle ADE\\).
[?]", "answer0": "31", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 197, "difficulty_id": 2, "explanation": "\\(\\angle DCE = \\angle ACB\\) (vertically opposite). In triangle ABC, \\(\\angle ACB = 180\\degree - 90\\degree - 55\\degree\\); in triangle DCE, \\(180\\degree - (\\angle ADE + 114\\degree) = 180\\degree - (90\\degree + 55\\degree)\\), so \\(\\angle ADE = 90\\degree + 55\\degree - 114\\degree = 31\\degree\\).", "hints": ["Mark the vertically opposite angles at C.", "Use the angle sums of the two triangles sharing vertex C."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q19", "image_needed": true, "image_options": false, "image_file": "q19.png", "image_page": 14, "image_bbox": [0.22, 0.14, 0.82, 0.32], "image_loc": "two triangles meeting at C with straight lines ACD and BCE, angles 55°, 114° and right angle marked, upper part of page", "notes": "Key working: ∠ADE = 90° + 55° − 114° = 31°. Key answer 31°." }, { "n": 20, "type_id": 2, "question": "WXYZ is a trapezium and WX is parallel to ZY. \\(\\angle WXY = 56\\degree\\) and \\(\\angle WZY = 68\\degree\\). Find \\(\\angle XWZ\\).
[?]", "answer0": "112", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 200, "difficulty_id": 2, "explanation": "WX is parallel to ZY, so \\(\\angle XWZ\\) and \\(\\angle WZY\\) are co-interior angles: \\(\\angle XWZ = 180\\degree - 68\\degree = 112\\degree\\).", "hints": ["WX parallel to ZY gives co-interior (supplementary) angles.", "∠XWZ = 180° − ∠WZY."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q20", "image_needed": true, "image_options": false, "image_file": "q20.png", "image_page": 14, "image_bbox": [0.34, 0.55, 0.62, 0.70], "image_loc": "trapezium WXYZ with 56° at X and 68° at Z, lower-middle of page", "notes": "Key answer 112°. The 56° at X is a distractor; only the parallel-side co-interior relation is needed." }, { "n": 21, "type_id": 2, "question": "JKN is an isosceles triangle and KLMN is a parallelogram. JNM is a straight line and JK = KN. \\(\\angle KNM = 110\\degree\\). Find \\(\\angle JKN\\).
[?]", "answer0": "40", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 200, "difficulty_id": 3, "explanation": "\\(\\angle KNJ = 180\\degree - 110\\degree = 70\\degree\\) (angles on straight line JNM). Triangle JKN is isosceles (JK = KN), so \\(\\angle KJN = \\angle KNJ = 70\\degree\\). \\(\\angle JKN = 180\\degree - 70\\degree - 70\\degree = 40\\degree\\).", "hints": ["Find ∠KNJ using the straight line JNM.", "JK = KN makes the base angles equal."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q21", "image_needed": true, "image_options": false, "image_file": "q21.png", "image_page": 15, "image_bbox": [0.30, 0.24, 0.66, 0.42], "image_loc": "isosceles triangle JKN attached to parallelogram KLMN, 110° marked at N, upper-middle of page", "notes": "Key working: ∠KNJ = 180° − 110° = 70°; ∠JKN = 180° − 70° − 70° = 40°. Key answer 40°." }, { "n": 22, "type_id": 2, "question": "Find the circumference of a circle of radius 5 cm. (Take \\(\\pi=3.14\\))
[?] cm", "answer0": "31.4", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 220, "difficulty_id": 1, "explanation": "Circumference = \\(2\\pi r = 2\\times3.14\\times5 = 31.4\\) cm.", "hints": ["Circumference = 2 × π × radius.", "2 × 3.14 × 5."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q22", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key answer 31.4 cm." }, { "n": 23, "type_id": 2, "question": "The figure is made up of 2 semicircles. XY is half of XZ. XY = 14 cm. Find the area of the shaded part. (Take \\(\\pi=\\dfrac{22}{7}\\))
[?] cm²", "answer0": "231", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 3, "explanation": "Big semicircle radius = XZ ÷ 2; small semicircle radius 7. Shaded area = \\(0.5\\times\\dfrac{22}{7}\\times(14^2-7^2) = 0.5\\times\\dfrac{22}{7}\\times(196-49) = 0.5\\times\\dfrac{22}{7}\\times147 = 231\\) cm².", "hints": ["The shaded area is the big half-circle minus the small half-circle.", "Use radii 14 cm and 7 cm: 0.5 × π × (14² − 7²)."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q23", "image_needed": true, "image_options": false, "image_file": "q23.png", "image_page": 16, "image_bbox": [0.26, 0.20, 0.66, 0.34], "image_loc": "large semicircle XZ with a small unshaded semicircle near Z, shaded region, upper-middle of page", "notes": "Key working: 0.5 × 22/7 × (14² − 7²) = 231 cm². Key answer 231 cm²." }, { "n": 24, "type_id": 0, "question": "A straight line EF is drawn on a square grid inside a box. G is one of the dots inside the box. Draw two lines FG and EG to complete triangle EFG with \\(\\angle EFG = 90\\degree\\) and EF = FG.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 195, "difficulty_id": 2, "explanation": "Choose the dot G so that FG is perpendicular to EF and FG equals EF, then draw FG and EG (see the answer-key figure).", "hints": ["FG must be at right angles to EF at F.", "FG must be the same length as EF."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q24", "image_needed": true, "image_options": false, "image_file": "q24.png", "image_page": 17, "image_bbox": [0.22, 0.12, 0.84, 0.46], "image_loc": "dot-grid box with line EF drawn near the centre, upper part of page", "notes": "Interactive draw-to-complete task with no app question type — type_id 0, skipped on insert. Key: see figure (right isosceles triangle EFG)." }, { "n": 25, "type_id": 2, "question": "A cuboid is 0.4 m long and 20 cm wide. It has a volume of 20 000 cm³. Find the height of the cuboid.
[?] cm", "answer0": "25", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 226, "difficulty_id": 2, "explanation": "0.4 m = 40 cm. Height = volume ÷ (length × width) = 20 000 ÷ (40 × 20) = 20 000 ÷ 800 = 25 cm.", "hints": ["Convert 0.4 m to centimetres (40 cm).", "Height = volume ÷ (length × width)."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q25", "image_needed": true, "image_options": false, "image_file": "q25.png", "image_page": 18, "image_bbox": [0.22, 0.18, 0.52, 0.34], "image_loc": "cuboid drawing labelled 0.4 m long and 20 cm wide, upper-left of page", "notes": "Key working: 0.4 m = 40 cm; 20000 ÷ 40 ÷ 20 = 25 cm. Key answer 25 cm." }, { "n": 26, "type_id": 2, "question": "Two numbers add up to 364. One of the numbers is a 2-digit number and the other is a 3-digit number. What is the smallest possible difference between the two numbers?
[?]", "answer0": "166", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "To make the difference smallest, make the 2-digit number as large as possible: 99. Then the 3-digit number = 364 − 99 = 265. Difference = 265 − 99 = 166.", "hints": ["The largest 2-digit number is 99.", "Subtract it from 364, then find the difference."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q26", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key working: 2-digit = 99, 3-digit = 265, difference = 166. Key answer 166." }, { "n": 27, "type_id": 2, "question": "Use all the digits 7, 0, 4 and 5 to form
(a) the smallest multiple of 10 [?]
(b) the even number closest to 5000 [?]", "answer0": "4570", "answer1": "5074", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 149, "difficulty_id": 2, "explanation": "(a) A multiple of 10 ends in 0. The smallest arrangement of 7, 4, 5 in front of 0 is 4570. (b) An even number ends in 0 or 4. The number closest to 5000 starts with 5: 5074 (ends in 4) is the closest even number to 5000.", "hints": ["(a) A multiple of 10 ends in 0; make the leading digits smallest.", "(b) Even means it ends in 0 or 4; get as close to 5000 as possible."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q27", "image_needed": false, "image_options": false, "image_file": null, "notes": "Two FIB blanks: (a) 4570, (b) 5074. Key answers match." }, { "n": 28, "type_id": 2, "question": "Shanice had a bottle of shampoo. She used an equal amount of shampoo each day. At the end of the 7th day, \\(\\dfrac{4}{5}\\) of the bottle was left. At the end of the 15th day, the amount of shampoo left was 280 ml. What was the amount of shampoo in the bottle at first?
[?] ml", "answer0": "490", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "In 7 days \\(\\dfrac{1}{5}\\) was used, so each day uses \\(\\dfrac{1}{35}\\) of the bottle. By the 15th day, \\(\\dfrac{15}{35}=\\dfrac{3}{7}\\) is used, leaving \\(1-\\dfrac{3}{7}=\\dfrac{4}{7}\\) = 280 ml. So the full bottle = \\(280\\times\\dfrac{7}{4} = 490\\) ml.", "hints": ["Find the fraction used per day from the 7-day information.", "Work out the fraction left at day 15, which equals 280 ml."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q28", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key working: 1/5 in 7 days → 1/35 per day; day 15 uses 15/35 = 3/7, left 4/7 = 280 ml, full = 490 ml. Key answer 490 ml." }, { "n": 29, "type_id": 2, "question": "The block of wood was dipped into a pail of paint. The block was then cut into 4 identical cubes along the dotted lines and taken apart. The total unpainted area of the 4 cubes was 150 cm². What was the volume of each cube?
[?] cm³", "answer0": "125", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 229, "difficulty_id": 3, "explanation": "Cutting into 4 cubes creates newly exposed (unpainted) square faces. The unpainted area per cube = 150 ÷ 6 = 25 cm² (one face). Side = \\(\\sqrt{25}=5\\) cm. Volume = \\(5^3 = 125\\) cm³.", "hints": ["Total unpainted area divided among the cubes gives one face area.", "From the face area find the side, then cube it for the volume."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q29", "image_needed": true, "image_options": false, "image_file": "q29.png", "image_page": 20, "image_bbox": [0.54, 0.62, 0.84, 0.80], "image_loc": "long shaded block of wood with dotted cut lines into 4 cubes, lower-right of page", "notes": "Key working: 150 ÷ 6 = 25 cm² per face; side = 5 cm; volume = 125 cm³. Key answer 125 cm³." }, { "n": 30, "type_id": 2, "question": "Three girls used the same number of beads to make necklaces. Devi used \\(\\dfrac{2}{5}\\) of her beads, Esther used \\(\\dfrac{3}{8}\\) of hers and Farah used \\(\\dfrac{2}{3}\\) of hers. They had a total of 1440 beads at first. How many beads did each girl use?
[?]", "answer0": "216", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "Each girl used the same number of beads. \\(\\dfrac{2}{5}=\\dfrac{6}{15}\\), \\(\\dfrac{3}{8}=\\dfrac{6}{16}\\), \\(\\dfrac{2}{3}=\\dfrac{6}{9}\\), so the girls had 15, 16 and 9 units. Total = 15 + 16 + 9 = 40 units = 1440, so 1 unit = 36. Beads used = 6 units = 6 × 36 = 216.", "hints": ["Make the numerators equal (6) so the 'used' amounts match.", "Total the unit counts, find 1 unit, then 6 units."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Q30", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key working: 2/5=6/15, 3/8=6/16, 2/3=6/9; 1440 ÷ 40 = 36; used = 6 × 36 = 216. Key answer 216 beads." }, { "n": 31, "type_id": 2, "question": "The bar graph shows the number of stamps collected by 4 boys. Greg collected 50 and Yusof collected 35. Complete the table: how many stamps did Lyndon and Leon collect?
Lyndon: [?]
Leon: [?]", "answer0": "80", "answer1": "95", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 147, "difficulty_id": 1, "explanation": "Reading the bar graph: Lyndon's bar reaches 80 and Leon's bar reaches 95.", "hints": ["Read the height of each boy's bar.", "Lyndon ≈ 80, Leon ≈ 95."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q1", "image_needed": true, "image_options": false, "image_file": "q31.png", "image_page": 24, "image_bbox": [0.20, 0.20, 0.82, 0.44], "image_loc": "bar graph of stamps collected by Greg, Lyndon, Yusof, Leon, upper part of page", "notes": "Paper 2 Q1. Two FIB blanks: Lyndon 80, Leon 95. Key answers 80, 95." }, { "n": 32, "type_id": 2, "question": "A bicycle wheel of diameter 80 cm made 3 complete turns. Find the distance covered. (Take \\(\\pi=3.14\\))
[?] cm", "answer0": "753.6", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 2, "explanation": "One turn = circumference = \\(3.14\\times80 = 251.2\\) cm. Three turns = 3 × 251.2 = 753.6 cm.", "hints": ["Distance for one turn equals the circumference (π × diameter).", "Multiply by 3 turns."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q2. Key working: 3 × 3.14 × 80 = 753.6 cm. Key answer 753.6 cm." }, { "n": 33, "type_id": 2, "question": "Mr Tan bought a laptop. The price of the laptop before GST was $2580. He had to pay GST of 9% on the price of the laptop. What was the amount of GST he had to pay?
$[?]", "answer0": "232.20", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 176, "difficulty_id": 1, "explanation": "GST = 9% of $2580 = \\(\\dfrac{9}{100}\\times2580 = $232.20\\).", "hints": ["GST is 9% of the price.", "Multiply $2580 by 0.09."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q3", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q3. Key answer $232.20." }, { "n": 34, "type_id": 2, "question": "A machine started printing brochures at 8 a.m. on Wednesday at a rate of 800 brochures per hour. After every 5 hours of printing, it would be stopped for an hour to cool down. How many brochures were printed by 6 a.m. the next day?
[?]", "answer0": "15200", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "Time from 8 a.m. Wednesday to 6 a.m. next day = 22 hours. Every cycle is 5 h printing + 1 h cooling. 22 ÷ 6 = 3 cycles remainder 4 h, so printing hours = 3 × 5 + 4 = 19 hours. Brochures = 19 × 800 = 15 200.", "hints": ["Find the total elapsed time (22 hours).", "Each 6-hour cycle has only 5 printing hours; count the printing hours."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q4", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q4. Key working: 22 ÷ 6 = 3 R4; (3×5 + 4) × 800 = 19 × 800 = 15200. Key answer 15200." }, { "n": 35, "type_id": 2, "question": "Kendrik bought 4 different storybooks. The first storybook cost $14 and the average cost of the remaining storybooks was \\(\\dfrac{3}{7}\\) of the cost of the first storybook. How much did he pay for all the storybooks?
$[?]", "answer0": "32", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 204, "difficulty_id": 2, "explanation": "Average cost of the 3 remaining books = \\(\\dfrac{3}{7}\\times14 = $6\\). Total for the 3 = 3 × $6 = $18. All 4 books = $14 + $18 = $32.", "hints": ["Find the average cost of the 3 remaining books (3/7 of $14).", "Total = first book + 3 × average of the rest."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q5", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q5. Key working: average = $6; total = $14 + $18 = $32. Key answer $32." }, { "n": 36, "type_id": 2, "question": "The figure is made up of a rectangle and a triangle overlapping each other. \\(\\dfrac{1}{4}\\) of the rectangle and \\(\\dfrac{2}{5}\\) of the triangle is unshaded. The area of the unshaded part of the figure is 57 cm². Find the area of the rectangle.
[?] cm²", "answer0": "228", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 138, "difficulty_id": 3, "explanation": "The unshaded part of the rectangle (\\(\\dfrac{1}{4}\\)) and the unshaded part of the triangle (\\(\\dfrac{2}{5}\\)) make up the 57 cm² unshaded region. From the key, \\(\\dfrac{1}{4}\\) of the rectangle = 57 cm², so the rectangle = 57 × 4 = 228 cm².", "hints": ["The unshaded region is 1/4 of the rectangle plus 2/5 of the triangle.", "Per the key, 1/4 of the rectangle equals 57 cm²."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q6", "image_needed": true, "image_options": false, "image_file": "q36.png", "image_page": 27, "image_bbox": [0.36, 0.26, 0.62, 0.40], "image_loc": "shaded rectangle with an overlapping triangle (white tip and grey base), upper-middle of page", "notes": "Paper 2 Q6 part (a). Key working: area of rectangle = 57 × 4 = 228 cm². Served blank is part (a) = 228. Part (b) 'What fraction of the figure is unshaded?' = 2/11 (fraction) recorded here in notes. Key answers (a) 228 cm², (b) 2/11." }, { "n": 37, "type_id": 1, "question": "PQRS is a rectangle and QRT is a right-angled triangle with sides measuring 30 cm, 40 cm and 50 cm. The perimeter of the shaded part is 174 cm. What is the ratio of the area of the triangle to the area of the shaded part? Give your answer in the simplest form.", "answer0": "4 : 5", "answer1": "3 : 4", "answer2": "5 : 4", "answer3": "3 : 5", "correct_answer": 0, "skill_id": 213, "difficulty_id": 3, "explanation": "Breadth of the rectangle = (174 − 30 − 40 − 50) ÷ 2 = 27 cm. Area of triangle = \\(0.5\\times30\\times40 = 600\\) cm². Area of shaded part = (50 × 27) − 600 = 1350 − 600 = 750 cm². Ratio triangle : shaded = 600 : 750 = 4 : 5.", "hints": ["Use the shaded perimeter to find the rectangle's breadth.", "Shaded area = rectangle area − triangle area; then form and simplify the ratio."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q7", "image_needed": true, "image_options": false, "image_file": "q37.png", "image_page": 28, "image_bbox": [0.26, 0.26, 0.76, 0.44], "image_loc": "rectangle PQRS with right-angled triangle QRT (30, 40, 50 cm) and shaded region, upper-middle of page", "notes": "Paper 2 Q7. Key working: breadth = 27 cm, triangle 600 cm², shaded 750 cm², ratio 600:750 = 4:5. Ratio answer so rendered as MCQ. Key answer 4 : 5." }, { "n": 38, "type_id": 2, "question": "Nurul and Peili went shopping together with a total sum of $60. Nurul spent twice as much as Peili. The amount Peili had left was $7 more than what she had spent. She had twice as much money left as Nurul. How much money did Nurul have at first?
$[?]", "answer0": "31", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 241, "difficulty_id": 3, "explanation": "Let Nurul have $u left. Then (from the key's table) Nurul at first = $(5u − 14), Peili at first = $(4u − 7), and these add to $60. So (5u − 14) + (4u − 7) = 60 → 9u = 81 → u = 9. Nurul at first = 5 × 9 − 14 = $31.", "hints": ["Let Nurul's leftover be u and express both girls' starting and spent amounts in terms of u.", "The two starting amounts add to $60."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q8", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q8. Key working: Nurul at first = 5u − 14, Peili = 4u − 7; sum = 60 gives u = 9; Nurul = $31. Key answer $31." }, { "n": 39, "type_id": 2, "question": "Suzi formed a solid using some 2-cm, 3-cm and 5-cm cubes. She used a total of 18 cubes to form the solid. The total volume of the solid was 707 cm³. How many 2-cm cubes did Suzi use?
[?]", "answer0": "9", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 229, "difficulty_id": 3, "explanation": "Volumes are 8, 27 and 125 cm³. Testing combinations to total 18 cubes and 707 cm³: with four 5-cm cubes (500 cm³), the remaining 14 cubes must total 207 cm³. Solving, five 3-cm cubes (135 cm³) and nine 2-cm cubes (72 cm³) give 207 cm³. Check: 4 × 125 + 5 × 27 + 9 × 8 = 707. So 9 two-cm cubes.", "hints": ["Find the volume of each cube size (8, 27, 125 cm³).", "Try numbers of 5-cm cubes, then split the rest into 3-cm and 2-cm cubes."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q9", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q9. Key answer 9 (solution: four 5-cm, five 3-cm, nine 2-cm cubes; check 500+135+72 = 707)." }, { "n": 40, "type_id": 2, "question": "The diagram shows a piece of paper QRST in the shape of a trapezium. \\(\\angle TQR = 115\\degree\\). The paper is folded along line UV which is parallel to QT.
(a) Find \\(\\angle x\\). [?]
(b) Find \\(\\angle y\\). [?]", "answer0": "245", "answer1": "50", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "(a) After folding, \\(\\angle x\\) is the reflex angle TQU: \\(\\angle x = 360\\degree - 115\\degree = 245\\degree\\) (angles at a point). (b) \\(\\angle QUV = 180\\degree - 115\\degree = 65\\degree\\) (co-interior angles, QT parallel to UV). On folding, \\(\\angle y = 180\\degree - 65\\degree - 65\\degree = 50\\degree\\) (angles on a straight line).", "hints": ["(a) ∠x is the reflex angle; use angles at a point (360°).", "(b) Use QT parallel to UV for the co-interior angle, then the fold symmetry."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q10", "image_needed": true, "image_options": false, "image_file": "q40.png", "image_page": 31, "image_bbox": [0.22, 0.18, 0.84, 0.34], "image_loc": "trapezium QRST with fold line UV before and after folding, 115° marked at Q, angles x and y, upper part of page", "notes": "Paper 2 Q10. Two FIB blanks: (a) ∠x = 245°, (b) ∠y = 50°. Key answers match." }, { "n": 41, "type_id": 1, "question": "EFGH is a square. \\(\\angle FKE = 106\\degree\\) and FE = EJ. FKH and JKE are straight lines. Find \\(\\angle KFJ\\).", "answer0": "30.5°", "answer1": "29°", "answer2": "45°", "answer3": "75.5°", "correct_answer": 0, "skill_id": 230, "difficulty_id": 3, "explanation": "\\(\\angle EFH = (180\\degree - 90\\degree)\\div2 = 45\\degree\\) (triangle EFH is right isosceles). \\(\\angle FEJ = 180\\degree - 45\\degree - 106\\degree = 29\\degree\\) (angle sum of triangle EFK). Triangle EFJ is isosceles (FE = EJ), so \\(\\angle EFJ = \\angle EJF = (180\\degree - 29\\degree)\\div2 = 75.5\\degree\\). \\(\\angle KFJ = \\angle EFJ - \\angle EFK = 75.5\\degree - 45\\degree = 30.5\\degree\\).", "hints": ["Find ∠EFH from the right-isosceles triangle in the square.", "Use the isosceles triangle EFJ (FE = EJ) to get ∠EFJ, then subtract."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q11", "image_needed": true, "image_options": false, "image_file": "q41.png", "image_page": 32, "image_bbox": [0.34, 0.16, 0.62, 0.36], "image_loc": "square EFGH with straight lines FKH and JKE crossing at K, 106° marked, upper-middle of page", "notes": "Paper 2 Q11. Key answer 30.5°. Non-integer degree answer so rendered as MCQ; distractors are the intermediate angles from the working." }, { "n": 42, "type_id": 2, "question": "At first, Lisa had a total of 66 blue and pink balloons. 17 pink balloons burst. She then increased the number of blue balloons by 75%. After that, Lisa had a total of 79 balloons. How many pink balloons did she have at first?
[?]", "answer0": "26", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 209, "difficulty_id": 3, "explanation": "After 17 pink burst, total = 66 − 17 = 49. The blue balloons increased by 75% added 79 − 49 = 30 balloons, which is 75% of the blue balloons, so blue at first = 30 ÷ 0.75 = 40. Pink at first = 66 − 40 = 26.", "hints": ["After the pink balloons burst, the new total is 66 − 17.", "The 75% increase in blue balloons equals the rise to 79; work back to the original blue count."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q12", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q12. Key working: 66 − 17 = 49; 79 − 49 = 30 = 75% of blue; blue = 40; pink = 66 − 40 = 26. Key answer 26." }, { "n": 43, "type_id": 2, "question": "A rectangular tank with a base area of 3500 cm² and a height of 80 cm was \\(\\dfrac{1}{4}\\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained from the tank at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off.
(a) How much water was drained from the tank? [?] ℓ
(b) After the tap was turned off, how much more water was needed to fill the tank completely? [?] ℓ", "answer0": "24", "answer1": "234", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 191, "difficulty_id": 3, "explanation": "(a) From 8.00 to 8.06 is 6 minutes. Drained = 4 × 6 = 24 ℓ. (b) Water left = \\(3500\\times80\\times\\dfrac{1}{4}\\div1000 - 24 = 70 - 24 = 46\\) ℓ. Full capacity = 3500 × 80 ÷ 1000 = 280 ℓ. More water needed = 280 − 46 = 234 ℓ.", "hints": ["(a) Drain rate × minutes (6 min).", "(b) Find the water left, then subtract from the full capacity."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q13", "image_needed": true, "image_options": false, "image_file": "q43.png", "image_page": 34, "image_bbox": [0.34, 0.22, 0.64, 0.36], "image_loc": "rectangular tank with a tap at the lower-left, upper-middle of page", "notes": "Paper 2 Q13. Two FIB blanks: (a) 24 ℓ, (b) 234 ℓ. Key answers match." }, { "n": 44, "type_id": 2, "question": "Six identical rectangular boxes can be stacked into a cupboard 0.9 m wide. The first arrangement (Figure A) leaves a 42-cm gap at the top. The second arrangement (Figure B) leaves a 10-cm gap at the top and another gap at the side.
(a) In the arrangement shown in Figure B, what is the width of the gap at the side? [?] cm
(b) What is the height of the cupboard in metres? [?] m", "answer0": "16", "answer1": "2.64", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 241, "difficulty_id": 3, "explanation": "Let the box be L long and B wide. Figure A: 3 boxes high → 3L + 42 = cupboard height. Figure B: a box rotated → L + 4B + 10 = same height, and 2B = cupboard width = 90 cm. From 3L + 42 = L + 4B + 10: 2L + 32 = 4B → L + 16 = 2B = 90, gap at side = 2B − L = 16 cm. (b) 2B = 90 so L = 90 − 16 = 74 cm; height = 3 × 74 + 42 = 264 cm = 2.64 m.", "hints": ["Set the two expressions for the cupboard height equal.", "Use 2B = 0.9 m = 90 cm to find the box length and then the height."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q14", "image_needed": true, "image_options": false, "image_file": "q44.png", "image_page": 35, "image_bbox": [0.22, 0.18, 0.72, 0.42], "image_loc": "Figure A (boxes with 42 cm top gap) and Figure B (10 cm top gap and side gap) side by side, upper part of page", "notes": "Paper 2 Q14. Two FIB blanks: (a) 16 cm, (b) 2.64 m. Key answers match." }, { "n": 45, "type_id": 2, "question": "Karl had clips of four different colours. \\(\\dfrac{1}{8}\\) of the clips were white and \\(\\dfrac{2}{7}\\) of the remaining clips were red. He had an equal number of blue clips and yellow clips. Karl had 35 blue clips.
(a) How many red clips did he have? [?]
(b) Karl packed all the blue clips into small, medium and large boxes. He filled each small box with 2 clips, each medium box with 3 clips and each large box with 6 clips. All the boxes were full and there were no clips left over. What was the least number of boxes used by Karl? [?]", "answer0": "28", "answer1": "7", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 209, "difficulty_id": 3, "explanation": "(a) Remaining after white = \\(1-\\dfrac{1}{8}=\\dfrac{7}{8}\\). Red = \\(\\dfrac{2}{7}\\times\\dfrac{7}{8}=\\dfrac{2}{8}\\). Blue + yellow = \\(1-\\dfrac{1}{8}-\\dfrac{2}{8}=\\dfrac{5}{8}\\), and these are equal, so 2 × 35 = 70 blue+yellow clips = \\(\\dfrac{5}{8}\\). Then \\(\\dfrac{2}{8}\\) (red) = 70 ÷ 5 × 2 = 28 red clips. (b) 35 blue clips: use as many large boxes (6 clips) as possible: 35 ÷ 6 = 5 R5; the leftover 5 clips fit in one box of 3 and one box of 2. Total = 5 + 1 + 1 = 7 boxes.", "hints": ["(a) Find the red fraction, then use the 35 blue clips to size each fraction.", "(b) Use the largest boxes first, then fill the remainder with fewer boxes."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q15", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q15. Two FIB blanks: (a) 28 red clips, (b) 7 boxes. Key answers match." }, { "n": 46, "type_id": 2, "question": "A symmetric figure is drawn on a rectangular piece of paper 20 cm by 15 cm. Its outline consists of a large semicircle, 2 smaller semicircles and 2 straight lines. (Take \\(\\pi=3.14\\))
(a) What is the area of the figure? [?] cm²
(b) What is its perimeter? [?] cm", "answer0": "178.5", "answer1": "72.8", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 3, "explanation": "(a) Area = large semicircle (radius 10) + rectangle strip − two small semicircles (radius 5): \\(0.5\\times3.14\\times10^2 + 20\\times5 - 3.14\\times5^2 = 157 + 100 - 78.5 = 178.5\\) cm². (b) Perimeter = large semicircle arc + two small semicircle arcs + two straight lines = \\(2\\times3.14\\times5 + \\dfrac{1}{2}\\times(2\\times3.14\\times10) + 5 + 5 = 31.4 + 31.4 + 10 = 72.8\\) cm.", "hints": ["(a) Big semicircle plus rectangle part minus the two small semicircles.", "(b) Add the arcs of the semicircles and the two straight edges."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q16", "image_needed": true, "image_options": false, "image_file": "q46.png", "image_page": 37, "image_bbox": [0.36, 0.20, 0.66, 0.36], "image_loc": "symmetric shaded figure: large semicircle on top with two small semicircles cut from the bottom, in a 20 cm by 15 cm box, upper-middle of page", "notes": "Paper 2 Q16. Two FIB blanks: (a) 178.5 cm², (b) 72.8 cm. Key answers match." }, { "n": 47, "type_id": 2, "question": "The amount of money Kathy had to the amount of money Alice had was 3 : 4. After Kathy spent $250 on a bag and gave $50 to Alice, the ratio became 1 : 2.
(a) How much money did Alice have at first? $[?]
(b) How much money did Kathy have at the end? $[?]", "answer0": "1300", "answer1": "675", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 214, "difficulty_id": 3, "explanation": "Let Kathy have $u at the end; Alice has $2u at the end (ratio 1 : 2). At first Kathy = u + 300 (spent $250, gave away $50) and Alice = 2u − 50. The first ratio is 3 : 4, so 4(u + 300) = 3(2u − 50): 4u + 1200 = 6u − 150 → 2u = 1350 → u = 675. (a) Alice at first = 2u − 50 = 1350 − 50 = $1300. (b) Kathy at the end = u = $675.", "hints": ["Let Kathy's end amount be u and Alice's be 2u; express their starting amounts.", "Apply the first ratio 3 : 4 to the starting amounts and solve."], "source": "Nanyang 2025 P6 Weighted Assessment 1 Paper 2 Q17", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q17. Two FIB blanks: (a) Alice at first $1300, (b) Kathy at the end $675. Key answers match." } ] }