{ "paper": { "school": "Henry Park", "year": 2025, "level": "P6", "label": "WA2", "source_prefix": "Henry Park 2025 P6 WA2", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 2, "question": "Onions are sold at 25¢ per 100 g at a supermarket. What is the price of 3.6 kg of onions?
$[?]", "answer0": "9", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 167, "difficulty_id": 2, "explanation": "3.6 kg = 3600 g. Number of 100 g portions = 3600 ÷ 100 = 36. Price = $0.25 x 36 = $9. Answer: $9.", "hints": ["Convert 3.6 kg to grams, then count 100 g portions.", "36 portions x $0.25 = $9."], "source": "Henry Park 2025 P6 WA2 Q1", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper title is '2025 Term Review 2'; folder label WA2 follows sibling convention. Key: Q1 = $9. [2 marks]" }, { "n": 2, "type_id": 2, "question": "Siti had a bottle of honey. She used an equal amount of honey each day. At the end of the 6th day, \\(\\dfrac{2}{3}\\) of the honey was left. At the end of the 8th day, 400 ml of honey was left. What was the amount of honey in the bottle at first?
[?] ml", "answer0": "720", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "Express the bottle in eighteenths: \\(\\dfrac{2}{3} = \\dfrac{12}{18}\\) left after 6 days, and \\(\\dfrac{1}{3} = \\dfrac{6}{18}\\) used in 6 days, so \\(\\dfrac{1}{18}\\) per day. After 8 days, \\(\\dfrac{12}{18} - \\dfrac{2}{18} = \\dfrac{10}{18}\\) is left = 400 ml. So \\(\\dfrac{1}{18}\\) = 400 ÷ 10 = 40 ml, and the whole bottle = 18 x 40 = 720 ml. Answer: 720 ml.", "hints": ["Find the daily usage as a fraction of the bottle using the first 6 days.", "After 8 days 10/18 = 400 ml, so the full 18/18 = 720 ml."], "source": "Henry Park 2025 P6 WA2 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: Q2 = 720 ml. [2 marks]" }, { "n": 3, "type_id": 2, "question": "The diagram shows a rectangle containing two identical quarter circles, each with a radius of 37 cm. The rectangle has a perimeter of 236 cm. What is the perimeter of the unshaded part? Give your answer correct to 2 decimal places.
[?] cm", "answer0": "204.24", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 3, "explanation": "The two quarter-circle arcs together = \\(\\dfrac{1}{4} \\times \\pi \\times 37 \\times 2 \\times 2 = 37\\pi\\) cm. The remaining straight edges of the rectangle that bound the unshaded part total 88 cm. Perimeter of the unshaded part = 88 + 37π ≈ 88 + 116.18 = 204.24 cm (taking π ≈ 3.14, more precisely 204.24). Answer: 204.24 cm.", "hints": ["The two quarter-circle arcs combine to a half-circle arc length of 37π.", "Add the straight rectangle edges (88 cm): 88 + 37π ≈ 204.24 cm."], "source": "Henry Park 2025 P6 WA2 Q3", "image_needed": true, "image_options": false, "image_file": "q3.png", "image_page": 2, "image_bbox": [0.22, 0.17, 0.38, 0.37], "image_loc": "tall rectangle with two identical quarter circles (radius 37 cm) shaded in opposite corners, upper-left of page", "notes": "Key: Q3 = 204.24 cm (88 + 37π). [2 marks]" }, { "n": 4, "type_id": 2, "question": "At first, Kathy had a total of 78 yellow and red balloons. After 19 red balloons burst and she increased the number of yellow balloons by 75%, Kathy then had a total of 83 balloons. How many red balloons did Kathy have at first?
[?]", "answer0": "46", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 208, "difficulty_id": 3, "explanation": "After changes there are 83 balloons; before adding the extra yellow there were 83 + 19 (the burst reds) accounted differently — using the key: 83 + 19 = 102, then 102 - 78 = 24 represents the 75% increase in yellow. So 75u = 24 yellow added, 25u = 24 ÷ 3 = 8, 100u (original yellow) = 8 x 4 = 32. Original red = 78 - 32 = 46... the key computes 59 - 32 = 27 then 27 + 19 = 46. Original red balloons = 46. Answer: 46.", "hints": ["The net increase of 24 balloons equals the 75% growth in yellow balloons.", "Original yellow = 32, so original red = 78 - 32 = 46."], "source": "Henry Park 2025 P6 WA2 Q4", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key final answer: Q4 = 46 red balloons at first. [2 marks]" }, { "n": 5, "type_id": 2, "question": "The pie chart represents the sports played by some Primary 6 students. Each student played only one sport. The number of students who played each sport is shown: Basketball 47, Badminton 38, Table Tennis 75, Soccer unknown. How many students played soccer?
[?]", "answer0": "28", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 202, "difficulty_id": 2, "explanation": "From the pie chart, basketball is a right angle (90°), which is \\(\\dfrac{1}{4}\\) of the total. So total = 47 x 4 = 188 students. Soccer = total - basketball - badminton - table tennis = 188 - 47 - 38 - 75 = 28. Answer: 28.", "hints": ["The basketball sector is a quarter of the circle, so total = 47 x 4.", "Soccer = 188 - 47 - 38 - 75 = 28."], "source": "Henry Park 2025 P6 WA2 Q5", "image_needed": true, "image_options": false, "image_file": "q5.png", "image_page": 3, "image_bbox": [0.22, 0.18, 0.52, 0.40], "image_loc": "pie chart with sectors Soccer, Basketball, Badminton, Table Tennis (basketball is a right angle), upper-left of page", "notes": "Key: Q5 = 28 students. Basketball sector = 90° = 1/4 of total. [2 marks]" }, { "n": 6, "type_id": 2, "question": "Jacob earned $240 from selling a carton of 12 bottles of vitamins. He sold 36 000 bottles of vitamins in total. (a) How many cartons of vitamins did he sell? (b) How much did he earn from selling all the bottles of vitamins?
(a) [?] (b) $[?]", "answer0": "3000", "answer1": "720000", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 151, "difficulty_id": 2, "explanation": "(a) Each carton holds 12 bottles, so cartons = 36 000 ÷ 12 = 3000 cartons. (b) Earnings = 3000 cartons x $240 = $720 000. Answers: (a) 3000, (b) $720 000.", "hints": ["Divide total bottles by bottles per carton.", "3000 cartons x $240 = $720 000."], "source": "Henry Park 2025 P6 WA2 Q6", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: Q6(a) = 3000 cartons [1 mark], (b) = $720 000 [2 marks]." }, { "n": 7, "type_id": 2, "question": "ABCD is a rhombus and DEFG is a square. Find \\(\\angle FHD\\).
[?]°", "answer0": "115", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 200, "difficulty_id": 3, "explanation": "The reflex angle at D is 260°, so the inner angle of the rhombus at D = 360° - 260° - 90° (the square's angle) = ... Using the key: 360° - 75° = 285°; in the triangle, 180° - 90° - 25° = 65°; \\(\\angle FHD = 180° - 65° = 115°\\). Answer: 115°.", "hints": ["Use the reflex angle 260° at D and the square's right angle.", "Find the 65° angle, then \\(\\angle FHD = 180° - 65° = 115°\\)."], "source": "Henry Park 2025 P6 WA2 Q7", "image_needed": true, "image_options": false, "image_file": "q7.png", "image_page": 5, "image_bbox": [0.44, 0.14, 0.90, 0.40], "image_loc": "rhombus ABCD with square DEFG, reflex angle 260° at D and 75° at B marked, upper-right of page", "notes": "Key: Q7 = 115°. [3 marks]" }, { "n": 8, "type_id": 2, "question": "5 identical tables cost as much as 7 identical chairs. Each table cost $54 more than the chair. Find the total cost of the 5 tables and 7 chairs.
$[?]", "answer0": "1890", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "5 tables = 7 chairs in cost. Each table = chair + $54, so 5 tables = 5 chairs + 5 x $54 = 5 chairs + $270. Setting equal to 7 chairs: 7 chairs = 5 chairs + $270, so 2 chairs = $270, 1 chair = $135, 7 chairs = $945. Total of 5 tables and 7 chairs = $945 (tables) + $945 (chairs) = $1890. Answer: $1890.", "hints": ["5 tables cost $270 more than 5 chairs, and that equals 7 chairs.", "2 chairs = $270, so 1 chair = $135; total = 2 x $945 = $1890."], "source": "Henry Park 2025 P6 WA2 Q8", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: Q8 = $1890. [3 marks]" }, { "n": 9, "type_id": 2, "question": "A hawker had some eggs. He used \\(\\dfrac{1}{3}\\) of them on Saturday and \\(\\dfrac{7}{12}\\) of the rest on Sunday. After that, he bought 104 eggs and then had as many eggs as he had at first. How many eggs did he have at first?
[?]", "answer0": "144", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "Let the start be 18u (LCM-friendly). Saturday used \\(\\dfrac{1}{3}\\) = 6u, leaving 12u. Sunday used \\(\\dfrac{7}{12}\\) of 12u = 7u, leaving 5u. To return to 18u he bought 104 eggs: 18u - 5u = 13u = 104, so 1u = 8 and 18u = 144. Answer: 144.", "hints": ["Use 18 units for the start so the fractions divide evenly.", "After both days 5u remain; 13u bought = 104, so start = 18u = 144."], "source": "Henry Park 2025 P6 WA2 Q9", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: Q9 = 144 eggs. [4 marks]" }, { "n": 10, "type_id": 2, "question": "Eagle Express Delivery Company charges $3.50 for parcels delivered on time and $1.80 for parcels delivered late. In May, the company could earn $382.50 more had all the parcels been delivered on time. (a) How many parcels were delivered late in May? (b) For every parcel that was delivered late in May, 9 parcels were delivered on time. How much did the company earn in May?
(a) [?] (b) $[?]", "answer0": "225", "answer1": "7492.50", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 167, "difficulty_id": 3, "explanation": "(a) Each late parcel loses $3.50 - $1.80 = $1.70 compared with on-time. Late parcels = $382.50 ÷ $1.70 = 225. (b) On-time parcels = 225 x 9 = 2025, earning 2025 x $3.50 = $7087.50. Late parcels earn 225 x $1.80 = $405. Total = $7087.50 + $405 = $7492.50. Answers: (a) 225, (b) $7492.50.", "hints": ["Each late parcel costs $1.70 less; divide $382.50 by $1.70.", "On-time = 225 x 9 = 2025; total = 2025 x $3.50 + 225 x $1.80 = $7492.50."], "source": "Henry Park 2025 P6 WA2 Q10", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: Q10(a) = 225 parcels [2 marks], (b) = $7492.50 [2 marks]." }, { "n": 11, "type_id": 1, "question": "Jacinta had two identical bottles each completely filled with mixtures of oil and water. The capacity of each bottle is 600 ml. The ratio of the amount of oil to the amount of water in the first bottle was 5 : 3 and in the second bottle it was 3 : 1. Jacinta emptied both bottles into an empty pail. What was the ratio of the amount of oil to the amount of water in the pail?", "answer0": "11 : 5", "answer1": "8 : 4", "answer2": "5 : 3", "answer3": "3 : 1", "correct_answer": 0, "skill_id": 183, "difficulty_id": 3, "explanation": "First bottle (5 : 3, 8 parts): oil = \\(\\dfrac{5}{8} \\times 600 = 375\\), water = 225. Second bottle (3 : 1, 4 parts): oil = \\(\\dfrac{3}{4} \\times 600 = 450\\), water = 150. Pail oil = 375 + 450 = 825, water = 225 + 150 = 375. Ratio = 825 : 375 = 11 : 5. Answer: 11 : 5.", "hints": ["Find the oil and water amounts in each bottle separately (both 600 ml).", "Combine and simplify: 825 : 375 = 11 : 5."], "source": "Henry Park 2025 P6 WA2 Q11", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper Q11(a). Answer is a ratio, so MCQ. Key: 11 : 5. Distractor options added. [2 marks]" }, { "n": 12, "type_id": 2, "question": "Jacinta had two identical 600 ml bottles of oil-water mixture; the pail combined them to oil : water = 11 : 5. After Jacinta poured more water into the pail, the amount of oil and water in the pail was in the ratio of 2 : 1. How much water did Jacinta pour into the pail?
[?] ml", "answer0": "37.5", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 183, "difficulty_id": 3, "explanation": "Before adding, oil : water = 11 : 5. The oil amount is fixed (825 ml). After adding, oil : water = 2 : 1 = 22 : 11. Matching the oil parts (11 → 22), the before ratio scales to 22 : 10. So water increases from 10 parts to 11 parts: 1 extra part of water added. With oil = 825 ml = 22 parts, 1 part = 825 ÷ 22 = 37.5 ml. Water added = 1 part = 37.5 ml. Answer: 37.5 ml.", "hints": ["The oil amount does not change; scale the ratios so the oil parts match (22).", "Before 22 : 10, after 22 : 11; the extra 1 part of water = 37.5 ml."], "source": "Henry Park 2025 P6 WA2 Q11", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper Q11(b). Key: 37.5 ml. [3 marks]" }, { "n": 13, "type_id": 2, "question": "25% of Lina's money was spent on 5 files and 10 erasers. The cost of each file was 4 times the cost of each eraser. Lina bought more files with 40% of her remaining money. How many files did she buy altogether?
[?]", "answer0": "14", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "Let an eraser = 1u, so a file = 4u. 5 files + 10 erasers = 5(4u) + 10(1u) = 20u + 10u = 30u = 25% of her money. Remaining 75% = 90u, and 40% of the remaining = 0.40 x 90u = 36u. Each file = 4u, so files bought with remaining = 36u ÷ 4u = 9. Total files = 9 + 5 = 14. Answer: 14.", "hints": ["Use units: eraser = 1u, file = 4u; the first purchase is 30u = 25% of money.", "40% of the remaining money = 36u buys 9 more files; total = 5 + 9 = 14."], "source": "Henry Park 2025 P6 WA2 Q12", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper Q12. Key: 14 files. [3 marks]" }, { "n": 14, "type_id": 2, "question": "The figure is made up of a square, a quarter circle and a semicircle. The area of the square is 196 cm². Find the total area of the shaded parts. (Take \\(\\pi = \\dfrac{22}{7}\\))
[?] cm²", "answer0": "28", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 223, "difficulty_id": 3, "explanation": "Square area 196 cm² means side 14 cm. Working through the overlapping quarter circle (radius 14) and semicircle (radius 7), the shaded parts compute (per the key) to 28 cm². Answer: 28 cm².", "hints": ["Side of the square = \\(\\sqrt{196}\\) = 14 cm; the semicircle radius is 7 cm.", "Combine the circular-segment areas to get 28 cm²."], "source": "Henry Park 2025 P6 WA2 Q13", "image_needed": true, "image_options": false, "image_file": "q14.png", "image_page": 10, "image_bbox": [0.38, 0.18, 0.62, 0.36], "image_loc": "square with a quarter circle and a semicircle drawn inside, shaded leaf-shaped regions, upper-middle of page", "notes": "Paper Q13. Key: 28 cm². [3 marks]" }, { "n": 15, "type_id": 2, "question": "At the start of a party, there were 70 children. Each boy was given 5 candies and each girl was given 3 candies. A total of 260 candies were given out. (a) Find the number of boys at the party. (b) Halfway through the party, 3 boys and some girls joined in. After that, the ratio of the number of boys to the number of girls at the party became 1 : 2. Find the number of girls who joined the party later.
(a) [?] (b) [?]", "answer0": "25", "answer1": "11", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 183, "difficulty_id": 3, "explanation": "(a) Assume all 70 were girls: 70 x 3 = 210 candies. Extra candies = 260 - 210 = 50, and each boy adds 5 - 3 = 2 extra, so boys = 50 ÷ 2 = 25. (b) Girls at start = 70 - 25 = 45. Boys become 25 + 3 = 28. New ratio boys : girls = 1 : 2, so girls now = 28 x 2 = 56. Girls who joined = 56 - 45 = 11. Answers: (a) 25, (b) 11.", "hints": ["(a) Assume all girls (210 candies); each boy adds 2 extra candies.", "(b) Boys become 28; at ratio 1 : 2, girls = 56, so 56 - 45 = 11 joined."], "source": "Henry Park 2025 P6 WA2 Q14", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper Q14. Key: (a) 25 boys [2 marks], (b) 11 girls joined [2 marks]." }, { "n": 16, "type_id": 1, "question": "The pie chart shows the number of $10, $20, $50 and $100 concert tickets sold by an event organiser. The $100 sector is 5%, the $50 sector is \\(\\dfrac{3}{20}\\), and AB is a straight line. What fraction of the tickets sold were $20 tickets?", "answer0": "\\(\\dfrac{3}{10}\\)", "answer1": "\\(\\dfrac{1}{5}\\)", "answer2": "\\(\\dfrac{3}{20}\\)", "answer3": "\\(\\dfrac{1}{2}\\)", "correct_answer": 0, "skill_id": 171, "difficulty_id": 3, "explanation": "AB is a straight line, so the $10 sector is half the circle = \\(\\dfrac{1}{2}\\) = 50%. The $100 sector is 5% and the $50 sector is \\(\\dfrac{3}{20}\\) = 15%. The $20 sector = 100% - 50% - 5% - 15% = 30% = \\(\\dfrac{30}{100} = \\dfrac{3}{10}\\). Answer: \\(\\dfrac{3}{10}\\).", "hints": ["AB straight means the $10 sector is half (50%).", "$20 = 100% - 50% - 5% - 15% = 30% = \\(\\dfrac{3}{10}\\)."], "source": "Henry Park 2025 P6 WA2 Q15", "image_needed": true, "image_options": false, "image_file": "q16.png", "image_page": 12, "image_bbox": [0.36, 0.15, 0.58, 0.30], "image_loc": "pie chart with sectors $10, $20, $50 (3/20), $100 (5%) and straight line AB, top of page", "notes": "Paper Q15(a). Answer is a fraction, so MCQ. Key: 6/20 = 3/10. Distractor options added. [1 mark]" }, { "n": 17, "type_id": 2, "question": "The pie chart shows the number of $10, $20, $50 and $100 concert tickets sold by an event organiser. The $100 sector is 5%, the $50 sector is \\(\\dfrac{3}{20}\\), and the $10 sector is half the circle. The organiser sold 160 more $50-tickets than $100-tickets. How many $10-tickets were sold?
[?]", "answer0": "800", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 174, "difficulty_id": 3, "explanation": "$50 sector = \\(\\dfrac{3}{20}\\) = 15% (3u where each u = 5%); $100 sector = 5% (1u). The difference = 3u - 1u = 2u = 160 tickets, so 1u (5%) = 80 tickets. The $10 sector is 50% = 10u = 10 x 80 = 800 tickets. Answer: 800.", "hints": ["$50 is 15% (3 units of 5%) and $100 is 5% (1 unit); their difference is 2 units = 160.", "1 unit (5%) = 80 tickets; $10 = 50% = 10 units = 800."], "source": "Henry Park 2025 P6 WA2 Q15", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper Q15(b). Key: 800 $10-tickets. [2 marks]" }, { "n": 18, "type_id": 2, "question": "In the figure, ABC is a triangle, BGF and ADJ are 2 identical quarter circles, CDEF and EGHJ are squares. The area of square CDEF is 25 cm² and the radius of the quarter circles is 18 cm. Find the total area of the unshaded parts of the figure. (Take \\(\\pi = 3.14\\))
[?] cm²", "answer0": "712.18", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 222, "difficulty_id": 3, "explanation": "Triangle ABC has legs 18 - 5 = 13 cm, area = \\(\\dfrac{1}{2} \\times 13 \\times 13 = 84.5\\) cm². Square EGHJ area = 13 x 13 = 169 cm². Each quarter circle (radius 18) area = \\(\\dfrac{1}{4} \\times 3.14 \\times 18 \\times 18 = 254.34\\) cm², and removing the small square (25 cm²) leaves 254.34 - 25 = 229.34 cm² per quarter circle. Total unshaded = 84.5 + 169 + (229.34 x 2) = 712.18 cm². Answer: 712.18 cm².", "hints": ["Find the triangle and square pieces (side 13 cm) and the two quarter circles (radius 18).", "84.5 + 169 + 2 x (254.34 - 25) = 712.18 cm²."], "source": "Henry Park 2025 P6 WA2 Q16", "image_needed": true, "image_options": false, "image_file": "q18.png", "image_page": 13, "image_bbox": [0.14, 0.20, 0.62, 0.46], "image_loc": "composite figure of triangle ABC, two quarter circles BGF and ADJ, squares CDEF and EGHJ, with small square shaded, middle of page", "notes": "Paper Q16. Key: 712.18 cm². [5 marks]" }, { "n": 19, "type_id": 2, "question": "The figures show trapezium ABCD and equilateral triangle EFG. The trapezium was folded along line DH to form a parallelogram HBCD. Triangle EFG was then pasted over parallelogram HBCD such that points E and B meet. Line DF is a straight line. \\(\\angle DHA = 36°\\) shown after folding. (a) Find \\(\\angle ADG\\). (b) Find \\(\\angle CEG\\).
(a) [?]° (b) [?]°", "answer0": "54", "answer1": "12", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 200, "difficulty_id": 3, "explanation": "(a) The fold creates \\(\\angle HAD = 90°\\) originally; after folding the marked angle is 36°. \\(\\angle GCE = 72° + 36° = 108°\\); the equilateral triangle base relations give \\((180° - 36°) ÷ 2 = 72°\\), and \\(\\angle ADG = 90° - 18° x 2 = 54°\\). (b) \\(\\angle CEG = 72° - 60° = 12°\\) (subtracting the 60° equilateral angle). Answers: (a) 54°, (b) 12°.", "hints": ["Use the folded right angle and the 36° mark; the triangle EFG is equilateral (60°).", "(a) 90° - 2 x 18° = 54°; (b) 72° - 60° = 12°."], "source": "Henry Park 2025 P6 WA2 Q17", "image_needed": true, "image_options": false, "image_file": "q19.png", "image_page": 14, "image_bbox": [0.18, 0.59, 0.72, 0.80], "image_loc": "final composite figure: parallelogram HBCD with equilateral triangle pasted on, 36° marked at H, lower part of page", "notes": "Paper Q17. Two parts. Key: (a) 54°, (b) 12°. The 36° appears after folding (second/third diagrams)." } ] }