{ "paper": { "school": "Nanyang", "year": 2024, "level": "P5", "label": "Term 2 Review", "source_prefix": "Nanyang 2024 P5 Term 2 Review", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 2, "question": "Find the area, in cm², of the shaded triangle. (Each grid square is 1 cm by 1 cm.) [?]", "answer0": "18", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 186, "difficulty_id": 2, "explanation": "The shaded triangle has base 6 cm and height 6 cm. Area = \\(\\dfrac{1}{2} \\times 6 \\times 6 = 18\\) cm².", "hints": ["Count the grid squares for the base and the height.", "Area of a triangle = \\(\\dfrac{1}{2} \\times \\text{base} \\times \\text{height}\\)."], "source": "Nanyang 2024 P5 Term 2 Review Q1", "image_needed": true, "image_options": false, "image_file": "fig-q1.png", "image_page": 1, "image_bbox": [0.3, 0.6, 0.62, 0.84], "image_loc": "shaded right-pointing triangle drawn on a 1 cm square grid, with 1 cm spacing marked top-right", "notes": "Key: 18 cm². Integer answer, FIB is fine." }, { "n": 2, "type_id": 2, "question": "In the figure, ABCD is a rectangle. AGB, BEC and DFC are straight lines. \\(DF = FC\\) and \\(BE = EC\\). AB = 22 cm and AD = 12 cm. Find the total area, in cm², of the shaded parts. [?]", "answer0": "165", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 186, "difficulty_id": 3, "explanation": "Whole rectangle area BCD-region = 22 \\times 12 = 264 cm². Unshaded triangle DGF = \\(\\dfrac{1}{2} \\times 11 \\times 12 = 66\\) cm² and unshaded triangle EFC = \\(\\dfrac{1}{2} \\times 6 \\times 11 = 33\\) cm². Shaded = 264 - 66 - 33 = 165 cm².", "hints": ["Find the full rectangle area first: 22 \\times 12.", "Subtract the two unshaded triangles (66 and 33)."], "source": "Nanyang 2024 P5 Term 2 Review Q2", "image_needed": true, "image_options": false, "image_file": "fig-q2.png", "image_page": 2, "image_bbox": [0.32, 0.13, 0.72, 0.32], "image_loc": "rectangle ABCD with G on AB, E on BC, F on DC, shaded polygon, 22 cm along top and 12 cm down the left side", "notes": "Key: 165 cm². Integer answer, FIB is fine." }, { "n": 3, "type_id": 2, "question": "The figure shows a solid made up of 1-m cubes. Find the volume, in m³, of the solid. [?]", "answer0": "33", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 190, "difficulty_id": 2, "explanation": "Counting the 1-m cubes in the stepped solid gives 33 cubes, so the volume is 33 m³.", "hints": ["Count cubes layer by layer, including hidden cubes.", "Each cube is 1 m³, so volume in m³ equals the cube count."], "source": "Nanyang 2024 P5 Term 2 Review Q3", "image_needed": true, "image_options": false, "image_file": "fig-q3.png", "image_page": 3, "image_bbox": [0.28, 0.22, 0.7, 0.45], "image_loc": "isometric stepped solid of 1-m cubes on a dot grid, with Top View / Front View / Side View arrows", "notes": "Key: 3a) 33 (printed '33 m^2' but the quantity is a volume of 1-m cubes, so the correct unit is m³ — flagged typo). Part (b) 'draw the front and side views' is a draw task with no value answer and is omitted (interactive)." }, { "n": 4, "type_id": 2, "question": "Sam, Abu and Ravi went jogging. They jogged 6100 m in total. Sam jogged 0.447 km more than Abu. Ravi jogged 0.8 km less than Sam.
(a) Express the total distance they jogged in kilometres. [?]
(b) How far, in km, did Ravi jog? [?]", "answer0": "6.1", "answer1": "1.649", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 184, "difficulty_id": 3, "explanation": "(a) 6100 m = 6100 \\div 1000 = 6.1 km. (b) Let Abu = A. Sam = A + 0.447; Ravi = Sam - 0.8 = A - 0.353. Total: A + (A + 0.447) + (A - 0.353) = 6.1 → 3A + 0.094 = 6.1 → 3A = 6.006 → A = 2.002. The key works it as: total 6.1, sum of the 'extra' relations gives Sam = (6.1 + 0.447 + 0.8) \\div 3 = 2.449, then Ravi = 2.449 - 0.8 = 1.649 km.", "hints": ["(a) Divide metres by 1000 to get kilometres.", "(b) Use the relations to find Sam first (2.449 km), then Ravi = Sam - 0.8."], "source": "Nanyang 2024 P5 Term 2 Review Q4", "image_needed": false, "image_options": false, "image_file": null, "notes": "Key: 4a) 6.1 km, 4b) 1.649 km. Both decimals, FIB is fine." }, { "n": 5, "type_id": 2, "question": "A rectangular tank, measuring 40 cm by 20 cm by 12 cm, contained 2400 ml of water.
(a) What was the volume, in cm³, of the tank? [?]
(b) A cubical container was filled with water to the brim. Some of the water was poured into the tank until the tank was \\(\\dfrac{3}{4}\\) full. There were 1032 ml of water left in the cubical container. What was the capacity, in litres, of the cubical container? [?]", "answer0": "9600", "answer1": "5.832", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 192, "difficulty_id": 3, "explanation": "(a) Tank volume = 40 \\times 20 \\times 12 = 9600 cm³. (b) \\(\\dfrac{3}{4}\\) full = \\(\\dfrac{3}{4} \\times 9600 = 7200\\) ml. Water poured in = 7200 - 2400 = 4800 ml. The cubical container held that 4800 ml plus the 1032 ml left = 5832 ml = 5832 \\div 1000 = 5.832 L.", "hints": ["(a) Multiply the three tank dimensions.", "(b) Water poured = \\(\\dfrac{3}{4}\\) full minus the 2400 ml already there; add the 1032 ml left over."], "source": "Nanyang 2024 P5 Term 2 Review Q5", "image_needed": true, "image_options": false, "image_file": "fig-q5.png", "image_page": 5, "image_bbox": [0.13, 0.47, 0.92, 0.62], "image_loc": "a cubical container plus two rectangular tanks (40 cm x 20 cm x 12 cm) showing water being poured from the cube into the tank", "notes": "Key: 5a) 9600 cm³, 5b) 5.832 L. (a) integer, (b) decimal — both FIB-fine." }, { "n": 6, "type_id": 2, "question": "The table shows the charges for sending parcels: up to 3 kg \\($10\\); up to 5 kg \\($18\\); up to 10 kg \\($35\\); more than 10 kg and up to 30 kg — first 10 kg \\($35\\) and every additional kilogram or less \\($4.50\\).
(a) Nancy sent two parcels with mass 4.6 kg and 12.5 kg. How much, in dollars, did she pay altogether? [?]
(b) Mrs Siva wants to send 35 kg of rice but cannot send any parcel above 30 kg, so she packs them into 2 smaller parcels. What is the least possible amount, in dollars, she must pay to send her rice? [?]", "answer0": "66.50", "answer1": "137.50", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "(a) 4.6 kg falls in 'up to 5 kg' = $18. 12.5 kg: first 10 kg = $35, then 2.5 kg → 3 extra kg-or-less at $4.50 = $13.50, so $35 + $13.50 = $48.50; total = $18 + $48.50 = $66.50. (b) Cheapest: one 30 kg parcel = $35 + 20 \\times $4.50 = $35 + $90 = $125 plus a 5 kg parcel $18 = $143; but the key splits to first-10 then 15 additional kg for the 25 kg part: 35 kg as a 25 kg + 10 kg pairing — key gives First 10 kg $35, next 15 kg = 15 \\times $4.50 = $67.50, total for that parcel $102.50, plus the other 10 kg parcel $35 → $102.50 + $35 = $137.50.", "hints": ["(a) Charge each parcel by its band; the 12.5 kg parcel needs first-10 plus extra kg.", "(b) Split 35 kg into two legal parcels to minimise cost; follow the key's $102.50 + $35 split."], "source": "Nanyang 2024 P5 Term 2 Review Q6", "image_needed": true, "image_options": false, "image_file": "fig-q6.png", "image_page": 6, "image_bbox": [0.22, 0.1, 0.82, 0.26], "image_loc": "parcel-charges table: Mass of parcel vs Charge, rows up to 3/5/10 kg and the more-than-10-kg band", "notes": "Key: 6a) $66.50, 6b) $137.50. Both money decimals, FIB is fine. The table is needed as the figure so the bands are visible." } ] }