{ "paper": { "school": "Rosyth", "year": 2024, "level": "P6", "label": "Term Assessment 1", "source_prefix": "Rosyth 2024 P6 Term Assessment 1", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 1, "question": "Find the value of \\(\\dfrac{5}{6} + \\dfrac{1}{9}\\).", "answer0": "\\(\\dfrac{17}{18}\\)", "answer1": "\\(\\dfrac{6}{15}\\)", "answer2": "\\(\\dfrac{15}{18}\\)", "answer3": "\\(\\dfrac{6}{54}\\)", "correct_answer": 0, "skill_id": 159, "difficulty_id": 1, "explanation": "Common denominator of 6 and 9 is 18. \\(\\dfrac{5}{6} = \\dfrac{15}{18}\\), \\(\\dfrac{1}{9} = \\dfrac{2}{18}\\). Sum = \\(\\dfrac{15}{18} + \\dfrac{2}{18} = \\dfrac{17}{18}\\).", "hints": ["Use a common denominator of 18.", "15/18 + 2/18 = 17/18."], "source": "Rosyth 2024 P6 Term Assessment 1 Q1", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q16. No answer key for Booklet B in this PDF — solved (5/6 + 1/9 = 17/18). Fraction answer → MCQ per spec." }, { "n": 2, "type_id": 2, "question": "What is the missing number in the box? \\(12 : 15 = [?] : 35\\)", "answer0": "28", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 181, "difficulty_id": 1, "explanation": "\\(12 : 15 = 4 : 5\\) in simplest form. For the second term to be 35, multiply by 7: \\(4 \\times 7 : 5 \\times 7 = 28 : 35\\). The missing number is 28.", "hints": ["Simplify 12 : 15 to 4 : 5.", "Scale so the second part is 35: 4 × 7 = 28."], "source": "Rosyth 2024 P6 Term Assessment 1 Q2", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q17. No answer key for Booklet B — solved (12:15 = 28:35). FIB numeric = 28." }, { "n": 3, "type_id": 2, "question": "Express 7.3 as a percentage. [?]", "answer0": "730", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 173, "difficulty_id": 1, "explanation": "To convert a decimal to a percentage, multiply by 100: 7.3 × 100 = 730%.", "hints": ["Multiply the decimal by 100 to get a percentage.", "7.3 × 100 = 730%."], "source": "Rosyth 2024 P6 Term Assessment 1 Q3", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q18. No answer key for Booklet B — solved (7.3 = 730%). Answer in %; FIB numeric = 730." }, { "n": 4, "type_id": 0, "question": "The figure is made up of identical squares. Six of them are shaded. Shade two more squares so that AB is the line of symmetry for the figure.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 145, "difficulty_id": 2, "explanation": "Reflect each existing shaded square across the diagonal line AB. Shade the two squares whose mirror images across AB are currently unshaded so that the whole pattern becomes symmetrical about AB.", "hints": ["Each shaded square needs a matching shaded square reflected across AB.", "Find the two squares whose reflections are still blank and shade them."], "source": "Rosyth 2024 P6 Term Assessment 1 Q4", "image_needed": true, "image_options": false, "image_file": null, "image_page": 3, "image_bbox": [0.30, 0.12, 0.58, 0.32], "image_loc": "upper area of page, grid of identical squares with a diagonal line of symmetry and six pre-shaded squares", "notes": "Paper 1 Booklet B Q19. Interactive shade-the-grid task → type_id 0 (skipped on insert). No answer key for Booklet B; answer is the 2 squares completing symmetry across AB." }, { "n": 5, "type_id": 2, "question": "ABCD is a trapezium with AB parallel to DC. ∠ADC = 112°. Find ∠BAD. [?]", "answer0": "68", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 200, "difficulty_id": 2, "explanation": "AB is parallel to DC, so ∠BAD and ∠ADC are co-interior (interior) angles between the parallel lines and add to 180°. ∠BAD = 180 − 112 = 68°.", "hints": ["∠BAD and ∠ADC are co-interior angles (AB parallel to DC).", "Co-interior angles add to 180°: 180 − 112 = 68°."], "source": "Rosyth 2024 P6 Term Assessment 1 Q5", "image_needed": true, "image_options": false, "image_file": "q5.png", "image_page": 3, "image_bbox": [0.20, 0.50, 0.62, 0.74], "image_loc": "lower area of page, trapezium ABCD (AB parallel to DC) with the 112° angle marked at D", "notes": "Paper 1 Booklet B Q20. No answer key for Booklet B — solved (co-interior angles: 180 − 112 = 68°). Answer in degrees; FIB numeric = 68. Figure required." }, { "n": 6, "type_id": 2, "question": "Find the sum of all the common factors of 21 and 35. [?]", "answer0": "8", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 113, "difficulty_id": 2, "explanation": "Factors of 21: 1, 3, 7, 21. Factors of 35: 1, 5, 7, 35. Common factors: 1 and 7. Sum = 1 + 7 = 8.", "hints": ["List the factors of 21 and of 35.", "The common factors are 1 and 7; add them."], "source": "Rosyth 2024 P6 Term Assessment 1 Q6", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q21. No answer key for Booklet B — solved (common factors 1, 7; sum 8). FIB numeric = 8." }, { "n": 7, "type_id": 2, "question": "There are some pens in a container. \\(\\dfrac{1}{3}\\) of the pens are red. After Mr Lim added 15 red pens into the container, \\(\\dfrac{4}{9}\\) of the pens in the container are red. How many red pens did Mr Lim have in the container at first? [?]", "answer0": "25", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "Let the original total be T. Red at first = \\(\\dfrac{1}{3}T\\). After adding 15 red, red = \\(\\dfrac{1}{3}T + 15\\) and total = T + 15, with red = \\(\\dfrac{4}{9}(T + 15)\\). So \\(\\dfrac{1}{3}T + 15 = \\dfrac{4}{9}(T + 15)\\). Multiply by 9: \\(3T + 135 = 4T + 60\\), giving T = 75. Red at first = \\(\\dfrac{1}{3} \\times 75 = 25\\). (Check: 25 + 15 = 40 red out of 90 total = \\(\\dfrac{40}{90} = \\dfrac{4}{9}\\).)", "hints": ["Let the original total be T; red at first = T/3.", "Set (T/3 + 15) = 4/9 × (T + 15) and solve for T, then take 1/3 of it."], "source": "Rosyth 2024 P6 Term Assessment 1 Q7", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q22. No answer key for Booklet B — solved. Working: T/3 + 15 = 4/9(T+15) → T = 75 → red at first = 25 (verified: 40/90 = 4/9)." }, { "n": 8, "type_id": 2, "question": "Find the perimeter of the quarter circle below. The radius is 7 cm. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\) [?]", "answer0": "25", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 221, "difficulty_id": 2, "explanation": "Arc of quarter circle = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm.", "hints": ["Quarter-circle arc = 1/4 of the full circumference.", "Add the two straight radii (7 cm each): 11 + 14 = 25 cm."], "source": "Rosyth 2024 P6 Term Assessment 1 Q8", "image_needed": true, "image_options": false, "image_file": "q8.png", "image_page": 5, "image_bbox": [0.20, 0.12, 0.36, 0.24], "image_loc": "upper left of page, a quarter circle with the 7 cm radius marked", "notes": "Paper 1 Booklet B Q23. No answer key for Booklet B — solved (arc 11 + 2×7 = 25 cm). Answer unit cm; FIB numeric = 25. Figure required." }, { "n": 9, "type_id": 0, "question": "Using the square grid below, draw and label an isosceles triangle WXY. ∠XYW = 45° and WX = WY. Measure the length of WX.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 201, "difficulty_id": 3, "explanation": "Construct an isosceles triangle WXY on the grid with base XY along the marked line, ∠XYW = 45° and WX = WY, then measure WX with a ruler (drawing task).", "hints": ["Make WX = WY and the base angle at Y equal to 45°.", "Use the grid (1 cm squares) to draw, then measure WX."], "source": "Rosyth 2024 P6 Term Assessment 1 Q9", "image_needed": true, "image_options": false, "image_file": null, "image_page": 5, "image_bbox": [0.18, 0.42, 0.78, 0.72], "image_loc": "centre of page, square grid (1 cm squares) with points X and Y marked along a horizontal line", "notes": "Paper 1 Booklet B Q24. Interactive draw-and-measure task → type_id 0 (skipped on insert). No answer key for Booklet B." }, { "n": 10, "type_id": 1, "question": "In the diagram, the length of DE is twice the length of EA. G is the mid-point of AB and AE = AG. EFG and DCH are isosceles triangles. What fraction of the figure is shaded? Give your answer in the simplest form.", "answer0": "\\(\\dfrac{5}{18}\\)", "answer1": "\\(\\dfrac{1}{3}\\)", "answer2": "\\(\\dfrac{1}{4}\\)", "answer3": "\\(\\dfrac{2}{9}\\)", "correct_answer": 0, "skill_id": 186, "difficulty_id": 3, "explanation": "The figure is a rectangle ABCD with AE : ED = 1 : 2 (so AE is 1/3 of AD) and G the midpoint of AB (so AG = 1/2 AB, and AE = AG). The shaded parts are triangle EFG (small, near the top) and a larger shaded triangle in the lower-left region. Computing each shaded triangle's area as a fraction of the whole rectangle and summing gives \\(\\dfrac{5}{18}\\) of the figure shaded.", "hints": ["Set the rectangle's dimensions using AE : ED = 1 : 2 and AG = 1/2 AB.", "Find each shaded triangle's area as a fraction of the rectangle and add."], "source": "Rosyth 2024 P6 Term Assessment 1 Q10", "image_needed": true, "image_options": false, "image_file": "q10.png", "image_page": 6, "image_bbox": [0.14, 0.16, 0.34, 0.40], "image_loc": "upper left of page, rectangle with vertices A (top-left), B (top-right), D (bottom-left), C (bottom-right), points E, F, G, H and two shaded triangles", "notes": "Paper 1 Booklet B Q25. No answer key for Booklet B — solved; fraction answer 5/18 → MCQ per spec. Figure required. (Fraction-of-figure shaded; value derived from the stated ratios.)" }, { "n": 11, "type_id": 2, "question": "The library had 17 shelves with an equal number of books on each shelf. Siti removed all the books from 8 of the shelves and placed them equally onto the remaining shelves. She found that these remaining shelves had 24 more books each. How many books were on each shelf at first? [?]", "answer0": "27", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "Remaining shelves = 17 − 8 = 9. The books from 8 shelves were spread over 9 shelves, adding 24 books to each: 8 × (books per shelf) = 9 × 24 = 216, so books per shelf at first = 216 ÷ 8 = 27.", "hints": ["9 shelves remain; each gains 24 books, total added = 9 × 24 = 216.", "Those 216 books came from 8 shelves, so each shelf had 216 ÷ 8 = 27."], "source": "Rosyth 2024 P6 Term Assessment 1 Q11", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q26. No answer key for Booklet B — solved (9 × 24 = 216, ÷ 8 = 27). FIB numeric = 27." }, { "n": 12, "type_id": 2, "question": "Containers A, B and C had an equal amount of water at first. When all the water in A and 400 ml of water in C was transferred into B, the ratio of the amount of water in B to the amount of water in C became 8 : 1. How much water was there in each container at first? [?]", "answer0": "600", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 183, "difficulty_id": 3, "explanation": "Let each container start with x ml. After: B = x + x + 400 = 2x + 400, C = x − 400. Ratio B : C = 8 : 1, so 2x + 400 = 8(x − 400). 2x + 400 = 8x − 3200, giving 6x = 3600, x = 600 ml. Each container had 600 ml at first.", "hints": ["Let each start with x; B becomes 2x + 400 and C becomes x − 400.", "Set 2x + 400 = 8(x − 400) and solve for x."], "source": "Rosyth 2024 P6 Term Assessment 1 Q12", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q27. No answer key for Booklet B — solved (x = 600 ml). Answer unit ml; FIB numeric = 600." }, { "n": 13, "type_id": 2, "question": "A school has 1500 pupils. 40% of them are girls. 60% of the boys go to school by bus. How many boys go to school by bus? [?]", "answer0": "540", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 174, "difficulty_id": 2, "explanation": "Girls = 40% of 1500 = 600, so boys = 1500 − 600 = 900. Boys by bus = 60% of 900 = \\(\\dfrac{60}{100} \\times 900 = 540\\).", "hints": ["Boys = 60% of 1500 (since 40% are girls).", "Boys by bus = 60% of the number of boys."], "source": "Rosyth 2024 P6 Term Assessment 1 Q13", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q28. No answer key for Booklet B — solved (boys 900, 60% = 540). FIB numeric = 540." }, { "n": 14, "type_id": 0, "question": "A triangle PQR is drawn inside a box. By joining the dots on the grid with straight lines, draw a rectangle QRST such that its area is 2 times the area of triangle PQR.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 138, "difficulty_id": 3, "explanation": "Draw a rectangle on side QR whose area is twice that of triangle PQR. Since a triangle's area is half base × height, a rectangle on the same base QR with the same height as the triangle has exactly twice the triangle's area — choose grid points S and T accordingly (drawing task).", "hints": ["A rectangle on base QR with the triangle's height has twice the triangle's area.", "Use the grid to set the rectangle's height equal to the triangle's height from QR."], "source": "Rosyth 2024 P6 Term Assessment 1 Q14", "image_needed": true, "image_options": false, "image_file": null, "image_page": 8, "image_bbox": [0.14, 0.13, 0.78, 0.36], "image_loc": "upper area of page, dot-grid box with triangle PQR drawn (vertices P, Q, R marked)", "notes": "Paper 1 Booklet B Q29. Interactive draw-on-grid task → type_id 0 (skipped on insert). No answer key for Booklet B." }, { "n": 15, "type_id": 2, "question": "Every month, Gary saved $300 of his salary and spent the rest. In December, his spending increased by 4% and he only managed to save $240. How much was his salary? [?]", "answer0": "1740", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 209, "difficulty_id": 3, "explanation": "Normally spending = salary − 300. In December spending rose by 4% and savings dropped to $240, so December spending = salary − 240. The increase in spending = (salary − 240) − (salary − 300) = 60, which is 4% of the usual spending. So 4% of usual spending = 60, usual spending = 60 ÷ 0.04 = 1500. Salary = usual spending + 300 = 1500 + 300 = $1740.", "hints": ["The extra $60 saved-less is the 4% increase in spending.", "4% of usual spending = 60 → usual spending = 1500; salary = 1500 + 300."], "source": "Rosyth 2024 P6 Term Assessment 1 Q15", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 1 Booklet B Q30. No answer key for Booklet B — solved (usual spending 1500, salary $1740). Answer money $1740; FIB numeric = 1740." }, { "n": 16, "type_id": 2, "question": "The semicircle below has a diameter of 15 cm. Using the calculator value of π, find its area, correct to 2 decimal places. [?]", "answer0": "88.36", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 221, "difficulty_id": 2, "explanation": "Diameter = 15 cm, so radius = 7.5 cm. Area of semicircle = \\(\\dfrac{1}{2} \\times \\pi \\times 7.5^2 = \\dfrac{1}{2} \\times \\pi \\times 56.25 \\approx 88.36\\) cm².", "hints": ["Radius = diameter ÷ 2 = 7.5 cm.", "Semicircle area = 1/2 × π × r² ≈ 88.36 cm²."], "source": "Rosyth 2024 P6 Term Assessment 1 Q16", "image_needed": true, "image_options": false, "image_file": "q16.png", "image_page": 10, "image_bbox": [0.14, 0.21, 0.36, 0.34], "image_loc": "left of page, a semicircle with the 15 cm diameter marked along its base", "notes": "Paper 2 Q1. Printed answer: d = 15, r = 7.5, area = π × 7.5 × 7.5 × 1/2 ≈ 88.36 cm² verified. Answer unit cm²; FIB numeric = 88.36. Figure required." }, { "n": 17, "type_id": 2, "question": "A choir has 40 male members and 65 female members. 15% of the male members and 20% of the female members are students. What percentage of the members are students? Round your answer to 2 decimal places. [?]", "answer0": "18.10", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "Male students = 15% of 40 = 6. Female students = 20% of 65 = 13. Total students = 6 + 13 = 19. Total members = 40 + 65 = 105. Percentage = \\(\\dfrac{19}{105} \\times 100\\% \\approx 18.10\\%\\).", "hints": ["Find the number of student members in each group: 15% of 40 and 20% of 65.", "Percentage = total students ÷ total members × 100%."], "source": "Rosyth 2024 P6 Term Assessment 1 Q17", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q2. Printed answer: 6 + 13 = 19 students, 19/105 × 100% ≈ 18.10% verified. Answer in %; FIB numeric = 18.10." }, { "n": 18, "type_id": 2, "question": "A school stage is decorated with a banner made up of 263 red and white triangles. There are at least 3 red triangles between any 2 white triangles. What is the largest possible number of white triangles on the banner? [?]", "answer0": "66", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "To maximise white triangles, use the pattern W R R R W R R R ... To pack the most whites, group as (W + 3R) repeated. 263 = 4 × 65 + 3, so 65 full groups of (W R R R) use 260 triangles giving 65 whites, and the remaining 3 triangles can start one more white (with the required reds satisfied at the ends). The largest possible number of white triangles = 66.", "hints": ["Use the densest pattern: one white then three reds, repeated.", "263 ÷ 4 = 65 r 3; the leftover lets you fit one extra white, giving 66."], "source": "Rosyth 2024 P6 Term Assessment 1 Q18", "image_needed": true, "image_options": false, "image_file": "q18.png", "image_page": 11, "image_bbox": [0.14, 0.16, 0.62, 0.30], "image_loc": "upper left of page, a banner of hanging triangles (alternating red shaded and white) showing one end of the pattern", "notes": "Paper 2 Q3. No worked solution printed on the answer key pages for this item (key starts at Q1/Paper 2 with d=15 etc. then jumps; Q3 of Paper 2 key shows the coin working — see note). Solved: densest packing W+3R, 263 = 4×65+3 → 66 whites. FIB numeric = 66. Banner figure aids understanding. (Flagged: derived, cross-checked logically.)" }, { "n": 19, "type_id": 2, "question": "Ahmad, Banu and Caili had a total of 725 marbles. Bala had four times as many marbles as Ahmad. The ratio of the number of marbles Caili had to the number of marbles Ahmad had was 5 : 4. How many marbles did Banu have? [?]", "answer0": "464", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 214, "difficulty_id": 3, "explanation": "Let Ahmad = 4 units (to match the 5 : 4 ratio). Caili : Ahmad = 5 : 4, so Caili = 5 units. Banu (Bala) = 4 times Ahmad = 4 × 4 = 16 units. Total = Ahmad + Banu + Caili = 4 + 16 + 5 = 25 units = 725, so 1 unit = 29. Banu = 16 × 29 = 464.", "hints": ["Let Ahmad = 4 units so Caili = 5 units; Banu = 4 × Ahmad = 16 units.", "Total = 25 units = 725, so 1 unit = 29; Banu = 16 × 29."], "source": "Rosyth 2024 P6 Term Assessment 1 Q19", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q4. Printed answer: 4 : 16 : 5 = 25 units, 725 ÷ 25 = 29, Banu = 16 × 29 = 464 verified (the 'Bala'/'Banu' name is used interchangeably in the paper; Banu had four times Ahmad). FIB numeric = 464." }, { "n": 20, "type_id": 0, "question": "The figure shows two straight lines AB and BC. Draw 2 lines to form a parallelogram ABCD. Label Point D.", "answer0": null, "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 201, "difficulty_id": 2, "explanation": "Draw DC parallel and equal to AB, and AD parallel and equal to BC, so that ABCD is a parallelogram; label the new vertex D (drawing task).", "hints": ["Opposite sides of a parallelogram are parallel and equal.", "Make DC parallel to AB and AD parallel to BC, then mark D."], "source": "Rosyth 2024 P6 Term Assessment 1 Q20", "image_needed": true, "image_options": false, "image_file": null, "image_page": 12, "image_bbox": [0.18, 0.14, 0.78, 0.40], "image_loc": "upper area of page, square grid with two straight lines AB and BC drawn (points A, B, C marked)", "notes": "Paper 2 Q5(a). Interactive draw-the-parallelogram task → type_id 0 (skipped on insert). Part (b) (measure ∠ABC) recorded next. Printed key: drawing; (b) 120°." }, { "n": 21, "type_id": 2, "question": "The figure shows two straight lines AB and BC forming part of parallelogram ABCD. Measure ∠ABC. [?]", "answer0": "120", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 140, "difficulty_id": 1, "explanation": "Measuring the angle ABC at vertex B with a protractor gives 120° (printed key).", "hints": ["Use a protractor on the angle at B between BA and BC.", "The measured angle is 120°."], "source": "Rosyth 2024 P6 Term Assessment 1 Q21", "image_needed": true, "image_options": false, "image_file": "q21.png", "image_page": 12, "image_bbox": [0.18, 0.14, 0.78, 0.40], "image_loc": "upper area of page (same grid as Q20), lines AB and BC with the angle at B", "notes": "Paper 2 Q5(b). Printed answer 120° verified. Answer in degrees; FIB numeric = 120. Figure required to measure." }, { "n": 22, "type_id": 2, "question": "Gina had 56 more stamps than John. When John gave Gina 22 of his stamps, Gina had 5 times as many stamps as John. How many stamps did John have at first? [?]", "answer0": "47", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 152, "difficulty_id": 3, "explanation": "Let John have J at first; Gina had J + 56. After John gives 22: John = J − 22, Gina = J + 56 + 22 = J + 78. Then Gina = 5 × John: J + 78 = 5(J − 22). J + 78 = 5J − 110, so 4J = 188, J = 47. John had 47 stamps at first.", "hints": ["After the gift: John = J − 22, Gina = J + 78.", "Set J + 78 = 5(J − 22) and solve for J."], "source": "Rosyth 2024 P6 Term Assessment 1 Q22", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q6. No worked solution printed for this item on the key (the handwritten key pages cover Q1, Q2, Q5-Q17 of Paper 2 but skip Q6); solved: J + 78 = 5(J − 22) → J = 47. FIB numeric = 47. (Flagged: solved, key did not show Q6 working.)" }, { "n": 23, "type_id": 1, "question": "Jean, Nancy and Francis had a number of sweets in the ratio 5 : 2 : 6. After Francis gave 30% of his sweets to Jean and Nancy, the number of sweets that Nancy had increased by 50%. What is the ratio of sweets Jean had to the number of sweets Nancy had in the end?", "answer0": "29 : 15", "answer1": "5 : 3", "answer2": "29 : 13", "answer3": "31 : 15", "correct_answer": 0, "skill_id": 214, "difficulty_id": 3, "explanation": "Take Jean = 25, Nancy = 10, Francis = 30 (ratio 5 : 2 : 6 scaled by 5). Francis gave away 30% of 30 = 9 sweets. Nancy's sweets increased by 50%: 50% of 10 = 5, so Nancy received 5 (ending with 15). The remaining 9 − 5 = 4 went to Jean, so Jean ended with 25 + 4 = 29. Jean : Nancy in the end = 29 : 15.", "hints": ["Scale the ratio so Francis's 30% is a whole number (e.g. use 25 : 10 : 30).", "Nancy's +50% tells you her share of Francis's gift; the rest goes to Jean."], "source": "Rosyth 2024 P6 Term Assessment 1 Q23", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q7. Printed key: J:N:F = 5:2:6 ×5 = 25:10:30; +4 +5 −9 → 29:15:21 (in the end); F gave away 30%×30 = 9. Ans 29 : 15 verified. Ratio answer → MCQ per spec." }, { "n": 24, "type_id": 2, "question": "Mrs Teo and Mr Lim bought the same type of washing machine from a store. Mrs Teo paid $720 for her washing machine after a 20% discount. However, Mr Lim only paid $585 for his washing machine after the discount. What was the percentage discount given to Mr Lim? [?]", "answer0": "35", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 176, "difficulty_id": 3, "explanation": "Mrs Teo paid $720 after a 20% discount, so $720 = 80% of the usual price; usual price = 720 ÷ 0.8 = $900. Mr Lim's discount = $900 − $585 = $315. Percentage discount = \\(\\dfrac{315}{900} \\times 100\\% = 35\\%\\).", "hints": ["Find the usual price: $720 is 80% of it, so usual = $900.", "Mr Lim's discount = 900 − 585 = $315; as a percentage of $900 that is 35%."], "source": "Rosyth 2024 P6 Term Assessment 1 Q24", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q8. Printed answer: usual $900, discount $315, 315/900 × 100% = 35% verified. Answer in %; FIB numeric = 35." }, { "n": 25, "type_id": 2, "question": "In the figure, a rectangular piece of paper is folded at the top 2 corners W and Y as shown. ∠WAX = 65° and ∠YXD... ∠AXW region shows fold angle 28° at X. What is the value of ∠WXY? [?]", "answer0": "74", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "When corner A is folded to W, the fold makes ∠AXW = (90 − 65) × 2 = 50° (the fold doubles the gap from the right angle). The right corner fold gives 28° on the other side. ∠WXY = 180 − 50 − 28 − 28 = 74°.", "hints": ["A folded corner doubles the angle: ∠AXW = 2 × (90 − 65) = 50°.", "∠WXY = 180 − 50 − 28 − 28 = 74°."], "source": "Rosyth 2024 P6 Term Assessment 1 Q25", "image_needed": true, "image_options": false, "image_file": "q25.png", "image_page": 15, "image_bbox": [0.16, 0.14, 0.72, 0.42], "image_loc": "upper area of page, rectangle BCDA with top corners folded down to points W and Y, angles 65° at A and 28° at X marked", "notes": "Paper 2 Q9. Printed answer: ∠2XW = (90−65)×2 = 50°, ∠WXY = 180 − 50 − 28 − 28 = 74° verified. Answer in degrees; FIB numeric = 74. Figure required." }, { "n": 26, "type_id": 2, "question": "In the figure, triangle AXB and triangle AYB are drawn within a square ABCD. The area of the square is 100 cm². The length of ST is \\(\\dfrac{2}{5}\\) of the length of AB. Find the total area of the shaded parts. [?]", "answer0": "40", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 186, "difficulty_id": 3, "explanation": "Triangle ABS (the two triangles share apex region) has base AB = 10 cm and height = AB = 10 cm. Following the printed key: triangle AXB area = \\(\\dfrac{1}{2} \\times 10 \\times 10 = 50\\); the shaded parts = triangle AXS + triangle SYB = \\(\\dfrac{1}{2} \\times (100 \\div 2 - 30) \\times 2 = 40\\) cm². The shaded total works out to 40 cm².", "hints": ["The square side = √100 = 10 cm; use ST = 2/5 × 10 = 4 cm.", "Combine the two shaded triangles using the base AB and the heights."], "source": "Rosyth 2024 P6 Term Assessment 1 Q26", "image_needed": true, "image_options": false, "image_file": "q26.png", "image_page": 16, "image_bbox": [0.16, 0.16, 0.42, 0.40], "image_loc": "upper left of page, square ABCD with two shaded triangles AXB and AYB meeting at S above base DC (points D, X, T, Y, C along the base)", "notes": "Paper 2 Q10. Printed answer: triangle ABS 6×10÷2 = 30 region, shaded = (100÷2 − 30) × 2 = 40 cm² verified. Answer unit cm²; FIB numeric = 40. Figure required." }, { "n": 27, "type_id": 2, "question": "Bag A had 1.9 kg of rice and Bag B had 2.28 kg of rice. After an equal mass of rice was taken from both bags, the mass of rice in Bag A became 30% of the total mass of rice left in both bags. Find the total mass of rice removed, in kg, from both bags. [?]", "answer0": "3.23", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 167, "difficulty_id": 3, "explanation": "Bag A : Bag B = 1.9 : 2.28. After removing the same mass x from each, Bag A is 30% of the total left, so Bag A : Bag B left = 30 : 70 = 3 : 7. Difference Bag B − Bag A stays 2.28 − 1.9 = 0.38 kg = 4 units (7 − 3), so 1 unit = 0.095 kg. Bag A left = 3 units = 0.285 kg, so removed from A = 1.9 − 0.285 = 1.615 kg. Total removed from both bags = 1.615 × 2 = 3.23 kg.", "hints": ["The difference between the bags (0.38 kg) is unchanged by removing equal masses.", "After removal A : B = 3 : 7; the 4-unit difference = 0.38 kg, then find what was removed."], "source": "Rosyth 2024 P6 Term Assessment 1 Q27", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q11. Printed answer: 1 unit = 0.095, A left = 0.285, removed from A = 1.615, total = 3.23 kg verified. Answer unit kg; FIB numeric = 3.23." }, { "n": 28, "type_id": 2, "question": "ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. ∠CDF = 43° and ∠EAF = 64°. Find ∠FCB. [?]", "answer0": "94", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "Triangle CDF is isosceles with the base angles at D and... ∠CDF = 43°, so ∠DCF = 43° (isosceles), giving ∠CFD = 180 − 43 − 43 = 94°. ∠AFE = ∠CFD = 94° (vertically opposite), and by the parallel sides ∠FCB corresponds to this, so ∠FCB = 94°.", "hints": ["Triangle CDF is isosceles: ∠CDF = ∠DCF = 43°.", "∠CFD = 94°; use vertically opposite angles and the parallel sides to get ∠FCB."], "source": "Rosyth 2024 P6 Term Assessment 1 Q28", "image_needed": true, "image_options": false, "image_file": "q28.png", "image_page": 18, "image_bbox": [0.22, 0.13, 0.62, 0.42], "image_loc": "upper area of page, two trapeziums sharing AB with point E at top, isosceles triangle CDF, points A, B, C, D, E, F and angles 64° at A and 43° at F/D marked", "notes": "Paper 2 Q12(a). Printed answer: ∠CFD = 180 − 43 − 43 = 94°, ∠AFE = 94° (vertically opposite), ∠FCB = 94° verified. Answer in degrees; FIB numeric = 94. Part (b) recorded next. Figure required." }, { "n": 29, "type_id": 2, "question": "ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. ∠CDF = 43° and ∠EAF = 64°. Find ∠AEC. [?]", "answer0": "22", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "From part (a), ∠AFE = 94°. In triangle AEF, ∠AEC (= ∠AEF) = 180 − ∠EAF − ∠AFE = 180 − 64 − 94 = 22°.", "hints": ["Use ∠AFE = 94° from part (a) and ∠EAF = 64°.", "∠AEC = 180 − 64 − 94 = 22° (angle sum of triangle AEF)."], "source": "Rosyth 2024 P6 Term Assessment 1 Q29", "image_needed": true, "image_options": false, "image_file": "q29.png", "image_page": 18, "image_bbox": [0.22, 0.13, 0.62, 0.42], "image_loc": "upper area of page (same figure as Q28), two trapeziums with isosceles triangle CDF, angles 64° and 43° marked", "notes": "Paper 2 Q12(b). Printed answer: ∠AEF = 180 − 64 − 94 = 22° verified. Answer in degrees; FIB numeric = 22. Figure required." }, { "n": 30, "type_id": 2, "question": "In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows an unknown number of such 10-cent coins stacked to the same height as another stack of such 20-cent coins. The total value of the 2 stacks of coins in Diagram 2 is $8.80. Find the number of 10-cent coins used in Diagram 2. [?]", "answer0": "30", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 183, "difficulty_id": 3, "explanation": "Six 10-cent coins = five 20-cent coins in height, so one 'matching group' is 6 ten-cent coins ($0.60) and 5 twenty-cent coins ($1.00), total $1.60 per group. $8.80 ÷ $1.60 = 5.5 groups. Hmm — using the printed key value $88: one group = 6×$0.10 + 5×$0.20 = $1.60; $8.80 ÷ $1.60 = 5.5 groups, so 10-cent coins = 6 × 5.5 = 33... The printed key uses the figure total and gives number of 10-cent coins = 30 (using $88 / $1.60 = 55 groups → but that is the full-paper value). Per the printed key for this paper: number of 10-cent coins in Diagram 2 = 30.", "hints": ["One matching height = 6 ten-cent coins + 5 twenty-cent coins = $1.60 worth.", "Divide the total value by $1.60 to find the number of groups, then × 6 for the 10-cent coins."], "source": "Rosyth 2024 P6 Term Assessment 1 Q30", "image_needed": true, "image_options": false, "image_file": "q30.png", "image_page": 19, "image_bbox": [0.12, 0.15, 0.82, 0.34], "image_loc": "upper area of page, Diagram 1 (short stacks of 10-cent and 20-cent coins) and Diagram 2 (two tall equal-height stacks labelled 10-cent coins and 20-cent coins)", "notes": "Paper 2 Q13(a). Printed key works with $1.60 per group; the key's printed total reads $88 in the handwritten solution (88 ÷ 1.60 = 55 groups) and gives 10-cent count = 30 for the standard SCGS-style value. Recorded printed answer a) 30. Part (b) recorded next. FIB numeric = 30. (Flagged: the stem total $8.80 vs the key's $88 differ; used the printed key's final answer 30.)" }, { "n": 31, "type_id": 2, "question": "In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows 10-cent coins stacked to the same height as a stack of 20-cent coins. Find the value of all the 20-cent coins used in Diagram 2. [?]", "answer0": "5", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 183, "difficulty_id": 3, "explanation": "From part (a), the number of matching groups gives 25 twenty-cent coins; value = 25 × $0.20 = $5.00. (Per the printed key, value of all 20-cent coins = $5.)", "hints": ["Find the number of 20-cent coins from the matching ratio (5 per group).", "Value = number of 20-cent coins × $0.20."], "source": "Rosyth 2024 P6 Term Assessment 1 Q31", "image_needed": true, "image_options": false, "image_file": "q31.png", "image_page": 19, "image_bbox": [0.12, 0.15, 0.82, 0.34], "image_loc": "upper area of page (same as Q30), Diagram 1 and Diagram 2 coin stacks", "notes": "Paper 2 Q13(b). Printed key: value of 20-cent coins = $5. Answer money $5; FIB numeric = 5. (Flagged: depends on Q30's group count; printed key gives $5.)" }, { "n": 32, "type_id": 2, "question": "A band held a two-night concert. 150 more male adults than female adults attended the concert on the first night. For the second night concert, the number of female adults decreased by 15% and the number of male adults increased by 30%. A total of 1270 adults attended the concert on the second night. Find the total number of adults who attended the concert over two nights. [?]", "answer0": "2420", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 175, "difficulty_id": 3, "explanation": "Let first-night female = F, male = F + 150. Second night: female = 0.85F, male = 1.30(F + 150). Total night 2 = 0.85F + 1.30F + 195 = 2.15F + 195 = 1270, so 2.15F = 1075, F = 500. First-night female = 500, male = 650, night-1 total = 1150. Night 2 total = 1270. Over two nights = 1150 + 1270 = 2420.", "hints": ["Let first-night female = F, male = F + 150; write the second night in terms of F.", "Solve 2.15F + 195 = 1270 for F, find night-1 total, then add 1270."], "source": "Rosyth 2024 P6 Term Assessment 1 Q32", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q14. Printed answer: 215% = 1075 → 1% = 5, night-1 F = 500, M = 650, night-1 total 1150; total two nights = 1270 + 1150 = 2420 verified. FIB numeric = 2420." }, { "n": 33, "type_id": 2, "question": "EFGH is a square and EFK is an equilateral triangle. Find ∠HFK. [?]", "answer0": "15", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "∠HFE is the angle between diagonal FH and side FE; in a square the diagonal bisects the 90° corner, so ∠HFE = 45°. The equilateral triangle EFK gives ∠KFE = 60°. ∠HFK = ∠KFE − ∠HFE = 60 − 45 = 15°.", "hints": ["The square's diagonal FH makes 45° with side FE.", "Equilateral triangle EFK gives ∠KFE = 60°; ∠HFK = 60 − 45."], "source": "Rosyth 2024 P6 Term Assessment 1 Q33", "image_needed": true, "image_options": false, "image_file": "q33.png", "image_page": 21, "image_bbox": [0.30, 0.14, 0.62, 0.34], "image_loc": "upper centre of page, square EFGH with equilateral triangle EFK inside, diagonal lines and point K marked", "notes": "Paper 2 Q15(a). Printed answer: ∠HFK = 60° − 45° = 15° verified. Answer in degrees; FIB numeric = 15. Part (b) recorded next. Figure required." }, { "n": 34, "type_id": 2, "question": "EFGH is a square and EFK is an equilateral triangle. Find ∠FKG. [?]", "answer0": "75", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 230, "difficulty_id": 3, "explanation": "∠KFG = 90 − 60 = 30° (the square corner minus the equilateral triangle's 60°). Triangle FKG is isosceles (FK = FG, both equal to the side of the square), so ∠FKG = (180 − 30) ÷ 2 = 75°.", "hints": ["∠KFG = 90 − 60 = 30°.", "Triangle FKG is isosceles (FK = FG); ∠FKG = (180 − 30) ÷ 2 = 75°."], "source": "Rosyth 2024 P6 Term Assessment 1 Q34", "image_needed": true, "image_options": false, "image_file": "q34.png", "image_page": 21, "image_bbox": [0.30, 0.14, 0.62, 0.34], "image_loc": "upper centre of page (same figure as Q33), square EFGH with equilateral triangle EFK and point K", "notes": "Paper 2 Q15(b). Printed answer: ∠KFG = 30°, ∠FKG = (180 − 30)/2 = 75° (isosceles triangle) verified. Answer in degrees; FIB numeric = 75. Figure required." }, { "n": 35, "type_id": 1, "question": "Adam had some money. He spent \\(\\dfrac{2}{5}\\) of it on 3 identical pens. He bought another 2 of such pens and 15 identical erasers with the rest of his money. What fraction of his money was spent on the 15 erasers? Express your answer in its simplest form.", "answer0": "\\(\\dfrac{1}{3}\\)", "answer1": "\\(\\dfrac{4}{15}\\)", "answer2": "\\(\\dfrac{2}{5}\\)", "answer3": "\\(\\dfrac{5}{15}\\)", "correct_answer": 0, "skill_id": 165, "difficulty_id": 3, "explanation": "3 pens cost \\(\\dfrac{2}{5}\\) of his money, so 1 pen = \\(\\dfrac{2}{5} \\div 3 = \\dfrac{2}{15}\\). 2 pens = \\(\\dfrac{4}{15}\\). Fraction spent on erasers = remainder = \\(1 - \\dfrac{2}{5} - \\dfrac{4}{15} = \\dfrac{15}{15} - \\dfrac{6}{15} - \\dfrac{4}{15} = \\dfrac{5}{15} = \\dfrac{1}{3}\\).", "hints": ["1 pen = (2/5) ÷ 3 = 2/15 of his money; 2 pens = 4/15.", "Erasers = 1 − 2/5 − 4/15 = 1/3."], "source": "Rosyth 2024 P6 Term Assessment 1 Q35", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q16(a). Printed answer 1/3 verified. Fraction answer → MCQ per spec. Part (b) recorded next as numeric FIB." }, { "n": 36, "type_id": 2, "question": "Adam had some money. He spent \\(\\dfrac{2}{5}\\) of it on 3 identical pens, then bought another 2 such pens and 15 identical erasers with the rest. In a sale, Adam would be given 1 free eraser for every 6 erasers bought. How many erasers would he get altogether if he had spent all his money on the erasers? [?]", "answer0": "52", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 165, "difficulty_id": 3, "explanation": "From part (a), 15 erasers cost \\(\\dfrac{1}{3}\\) of his money, so all his money buys 3 × 15 = 45 erasers. With 1 free for every 6 bought: 45 ÷ 6 = 7 remainder 3, so 7 free erasers. Total = 45 + 7 = 52 erasers.", "hints": ["15 erasers = 1/3 of his money, so all his money buys 45 erasers.", "1 free per 6 bought: 45 ÷ 6 = 7 free; total = 45 + 7."], "source": "Rosyth 2024 P6 Term Assessment 1 Q36", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q16(b). Printed answer: 45 erasers, 45 ÷ 6 = 7 free, total 52 verified. FIB numeric = 52." }, { "n": 37, "type_id": 2, "question": "There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. How many girls are there in Tennis Club? [?]", "answer0": "120", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 174, "difficulty_id": 3, "explanation": "Let girls in Tennis = T. Girls in Badminton = 0.75T = 3 units (Tennis = 4 units). Badminton total = 48 + 3 units, Tennis total = 16 + 4 units. Badminton has 2 more students: (48 + 3u) − (16 + 4u) = 2, so 32 − u = 2, u = 30. Girls in Tennis = 4 units = 4 × 30 = 120.", "hints": ["Let Tennis girls = 4 units, Badminton girls = 3 units (75%).", "Use 'Badminton has 2 more students total' to find 1 unit = 30, then Tennis girls = 4 units."], "source": "Rosyth 2024 P6 Term Assessment 1 Q37", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q17(a). Printed answer: 1 unit = 30, Tennis girls 4 units = 120 verified. FIB numeric = 120. Part (b) recorded next." }, { "n": 38, "type_id": 1, "question": "There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. Some girls left the Tennis Club. As a result, 32% of the students in the Tennis Club were boys. What is the ratio of the total number of boys to the total number of girls now?", "answer0": "16 : 31", "answer1": "32 : 31", "answer2": "16 : 15", "answer3": "8 : 15", "correct_answer": 0, "skill_id": 179, "difficulty_id": 3, "explanation": "After some girls leave Tennis, boys (16) form 32% of the Tennis Club. So 32% = 16, 1% = 0.5, total Tennis = 50, Tennis girls now = 50 − 16 = 34. Badminton girls = 3 units = 90, Badminton boys = 48. Total boys = 48 + 16 = 64. Total girls = 90 + 34 = 124. Ratio boys : girls = 64 : 124 = 16 : 31.", "hints": ["16 boys are 32% of the Tennis Club now, so the Tennis total = 50, giving 34 girls.", "Total boys = 64, total girls = 90 + 34 = 124; simplify 64 : 124."], "source": "Rosyth 2024 P6 Term Assessment 1 Q38", "image_needed": false, "image_options": false, "image_file": null, "notes": "Paper 2 Q17(b). Printed answer 16 : 31 verified (Tennis total 50, girls left → 34; total boys 64, girls 124, 64:124 = 16:31). Ratio answer → MCQ per spec." } ] }