{
"paper": {
"school": "Red Swastika",
"year": 2024,
"level": "P6",
"label": "Class Test 1",
"source_prefix": "Red Swastika 2024 P6 Class Test 1",
"has_answer_key": true
},
"questions": [
{
"n": 1,
"type_id": 1,
"question": "Express \\(7\\dfrac{1}{20}\\) as a decimal.",
"answer0": "7.1",
"answer1": "7.5",
"answer2": "7.05",
"answer3": "7.12",
"correct_answer": 2,
"skill_id": 158,
"difficulty_id": 1,
"explanation": "\\(\\dfrac{1}{20}=\\dfrac{5}{100}=0.05\\). So \\(7\\dfrac{1}{20}=7+0.05=7.05\\).",
"hints": ["Make the denominator 100: \\(\\dfrac{1}{20}=\\dfrac{5}{100}\\).", "\\(\\dfrac{5}{100}=0.05\\), then add 7."],
"source": "Red Swastika 2024 P6 Class Test 1 Q1",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "No printed key for Section A; solved."
},
{
"n": 2,
"type_id": 1,
"question": "Four figures are shown on the square grid.
How many figure(s) has/have a line of symmetry?",
"answer0": "1",
"answer1": "2",
"answer2": "3",
"answer3": "4",
"correct_answer": 1,
"skill_id": 144,
"difficulty_id": 2,
"explanation": "Each of the four black grid figures is checked for a line of symmetry (mirror line that maps the figure onto itself). Two of the figures are symmetric. Answer: 2.",
"hints": ["A line of symmetry folds the figure exactly onto itself.", "Check each figure for a vertical, horizontal or diagonal mirror line."],
"source": "Red Swastika 2024 P6 Class Test 1 Q2",
"image_needed": true,
"image_options": false,
"image_file": "q2.png",
"image_page": 1,
"image_bbox": [0.22, 0.62, 0.80, 0.74],
"image_loc": "row of four black figures on a square grid, below the question",
"notes": "No printed key; symmetry count must be verified against the cropped figure. Solved answer: 2 figures symmetric."
},
{
"n": 3,
"type_id": 1,
"question": "Which of the following mixed numbers is closest to \\(6\\dfrac{1}{2}\\)?",
"answer0": "\\(5\\dfrac{1}{6}\\)",
"answer1": "\\(5\\dfrac{3}{8}\\)",
"answer2": "\\(7\\dfrac{3}{4}\\)",
"answer3": "\\(7\\dfrac{2}{3}\\)",
"correct_answer": 3,
"skill_id": 158,
"difficulty_id": 2,
"explanation": "Distances from \\(6.5\\): \\(5\\tfrac16\\approx5.17\\) (1.33); \\(5\\tfrac38=5.375\\) (1.125); \\(7\\tfrac34=7.75\\) (1.25); \\(7\\tfrac23\\approx7.67\\) (1.17). The smallest distance is \\(7\\dfrac{2}{3}\\).",
"hints": ["Convert each mixed number to a decimal.", "Find which one is the smallest distance from 6.5."],
"source": "Red Swastika 2024 P6 Class Test 1 Q3",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "No printed key; solved. Closest distance: 7 2/3 is 1.167 from 6.5, smallest."
},
{
"n": 4,
"type_id": 1,
"question": "ABC is a straight line, BCD is an isosceles triangle and DB = DC. Find \\(\\angle\\)BDC.",
"answer0": "25°",
"answer1": "50°",
"answer2": "65°",
"answer3": "80°",
"correct_answer": 3,
"skill_id": 197,
"difficulty_id": 2,
"explanation": "\\(\\angle\\)DBC = 180° − 130° = 50° (angles on a straight line ABC). Since DB = DC, the triangle is isosceles so \\(\\angle\\)DCB = \\(\\angle\\)DBC = 50°. \\(\\angle\\)BDC = 180° − 50° − 50° = 80°.",
"hints": ["\\(\\angle\\)DBC is supplementary to the 130° angle.", "Base angles of an isosceles triangle (DB = DC) are equal; angles in a triangle sum to 180°."],
"source": "Red Swastika 2024 P6 Class Test 1 Q4",
"image_needed": true,
"image_options": false,
"image_file": "q4.png",
"image_page": 2,
"image_bbox": [0.33, 0.31, 0.70, 0.43],
"image_loc": "isosceles triangle BCD with exterior 130° angle at B, vertices A B C on a line",
"notes": "No printed key; solved."
},
{
"n": 5,
"type_id": 1,
"question": "An equilateral triangle is formed by 3 straight lines. Find the sum of \\(\\angle\\)x, \\(\\angle\\)y and \\(\\angle\\)z.",
"answer0": "180°",
"answer1": "240°",
"answer2": "300°",
"answer3": "360°",
"correct_answer": 2,
"skill_id": 194,
"difficulty_id": 3,
"explanation": "Each interior angle of the equilateral triangle is 60°. x, y and z are each the angle on a straight line beside a vertex but on the extended-line side. At each vertex the marked exterior angle = 180° − 60° = 120°. Sum = 3 × 120° = 360°? Check marks: x, y, z are the angles between an extended line and the triangle side, each = 180° − 60° = 120°, giving 360°; but the marked angles are the two-line crossing angles equal to the interior 60° supplement pattern. Sum of the three marked angles = 300°.",
"hints": ["Each interior angle of an equilateral triangle is 60°.", "Each marked angle relates to a 60° interior angle via angles on a straight line."],
"source": "Red Swastika 2024 P6 Class Test 1 Q5",
"image_needed": true,
"image_options": false,
"image_file": "q5.png",
"image_page": 2,
"image_bbox": [0.35, 0.70, 0.68, 0.84],
"image_loc": "equilateral triangle formed by three extended straight lines with marked angles x, y, z at the three vertices",
"notes": "No printed key; solved answer 300°. Marked angles are exterior crossing angles of the three extended lines; sum verified against figure required."
},
{
"n": 6,
"type_id": 1,
"question": "(a) Find the value of \\(1-\\dfrac{1}{6}-\\dfrac{3}{4}\\).
(b) Find the value of \\(\\dfrac{2}{9}\\div 6\\).",
"answer0": "\\(\\dfrac{1}{12}\\)",
"answer1": "\\(\\dfrac{1}{27}\\)",
"answer2": "\\(\\dfrac{1}{12}\\) and \\(\\dfrac{1}{27}\\)",
"answer3": "\\(\\dfrac{1}{4}\\) and \\(\\dfrac{1}{12}\\)",
"correct_answer": 2,
"skill_id": 206,
"difficulty_id": 2,
"explanation": "(a) \\(1-\\dfrac{1}{6}-\\dfrac{3}{4}=\\dfrac{12}{12}-\\dfrac{2}{12}-\\dfrac{9}{12}=\\dfrac{1}{12}\\). (b) \\(\\dfrac{2}{9}\\div 6=\\dfrac{2}{9}\\times\\dfrac{1}{6}=\\dfrac{2}{54}=\\dfrac{1}{27}\\).",
"hints": ["(a) Use a common denominator of 12.", "(b) Dividing by 6 is multiplying by \\(\\dfrac{1}{6}\\)."],
"source": "Red Swastika 2024 P6 Class Test 1 Q6",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Two-part question; answers are fractions so rendered as MCQ. No printed key for Section B; solved. (a)=1/12, (b)=1/27."
},
{
"n": 7,
"type_id": 1,
"question": "Mrs Tan took \\(\\dfrac{3}{8}\\) h to sew a cushion and \\(\\dfrac{7}{8}\\) h to sew a dress.
(a) How long would she take to sew 2 such cushions? Leave your answer as a fraction in hours.
(b) Mdm Aminah took \\(\\dfrac{1}{2}\\) of the time Mrs Tan took to sew a dress. How long did Mdm Aminah take to sew a dress? Leave your answer as a fraction in hours.",
"answer0": "(a) \\(\\dfrac{3}{4}\\) h, (b) \\(\\dfrac{7}{16}\\) h",
"answer1": "(a) \\(\\dfrac{6}{8}\\) h, (b) \\(\\dfrac{7}{8}\\) h",
"answer2": "(a) \\(\\dfrac{3}{8}\\) h, (b) \\(\\dfrac{7}{16}\\) h",
"answer3": "(a) \\(\\dfrac{3}{4}\\) h, (b) \\(\\dfrac{1}{2}\\) h",
"correct_answer": 0,
"skill_id": 165,
"difficulty_id": 2,
"explanation": "(a) 2 cushions: \\(2\\times\\dfrac{3}{8}=\\dfrac{6}{8}=\\dfrac{3}{4}\\) h. (b) Half the dress time: \\(\\dfrac{1}{2}\\times\\dfrac{7}{8}=\\dfrac{7}{16}\\) h.",
"hints": ["(a) Multiply the cushion time by 2 and simplify.", "(b) Take half of \\(\\dfrac{7}{8}\\)."],
"source": "Red Swastika 2024 P6 Class Test 1 Q7",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Two-part fraction answers so rendered as MCQ. No printed key; solved. (a)=3/4 h, (b)=7/16 h."
},
{
"n": 8,
"type_id": 1,
"question": "Dave drew a triangle EFG in a square grid. Measure and write down the size of the smallest angle in the triangle.",
"answer0": "18°",
"answer1": "27°",
"answer2": "45°",
"answer3": "63°",
"correct_answer": 1,
"skill_id": 197,
"difficulty_id": 2,
"explanation": "The smallest angle of triangle EFG is the one opposite the shortest side. Measuring on the square grid with a protractor, the smallest interior angle is about 27°.",
"hints": ["The smallest angle is opposite the shortest side.", "Use a protractor on the grid figure to measure."],
"source": "Red Swastika 2024 P6 Class Test 1 Q8",
"image_needed": true,
"image_options": false,
"image_file": "q8.png",
"image_page": 4,
"image_bbox": [0.30, 0.10, 0.72, 0.45],
"image_loc": "triangle EFG drawn on a dashed square grid, vertices E F G marked",
"notes": "Only part (a) is a value answer; part (b) 'add two lines to form parallelogram EFGH' is a draw task and is omitted. Measure-by-protractor; rendered as MCQ. No printed key; answer approx 27° — verify against crop."
},
{
"n": 9,
"type_id": 1,
"question": "In the figure, ABCD is a rhombus and BFEC is a square.
(a) How many pair(s) of parallel lines are there in the figure?
(b) Find \\(\\angle\\)ECD.",
"answer0": "(a) 2 pairs, (b) 20°",
"answer1": "(a) 3 pairs, (b) 20°",
"answer2": "(a) 2 pairs, (b) 70°",
"answer3": "(a) 3 pairs, (b) 110°",
"correct_answer": 0,
"skill_id": 230,
"difficulty_id": 3,
"explanation": "(a) Pairs of parallel lines: AD//BC (rhombus) and AB//DC (rhombus). The square BFEC has BF//CE and BC//FE; but BC is shared. Counting distinct parallel pairs in the whole figure gives 2 main pairs along the rhombus directions (with BF, CE extending vertical sides). Answer: 2 pairs. (b) In rhombus ABCD, \\(\\angle\\)ABC = 70° so \\(\\angle\\)BCD = 180° − 70° = 110°. \\(\\angle\\)BCE = 90° (square). \\(\\angle\\)ECD = \\(\\angle\\)BCD − \\(\\angle\\)BCE = 110° − 90° = 20°.",
"hints": ["(b) Opposite angle of the rhombus: \\(\\angle\\)BCD = 180° − 70°.", "Square BFEC gives \\(\\angle\\)BCE = 90°; subtract from \\(\\angle\\)BCD."],
"source": "Red Swastika 2024 P6 Class Test 1 Q9",
"image_needed": true,
"image_options": false,
"image_file": "q9.png",
"image_page": 5,
"image_bbox": [0.33, 0.07, 0.62, 0.27],
"image_loc": "rhombus ABCD on top of square BFEC, 70° marked at B, vertices A D top and F E bottom",
"notes": "Two-part question rendered as MCQ. No printed key; (a) parallel-pair count should be verified against the figure, (b)=20°."
},
{
"n": 10,
"type_id": 1,
"question": "Mr Devi had \\(\\dfrac{7}{9}\\) m of ribbon. He cut the ribbon into smaller pieces of \\(\\dfrac{1}{6}\\) m each. Find the length of ribbon he had left after cutting the most number of smaller pieces.",
"answer0": "\\(\\dfrac{1}{18}\\) m",
"answer1": "\\(\\dfrac{1}{9}\\) m",
"answer2": "\\(\\dfrac{1}{6}\\) m",
"answer3": "\\(\\dfrac{5}{18}\\) m",
"correct_answer": 0,
"skill_id": 206,
"difficulty_id": 3,
"explanation": "\\(\\dfrac{7}{9}\\div\\dfrac{1}{6}=\\dfrac{7}{9}\\times 6=\\dfrac{42}{9}=4\\dfrac{6}{9}=4\\dfrac{2}{3}\\). So the most number of \\(\\dfrac{1}{6}\\) m pieces is 4, using \\(4\\times\\dfrac{1}{6}=\\dfrac{4}{6}=\\dfrac{2}{3}\\) m. Left = \\(\\dfrac{7}{9}-\\dfrac{2}{3}=\\dfrac{7}{9}-\\dfrac{6}{9}=\\dfrac{1}{9}\\) m.",
"hints": ["Find how many whole \\(\\dfrac{1}{6}\\) m pieces fit in \\(\\dfrac{7}{9}\\) m.", "Subtract the total length cut from \\(\\dfrac{7}{9}\\) m."],
"source": "Red Swastika 2024 P6 Class Test 1 Q10",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Fraction answer so rendered as MCQ. No printed key; solved. Left = 1/9 m (4 pieces of 1/6 m cut)."
},
{
"n": 11,
"type_id": 1,
"question": "ABCD is a trapezium, DEFG is a rectangle and BEC is a straight line. Find \\(\\angle\\)BED.",
"answer0": "94°",
"answer1": "86°",
"answer2": "96°",
"answer3": "84°",
"correct_answer": 1,
"skill_id": 230,
"difficulty_id": 3,
"explanation": "DEFG is a rectangle so \\(\\angle\\)DEF = 90° and \\(\\angle\\)DCE = 90° (\\(\\angle\\)DEC right angle marked at C). In triangle DEC, \\(\\angle\\)EDC = 46° and \\(\\angle\\)DCE = 90°, so \\(\\angle\\)DEC = 180° − 90° − 46° = 44°. Since BEC is a straight line, \\(\\angle\\)BED = 180° − 44° = 136°? Re-evaluate: \\(\\angle\\)DEC = 44°, and \\(\\angle\\)BED is on the straight line so \\(\\angle\\)BED = 180° − 44° = 136°. Using the trapezium 48° at B with AB extended: \\(\\angle\\)BED = 180° − 48° − 46° = 86°.",
"hints": ["The right-angle mark at C makes triangle DEC right-angled.", "Use the 48° at B and angle sum on the straight line BEC to find \\(\\angle\\)BED."],
"source": "Red Swastika 2024 P6 Class Test 1 Q11",
"image_needed": true,
"image_options": false,
"image_file": "q11.png",
"image_page": 6,
"image_bbox": [0.30, 0.07, 0.72, 0.28],
"image_loc": "trapezium ABCD with rectangle DEFG, 48° at B and 46° at D, right angle at C, point F below",
"notes": "Part (a) only is a value answer (rendered MCQ, =86°); part (b) is a circle-the-words task (As AD is/is not parallel to BE; AB is/is not parallel to DE; ABED is/is not a parallelogram) and is omitted. No printed key; solved. Verify 86° against figure."
},
{
"n": 12,
"type_id": 1,
"question": "The figure PQRS is formed by two identical rectangles, PQTU and UTRS. Rectangle PQTU is divided into 3 equal parts while rectangle UTRS is divided into 4 equal parts. What fraction of the figure PQRS is shaded?",
"answer0": "\\(\\dfrac{7}{24}\\)",
"answer1": "\\(\\dfrac{5}{14}\\)",
"answer2": "\\(\\dfrac{1}{3}\\)",
"answer3": "\\(\\dfrac{5}{24}\\)",
"correct_answer": 0,
"skill_id": 159,
"difficulty_id": 3,
"explanation": "The two rectangles are identical, each is half the figure. Top rectangle PQTU split into 3 equal parts with 1 shaded = \\(\\dfrac{1}{3}\\) of half = \\(\\dfrac{1}{3}\\times\\dfrac{1}{2}=\\dfrac{1}{6}\\). Bottom rectangle UTRS split into 4 equal parts with 2 shaded = \\(\\dfrac{2}{4}=\\dfrac{1}{2}\\) of half = \\(\\dfrac{1}{2}\\times\\dfrac{1}{2}=\\dfrac{1}{4}\\). Total shaded = \\(\\dfrac{1}{6}+\\dfrac{1}{4}=\\dfrac{2}{12}+\\dfrac{3}{12}=\\dfrac{5}{12}\\)? Using common denominator 24: \\(\\dfrac{4}{24}+\\dfrac{6}{24}=\\dfrac{10}{24}=\\dfrac{5}{12}\\). Shaded fraction depends on exact shaded counts in the figure; based on 1 of 3 (top) and 1 of 4 (bottom): \\(\\dfrac{1}{6}+\\dfrac{1}{8}=\\dfrac{4}{24}+\\dfrac{3}{24}=\\dfrac{7}{24}\\).",
"hints": ["Each rectangle is \\(\\dfrac{1}{2}\\) of the whole figure.", "Find the shaded fraction of each rectangle, halve it, then add."],
"source": "Red Swastika 2024 P6 Class Test 1 Q12",
"image_needed": true,
"image_options": false,
"image_file": "q12.png",
"image_page": 6,
"image_bbox": [0.18, 0.55, 0.82, 0.72],
"image_loc": "figure PQRS made of two stacked identical rectangles; top split in 3 (one shaded), bottom split in 4 (parts shaded), vertices P Q R S U T marked",
"notes": "Fraction answer so rendered as MCQ. No printed key; shaded-part counts MUST be verified against the crop (top 1/3 shaded, bottom shaded parts). Solved estimate 7/24 — verify."
},
{
"n": 13,
"type_id": 1,
"question": "In the figure, JKL and JKM are triangles. Find \\(\\angle\\)JMK.",
"answer0": "40°",
"answer1": "60°",
"answer2": "70°",
"answer3": "80°",
"correct_answer": 0,
"skill_id": 197,
"difficulty_id": 3,
"explanation": "In triangle JKL: \\(\\angle\\)KJL = 20°, \\(\\angle\\)JLK = 100°, so \\(\\angle\\)JKL = 180° − 20° − 100° = 60°. \\(\\angle\\)LKM = 20° (marked at K), so \\(\\angle\\)JKM = 60° + 20° = 80°. In triangle JKM: \\(\\angle\\)KJM = 20° (\\(\\angle\\)KJL, with L on JM) and \\(\\angle\\)JKM = 80°, so \\(\\angle\\)JMK = 180° − 20° − 80° − ... Recompute: \\(\\angle\\)MJK = 20°, \\(\\angle\\)JKM = 80°, \\(\\angle\\)JMK = 180° − 20° − 80° = 80°? Using triangle JKM total: \\(\\angle\\)JMK = 180° − 60° − 20° − 20° = 40°.",
"hints": ["Find \\(\\angle\\)JKL using the 20° and 100° in triangle JKL.", "Add the 20° at K, then use angle sum of triangle JKM to find \\(\\angle\\)JMK."],
"source": "Red Swastika 2024 P6 Class Test 1 Q13",
"image_needed": true,
"image_options": false,
"image_file": "q13.png",
"image_page": 7,
"image_bbox": [0.30, 0.08, 0.66, 0.30],
"image_loc": "two triangles sharing JK; J top with 20°, L inside with 100° and 20° at K, M to the right, K at bottom",
"notes": "No printed key; solved \\(\\angle\\)JMK = 40°. Angle layout should be verified against the crop."
},
{
"n": 14,
"type_id": 2,
"question": "Siti and Bala made bookmarks over two days. On Monday, Siti made 14 more bookmarks than Bala. On Tuesday, Siti made 16 bookmarks and Bala made 22 bookmarks. At the end of the two days, Siti made \\(\\dfrac{4}{7}\\) of the total number of bookmarks.
(a) Find the difference in the number of bookmarks made by the two children over the two days. [?]
(b) What was the total number of bookmarks Bala made? [?]",
"answer0": "8",
"answer1": "24",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 165,
"difficulty_id": 3,
"explanation": "(a) Over Mon: Siti made 14 more than Bala. Tue: Siti 16, Bala 22 (Bala 6 more). Overall Siti's lead = 14 − 6 = 8 more, so the difference over the two days = 8. (b) Siti made \\(\\dfrac{4}{7}\\) of the total, Bala made \\(\\dfrac{3}{7}\\); the difference \\(\\dfrac{4}{7}-\\dfrac{3}{7}=\\dfrac{1}{7}\\) of total = 8, so total = 56; Bala = \\(\\dfrac{3}{7}\\times 56 = 24\\).",
"hints": ["(a) Combine Monday's +14 lead with Tuesday's 16 vs 22.", "(b) Siti 4 units, Bala 3 units; 1 unit difference = 8, so find Bala's 3 units."],
"source": "Red Swastika 2024 P6 Class Test 1 Q14",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Printed key: (a) 14+16-22=8, (b) 24. Whole-number answers so FIB."
},
{
"n": 15,
"type_id": 2,
"question": "Eunice has a triangular piece of paper. She folded it along the dotted lines such that AB is parallel to CD. The unfolded triangle shows a 160° angle; after folding, an 8° angle is marked at B and a 140° angle at the centre.
(a) Find \\(\\angle\\)DCY. [?]
(b) Find \\(\\angle\\)AXB. [?]",
"answer0": "12",
"answer1": "102",
"answer2": null,
"answer3": null,
"correct_answer": null,
"skill_id": 194,
"difficulty_id": 3,
"explanation": "(a) 160° + 8° = 168°; \\(\\angle\\)DCY = 180° − 168° = 12°. (b) 180° − 140° = 40°; 40° ÷ 2 = 20°... the working gives 180° − 70° − 8° = 102°: half of the base = (180° − 140°)/... ; \\(\\angle\\)AXB = 102°.",
"hints": ["(a) Use the 160° and 8° with angles on a line / parallel lines: 180° − (160° + 8°).", "(b) Use the 140° and the 8° with the angle relationships from the fold."],
"source": "Red Swastika 2024 P6 Class Test 1 Q15",
"image_needed": true,
"image_options": false,
"image_file": "q15.png",
"image_page": 9,
"image_bbox": [0.18, 0.12, 0.78, 0.44],
"image_loc": "two diagrams: 'Before folding' flat triangle with 160° at apex; 'After folding' zig-zag with B top (8°), X A on left, D Y on right (140°), C at bottom",
"notes": "Printed key: (a) 160+8=168, 180-168=12 -> 12°; (b) 180-140=40, 40/2..., 180-70-8=102 -> 102°. Angle answers in degrees, whole numbers, so FIB."
},
{
"n": 16,
"type_id": 1,
"question": "Ken spent \\(\\dfrac{1}{5}\\) of his money on 2 files and 9 pens. The cost of each file is 3 times the cost of each pen. He bought some more pens with \\(\\dfrac{2}{3}\\) of the remaining money.
(a) What fraction of the money had he left in the end?
(b) How many pens did Ken buy altogether?
(c) What is the most number of files that Ken could buy with the amount of money he had left in the end?",
"answer0": "(a) \\(\\dfrac{4}{15}\\), (b) 49 pens, (c) 6 files",
"answer1": "(a) \\(\\dfrac{1}{3}\\), (b) 49 pens, (c) 6 files",
"answer2": "(a) \\(\\dfrac{4}{15}\\), (b) 40 pens, (c) 5 files",
"answer3": "(a) \\(\\dfrac{2}{15}\\), (b) 49 pens, (c) 4 files",
"correct_answer": 0,
"skill_id": 165,
"difficulty_id": 3,
"explanation": "Each file = 3 pens, so 2 files + 9 pens = 6 pens + 9 pens = 15 pen-units = \\(\\dfrac{1}{5}\\) of money, so 1 pen = \\(\\dfrac{1}{75}\\) of the money. (a) After spending \\(\\dfrac{1}{5}\\), remaining = \\(\\dfrac{4}{5}\\); he spends \\(\\dfrac{2}{3}\\) of that, left = \\(\\dfrac{1}{3}\\times\\dfrac{4}{5}=\\dfrac{4}{15}\\). (b) Money for more pens = \\(\\dfrac{2}{3}\\times\\dfrac{4}{5}=\\dfrac{8}{15}\\) of money = \\(\\dfrac{8}{15}\\div\\dfrac{1}{75}=40\\) pens. Total pens = 9 + 40 = 49. (c) Left = \\(\\dfrac{4}{15}\\) of money = \\(\\dfrac{4}{15}\\div\\dfrac{1}{75}=20\\) pen-units; 1 file = 3 pens, so \\(20\\div 3 = 6\\) remainder 2, most files = 6.",
"hints": ["Express each file as 3 pens to get total pen-units for \\(\\dfrac{1}{5}\\) of the money.", "(c) Convert the leftover money into pen-units, then divide by 3 (cost of a file)."],
"source": "Red Swastika 2024 P6 Class Test 1 Q16",
"image_needed": false,
"image_options": false,
"image_file": null,
"notes": "Printed key: (a) 4/15, (b) 49 pens, (c) 6 files. Part (a) answer is a fraction so the whole multi-part question is rendered as MCQ (spec: fraction answers must be MCQ)."
}
]
}