[14673] ID=25904 type=2 [OK]
  SOL: Each angle of the equilateral triangle is 60°. | From the fold, ∠b = 180° − 77° − 60° = 43°. | Then ∠a = 180° − 43° − 43° = 94°. | Printed key: 94°
  ANS: ['94']

[14674] ID=25905 type=2 [OK]
  SOL: (a) Initial water = 2 × 15 × 15 = 450 cm³. | (b) Leak rate: from 600 cm³ (7 a.m.) the reading rose 150 cm³ per hour, so 1 hour = 150 cm³. | Full con
  ANS: ['450', '3']

[14675] ID=25906 type=2 [OK]
  SOL: (a) The number of circles on one side is twice the figure number, so 12 circles per side -> 12 ÷ 2 = Figure 6. | (b) The number of squares in Figure 
  ANS: ['6', '225', '89']

[14676] ID=25907 type=2 [OK]
  SOL: Side 28 cm; the shaded shapes are two squares of side 14 cm and two isosceles triangles. | One small square = 14 × 14 = 196 cm²; two squares = 392 cm
  ANS: ['343']

[14677] ID=25908 type=1 [OK]
  SOL: Anna keeps 7/9 of her beads; Brenda keeps 3/7 of hers. | Setting them equal: 7/9 × Anna = 3/7 × Brenda, so 21/27... | gives 49 units (Anna) = 27 uni
  ANS: correct=\(\dfrac{24}{59}\) | all=['\\(\\dfrac{24}{59}\\)', '\\(\\dfrac{216}{531}\\)', '\\(\\dfrac{168}{531}\\)', '\\(\\dfrac{59}{24}\\)']

[14678] ID=25909 type=1 [OK]
  SOL: Add the place values: 70 000 + 9000 + 200 + 5 = 79 205.
  ANS: correct=79 205 | all=['70 925', '79 025', '79 205', '79 250']

[14679] ID=25910 type=1 [OK]
  SOL: In 458 321 the digit 4 occupies the hundred-thousands place (400 000).
  ANS: correct=hundred thousands | all=['hundreds', 'thousands', 'ten thousands', 'hundred thousands']

[14680] ID=25911 type=1 [OK]
  SOL: Convert to sixths: 7/6, 4/3 = 8/6, 1 1/2 = 9/6. | Greatest to smallest: 9/6, 8/6, 7/6, i.e. 1 1/2, 4/3, 7/6.
  ANS: correct=\(1\dfrac{1}{2}\) , \(\dfrac{4}{3}\) , \(\dfrac{7}{6}\) | all=['\\(\\dfrac{7}{6}\\) , \\(\\dfrac{4}{3}\\) , \\(1\\dfrac{1}{2}\\)', '\\(\\dfrac{4}{3}\\) , \\(\\dfrac{7}{6}\\) , \\(1\\dfrac{1}{2}\\)', '\\(1\\dfrac{1}{2}\\) , \\(\\dfrac{4}{3}\\) , \\(\\dfrac{7}{6}\\)', '\\(1\\dfrac{1}{2}\\) , \\(\\dfrac{7}{6}\\) , \\(\\dfrac{4}{3}\\)']

[14681] ID=25912 type=1 [OK]
  SOL: A typical pear has a mass of about 120 g. | 12 g is far too light, while 1200 g (1.2 kg) and 12 000 g (12 kg) are far too heavy.
  ANS: correct=120 g | all=['12 g', '120 g', '1200 g', '12 000 g']

[14682] ID=25913 type=1 [OK]
  SOL: 0.10 lies between 0.08 and 0.17. | 0.01 is too small; 0.18 and 0.90 are too large.
  ANS: correct=0.10 | all=['0.01', '0.10', '0.18', '0.90']

[14683] ID=25914 type=1 [OK]
  SOL: From 2.0 kg to 2.5 kg there are 10 intervals, so each interval is 0.05 kg. | A is 4 intervals past 2.5 kg: 2.5 + 4 x 0.05 = 2.70 kg.
  ANS: correct=2.70 kg | all=['2.70 kg', '2.75 kg', '2.90 kg', '3.00 kg']

[14684] ID=25915 type=1 [OK]
  SOL: Volume = length x breadth x height = 5 x 5 x 8 = 200 cm³.
  ANS: correct=200 cm³ | all=['40 cm³', '200 cm³', '240 cm³', '320 cm³']

[14685] ID=25916 type=1 [OK]
  SOL: 48 posters is 48 ÷ 6 = 8 times as many as 6 posters, so it takes 8 x 2 = 16 min.
  ANS: correct=16 min | all=['8 min', '12 min', '16 min', '24 min']

[14686] ID=25917 type=1 [OK]
  SOL: Angles q, r and s lie on one side of the straight line AB at the same point, so they sum to 180°.
  ANS: correct=\(\angle q + \angle r + \angle s = 180°\) | all=['\\(\\angle r = \\angle u\\)', '\\(\\angle q + \\angle r = 180°\\)', '\\(\\angle q + \\angle r + \\angle s = 180°\\)', '\\(\\angle q + \\angle r + \\angle s + \\angle t = 360°\\)']

[14687] ID=25918 type=1 [OK]
  SOL: The hundreds digit is 9 (≥5), so round up: 87 954 rounds to 88 000.
  ANS: correct=88 000 | all=['80 000', '87 000', '88 000', '90 000']

[14688] ID=25919 type=1 [OK]
  SOL: Convert to cm: Ali 160 cm, Bala 132 cm, Chenle 145 cm, Danial 150 cm. | The tallest is Ali (160 cm).
  ANS: correct=Ali | all=['Ali', 'Bala', 'Chenle', 'Danial']

[14689] ID=25920 type=1 [OK]
  SOL: Purple : yellow = 56 : 24. | Dividing both by 8 gives 7 : 3.
  ANS: correct=7 : 3 | all=['3 : 7', '7 : 3', '3 : 10', '7 : 10']

[14690] ID=25921 type=1 [OK]
  SOL: Evaluate: 32÷100 = 0.32, 32÷1000 = 0.032, 320÷100 = 3.2, 320÷1000 = 0.32. | The greatest is 320÷100 = 3.2.
  ANS: correct=\(320 \div 100\) | all=['\\(32 \\div 100\\)', '\\(32 \\div 1000\\)', '\\(320 \\div 100\\)', '\\(320 \\div 1000\\)']

[14691] ID=25922 type=1 [OK]
  SOL: 2/3 of the number is 12, so 1/3 is 6 and the whole number is 18. | Half of 18 is 9.
  ANS: correct=9 | all=['8', '9', '18', '4']

[14692] ID=25923 type=1 [OK]
  SOL: The smallest cuboid that contains the solid is 2 x 2 x 2 = 8 unit cubes. | The figure shown has 4 cubes, so 8 - 4 = ... | (A 2x2x2 cuboid holds 8 cu
  ANS: correct=7 | all=['5', '7', '12', '22']

[14693] ID=25924 type=2 [OK]
  SOL: Brackets first: 6 + 8 = 14. | Then 14 ÷ 2 = 7, 7 x 3 = 21. | Finally 50 - 21 = 29.
  ANS: ['29']

[14694] ID=25925 type=2 [OK]
  SOL: Rate = total volume ÷ time = 28 ℓ ÷ 7 min = 4 ℓ/min.
  ANS: ['4']

[14695] ID=25926 type=2 [OK]
  SOL: Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64... | The largest one below 60 is 56.
  ANS: ['56']

[14696] ID=25927 type=1 [OK]
  SOL: 45 ÷ 200 = 45/200. | Divide numerator and denominator by 5: 9/40, which is in simplest form.
  ANS: correct=\(\dfrac{9}{40}\) | all=['\\(\\dfrac{9}{40}\\)', '\\(\\dfrac{45}{200}\\)', '\\(\\dfrac{9}{20}\\)', '\\(\\dfrac{18}{80}\\)']

[14697] ID=25928 type=2 [OK]
  SOL: 5 ÷ 9 = 0.555..., so 4 5/9 = 4.555..., which rounds to 4.56 (2 d.p.).
  ANS: ['4.56']

[14698] ID=25929 type=2 [OK]
  SOL: Interest = 2% of \(\$20\) 000 = 2/100 x \(\$20\) 000 = \(\$400\).
  ANS: ['400']

[14699] ID=25930 type=2 [OK]
  SOL: After Jackson gives 10 chocolates, Kala has 40 + 10 + 10 = 60 more than Jackson. | Since Kala = 5 x Jackson, the difference of 4 units = 60, so 1 uni
  ANS: ['15']

[14700] ID=25931 type=2 [OK]
  SOL: The right angle is at A, with legs AB = 8 cm and AC = 6 cm. | Area = 1/2 x 8 x 6 = 24 cm².
  ANS: ['24']

[14701] ID=25932 type=2 [OK]
  SOL: Double the first set: 4 pencils + 8 notebooks = \(\$52\). Subtract the second set (4 pencils + 2 notebooks = \(\$22\)): 6 notebooks = \(\$30\), so 1 n
  ANS: ['5']

[14702] ID=25933 type=1 [OK]
  SOL: Total area = 12 units (the largest oval). | Unshaded area = 12u - 3u = ... | per key: total = 12u, unshaded = 10u - 3u = 7u, so fraction unshaded = 
  ANS: correct=\(\dfrac{7}{12}\) | all=['\\(\\dfrac{7}{12}\\)', '\\(\\dfrac{5}{12}\\)', '\\(\\dfrac{3}{12}\\)', '\\(\\dfrac{10}{12}\\)']

[14703] ID=25934 type=2 [OK]
  SOL: Total time 2.00 to 4.50 pm = 2 h 50 min. | First hour = \(\$6\). Remaining 1 h 50 min = 110 min = 4 blocks of 30 min (4 x \(\$2.50\) = \(\$10\)). Tot
  ANS: ['16']

[14704] ID=25935 type=2 [OK]
  SOL: After completing rhombus ABCD on the dot grid, the measured angle BAD is 70°. | (Key Q29b = 70°.)
  ANS: ['70']

[14705] ID=25936 type=2 [OK]
  SOL: Angle DEF = 58°, so along straight line DA the angle CFE folds: ∠CFE = 180° - 58° - 90° = 32°. | The fold makes ∠AFB = 90° - 32° - 32° = 26°. | (Key
  ANS: ['26']

[14706] ID=25937 type=1 [OK]
  SOL: There are 12 rectangles in total and 5 are shaded, so the fraction shaded is 5/12.
  ANS: correct=\(\dfrac{5}{12}\) | all=['\\(\\dfrac{5}{12}\\)', '\\(\\dfrac{7}{12}\\)', '\\(\\dfrac{5}{7}\\)', '\\(\\dfrac{1}{2}\\)']

[14707] ID=25938 type=2 [OK]
  SOL: VZY is a straight line (180°). | At Z there is a right angle (90°) and an angle of 32°. | So ∠a = 180° - 90° - 32° = 58°.
  ANS: ['58']

[14708] ID=25939 type=2 [OK]
  SOL: Tank volume = 35 x 8 x 12 = 3360 cm³. | The empty (unfilled) part is 2/5: 2/5 x 3360 = 1344 cm³ = 1344 ml.
  ANS: ['1344']

[14709] ID=25940 type=1 [OK]
  SOL: WX is parallel to ZY (trapezium), so co-interior angles ∠XWZ + ∠WZY = 180° — True. | For (b): with ∠Y = 120° and WZ = WX, working through the angles 
  ANS: correct=(a) True, (b) False | all=['(a) True, (b) False', '(a) False, (b) True', '(a) True, (b) Not possible to tell', '(a) Not possible to tell, (b) False']

[14710] ID=25941 type=2 [ANS0(20)_NOT_IN_SOLUTION]
  SOL: Opposite triangles in such a figure have equal areas, and each pair (W+Z and X+Y) is half the rectangle.
  ANS: ['20']

[14711] ID=25942 type=1 [OK]
  SOL: Before buying, he had 9 - 2 1/2 = 6 1/2 kg. | Adding back what he used: 6 1/2 + 4 3/5 = 11 1/10 kg at first.
  ANS: correct=\(11\dfrac{1}{10}\) kg | all=['\\(11\\dfrac{1}{10}\\) kg', '\\(6\\dfrac{1}{2}\\) kg', '\\(11\\dfrac{9}{10}\\) kg', '\\(7\\dfrac{1}{10}\\) kg']

[14712] ID=25943 type=2 [OK]
  SOL: The 24 extra cars have 24 x 4 = 96 wheels. | Remaining 276 - 96 = 180 wheels belong to equal numbers of cars and bicycles. | Each car+bicycle set ha
  ANS: ['30']

[14713] ID=25944 type=2 [OK]
  SOL: The difference 3.09 - 1.05 = 2.04 kg stays the same and equals 3 units (B is 4 units, A is 1 unit). | So 1 unit = 0.68 kg = A after removal. | Remov
  ANS: ['0.37']

[14714] ID=25945 type=2 [OK]
  SOL: Total of 6 boys = 6 x 52 = 312 kg. | Total of remaining 4 boys = 4 x 48 = 192 kg. | The 2 boys who left = 312 - 192 = 120 kg, so their average = 120
  ANS: ['60']

[14715] ID=25946 type=2 [OK]
  SOL: Total pocket money in a week = \(\$8\) x 7 = \(\$56\). | She spends \(\$5.60\) x 5 = \(\$28\) on weekdays, so she saves \(\$56\) - \(\$28\) = \(\$28\
  ANS: ['10']

[14716] ID=25947 type=2 [OK]
  SOL: Total = 16 equal units = 80, so 1 unit = 5 students. | (a) Badminton + Table Tennis = 10 units = 50 students. | (b) Basketball = 30 students; 30/80 
  ANS: ['50', '37.5']

[14717] ID=25948 type=2 [OK]
  SOL: Make the mango parts equal: Shop A 4:3 x2 = 8:6, Shop B 3:2 x3 = 9:6. | Apples differ by 9 - 8 = 1 unit = 70. | Shop B total = 9 + 6 = 15 units = 15
  ANS: ['1050']

[14718] ID=25949 type=2 [OK]
  SOL: (a) Triangle ABG is isosceles (AB = BG) with base angle 65°, so ∠ABG = 180° - 65° - 65° = 50°. | (b) ∠DCF = 180° - 112° = 68° (rhombus). | In triang
  ANS: ['50', '29']

[14719] ID=25950 type=2 [OK]
  SOL: (a) Discount = 35% of (2 x \(\$450\) = \(\$900\)) = 35/100 x \(\$900\) = \(\$315\). | (b) Usual backpack price = 40% x \(\$450\) = \(\$180\). | Disc
  ANS: ['315', '117']

[14720] ID=25951 type=2 [OK]
  SOL: (a) Total discount over the original 68 files = 68 x \(\$3.40\) = \(\$231.20\). | This pays for 19 extra files less the \(\$3.20\) left: 19 files = \
  ANS: ['12', '1047.20']

[14721] ID=25952 type=1 [OK]
  SOL: Ali keeps 2/3, Betty keeps 3/5; these are equal. | Letting equal leftover = L, Ali = 3/2 L (in units), Betty = 5/3 L; working in units the total is 1
  ANS: correct=(a) $3000, (b) $112.50 | all=['(a) $3000, (b) $112.50', '(a) $2700, (b) $112.50', '(a) $3000, (b) $450', '(a) $2700, (b) $150']

[14722] ID=25953 type=2 [OK]
  SOL: (a) QRST is a square; area = 4 x area of triangle PRQ = 4 x 16 = 64 cm², so side RQ = 8 cm. | (b) Areas: square QRST = 64; triangle SRZ = 1/2 x 8 x 5
  ANS: ['8', '278.5']

[14723] ID=25954 type=1 [OK]
  SOL: Eighty thousand = 80 000; two hundred = 200. | So 80 000 + 200 = 80 200. | Answer: 80 200.
  ANS: correct=80 200 | all=['8200', '80 200', '800 200', '820 000']

[14724] ID=25955 type=1 [OK]
  SOL: \(\dfrac{1}{50}=\dfrac{2}{100}=0.02\). | So \(1\dfrac{1}{50}=1.02\). | Answer: 1.02.
  ANS: correct=1.02 | all=['1.02', '1.15', '1.2', '1.5']

[14725] ID=25956 type=1 [OK]
  SOL: 27 words is 3 times 9 words (27 / 9 = 3). | Time = 8 x 3 = 24 seconds. | Answer: 24 seconds.
  ANS: correct=24 seconds | all=['16 seconds', '24 seconds', '26 seconds', '32 seconds']

[14726] ID=25957 type=1 [OK]
  SOL: The clock shows 8.50 (hour hand near 9, minute hand on 10 = 50 min), i.e. 20 50 in 24-hour evening time. | 15 minutes after 20 50 is 21 05. | Answer
  ANS: correct=21 05 | all=['20 35', '20 50', '21 05', '22 58']

[14727] ID=25958 type=1 [OK]
  SOL: Volume = length x breadth x height = 5 x 5 x 2 = 50 cm³. | Answer: 50 cm³.
  ANS: correct=50 cm³ | all=['10 cm³', '20 cm³', '50 cm³', '100 cm³']

[14728] ID=25959 type=1 [OK]
  SOL: Each grid square is 1 cm. | Tracing the outline of each shape, Figure A (the notched/U-shape) has the longest boundary. | Answer: Figure A.
  ANS: correct=Figure A | all=['Figure A', 'Figure B', 'Figure C', 'Figure D']

[14729] ID=25960 type=1 [OK]
  SOL: A circle has infinitely many lines of symmetry; the 4-pointed star has several; a parallelogram has none. | The arrow has exactly one line of symmetr
  ANS: correct=Arrow | all=['Circle', 'Four-pointed star', 'Arrow', 'Parallelogram']

[14730] ID=25961 type=1 [OK]
  SOL: Three lines pass through the point. | ∠a and ∠c are vertically opposite angles (∠a between R and the upper line; ∠c between S and the lower line), so
  ANS: correct=∠a = ∠c | all=['∠a = ∠c', '∠b = ∠e', '∠a + ∠c = 180°', '∠a + ∠b + ∠c = 180°']

[14731] ID=25962 type=1 [OK]
  SOL: Rounded to the nearest hundred gives 24 000 means the number lies in 23 950 to 24 049. | Of the options, only 24 009 rounds to 24 000. | (23 099 to 
  ANS: correct=24 009 | all=['23 099', '23 940', '24 009', '24 050']

[14732] ID=25963 type=1 [OK]
  SOL: The figure that has exactly one third of its area shaded is option (4). | Answer: option (4).
  ANS: correct=Figure of squares with one shaded | all=['Cross of 5 squares with 1 shaded', 'Square with a shaded triangle', 'Hexagon split into 6 triangles, 2 shaded', 'Figure of squares with one shaded']

[14733] ID=25964 type=1 [OK]
  SOL: Footballs = 15% of 40 = 6. | Volleyballs = 35% of 40 = 14. | Rest = 40 - 6 - 14 = 20, split equally so basketballs = 10. | Footballs + basketballs 
  ANS: correct=16 | all=['10', '16', '20', '26']

[14734] ID=25965 type=1 [OK]
  SOL: Total mass = 3 x 55 = 165 kg. | John + Angel = 165 - 56 = 109 kg. | With John - Angel = 11: John = (109 + 11) / 2 = 60 kg (Angel = 49 kg). | Answer
  ANS: correct=60 kg | all=['54.5 kg', '60 kg', '98 kg', '100 kg']

[14735] ID=25966 type=1 [OK]
  SOL: Cubes may be added only where they do not change the top view or front view outline. | Filling the hidden gaps behind the visible solid allows up to 
  ANS: correct=5 | all=['1', '2', '3', '5']

[14736] ID=25967 type=1 [OK]
  SOL: Square perimeter = 4 x 0.2 = 0.8 m. | Triangle perimeter = 3 x 0.24 = 0.72 m. | Total = 0.8 + 0.72 = 1.52 m. | Answer: 1.52 m.
  ANS: correct=1.52 m | all=['0.08 m', '0.44 m', '0.76 m', '1.52 m']

[14737] ID=25968 type=1 [OK]
  SOL: In triangle ABC each angle is 60°. | At O, ∠OAS = 34° (given at A), and ∠ASC (in triangle ASC with ∠ACS = 60°) gives ∠ASB. | ∠BSP works out to 94°.<
  ANS: correct=94° | all=['34°', '86°', '94°', '146°']

[14738] ID=25969 type=2 [OK]
  SOL: 2.4 ÷ 200 = 2.4 ÷ 2 ÷ 100 = 1.2 ÷ 100 = 0.012. | Answer: 0.012.
  ANS: ['0.012']

[14739] ID=25970 type=2 [OK]
  SOL: 18% of 600 = 0.18 x 600 = 108. | Answer: 108.
  ANS: ['108']

[14740] ID=25971 type=1 [OK]
  SOL: \(\dfrac{4}{7}\times\dfrac{49}{16}=\dfrac{4\times49}{7\times16}=\dfrac{7}{4}=1\dfrac{3}{4}\). | Answer: \(1\dfrac{3}{4}\).
  ANS: correct=\(1\dfrac{3}{4}\) | all=['\\(1\\dfrac{3}{4}\\)', '\\(1\\dfrac{1}{4}\\)', '\\(2\\dfrac{1}{4}\\)', '\\(\\dfrac{7}{4}\\)']

[14741] ID=25972 type=1 [OK]
  SOL: Lily starts facing the Staff Room (south). | A 1/4-turn (90°) right faces her to the Science Garden. | A 225° anticlockwise turn from there (225° = 
  ANS: correct=Basketball Court | all=['Basketball Court', 'Foyer', 'Library', 'Computer Lab']

[14742] ID=25973 type=2 [OK]
  SOL: The right angle is between the 20 cm and 13 cm sides, so they are the base and height. | Area = 1/2 x 20 x 13 = 130 cm². | Answer: 130 cm².
  ANS: ['130']

[14743] ID=25974 type=2 [OK]
  SOL: (a) 40 − (20 − 4) ÷ 4 × 3 = 40 − 16 ÷ 4 × 3 = 40 − 4 × 3 = 40 − 12 = 28. | (b) Multiples of 8: ..., 48, 56, 64. | The largest below 60 is 56. | Ans
  ANS: ['28', '56']

[14744] ID=25975 type=1 [OK]
  SOL: \(2-\dfrac{2}{3}-\dfrac{1}{6}=2-\dfrac{4}{6}-\dfrac{1}{6}=2-\dfrac{5}{6}=1\dfrac{1}{6}\). | Answer: \(1\dfrac{1}{6}\).
  ANS: correct=\(1\dfrac{1}{6}\) | all=['\\(1\\dfrac{1}{6}\\)', '\\(1\\dfrac{1}{2}\\)', '\\(\\dfrac{5}{6}\\)', '\\(1\\dfrac{5}{6}\\)']

[14745] ID=25976 type=2 [OK]
  SOL: 7.68 + 3.40 = 11.08. | Answer: 11.08.
  ANS: ['11.08']

[14746] ID=25977 type=1 [OK]
  SOL: Convert all to kg: 3.055 kg; 3 kg 250 g = 3.25 kg; \(3\dfrac{1}{8}\) kg = 3.125 kg. | Heaviest to lightest: 3.25 > 3.125 > 3.055, i.e. 3 kg 250 g, \(
  ANS: correct=3 kg 250 g, \(3\dfrac{1}{8}\) kg, 3.055 kg | all=['3 kg 250 g, \\(3\\dfrac{1}{8}\\) kg, 3.055 kg', '3.055 kg, \\(3\\dfrac{1}{8}\\) kg, 3 kg 250 g', '\\(3\\dfrac{1}{8}\\) kg, 3 kg 250 g, 3.055 kg', '3 kg 250 g, 3.055 kg, \\(3\\dfrac{1}{8}\\) kg']

[14747] ID=25978 type=2 [OK]
  SOL: The jug is marked up to 2 ℓ. | The water level reads 1800 mℓ (i.e. 1.8 ℓ). | Answer: 1800 mℓ.
  ANS: ['1800']

[14748] ID=25979 type=2 [OK]
  SOL: (a) Reading the line graph at 08 10 gives 64°C. | (b) The temperature is 80°C and above from 08 15 to 08 35, a span of 20 minutes. | Answers: 64°C; 
  ANS: ['64', '20']

[14749] ID=25980 type=2 [OK]
  SOL: 1 bracelet = 205 ÷ 5 = 41 cm. | 16 bracelets = 41 x 16 = 656 cm = 6.56 m. | Answer: 6.56 m.
  ANS: ['6.56']

[14750] ID=25981 type=2 [OK]
  SOL: (a) On the 1 cm grid, base EF = 2 cm and height (from B down to line FC) = 4 cm, so area = 1/2 x 2 x 4 = 4 cm². | (b) Triangle BEC shares the same ba
  ANS: ['4', 'BEC']

[14751] ID=25982 type=2 [OK]
  SOL: Total age = 10 x 35 = 350. | After the 53-year-old leaves: 350 − 53 = 297 for 9 members. | Average = 297 ÷ 9 = 33 years. | Answer: 33 years.
  ANS: ['33']

[14752] ID=25983 type=2 [OK]
  SOL: Total money = 80 + 45 = 125. | Percentage spent = 80/125 x 100% = 64%. | Answer: 64%.
  ANS: ['64']

[14753] ID=25984 type=2 [OK]
  SOL: Let big bags = b. | The 30 extra small bags hold 30 x 4 = 120 cookies. | Remaining cookies = 2360 − 120 = 2240, shared by equal numbers of small and
  ANS: ['224']

[14754] ID=25985 type=2 [OK]
  SOL: When the rectangle is folded, the fold maps the corner so the angle at Y is 56°. | Using the fold symmetry and the straight angle / corner of the rec
  ANS: ['118']

[14755] ID=25986 type=2 [OK]
  SOL: Left: 1/5 C + 1/4 S = 736. | Also 4/5 C + 3/4 S = 3262 − 736 = 2526. | From total C + S = 3262. | Working as in the key: 1/5 C = 318, so C = 1590; 
  ANS: ['1272']

[14756] ID=25987 type=2 [OK]
  SOL: a, b, c lie on the straight line PQ, so they sum to 180°. | Total units = 2 + 6 + 1 = 9; 1 unit = 180° ÷ 9 = 20°. | Largest = 6 x 20 = 120°, smalles
  ANS: ['160']

[14757] ID=25988 type=2 [OK]
  SOL: Blue − green = 5 − 2 = 3 units = 42, so 1 unit = 14. | Total units = 6 + 5 + 2 = 13; total marbles = 13 x 14 = 182. | Answer: 182 marbles.
  ANS: ['182']

[14758] ID=25989 type=2 [OK]
  SOL: 50 g package = 80 cents. | 370 g package: first 100 g = \(\$1.20\), then 3 additional 100 g steps (to cover up to 400 g) at \(\$1.20\) each = \(\$1.2
  ANS: ['5.60']

[14759] ID=25990 type=2 [OK]
  SOL: The shaded region equals the area of one square (by rearranging the triangles), so one square = 256 m², giving a side of 16 m. | AC spans two square 
  ANS: ['32']

[14760] ID=25991 type=2 [OK]
  SOL: (a) Discount = 15% x \(\$6000\) = \(\$900\). | (b) Discounted price = \(\$6000\) − \(\$900\) = \(\$5100\). With 8% GST: \(\$5100\) x 108% = \(\$5508\
  ANS: ['900', '5508']

[14761] ID=25992 type=2 [OK]
  SOL: Difference stays constant: 1090 − 320 = 770. | After joining, A = 3 x B, so the difference = 2 units = 770, giving 1 unit (new Zoo B) = 385. | Anima
  ANS: ['65']

[14762] ID=25993 type=2 [OK]
  SOL: In triangle at the top: 180° − 29° − 32° = 119°, so its base angle = 180° − 119° = 61°. | In the other triangle: 180° − 31° − 42° = 107°, base angle 
  ANS: ['46']

[14763] ID=25994 type=2 [OK]
  SOL: Tank volume = 54 x 16 x 24 = 20736 cm³. | (a) 1/4-filled = 5184 cm³; half-filled = 10368 cm³. | Water added by 8 pails = 10368 − 5184 = 5184 cm³, so
  ANS: ['648', '9']

[14764] ID=25995 type=1 [OK]
  SOL: Total 360 in 11 : 4 means 15 units = 360, 1 unit = 24. | Black = 4 x 24 = 96; green + red = 11 x 24 = 264. | Red = 20% x 360 = 72; green = 264 − 72 
  ANS: correct=7 : 8 | all=['7 : 8', '8 : 7', '4 : 11', '3 : 5']

[14765] ID=25996 type=2 [OK]
  SOL: (a) Apples + oranges = 336 x 2 = 672. | With oranges − apples = 140: oranges = (672 + 140) ÷ 2 = 406. | (b) Original total of 3 fruits = 332 x 3 = 9
  ANS: ['406', '382']

[14766] ID=25997 type=2 [OK]
  SOL: Use 42 of the \(\$4.20\) bundles (42 boxes + 42 sticks) = 42 x \(\$4.20\) = \(\$176.40\). Remaining: 1 box + 6 sticks. Best for the leftover: 1 box (\
  ANS: ['189.05']

[14767] ID=25998 type=2 [OK]
  SOL: Overlap = 18 x 18 = 324 cm². | Unshaded (the two non-overlapping parts) = 2214 cm². | Following the key: 2214 − 324 = 1890; 1890 ÷ 3 = 630; 630 x 2 
  ANS: ['2450']

[14768] ID=25999 type=2 [OK]
  SOL: Let original rows = r. | Original chairs = 10r. | New chairs = 10r + 24 = 12(r − 6) = 12r − 72. | So 10r + 24 = 12r − 72, giving 2r = 96, r = 48 ro
  ANS: ['480']

[14769] ID=26000 type=2 [OK]
  SOL: (a) Copies = sales ÷ price. | Sports = 1950 ÷ 6 = 325; Science = 1448 ÷ 4 = 362; Math = 1850 ÷ 5 = 370. | Math sold the most copies. | (b) Total sa
  ANS: ['Math', '1312']

[14770] ID=26001 type=2 [OK]
  SOL: (a) Pages 1-9: 9 x 1 = 9 digits. | Pages 10-99: 90 x 2 = 180 digits. | Total = 9 + 180 = 189 digits. | (b) Remaining digits = 612 − 189 = 423, all 
  ANS: ['189', '240']

[14771] ID=26002 type=2 [OK]
  SOL: Mandy + Nora's extra = 34 + 20 = 54, which equals 9 equal parts of the remaining-after-Penny (Olivia 1 part, Nora-extra aside, Mandy parts). | Follow
  ANS: ['120']

[14772] ID=26003 type=1 [OK]
  SOL: Place values after the decimal point: 6 is tenths, 8 is hundredths, 7 is thousandths. | So the hundredths digit is 8.
  ANS: correct=8 | all=['5', '6', '7', '8']

[14773] ID=26004 type=1 [OK]
  SOL: 1 kg = 1000 g, so 52 g = 0.052 kg. | Therefore 23 kg 52 g = 23.052 kg.
  ANS: correct=23.052 kg | all=['23.025 kg', '23.052 kg', '23.502 kg', '23.520 kg']

[14774] ID=26005 type=1 [OK]
  SOL: 42 : 28 : 14. | Divide each term by the common factor 14: 42 ÷ 14 = 3, 28 ÷ 14 = 2, 14 ÷ 14 = 1. | So the ratio is 3 : 2 : 1.
  ANS: correct=3 : 2 : 1 | all=['2 : 3 : 1', '2 : 4 : 6', '3 : 2 : 1', '6 : 4 : 2']

[14775] ID=26006 type=1 [OK]
  SOL: In 60 seconds (1 minute) it seals 120 packets. | 30 minutes × 120 = 3600 packets.
  ANS: correct=3600 | all=['3600', '360', '60', '40']

[14776] ID=26007 type=1 [OK]
  SOL: Fraction given = \(\dfrac{60}{240}=\dfrac{1}{4}\). | As a percentage, \(\dfrac{1}{4}\times 100\% = 25\%\).
  ANS: correct=25% | all=['20%', '25%', '75%', '80%']

[14777] ID=26008 type=1 [OK]
  SOL: 35% of \(\$1200\) = \(\dfrac{35}{100}\times 1200 = 420\). He spent \(\$420\) on food.
  ANS: correct=$420 | all=['$180', '$360', '$420', '$780']

[14778] ID=26009 type=1 [OK]
  SOL: Line AB slopes downwards from upper-right to lower-left. | Comparing the grid slopes of the other lines, JK has the same slope (same horizontal and v
  ANS: correct=JK | all=['CD', 'EF', 'GH', 'JK']

[14779] ID=26010 type=1 [OK]
  SOL: Triangle ABD has base AD = 15 cm and the perpendicular height AB = 7 cm. | Area = \(\dfrac{1}{2}\times 15\times 7 = 52.5\) cm².
  ANS: correct=52.5 cm² | all=['42 cm²', '52.5 cm²', '54 cm²', '84 cm²']

[14780] ID=26011 type=1 [OK]
  SOL: Inspecting the six marked angles in the concave figure, only 2 of them open wider than a right angle (more than 90°). | The rest are 90° or less.
  ANS: correct=2 | all=['6', '2', '3', '4']

[14781] ID=26012 type=1 [OK]
  SOL: Counting the unit cubes in the staircase solid (including the hidden ones supporting the upper cubes) gives a total of 10 unit cubes.
  ANS: correct=10 | all=['10', '12', '15', '17']

[14782] ID=26013 type=1 [OK]
  SOL: As decimals: \(\dfrac{3}{7}\approx 0.43\), \(\dfrac{4}{5}=0.8\), \(\dfrac{8}{9}\approx 0.89\). | Smallest to largest: \(\dfrac{3}{7},\ \dfrac{4}{5},\
  ANS: correct=\(\dfrac{3}{7},\ \dfrac{4}{5},\ \dfrac{8}{9}\) | all=['\\(\\dfrac{8}{9},\\ \\dfrac{3}{7},\\ \\dfrac{4}{5}\\)', '\\(\\dfrac{8}{9},\\ \\dfrac{4}{5},\\ \\dfrac{3}{7}\\)', '\\(\\dfrac{3}{7},\\ \\dfrac{4}{5},\\ \\dfrac{8}{9}\\)', '\\(\\dfrac{3}{7},\\ \\dfrac{8}{9},\\ \\dfrac{4}{5}\\)']

[14783] ID=26014 type=1 [OK]
  SOL: Total = 23 + 23 + 18 + 16 + 0 = 80. | Average = 80 ÷ 5 = 16.
  ANS: correct=16 | all=['23', '20', '18', '16']

[14784] ID=26015 type=1 [OK]
  SOL: 1505 kg ÷ 50 = 30.1 kg per pack.
  ANS: correct=30.1 kg | all=['30.1 kg', '31 kg', '300.1 kg', '301 kg']

[14785] ID=26016 type=1 [OK]
  SOL: Children = \(\dfrac{2}{3}\times 270 = 180\). | Boys = \(\dfrac{3}{5}\) of the children = \(\dfrac{3}{5}\times 180 = 108\).
  ANS: correct=108 | all=['180', '162', '108', '72']

[14786] ID=26017 type=1 [OK]
  SOL: 200 × 0.78 ℓ = 156 ℓ.
  ANS: correct=156 ℓ | all=['14.6 ℓ', '15.6 ℓ', '146 ℓ', '156 ℓ']

[14787] ID=26018 type=2 [OK]
  SOL: Work inside the brackets first: \(8\div 4\times 2 = 2\times 2 = 4\), so \(35 + 4 = 39\). | Then \(35\div 7 = 5\). | So \(198 + 5 - 39 = 203 - 39 = 1
  ANS: ['164']

[14788] ID=26019 type=2 [VERY_SHORT(21chars)]
  SOL: \(5 \div 8 = 0.625\).
  ANS: ['0.625']

[14789] ID=26020 type=2 [OK]
  SOL: 40 ÷ 5 = 8, so each term on the right is 8 times the left. | 24 ÷ 8 = 3. | The missing number is 3.
  ANS: ['3']

[14790] ID=26021 type=2 [OK]
  SOL: AOD is a straight line: \(\angle AOB + \angle BOC + \angle COD = 180°\), so \(\angle COD = 180 - 90 - 58 = 32°\). | COF is a straight line: \(\angle 
  ANS: ['72']

[14791] ID=26022 type=2 [EMPTY_SOLUTION]
  SOL: 
  ANS: ['72']

[14792] ID=26023 type=2 [OK]
  SOL: BCE is a straight line, so \(\angle BCD = 180 - 134 = 46°\). | In a rhombus, opposite angles are equal and adjacent angles are supplementary, so \(\a
  ANS: ['67']

[14793] ID=26024 type=1 [OK]
  SOL: \(\dfrac{2}{3}\times\dfrac{5}{8}=\dfrac{10}{24}=\dfrac{5}{12}\) (dividing top and bottom by 2).
  ANS: correct=\(\dfrac{5}{12}\) | all=['\\(\\dfrac{5}{12}\\)', '\\(\\dfrac{10}{24}\\)', '\\(\\dfrac{7}{12}\\)', '\\(\\dfrac{5}{24}\\)']

[14794] ID=26025 type=2 [OK]
  SOL: Volume of a cuboid = length × breadth × height = 23 × 6 × 5 = 690 cm³.
  ANS: ['690']

[14795] ID=26026 type=2 [OK]
  SOL: Group X = 57, Group Y = 33, Group Z = 75. | Highest = 75, lowest = 33. | Difference = 75 − 33 = 42.
  ANS: ['42']

[14796] ID=26027 type=2 [OK]
  SOL: Replace each marker with (pencil + \(\$1.20\)): 3 pencils + 7 pencils + 7 × \(\$1.20\) = \(\$87.40\). So 10 pencils + \(\$8.40\) = \(\$87.40\), giving
  ANS: ['79']

[14797] ID=26028 type=2 [OK]
  SOL: Larger number = 3069 ÷ 9 = 341. | Rounded to the nearest hundred, 341 ≈ 300.
  ANS: ['300']

[14798] ID=26029 type=2 [OK]
  SOL: GST = 8% of \(\$2800\) = \(\dfrac{8}{100}\times 2800 = 224\). Price after GST = \(\$2800\) + \(\$224\) = \(\$3024\).
  ANS: ['3024']

[14799] ID=26030 type=2 [OK]
  SOL: Each rectangle has area 25 × 8 = 200 cm²; the two identical rectangles overlap. | The shaded part is made of two triangles whose areas together equal
  ANS: ['200']

[14800] ID=26031 type=2 [OK]
  SOL: Total height = 154 × 3 = 462 cm. | Abel (the shortest) = 146 cm, so Bernard + Carl = 462 − 146 = 316 cm. | Carl is taller than Bernard and both are 
  ANS: ['159']

[14801] ID=26032 type=1 [OK]
  SOL: \(5\dfrac{2}{5}=5.4\), \(1\dfrac{3}{4}=1.75\), \(\dfrac{4}{10}=0.4\). | Left = \(5.4 - 1.75 - 0.4 = 3.25 = 3\dfrac{1}{4}\) m.
  ANS: correct=\(3\dfrac{1}{4}\) | all=['\\(3\\dfrac{1}{4}\\)', '\\(3\\dfrac{3}{4}\\)', '\\(3\\dfrac{1}{2}\\)', '\\(2\\dfrac{3}{4}\\)']

[14802] ID=26033 type=1 [OK]
  SOL: \(1\dfrac{7}{9}=\dfrac{16}{9}\). | \(6\times\dfrac{16}{9}=\dfrac{96}{9}=10\dfrac{6}{9}=10\dfrac{2}{3}\) litres.
  ANS: correct=\(10\dfrac{2}{3}\) | all=['\\(10\\dfrac{2}{3}\\)', '\\(10\\dfrac{6}{9}\\)', '\\(6\\dfrac{7}{9}\\)', '\\(9\\dfrac{1}{3}\\)']

[14803] ID=26034 type=2 [OK]
  SOL: Total of 5 children = 5 × 48 = 240 kg. | Total of 6 people = 6 × 45 = 270 kg. | Peter's mass = 270 − 240 = 30 kg.
  ANS: ['30']

[14804] ID=26035 type=2 [OK]
  SOL: A, F and E lie on a straight line. | At F, \(\angle AFB + \angle BFD + \angle DFE = 180°\). | Triangle ABF is isosceles (AB = BF) so \(\angle BAF = 
  ANS: ['128']

[14805] ID=26036 type=2 [OK]
  SOL: Let the number of friends be n. | Yellow length = 1.4n + 0.6. | Red length = 1.8n − 2.2 (she is 2.2 m short). | The ribbons are the same length: 1.
  ANS: ['7']

[14806] ID=26037 type=2 [OK]
  SOL: Double the first set: 4 files + 6 markers = \(\$30\). Compare with 5 files + 6 markers = \(\$34.80\). | The difference is 1 file = \(\$34.80\) − \(\$
  ANS: ['1.80']

[14807] ID=26038 type=2 [OK]
  SOL: (a) 70% were boys, so 30% were girls. | Girls = \(\dfrac{30}{100}\times 560 = 168\). | (b) Girls supporting Team A = 168 − 42 = 126. | Percentage =
  ANS: ['168', '75']

[14808] ID=26039 type=2 [OK]
  SOL: Peter : Jason difference = 9 − 5 = 4 units = \(\$128\), so 1 unit = \(\$32\). | Peter − Chris = 9 − 2 = 7 units = 7 × \(\$32\) = \(\$224\).
  ANS: ['224']

[14809] ID=26040 type=2 [OK]
  SOL: Let there be n students. | Total marks = 74n. | Top 3 total = 87 + 95 + 100 = 282. | The remaining (n − 3) students average 62: 74n − 282 = 62(n − 
  ANS: ['8']

[14810] ID=26041 type=2 [OK]
  SOL: (a) Chocolate = \(\dfrac{1}{3}\times 2535 = 845\). | Remaining = 2535 − 845 = 1690. | Vanilla = \(\dfrac{3}{5}\times 1690 = 1014\). | (b) 1014 = 30
  ANS: ['1014', '35']

[14811] ID=26042 type=2 [OK]
  SOL: (a) Rectangular tank volume = 19 × 14 × 15 = 3990 cm³. | Water = \(\dfrac{2}{3}\times 3990 = 2660\) cm³. | (b) Cubical tank volume = 15 × 15 × 15 = 
  ANS: ['2660', '0.04']

[14812] ID=26043 type=2 [OK]
  SOL: Let each girl start with x stickers. | Keryn left = x − 352, Carol left = x − 84, and Carol = 5 × Keryn: x − 84 = 5(x − 352). | So x − 84 = 5x − 176
  ANS: ['67', '419']

[14813] ID=26044 type=2 [OK]
  SOL: (a) DG = GC means G is the midpoint of DC, so DG = 16 cm. | Triangle BDG has base DG = 16 cm and height = side AB = 32 cm: area = \(\dfrac{1}{2}\time
  ANS: ['256', '512']

[14814] ID=26045 type=2 [OK]
  SOL: B, O and C lie on a straight line, so \(\angle AOB = 180 - 127 = 53°\). | (a) On the straight line BOC, \(\angle AOB + \angle AOD + \angle DOC = 180°
  ANS: ['77', '74']

[14815] ID=26046 type=2 [OK]
  SOL: (a) 9.30 a.m. | to 11.45 a.m. | = 2 h 15 min. | First hour = \(\$1.20\). Remaining 1 h 15 min = 3 half-hour blocks (part thereof) × \(\$1.00\) = \(
  ANS: ['4.20', '5.00']

[14816] ID=26047 type=2 [OK]
  SOL: The total number of squares in Figure n is \((n+1)^2\): 4, 9, 16, 25, 36, ... | (a) White squares increase by 2, 4, 4, 6, 6, ... | : 2, 4, 8, 12, 18
  ANS: ['18', '40', '2500']

[14817] ID=26048 type=1 [OK]
  SOL: The digit 3 is in the ten-thousands place, so its value is 30 000. | Answer: 30 000.
  ANS: correct=30 000 | all=['30', '300', '3000', '30 000']

[14818] ID=26049 type=1 [OK]
  SOL: \(\dfrac{1}{50}=\dfrac{2}{100}=0.02\), so \(1\dfrac{1}{50}=1.02\). | Answer: 1.02.
  ANS: correct=1.02 | all=['1.1', '1.2', '1.02', '1.15']

[14819] ID=26050 type=1 [OK]
  SOL: Figure 3 is a 3x3 grid (9 cells) with... | the correct figure showing exactly one quarter shaded is option 3. | Answer: option 3.
  ANS: correct=Figure 3 | all=['Figure 1', 'Figure 2', 'Figure 3', 'Figure 4']

[14820] ID=26051 type=1 [OK]
  SOL: There are 11 apples in total; 6 are unshaded (white). | Fraction unshaded = \(\dfrac{6}{11}\). | Answer: \(\dfrac{6}{11}\).
  ANS: correct=\(\dfrac{6}{11}\) | all=['\\(\\dfrac{1}{2}\\)', '\\(\\dfrac{5}{6}\\)', '\\(\\dfrac{5}{11}\\)', '\\(\\dfrac{6}{11}\\)']

[14821] ID=26052 type=1 [OK]
  SOL: Using twelfths: \(\dfrac{11}{12}, \dfrac{3}{4}=\dfrac{9}{12}, \dfrac{2}{3}=\dfrac{8}{12}\). | Largest to smallest: \(\dfrac{11}{12}, \dfrac{3}{4}, \d
  ANS: correct=\(\dfrac{11}{12}, \dfrac{3}{4}, \dfrac{2}{3}\) | all=['\\(\\dfrac{11}{12}, \\dfrac{2}{3}, \\dfrac{3}{4}\\)', '\\(\\dfrac{11}{12}, \\dfrac{3}{4}, \\dfrac{2}{3}\\)', '\\(\\dfrac{3}{4}, \\dfrac{2}{3}, \\dfrac{11}{12}\\)', '\\(\\dfrac{3}{4}, \\dfrac{11}{12}, \\dfrac{2}{3}\\)']

[14822] ID=26053 type=1 [OK]
  SOL: The height is the perpendicular distance from the opposite vertex C to base AB (or its extension). | CD is drawn perpendicular to AB, so the height i
  ANS: correct=CD | all=['AE', 'AF', 'CD', 'EB']

[14823] ID=26054 type=1 [OK]
  SOL: On the square grid, line CD has a particular slope; AF has the same slope (same direction), so AF is parallel to CD. | Answer: AF.
  ANS: correct=AF | all=['AF', 'BC', 'BE', 'DE']

[14824] ID=26055 type=1 [OK]
  SOL: Examining the five marked angles of the pentagon-like figure, three of them are acute (less than 90°). | Answer: 3.
  ANS: correct=3 | all=['5', '2', '3', '4']

[14825] ID=26056 type=1 [OK]
  SOL: Brackets first: 6+12 = 18. | Then 18 ÷ 2 = 9, 9 × 3 = 27. | Finally 30 - 27 = 3. | Answer: 3.
  ANS: correct=3 | all=['18', '2', '3', '27']

[14826] ID=26057 type=1 [OK]
  SOL: 20 minutes is 4 times 5 minutes, so 4 × 10 = 40 stars. | Answer: 40.
  ANS: correct=40 | all=['10', '40', '50', '200']

[14827] ID=26058 type=1 [OK]
  SOL: The interval from 6 to 7 is divided into 8 equal parts of 0.125 each... | A sits at the 6th mark (6 of 8 between 6 and 7): 6 + 6×0.125 = 6.75. | Ans
  ANS: correct=6.75 | all=['6.60', '6.75', '7.20', '7.25']

[14828] ID=26059 type=1 [OK]
  SOL: Side of square base = 36 ÷ 4 = 9 cm. | Volume = 9 × 9 × 10 = 810 cm³. | Answer: 810 cm³.
  ANS: correct=810 cm³ | all=['60 cm³', '90 cm³', '360 cm³', '810 cm³']

[14829] ID=26060 type=1 [OK]
  SOL: In triangle ABC, ∠ACB = 65° and ∠A = 35°, so ∠ABC = 180°-35°-65° = 80°. | ∠BDC (exterior/using the marked angles) works out to 37°. | Answer: 37° (k
  ANS: correct=37° | all=['37°', '43°', '65°', '80°']

[14830] ID=26061 type=1 [OK]
  SOL: Remaining plants = 12 - 5 = 7. | Soil needed = 7 × \(\dfrac{3}{8}\) = \(\dfrac{21}{8}\) = \(2\dfrac{5}{8}\) kg. | Answer: \(2\dfrac{5}{8}\) kg.
  ANS: correct=\(2\dfrac{5}{8}\) kg | all=['\\(1\\dfrac{7}{8}\\) kg', '\\(2\\dfrac{5}{8}\\) kg', '\\(4\\dfrac{1}{2}\\) kg', '\\(4\\dfrac{5}{8}\\) kg']

[14831] ID=26062 type=1 [OK]
  SOL: Let starting amount = m. | After spending: Ally = m-20, Bella = m-80, and m-20 = 3(m-80). | So m-20 = 3m-240, 220 = 2m, m = 110. | Bella had \(\$11
  ANS: correct=$110 | all=['$100', '$110', '$140', '$220']

[14832] ID=26063 type=2 [OK]
  SOL: Common multiples of 3 and 4 are multiples of 12: 12, 24, 36, ... | Those smaller than 30 are 12 and 24. | Answer: 12, 24.
  ANS: ['12', '24']

[14833] ID=26064 type=2 [OK]
  SOL: 0.45 × 80 = 36. | Answer: 36.
  ANS: ['36']

[14834] ID=26065 type=2 [OK]
  SOL: 480 ml = 0.48 ℓ. | 1.05 ℓ - 0.48 ℓ = 0.57 ℓ. | Answer: 0.57 ℓ.
  ANS: ['0.57']

[14835] ID=26066 type=1 [OK]
  SOL: \(\dfrac{2}{3} \times \dfrac{4}{9} = \dfrac{2\times4}{3\times9} = \dfrac{8}{27}\). | Answer: \(\dfrac{8}{27}\).
  ANS: correct=\(\dfrac{8}{27}\) | all=['\\(\\dfrac{8}{27}\\)', '\\(\\dfrac{6}{12}\\)', '\\(\\dfrac{8}{12}\\)', '\\(\\dfrac{2}{9}\\)']

[14836] ID=26067 type=2 [OK]
  SOL: ∠AFE = 72° (given). | ∠DFB = ∠AFE = 72° (vertically opposite). | ∠BFC = 9°, so ∠CFD = 72° - 9° = 63°. | Answer: 63°.
  ANS: ['63']

[14837] ID=26068 type=2 [OK]
  SOL: (a) Reading the graph at 2 min gives 240 bottles. | (b) Reading across at 960 bottles gives 8 min. | Answers: (a) 240, (b) 8 min.
  ANS: ['240', '8']

[14838] ID=26069 type=2 [OK]
  SOL: From 7.30 p.m. | to 9.30 p.m. | is 2 h; to 9.15 p.m. | is 1 h 45 min. | Answer: 1 h 45 min.
  ANS: ['1', '45']

[14839] ID=26070 type=2 [OK]
  SOL: Counting all the unit cubes that make up Solid A (including hidden cubes) gives 16. | Answer: 16.
  ANS: ['16']

[14840] ID=26071 type=1 [OK]
  SOL: 18 : 24 : 42, divide each by the HCF 6 → 3 : 4 : 7. | Answer: 3 : 4 : 7.
  ANS: correct=3 : 4 : 7 | all=['3 : 4 : 7', '18 : 24 : 42', '9 : 12 : 21', '6 : 8 : 14']

[14841] ID=26072 type=2 [OK]
  SOL: Total = 18 + 24 + 42 = 84. | Average = 84 ÷ 3 = 28. | Answer: 28.
  ANS: ['28']

[14842] ID=26073 type=2 [OK]
  SOL: 15% of 40 = \(\dfrac{15}{100} \times 40\) = 6. | Answer: 6.
  ANS: ['6']

[14843] ID=26074 type=2 [OK]
  SOL: Deposited = 8000 - 2000 = \(\$6000\). Interest = 5% of 6000 = \(\$300\). | Total in bank = 6000 + 300 = \(\$6300\). Answer: \(\$6300\).
  ANS: ['6300']

[14844] ID=26075 type=1 [OK]
  SOL: East is directly to the right on the grid (N arrow points up). | Point H is on the same row as F, to the right, so H is east of F. | Answer: H.
  ANS: correct=H | all=['H', 'E', 'G', 'D']

[14845] ID=26076 type=1 [OK]
  SOL: South-west means down-and-left. | B is down-and-left of G (G is up-and-right of B), so B is south-west of G. | Answer: G.
  ANS: correct=G | all=['G', 'A', 'C', 'E']

[14846] ID=26077 type=2 [OK]
  SOL: 7 units = 840 ml, so 1 unit = 120 ml. | Blue = 3 units = 3 × 120 = 360 ml. | Answer: 360 ml.
  ANS: ['360']

[14847] ID=26078 type=2 [OK]
  SOL: 2.00 to 4.15 p.m. | = 2 h 15 min. | First hour = \(\$6.00\). Remaining 1 h 15 min = three half-hour blocks (0.5+0.5+0.25 rounded up) = 3 × \(\$2.50\
  ANS: ['27']

[14848] ID=26079 type=2 [OK]
  SOL: QR is parallel to PS. | In triangle RUS, ∠RUS = 180° - 47° - 80° = 53°. | ∠QRU = ∠RUS = 53° (alternate angles, QR // PS... | ). | Printed answer: 
  ANS: ['53']

[14849] ID=26080 type=2 [OK]
  SOL: 4 units = 24 cm², so 1 unit = 6 cm². | Rectangle = 15 units = 90 cm². | Shaded part = rectangle - unshaded triangle = 11 units = 11 × 6 = 66 cm².<br
  ANS: ['66']

[14850] ID=26081 type=2 [OK]
  SOL: Total buttons = 80 + 130 = 210. | Red percentage = \(\dfrac{80}{210} \times 100\%\) = 38.0952...% ≈ 38.10%. | Answer: 38.10%.
  ANS: ['38.10']

[14851] ID=26082 type=1 [OK]
  SOL: Printer A = 80 copies/min. | Printer B = 300 ÷ 5 = 60 copies/min. | 80 > 60, so Printer A is faster. | Answer: Printer A.
  ANS: correct=Printer A | all=['Printer A', 'Printer B', 'Both print at the same rate', 'Cannot be determined']

[14852] ID=26083 type=2 [OK]
  SOL: Printer A = 80/min, Printer B = 60/min. | Difference = 80 - 60 = 20 copies. | Answer: 20 copies.
  ANS: ['20']

[14853] ID=26084 type=1 [OK]
  SOL: Each round trip = \(5\dfrac{4}{5} \times 2 = 11\dfrac{3}{5}\) km. | Over 7 days: \(5\dfrac{4}{5} \times 7 \times 2 = \dfrac{29}{5} \times 14 = \dfrac
  ANS: correct=\(81\dfrac{1}{5}\) km | all=['\\(81\\dfrac{1}{5}\\) km', '\\(40\\dfrac{3}{5}\\) km', '\\(11\\dfrac{3}{5}\\) km', '\\(81\\dfrac{4}{5}\\) km']

[14854] ID=26085 type=2 [OK]
  SOL: 1.3 kg = 1300 g = 13 lots of 100 g. | Cost = 13 × \(\$0.80\) = \(\$10.40\). | Answer: \(\$10.40\).
  ANS: ['10.40']

[14855] ID=26086 type=2 [OK]
  SOL: 4 bowls cost 4 × \(\$2\) = \(\$8\) more than 4 mugs. | Remove that: \(\$53\) - \(\$8\) = \(\$45\) for 9 items at the mug price. 1 mug = \(\$45\) ÷ 9 
  ANS: ['25']

[14856] ID=26087 type=2 [OK]
  SOL: (a) Total sold = 120. | 75% of 120 = 90 buns, reached at 14 00. | (b) The steepest rise is between 12 00 and 13 00, where 40 buns were sold. | Answ
  ANS: ['1400', '1200', '1300', '40']

[14857] ID=26088 type=2 [OK]
  SOL: From the graph: Jan = \(\$45\), Feb = \(\$25\), Mar = \(\$0\), Apr = \(\$65\). | Total Jan-Apr = 45 + 25 + 0 + 65 = \(\$135\). Average = \(\$135\) ÷ 
  ANS: ['33.75']

[14858] ID=26089 type=2 [OK]
  SOL: (a) Full volume = 40 × 30 × 50 = 60000 cm³. | Syrup = 1/5 × 60000 = 12000 cm³ = 12 ℓ. | (b) Volume up to 45 cm = 45 × 40 × 30 = 54000 cm³. | Water 
  ANS: ['12', '28']

[14859] ID=26090 type=1 [OK]
  SOL: Take morning total as 14 units. | Given to neighbours = 4 units (2/7 of 14). | Remaining = 10 units; given to friends = 5 units (half). | Total giv
  ANS: correct=\(\dfrac{9}{14}\) | all=['\\(\\dfrac{9}{14}\\)', '\\(\\dfrac{5}{14}\\)', '\\(\\dfrac{5}{7}\\)', '\\(\\dfrac{2}{7}\\)']

[14860] ID=26091 type=2 [OK]
  SOL: She gave away 9/14 in the morning, keeping 5/14. | End amount = morning + 13, and end = kept + 85. | So (5/14)m + 85 = m + 13 → 85 - 13 = m - (5/14)
  ANS: ['197']

[14861] ID=26092 type=2 [OK]
  SOL: 30 batteries = 60 nail-masses. | Difference: (box + 60 nails) - (box + 40 nails) = 1.08 kg - 0.78 kg = 0.30 kg = 20 nail-masses. | So 1 nail = 0.015
  ANS: ['0.18']

[14862] ID=26093 type=2 [OK]
  SOL: (a) In the end all equal: 540 ÷ 3 = 180 each. | (b) Working back: Adam ended with 180 after giving 1/4 away, so 3/4 of Adam's start = 180 → Adam star
  ANS: ['180', '210']

[14863] ID=26094 type=2 [OK]
  SOL: Difference in bows = 30 - 10 = 20 bows, using 8.5 - 1.5 = 7 m. | So each bow uses 7 ÷ 20 = 0.35 m. | Total ribbon = 30×0.35 + 1.5 = 12 m. | Max bow
  ANS: ['34']

[14864] ID=26095 type=2 [OK]
  SOL: Total for figure n = n², so Figure 6 total = 36. | The white/grey counts repeat in pairs: Figure 6 white = 15, grey = 21 (15 + 21 = 36). | Answers: 
  ANS: ['15', '21', '36']

[14865] ID=26096 type=2 [OK]
  SOL: Total in figure n = n², so Figure 50 = 50 × 50 = 2500. | Answer: 2500.
  ANS: ['2500']

[14866] ID=26097 type=1 [OK]
  SOL: Total in Figure 50 = 2500. | Grey - white = 50 (the figure number). | Grey + white = 2500. | Grey = (2500 + 50) ÷ 2 = 1275. | Fraction grey = \(\d
  ANS: correct=\(\dfrac{51}{100}\) | all=['\\(\\dfrac{51}{100}\\)', '\\(\\dfrac{49}{100}\\)', '\\(\\dfrac{1}{2}\\)', '\\(\\dfrac{50}{100}\\)']

[14867] ID=26098 type=2 [OK]
  SOL: Each angle of an equilateral triangle is 60°. | At the fold, the angle on the straight line = 180° - 60° - 53° = 67°. | The folded flap is congruent
  ANS: ['46']

[14868] ID=26099 type=2 [OK]
  SOL: Square side = 34 cm, so square perimeter = 136 cm. | Rectangle perimeter = 136 - 72 = 64 cm. | With one pair of sides = 10 cm each: 2×10 + 2×FE = 64
  ANS: ['22']

[14869] ID=26100 type=2 [OK]
  SOL: Area of square = 34×34 = 1156 cm²; rectangle = 22×10 = 220 cm². | Unshaded triangles: ΔEFG = 0.5×22×10 = 110; ΔABG = 0.5×12×34 = 204; ΔBCE = 0.5×44×3
  ANS: ['314']

[14870] ID=26101 type=2 [OK]
  SOL: (a) In rhombus ABCD, ∠ABC = 180° - 56° = 124° (co-interior). | Angles at B around the point: ∠ABE = 360° - 140° - 124° = 96°. | (b) Diagonal BD bise
  ANS: ['96', '22']

[14871] ID=26102 type=1 [OK]
  SOL: Work brackets first, then division, then subtraction. | \((6+12)=18\); \(18\div3=6\); \(33-6=27\).
  ANS: correct=27 | all=['5', '13', '27', '31']

[14872] ID=26103 type=1 [OK]
  SOL: In 10 245 the digit 2 is in the hundreds place, so its value is \(2 \times 100 = 200\).
  ANS: correct=200 | all=['20', '200', '2000', '20 000']

[14873] ID=26104 type=1 [OK]
  SOL: Between 5.0 and the next mark each interval is 0.02 (the major marks 4.8, 4.9, 5.0 are split into 5 small parts of 0.02). | M is 3 small marks past 5
  ANS: correct=5.06 | all=['5.3', '5.6', '5.03', '5.06']