Question 1Topic – E1.1From the list of numbers, write down(a) a cube number(b) a prime number▶️ Answer/ExplanationSolution(a) Ans: 64 (since 4³ = 64)(b) Ans: 61 or 67 (both are prime numbers)For part (a), we identify 64 as it’s 4 cubed.For part (b), we check divisibility: 61 and 67 have no divisors other than 1 and themselves. Question 2Topic – E1.15A train journey starts at 23:30 and finishes at 07:15 the next day.Find the time taken for this journey.▶️ Answer/ExplanationSolutionAns: 7 hours 45 minutesFrom 23:30 to 00:30 is 1 hourFrom 00:30 to 07:30 is 7 hours07:30 to 07:15 is -15 minutesTotal = 7 hours + 45 minutes = 7h 45m Question 3Topic – E2.2Simplify: 3p – t – p – 4t▶️ Answer/ExplanationSolutionAns: 2p – 5tCombine like terms:3p – p = 2p-t – 4t = -5tFinal simplified form: 2p – 5t Question 4Topic – E4.3The scale drawing shows the positions of town K and town L. The scale is 1 cm represents 10 km.(a) Find the actual distance between town K and town L.(b) Measure the bearing of town L from town K.▶️ Answer/ExplanationSolution(a) Ans: 85 kmFrom the diagram, distance is 8.5 cmActual distance = 8.5 × 10 = 85 km(b) Ans: 065°Using a protractor, measure the angle from north line clockwise to KL Question 5Topic – E9.3Each student in a class of 20 students records the number of coins in their pockets.The table shows the results:Number of coins0123456Frequency3178001(a) Find the median.(b) Calculate the mean.▶️ Answer/ExplanationSolution(a) Ans: 2Median position = (20+1)/2 = 10.5th valueCumulative frequencies: 3, 4, 11, 19 → median is 2(b) Ans: 2.25Total coins = (0×3)+(1×1)+(2×7)+(3×8)+(6×1) = 45Mean = 45 ÷ 20 = 2.25 Question 6Topic – E4.6The diagram shows three lines meeting at a point.Find the value of w.▶️ Answer/ExplanationSolutionAns: 190Angles on a straight line add up to 180°.The angle opposite 80° is also 80° (vertically opposite angles).w = 360° – (80° + 80° + 10°) = 190° (angles around a point). Question 7Topic – E2.5Solve the equation.$7 – h = 3 – 5h$▶️ Answer/ExplanationSolutionAns: -1$7 – h = 3 – 5h$$-h + 5h = 3 – 7$$4h = -4$$h = -1$ Question 8Topic – E2.1Sacha buys b books and m magazines.The cost of each book is \$12 and the cost of each magazine is \$5.Write an expression, in terms of b and m, for the total cost of the books and the magazines.▶️ Answer/ExplanationSolutionAns: $12b + 5m$Cost of books = number of books × price per book = $12 × b$Cost of magazines = number of magazines × price per magazine = $5 × m$Total cost = Cost of books + Cost of magazines = $12b + 5m$ Question 9Topic – E4.6Find the size of an interior angle of a regular 15-sided polygon.▶️ Answer/ExplanationSolutionAns: 156°Sum of interior angles = (n-2) × 180° = (15-2) × 180° = 2340°Each interior angle = Total sum ÷ number of sides = 2340° ÷ 15 = 156°Alternatively: 180° – (360° ÷ 15) = 180° – 24° = 156° Question 10Topic – E1.4Without using a calculator, work out $2\frac{1}{4} – 1\frac{11}{12}$.You must show all your working and give your answer as a fraction in its simplest form.▶️ Answer/ExplanationSolutionAns: $\frac{1}{3}$Convert to improper fractions: $\frac{9}{4} – \frac{23}{12}$Find common denominator (12): $\frac{27}{12} – \frac{23}{12}$Subtract numerators: $\frac{4}{12}$Simplify: $\frac{1}{3}$ Question 11Topic – E2.5Solve the simultaneous equations.$3p – 2q = 7$$p + 2q = 1$▶️ Answer/ExplanationSolutionAns: p = 2, q = -½Add both equations to eliminate q:$(3p – 2q) + (p + 2q) = 7 + 1$ → $4p = 8$ → $p = 2$Substitute p=2 into second equation:$2 + 2q = 1$ → $2q = -1$ → $q = -½$ Question 12Topic – E2.2$V = \sqrt[3]{\frac{x}{y}}$Rearrange the formula to write x in terms of V and y.▶️ Answer/ExplanationSolutionAns: x = V³yStart with V = ∛(x/y)Cube both sides: V³ = x/yMultiply both sides by y: x = V³y Question 13Topic – E2.7Find the nth term of each sequence:(a) 21, 13, 5, -3, -11, …(b) 2.5, 5, 10, 20, 40, …▶️ Answer/ExplanationSolutionAns:(a) 29 – 8nCommon difference is -8, first term is 21, so nth term = 21 – 8(n-1) = 29 – 8n(b) 5 × 2ⁿ⁻²Geometric sequence with ratio 2, first term 2.5, so nth term = 2.5 × 2ⁿ⁻¹ = 5 × 2ⁿ⁻² Question 14Topic – E4.7(a) A, B, C and D lie on the circle. TAS is a tangent to the circle at A.(i) Find the value of x.(ii) Find the value of y.(b) P, Q and R lie on the circle, center O. Find the value of w.▶️ Answer/ExplanationSolutionAns:(a)(i) x = 41° (angle between tangent and chord equals angle in alternate segment)(a)(ii) y = 37° (angles in same segment are equal)(b) w = 130° (angle at center is twice angle at circumference) Question 15Topic – E9.3The box-and-whisker diagram shows information about the heights of some plants.(a) Find the median height.(b) Find the interquartile range of the heights.▶️ Answer/ExplanationSolutionAns:(a) 28 cm (middle line of box plot)(b) 33 cm (upper quartile – lower quartile = 45 – 12 = 33) Question 16Topic – E6.5Calculate the shortest distance from C to AB.▶️ Answer/ExplanationSolutionAns: 9.40 cmShortest distance is perpendicular from C to AB.Using right triangle trigonometry: distance = BC × sin(33.14°).Calculation: 17.2 × sin(33.14°) ≈ 9.40 cm. Question 17Topic – E2.3Simplify:(a) $18x^{18} \div 3x^{3}$(b) $(125y^{75})^{\frac{2}{3}}$▶️ Answer/ExplanationSolutionAns:(a) $6x^{15}$Divide coefficients: $18 ÷ 3 = 6$Subtract exponents: $x^{18-3} = x^{15}$(b) $25y^{50}$Cube root first: $125^{\frac{1}{3}} = 5$ and $y^{75×\frac{1}{3}} = y^{25}$Then square: $5^2 = 25$ and $(y^{25})^2 = y^{50}$ Question 18Topic – E4.4Two mathematically similar solids have volumes 81 cm³ and 24 cm³.The height of the smaller solid is 4.8 cm.Calculate the height of the larger solid.▶️ Answer/ExplanationSolutionAns: 7.2 cmVolume ratio is 81:24 = 27:8.Linear scale factor is cube root of volume ratio: 3:2.Height of larger solid = 4.8 × (3/2) = 7.2 cm. Question 19Topic – E2.8y is inversely proportional to √(x+2).When x = 2, y = 3.Find y in terms of x.▶️ Answer/ExplanationSolutionAns: y = 6/√(x+2)General form: y = k/√(x+2).Substitute x=2, y=3: 3 = k/√4 → k = 6.Final equation: y = 6/√(x+2). Question 20Topic – E6.4Solve the equation tanx + 2 = 0 for 0° ≤ x ≤ 360°.▶️ Answer/ExplanationSolutionAns: 116.6°, 296.6°Rearrange: tanx = -2.First solution: x = tan⁻¹(-2) ≈ -63.4° → 180-63.4=116.6°.Second solution: 116.6° + 180° = 296.6°. Question 21Topic – E1.2The Venn diagram shows the number of elements in each region.(a) Use set notation to describe the shaded region.(b) Find n(A ∩ B ∩ C).▶️ Answer/ExplanationSolution(a) Ans: (B ∪ C) ∩ A’The shaded region represents elements in B or C but not in A.(b) Ans: 5This is the intersection of all three sets A, B, and C. Question 22Topic – E6.6The diagram shows a cuboid with a diagonal PB.Calculate the angle between the diagonal PB and the base ABCD.▶️ Answer/ExplanationSolutionAns: 17.1°First find diagonal BD using Pythagoras: √(12² + 5²) = 13 cm.The angle is between PB and BD. Use tanθ = opposite/adjacent = height/BD = 4/13.θ = tan⁻¹(4/13) ≈ 17.1°. Question 23Topic – E2.2Write x² + 8x – 7 in the form (x + a)² + b.▶️ Answer/ExplanationSolutionAns: (x + 4)² – 23Take coefficient of x: 8 → half is 4.Square it: 4² = 16.Rewrite: x² + 8x as (x + 4)² – 16.Final form: (x + 4)² – 16 – 7 = (x + 4)² – 23. Question 24Topic – E1.10A rectangle has an area of 150 m², correct to the nearest square metre.The length of the rectangle is 22 m, correct to the nearest metre.Calculate the upper bound of the width of the rectangle.▶️ Answer/ExplanationSolutionAns: 7 mUpper bound of area = 150.5 m².Lower bound of length = 21.5 m.Upper bound width = max area/min length = 150.5/21.5 ≈ 7 m. Question 25Topic – E2.3Simplify:$\frac{3x – 2 – 3xy + 2y}{1 – y^2}$▶️ Answer/ExplanationSolutionAns: $\frac{3x – 2}{1 + y}$Factor numerator: (3x – 2)(1 – y).Factor denominator: (1 – y)(1 + y).Cancel common factor (1 – y).Final simplified form: (3x – 2)/(1 + y). Question 26Topic – E7.4In the diagram, $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$.AK : KB = 2 : 1 OK = KC.Find $\overrightarrow{AC}$ in terms of a and b.Give your answer in its simplest form.▶️ Answer/ExplanationSolutionAns: $-\frac{1}{3}a + \frac{4}{3}b$Find K’s position: $\overrightarrow{OK} = \frac{2}{3}b + \frac{1}{3}a$.Since OK = KC, C’s position is $\overrightarrow{OC} = 2\overrightarrow{OK} = \frac{2}{3}a + \frac{4}{3}b$.$\overrightarrow{AC} = \overrightarrow{OC} – \overrightarrow{OA} = (\frac{2}{3}a + \frac{4}{3}b) – a = -\frac{1}{3}a + \frac{4}{3}b$.
