Question 1Topic – E1.15Tara goes on a journey by train. The train leaves at 06 48. The journey takes 12 hours and 35 minutes.Find the time when Tara arrives.▶️ Answer/ExplanationSolutionAns: 19:23 or 7:23 pmStart time: 06:48Add 12 hours: 18:48Add 35 minutes: 19:23 Question 2Topic – E1.3From the list, write down:(a) a cube number(b) a prime number.▶️ Answer/ExplanationSolutionAns:(a) 64 (since 4×4×4 = 64)(b) 61 (only number divisible by 1 and itself) Question 3Topic – E9.3The stem-and-leaf diagram shows the heights, in centimetres, of some plants:(a) Find the median height.(b) Work out the mean height.▶️ Answer/ExplanationSolutionAns:(a) 12.2 cm (7th value in ordered list of 13 values)(b) 12.1 cm (sum of all heights ÷ 13 = 157.3 ÷ 13) Question 4Topic – E1.13Shubhu invests \($750\) in a savings account for 5 years. The account pays simple interest at a rate of 1.8% per year.Calculate the total interest she earns during the 5 years.▶️ Answer/ExplanationSolutionAns: \$67.50Simple Interest = Principal × Rate × Time= \$750 × 0.018 × 5= \$67.50 Question 5Topic – E4.1The diagram shows triangle ABC. M is the midpoint of AC.Triangle ABC is rotated 180° about centre M. The image and the original triangle together form a quadrilateral ABCD.(a) Write down the mathematical name of the quadrilateral ABCD.(b) Find angle BAD.▶️ Answer/ExplanationSolutionAns:(a) Parallelogram (opposite sides equal and parallel)(b) 68° (180° – 112° from rotation properties) Question 6Topic – E8.1Rama asks a group of students how they travel to school. The table shows the probability of how a student, chosen at random, travels to school.(a) Complete the table.(b) There are 1800 students at the school. Find the expected number of students that walk to school.▶️ Answer/ExplanationSolutionAns:(a) Other = 1 – (0.4 + 0.32 + 0.17) = 0.11(b) Expected number = 1800 × 0.32 = 576 studentsFor part (a), probabilities must sum to 1. For part (b), multiply total students by walk probability. Question 7Topic – E1.4Without using a calculator, work out $1\frac{5}{6} \div \frac{11}{15}$.You must show all your working and give your answer as a mixed number in its simplest form.▶️ Answer/ExplanationSolutionAns: $2\frac{1}{2}$First convert mixed number: $1\frac{5}{6} = \frac{11}{6}$Division becomes multiplication by reciprocal: $\frac{11}{6} \times \frac{15}{11} = \frac{165}{66}$Simplify: $\frac{165}{66} = \frac{5}{2} = 2\frac{1}{2}$The 11’s cancel out, then divide numerator and denominator by 33. Question 8Topic – E1.1Find the highest common factor (HCF) of 48 and 80.▶️ Answer/ExplanationSolutionAns: 16Prime factors of 48: $2^4 \times 3$Prime factors of 80: $2^4 \times 5$Common factors: $2^4 = 16$Therefore, HCF is 16. Question 9Topic – E2.2$P = \frac{2w y^2}{3}$Find the positive value of $y$ when $P = 108$ and $w = 8$.▶️ Answer/ExplanationSolutionAns: 4.5 or $\frac{9}{2}$$108 = \frac{2 \times 8 \times y^2}{3}$$108 = \frac{16y^2}{3}$$324 = 16y^2$$y^2 = 20.25$ → $y = 4.5$ (positive value) Question 10Topic – E7.1$\overrightarrow{AB} = \begin{pmatrix} 7 \\ -3 \end{pmatrix}$(a) Find $3\overrightarrow{AB}$.(b) Find $|\overrightarrow{AB}|$.▶️ Answer/ExplanationSolutionAns:(a) $3\begin{pmatrix} 7 \\ -3 \end{pmatrix} = \begin{pmatrix} 21 \\ -9 \end{pmatrix}$(b) $|\overrightarrow{AB}| = \sqrt{7^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} ≈ 7.62$For (a), multiply each component by 3. For (b), use Pythagoras’ theorem. Question 11Topic – E5.4A bronze sphere has radius 3.6 cm. The density of bronze is 8.05 g/cm3.Find the mass of the sphere. Give your answer in kilograms, correct to the nearest gram.[The volume, V, of a sphere with radius r is V = $\frac{4}{3}\pi r^3$.][Density = mass ÷ volume.]▶️ Answer/ExplanationSolutionAns: 1.573 kg1. Calculate volume: $\frac{4}{3}\pi(3.6)^3 ≈ 195.432 cm^3$2. Find mass: 195.432 × 8.05 ≈ 1573.23 g3. Convert to kg: 1573.23 g = 1.57323 kg4. Round to nearest gram: 1.573 kg Question 12Topic – E1.13Oliver sent 22% more messages in June than in May. He sent 305 messages in June.Find how many more messages he sent in June than in May.▶️ Answer/ExplanationSolutionAns: 551. Let May messages = M2. June messages = 1.22 × M = 3053. Solve for M: M = 305 ÷ 1.22 = 2504. Difference: 305 – 250 = 55 more messages Question 13Topic – E2.6The graph of y = 2x + 1 is drawn on the grid.By shading the unwanted regions of the grid, find and label the region R which satisfies these inequalities:y ≥ 2x + 1 y ≥ 1 4x + 3y < 12▶️ Answer/ExplanationSolutionAns: 1. Draw y=2x+1 as solid line, shade above it2. Draw y=1 as solid line, shade above it3. Draw 4x+3y=12 as dashed line, shade below it4. The region satisfying all three is the triangular area where all shadings overlap Question 14Topic – E9.4The box-and-whisker plot shows information about the mass, in kg, of some parcels.(a) Find the mass of the heaviest parcel.(b) Find the interquartile range.▶️ Answer/ExplanationSolutionAns: (a) 3.3 kg (b) 0.55 kg(a) The rightmost point on the plot shows the maximum mass = 3.3 kg(b) IQR = Q3 – Q1 = 2.75 – 2.2 = 0.55 kg (read from the box ends) Question 15Topic – E2.2T = $\sqrt{3d – e}$Rearrange the formula to make d the subject.▶️ Answer/ExplanationSolutionAns: d = $\frac{T^2 + e}{3}$1. Square both sides: T² = 3d – e2. Add e to both sides: T² + e = 3d3. Divide by 3: d = (T² + e)/3 Question 16Topic – E5.4A cylinder with height 12.5 cm has a curved surface area of 105π cm².Calculate the volume of the cylinder.▶️ Answer/ExplanationSolutionAns: 693 or 692.7 to 692.8 cm³1. Find radius using curved surface area formula: 2πrh = 105π → r = 105/(2×12.5) = 4.2 cm2. Volume formula: πr²h = π×(4.2)²×12.53. Calculate: π×17.64×12.5 ≈ 693 cm³ Question 17Topic – E2.4(a) Simplify $(64y^{27})^{\frac{2}{3}}$(b) Simplify $\frac{x – 5}{x^2 – 25}$▶️ Answer/ExplanationSolutionAns:(a) $16y^{18}$1. Cube root first: $(64y^{27})^{1/3} = 4y^9$2. Then square: $(4y^9)^2 = 16y^{18}$(b) $\frac{1}{x+5}$1. Factor denominator: $x^2-25 = (x+5)(x-5)$2. Cancel (x-5) terms Question 18Topic – E1.11$F$ is proportional to the product of $m$ and $a$.Calculate the percentage change in $F$ when $m$ is increased by 40% and $a$ is decreased by 15%.▶️ Answer/ExplanationSolutionAns: 19%1. New m = 1.4m (40% increase)2. New a = 0.85a (15% decrease)3. New F = k×(1.4m)×(0.85a) = 1.19kma4. Percentage change = (1.19 – 1)×100 = 19% Question 19Topic – E6.2Calculate the obtuse angle $PRQ$.▶️ Answer/ExplanationSolutionAns: 116.9°1. Use sine rule: $\frac{QR}{\sin P} = \frac{PQ}{\sin R}$2. $\sin R = \frac{PQ \sin P}{QR} = \frac{13.5 \sin 42°}{18}$ ≈ 0.53. Acute angle would be ≈63.1°4. Obtuse angle = 180° – 63.1° = 116.9° Question 20Topic – E2.2$(x+a)(x+2)(2x+3)$ is equivalent to $2x^3+bx^2+cx-18$.Find the value of each of $a$, $b$ and $c$.▶️ Answer/ExplanationSolutionAns: a = -3, b = 1, c = -151. Constant term: a×2×3 = -18 → a = -32. Expand first two brackets: (x² – x – 6)(2x + 3)3. Final expansion: 2x³ + x² – 15x – 184. Compare coefficients: b=1, c=-15 Question 21Topic – E6.1The diagram shows a cuboid ABCDEFGH. AB = 14 cm, BC = 5 cm and CG = 8 cm. M is the midpoint of HG.(a) Calculate BM.(b) Calculate the angle that BM makes with the base ABCD.▶️ Answer/ExplanationSolution(a) Ans: 11.7 cmFirst find horizontal distance from B to M: 7 cm (half AB) and 5 cm (BC).Then use 3D Pythagoras: √(7² + 5² + 8²) = √(49 + 25 + 64) = √138 ≈ 11.7 cm(b) Ans: 43.0°Angle is between BM and its projection on base. Vertical component is 8 cm.tanθ = 8/√(7² + 5²) → θ = tan⁻¹(8/√74) ≈ 43.0° Question 22 Topic – E2.5 Find the coordinates of the point where the line 4x + y = 9 intersects the curve y + x² = 5. ▶️ Answer/Explanation Solution Ans: (2, 1) From line equation: y = 9 – 4x Substitute into curve: (9 – 4x) + x² = 5 → x² – 4x + 4 = 0 Solve: (x – 2)² = 0 → x = 2 Find y: y = 9 – 4(2) = 1
