S2 EOY Practice — Mock Paper B1

Compiled from the 2023–2025 end-of-year Paper 1 papers
Prepared by Miss Clarissa Ng
www.clartutors.com

END-OF-YEAR PRACTICE — SECONDARY 2

Integrated Mathematics · Paper 1 · 1 hour 15 minutes · 50 marks
Class: ______________ Register Number: ______________ Name: ____________________________

INSTRUCTIONS

  1. Write your name, register number and class on all the work you hand in.
  2. Write in dark blue or black ink.
  3. You may use an HB pencil for any diagrams or graphs.
  4. Do not use staples, paper clips, glue or correction tape/fluid.
  5. Write your answers and working in the blank spaces provided.
  6. Answer all questions.
  7. Omission of essential working will result in loss of marks.
  8. The use of an approved scientific calculator is expected, where appropriate.
  9. If the degree of accuracy is not specified in a question and the answer is not exact, give your answer to three significant figures. For π, use either your calculator value or 3.142.
  10. The number of marks is given in brackets [ ] at the end of each question or part question.
  11. The total number of marks for this paper is 50.

INFORMATION

Questions

Q1. A grain of rice has a mass of 2.5 × 10−5 kg. The mass of one carbon atom is 2.0 × 10−26 kg.

(a) Express the mass of the grain of rice as a multiple of the mass of one carbon atom, giving your answer in standard form. [2]

(b) A water molecule is made up of 2 hydrogen atoms and 1 oxygen atom. The mass of an oxygen atom is 16 times the mass of a hydrogen atom, and one hydrogen atom has a mass equal to one twelfth of the mass of a carbon atom. Find the number of water molecules in the grain of rice, giving your answer in standard form. [2]

Q2. y is inversely proportional to the square of x. When x = 2, y = 45.

(a) Express y in terms of x. [2]

(b) Find the value of y when x = 3. [1]

Q3. (a) Solve the inequality 5 − 2x ≤ 3x − 5 < 2x + 7 and illustrate the solution set on a number line. [3]

(b) Hence state the largest integer value of x. [1]

Q4. (a) Solve the equation 42x−1 = 8x+2. [2]

(b) Simplify (16a8b−4)½ × (2a−1b3)3, leaving your answer in positive index form. [3]

Q5. (a) Factorise 6x2 + x − 12. [2]

(b) Hence express 53x − 4 − 26x2 + x − 12 as a single fraction in its simplest form. [2]

Q6. (a) The pair of simultaneous equations

3x + ky = 6     and     6x − 4y = 9

has no solution. Find the value of k. [2]

(b) Solve the simultaneous equations y = x + 1 and x2 + y2 = 25. [3]

Q7. (a) Express 2x2 − 12x + 7 in the form a(x + p)2 + q. [3]

(b) Hence write down the coordinates of the turning point of the graph of y = 2x2 − 12x + 7. [1]

(c) State the equation of the line of symmetry of the same graph. [1]

Q8. The cumulative frequency curve below shows the marks scored by 160 students in a weighted assessment.

Marks scored by 160 studentsread the values off the curve010203040506070020406080100120140160MarksCumulative frequencyestimate from the curve; answers correct to the nearest mark are accepted

(a) Use the curve to estimate

(i) the median mark, [1]

(ii) the interquartile range. [2]

(b) The marks scored by Class 2 in the same assessment are shown in the box-and-whisker plot below.

Class 2 — marks on the same paperminimum 15, lower quartile 30, median 42, upper quartile 48, maximum 62010203040506070Marks

Compare the performance of the two classes. [2]

Q9. In triangle ABC, AB = 9 cm, BC = 15 cm and AC = 12 cm. E lies on BC such that AE is perpendicular to BC.

Triangle ABCABCE9 cm12 cm15 cmE lies on BC and AE is perpendicular to BC

(a) Show that angle BAC is a right angle. [2]

(b) Write down the value of cos BCA. [1]

(c) Calculate the length of AE. [2]

Q10. Two similar cylindrical jars, S and L, have heights 8 cm and 12 cm respectively. The volume of jar S is 320 cm3.

(a) Calculate the volume of jar L. [2]

(b) The curved outside surface of both jars is painted. Find the area painted on jar L as a percentage of the area painted on jar S. [2]

(c) Jar L is filled completely with water and all of the water is then poured into jar S, which is empty. Explain whether jar S overflows. [1]

Q11. A rectangular hall is 5 m longer than it is wide. The area of the floor of the hall is 128 m2.

(a) Taking the width of the hall as x m, form an equation in x and show that it reduces to x2 + 5x − 128 = 0. [2]

(b) Solve the equation x2 + 5x − 128 = 0, giving your answers correct to 3 significant figures. [2]

(c) Find the length of the hall. [1]

End of Paper 1. Check your work — every answer in its simplest form, units where they are needed, and all working shown.

Answer Key — S2 EOY Practice, Mock Paper B1

Total: 50 marks · 11 questions. Method marks are awarded for a correct method even where the final answer is wrong; accuracy marks only for a correct answer. Values read from the graph in Q8 are accepted within about one mark of the value given.
Q1 Standard form [4]
(a) 2.5 × 10−5 ÷ 2.0 × 10−26 = 1.25 × 1021
(b) One hydrogen atom = 2.0 × 10−26 ÷ 12 = 1.67 × 10−27 kg, so one water molecule = 18 × 1.67 × 10−27 = 3.0 × 10−26 kg. Number of molecules = 2.5 × 10−5 ÷ 3.0 × 10−26 = 8.33 × 1020
Q2 Inverse proportion [3]
(a) y = k/x2; 45 = k/4 so k = 180, giving y = 180/x2
(b) y = 180 ÷ 9 = 20
Q3 Linear inequalities [4]
(a) 5 − 2x ≤ 3x − 5 gives 10 ≤ 5x, so x ≥ 2. 3x − 5 < 2x + 7 gives x < 12. Solution set: 2 ≤ x < 12, drawn with a solid circle at 2 and an open circle at 12
(b) largest integer value of x = 11
Q4 Indices [5]
(a) 22(2x−1) = 23(x+2), so 4x − 2 = 3x + 6 and x = 8
(b) (16a8b−4)½ = 4a4b−2; (2a−1b3)3 = 8a−3b9. Product = 32ab7
Q5 Algebraic fractions [4]
(a) 6x2 + x − 12 = (3x − 4)(2x + 3)
(b) Common denominator (3x − 4)(2x + 3). The numerator is 5(2x + 3) − 2 = 10x + 13, so the answer is (10x + 13) / ((3x − 4)(2x + 3))
Q6 Simultaneous equations [5]
(a) The gradients are −3/k and 6/4 = 1.5. No solution means the lines are parallel, so −3/k = 1.5 and k = −2
(b) Substituting: x2 + (x + 1)2 = 25 gives 2x2 + 2x − 24 = 0, that is (x + 4)(x − 3) = 0. x = −4, y = −3 or x = 3, y = 4
Q7 Completing the square [5]
(a) 2(x2 − 6x) + 7 = 2[(x − 3)2 − 9] + 7 = 2(x − 3)2 − 11
(b) turning point (3, −11)
(c) line of symmetry x = 3
Q8 Statistics [5]
(a)(i) the median is the 80th value; read from the curve, ≈ 37 marks
(a)(ii) lower quartile (40th value) ≈ 24 and upper quartile (120th value) = 50, so the interquartile range ≈ 26 marks
(b) Class 2 has the higher median (42 compared with about 37), so it performed better on average; it also has the smaller interquartile range (18 compared with about 26), so its marks were more consistent
Q9 Pythagoras and area [5]
(a) AB2 + AC2 = 81 + 144 = 225 and BC2 = 225. Since AB2 + AC2 = BC2, the converse of Pythagoras’ theorem gives angle BAC = 90°
(b) cos BCA = AC/BC = 12/15 = 0.8
(c) Area of ABC = ½ × 9 × 12 = 54 cm2, and also ½ × 15 × AE = 54, so AE = 7.2 cm
Q10 Similar solids [5]
(a) length ratio 8 : 12 = 2 : 3, so volume ratio = 23 : 33 = 8 : 27. Volume of L = 320 × 27/8 = 1080 cm3
(b) area ratio = (3/2)2 = 2.25, so the area painted on L is 225 % of the area painted on S
(c) the volume of L (1080 cm3) is greater than the volume of S (320 cm3), so the water cannot all fit and jar S overflows
Q11 Quadratic word problem [5]
(a) x(x + 5) = 128, which expands to x2 + 5x − 128 = 0
(b) x = [−5 ± √(25 + 512)] / 2 = [−5 ± √537] / 2, so x = 9.09 or −14.1; the negative value is rejected because a width cannot be negative
(c) length = 9.09 + 5 = 14.1 m (3 s.f.)
Miss Clarissa Ng | www.clartutors.com