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Transcript of The time in minutes (T) for meals to be served at a busy … · · 2018-04-19... AD = 6 cm and BC...
![Page 1: The time in minutes (T) for meals to be served at a busy … · · 2018-04-19... AD = 6 cm and BC = 9 cm. Angle ABC = 75° and angle ADC = 90° ... cm3 (5) (Total 8 marks) € €](https://reader034.fdocuments.net/reader034/viewer/2022042510/5b0256a37f8b9a6a2e8fa372/html5/thumbnails/1.jpg)
Q1. The time in minutes (T) for meals to be served at a busy restaurant is inversely proportional to the square of the number of waiters (W) working at that time.
It takes 20 minutes for meals to be served when 12 waiters are working.
(a) Find an equation connecting T and W.
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Answer ................................................. (3)
(b) What is the minimum number of waiters that must be working for a meal to be served within 30 minutes?
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Answer ................................................. (3)
(Total 6 marks)
Q2. In the diagram, O is the centre of the circle. A, B, C and D are points on the circumference. Angle AOC = 130°
(a) Calculate the value of x. Give a reason for your answer.
Answer x = ..........................degrees
Reason ............................................................................................................
......................................................................................................................... (2)
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(b) Calculate the value of y. Give a reason for your answer.
Answer y = ..........................degrees
Reason ............................................................................................................. (2)
(Total 4 marks)
Q3. (a) Enlarge the shaded shape by a scale factor of 3.
(2)
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(b) How many times bigger is the area of the enlarged shape than the area of the small shape?
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Answer ................................................. (2)
(Total 4 marks)
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Q4. ABCD is a quadrilateral. AB = 7 cm, AD = 6 cm and BC = 9 cm. Angle ABC = 75° and angle ADC = 90°
Calculate the perimeter of ABCD.
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Answer ................................................. cm (Total 5 marks)
Q5. OACB is a parallelogram and M is the mid-point of BC.
= a and = b
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(a) Express the following vectors in terms of a and b
(i)
Answer ................................................. (1)
(ii)
Answer ................................................. (1)
(b) AM is extended to N, where .
Show that = b
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.......................................................................................................................... (2)
(c) What does this tell you about the position of N?
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(Total 5 marks)
Q6. (a) Calculate the size of an interior angle of a regular octagon.
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Answer ................................................. degrees (3)
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(b) Part of a tiled floor is shown.
The tiles labelled P, Q, R and S are regular octagons.
Explain why the tile labelled X is a square.
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.......................................................................................................................... (3)
(Total 6 marks)
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Q7. The diagrams show a rectangle and an L shape All the angles are right angles. All lengths are in centimetres. The shapes are equal in area.
Calculate the value of y.
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Answer ................................................. cm (Total 6 marks)
Q8. (a) Complete the table of values for y = 3x2 – 6
(1)
x –3 –2 –1 0 1 2 3 4
y 21 6 –3 –6 –3 21 42
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(b) On the grid below, draw the graph of y = 3x2 – 6 for values of x between –3 and +4.
(2)
(c) Use your graph to write down the solutions of 3x2 – 6 = 0
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Answer ...................... and ...................... (1)
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(d) By drawing an appropriate linear graph, write down the solutions of
3x2 – 5x – 6 = 0
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Answer ................................................. (3)
(Total 7 marks)
Q9. The area of the screen of a television set is A square inches. The length of the diagonal of the screen is d inches. A is directly proportional to the square of d.
A television set with an area of 90 square inches has a diagonal of length 15 inches.
(a) Find an equation connecting A and d.
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Answer ................................................. (3)
(b) Find the area of the screen of a television set with a diagonal of length 20 inches.
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Answer ................................................. square inches (1)
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(c) Another television set has a screen with an area of 250 square inches.
Find the length of its diagonal.
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Answer ................................................. inches (3)
(Total 7 marks)
Q10. The graph shows the function
(a) Write down the coordinates of the point where the graph intersects with the y-axis.
Answer ( ........................ , ....................... ) (1)
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(b) Find the value of a.
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Answer ................................................. (2)
(Total 3 marks)
Q11. Two spheres of radius 5 cm just fit inside a tube.
Calculate the volume inside the tube not filled by the spheres.
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Answer ................................................. cm2
(Total 5 marks)
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Q12. A sign maker designs a letter L. All arcs are quarter circles of radius 2 cm.
Calculate the area of the L.
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Answer ................................................. cm2
(Total 4 marks)
Not drawn accurately
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Q13. (a) P is inversely proportional to Q. When P = 100, Q = 32
Express P in terms of Q.
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Answer ................................................. (3)
(b) P and Q are positive quantities. Sketch a graph of the relationship between P and Q on this diagram.
(1)
(c) Calculate the value of Q when P is twice as big as Q.
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Answer ................................................. (2)
(Total 6 marks)
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Q14. A(1, 1) and B(–2, 4) are two points on the graph of y = x2
Here are three transformations of the graph y = x2.
On each diagram the graph of y = x2 is shown dotted. The images A′ and B′ of A and B are shown. Write down the equation of the transformed graph in each case.
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(a)
y = ........................................ (1)
(b)
y = ........................................ (1)
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(c)
y = ........................................ (1)
(Total 3 marks)
Q15. The diagram shows a solid made from a cone and a hemisphere. The radius of both shapes is r. The slant height of the cone is l. The perpendicular height of the cone is h.
The curved surface area of the cone and the curved surface area of the hemisphere are equal.
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(a) Show that l = 2r
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(b) Find the perpendicular height, h, of the cone in terms of r.
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Answer h = .............................................. (2)
(c) Find the ratio of the volumes of the cone and the hemisphere. Give your answer in surd form.
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Answer ................................................. (2)
(Total 6 marks)
Q16. (a) Show that can be written as 2x2 – 9x + 4 = 0
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(b) Part of the graph of y = is shown on the grid below.
Draw a straight line on the grid which will enable you to solve the equation
2x2– 9x + 4 = 0
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(c) Hence, or otherwise, solve the equation 2x2– 9x + 4 = 0
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Answer ................................................. (2)
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(Total 7 marks)
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Q17. A circle fits exactly inside a semi-circle of diameter 20 cm.
The shaded area is a × π square centimetres. Work out the value of a. You must show your working.
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Answer a = .............................................. (Total 4 marks)
Q18. This is the graph of y = cos x for 0° ≤ x ≤ 360°
Not drawn accurately
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Write the equation of each of the transformed graphs. In each case the graph of y = cos x is shown dotted to help you.
(a)
Equation y = .............................................. (1)
(b)
Equation y = .............................................. (1)
(c)
Equation y = .............................................. (1)
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(d)
Equation y = .............................................. (1)
(Total 4 marks)
Q19. ABCD is a cyclic quadrilateral. PAQ is a tangent to the circle at A. BC = CD Angle QAB = 38° and angle BAD = 76°
Not drawn accurately
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Show that AD is parallel to BC. Give reasons to justify any values you write down or calculate.
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Q20. (a) A calculator displays a number in standard form as
Which of the following numbers does the display show? Circle the correct answer.
7000 0.700 0.007 700 0.0007 (1)
(b) Use your calculator to work out
cos (tan–10.45)
(i) Give all the figures in your calculator display.
Answer ................................................. (1)
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(ii) Write your answer to an appropriate degree of accuracy.
Answer ................................................. (1)
(c) Use your calculator to work out
Answer ................................................. (1)
(Total 4 marks)
Q21. A sphere has radius r. A cone has base radius r and perpendicular height x. The volume of the sphere is double the volume of the cone.
Not drawn accurately
(a) Show that x =2r
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(b) Calculate the ratio of the surface area of the sphere to the curved surface area of the cone. Give your answer in surd form.
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Answer ................................................. (4)
(Total 6 marks)
Q22. The diagram shows a cuboid. AB = 3 cm, AE = 4 cm, BC = 12 cm.
(a) Find the length of BH.
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Answer ............................................ cm (2)
Not drawn accurately
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(b) The angle between BH and BD is x and the angle between BH and BC is y.
Which angle is bigger, x or y? You must show your working.
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Answer ................................................. (3)
(Total 5 marks)
Q23. XYZ is an isosceles triangle in which XZ = XY M and N are points on XZ and XY such that angle MYZ = angle NZY.
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Prove that triangles YMZ and ZNY are congruent.
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Q24. In the diagram SR is parallel to PT. SQT and RQP are straight lines. SR = 20 cm and PT = 30 cm The total height of the two triangles is 40 cm.
Not drawn accurately
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Use similar triangles to calculate the height, h cm, of triangle PQT.
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Answer h = .................................................. cm (Total 3 marks)
Q25. (a) A circle has a radius of 6 cm. A sector has an arc length of 8.4 cm. The angle at the centre of the sector is θ.
Calculate the value of θ.
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Answer ................................................. degrees (3)
Not drawn accurately
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(b) A cone has base radius 6 cm and height h cm. A smaller cone of base radius 2 cm and height 3 cm is cut from the top. The remaining frustum has dimensions as shown.
Calculate the volume of the frustum.
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Answer ................................................. cm3
(5) (Total 8 marks)
Not drawn accurately
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Q26. The grid below shows graphs of a curve
y = x2 + 2x – 3
and 3 straight lines
y = x + 1
y = – x – 2
and y = – x + 2
You must use the graphs to answer the following questions.
(a) Write down a pair of simultaneous linear equations that have a solution
x = – , y =
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Answer ................................................. (1)
(b) Write down and simplify a quadratic equation whose solutions are approximately – 3.3 or 0.3. You must show clearly how you obtain your answer.
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Answer ................................................. (2)
(c) Write down the approximate solutions to the equation x2 + x – 4 = 0.
You must show clearly how you obtain your answer.
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Answer ................................................. (2)
(Total 5 marks)
Q27. A square of side x and a quarter-circle of radius r have the same area.
Not to scale
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Express r in terms of x. Simplify your answer.
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Answer r = ............................................ (Total 3 marks)
Q28. (a) ABC is a triangle. AC = 19 cm, BC = 17 cm and angle BAC = 60°
Calculate the size of angle ABC.
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Answer ................................................. degrees (3)
Not to scale
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(b) PQR is a triangle. PR = 23 cm, PQ = 22 cm and angle QPR = 48°
Calculate the length of QR. Give your answer to an appropriate degree of accuracy.
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Answer ................................................. cm (4)
(Total 7 marks)
Q29. ABC is an isosceles triangle. The lengths, in cm, of the sides are
AB = 4a + 3, BC = 2b + 5 and AC = 2a + b
Not to scale
Not to scale
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(a) AB = BC
Show that 2a – b = 1
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(b) The perimeter of the triangle is 32 cm. Find the values of a and b.
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Answer a = ..................... cm, b =....................... cm (4)
(Total 6 marks)
Q30. For a ladder to be safe it must be inclined at between 70° and 80° to the ground.
(a) The diagram shows a ladder resting against a wall.
Not to scale
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Is it safe? You must show your working.
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(b) Another ladder rests against a wall.
Work out the closest distance that the bottom of the ladder can be from the wall so that it is safe.
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Answer ................................................. m (3)
(Total 6 marks)
Not to scale
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Q31. A hemispherical bowl of radius 6 cm has the same volume as a cone of perpendicular height 27 cm.
Not drawn accurately
Calculate the base radius, r, of the cone.
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Answer ................................................. cm (Total 4 marks)
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Q32. In the diagram below points Q and S lie on a circle centre O. SR is a tangent to the circle at S. Angle QRS = 40° and angle SOQ = 80°
Prove that triangle QSR is isosceles.
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Not drawn accurately
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Q33. Match each of the sketch graphs to one of these equations.
A y = 2 – 2x B y = 2x + 2 C y = 3 – x2 D y = x3 + 4 E y =
Graph 1 represents equation ……................
Graph 2 represents equation ……................
Graph 3 represents equation ……................
Graph 4 represents equation ……................ (Total 4 marks)
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Q34. The grid below shows the graph of y = x2 + 3x – 2
(a) By drawing an appropriate straight line on the graph solve the equation
x2 + 3x – 3 = 0
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Answer ................................................. (2)
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(b) By drawing an appropriate straight line on the graph solve the equation
x2 + 2x – 1 = 0
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Answer ................................................. (3)
(Total 5 marks)
Q35. (a) Explain why the volume of a cube increases by a factor of 8 when the side length is doubled.
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(b) June recently bought a small toy in the local shop.
It was originally 8 cm tall. After she placed it in water it grew to a similarly shaped alien. The height was then 14.5 cm. Is the claim on the pack justified?
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(Total 5 marks)
ALIEN
Place in
water and it becomes
6 times bigger!
Q36. A marble paperweight consists of a cuboid and a hemisphere as shown in the diagram. The hemisphere has a radius of 4 cm.
Not to scale
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Calculate the volume of the paperweight.
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Answer ................................................. (Total 4 marks)
Q37. A circle fits inside a semicircle of diameter 10 cm as shown.
Not drawn accurately
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Calculate the shaded area.
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Answer ................................................. cm2
(Total 3 marks)
Q38. y is directly proportional to the square of x. When y = 5, x = 4. Find the value of y when x = 8.
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Answer ................................................. (Total 3 marks)
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Q39. A giant paper clip is placed alongside a centimetre ruler. The curved ends are semicircles.
Calculate the length of wire used to make the clip.
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Answer ................................................. cm (Total 5 marks)
Q40. (a) ABC is a right-angled triangle. AC = 19 cm and AB = 9 cm.
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Calculate the length of BC.
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Answer ................................................. cm (3)
(b) PQR is a right-angled triangle. PQ = 11 cm and QR = 24 cm.
Calculate the size of angle PRQ.
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Answer ................................................. degrees (3)
(Total 6 marks)
Q41. Two towns, A and B, are connected by a motorway of length 100 miles and a dual carriageway of length 80 miles as shown.
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Jack travels from A to B along the motorway at an average speed of 60 mph. Fred travels from A to B along the dual carriageway at an average speed of 50 mph. What is the difference in time between the two journeys? Give your answer in minutes.
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Answer ................................................. minutes (Total 4 marks)
Q42. A straight line has the equation y = 2x – 3
A curve has the equation y2 = 8x – 16
(a) Solve these simultaneous equations to find any points of intersection of the line and the curve. Do not use trial and improvement. You must show all your working.
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Answer ................................................. (5)
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(b) Here are three sketches showing the curve y2 = 8x – 16 and three possible positions of the line y = 2x – 3
Sketch 1
Sketch 2
Sketch 3
Which is the correct sketch?
You must explain your answer.
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(Total 7 marks)
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Q43. The sketch shows the graph of y = sin x for 0° ≤ x ≤ 360°
You are given that sin 70° = 0.9397
(a) Write down another solution of the equation sin x = 0.9397
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Answer ................................................. degrees (1)
(b) Solve the equation sin x = –0.9397 for 0° ≤ x ≤ 360°
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Answer ................................................. degrees
................................................. degrees (2)
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(c) On the axes below sketch the graph of y = sin 2x for 0° ≤ x ≤ 360°
(2)
(d) Hence write down the four solutions of the equation sin 2x = 0.9397
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Answer ................................................. degrees
................................................. degrees
................................................. degrees
................................................. degrees (3)
(Total 8 marks)
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Q44. The diagram shows the graph of the equation y = x2 + px + q
The graph crosses the x-axis at A and B (2,0).
C (–3, –5) also lies on the graph.
(a) Find the values of p and q.
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Answer p = ...................... q = ...................... (4)
(b) Hence work out the coordinates of A.
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Answer ( ........................ , ........................ ) (2)
(Total 6 marks)
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Q45. The diagram shows a cylinder. The diameter of the cylinder is 10 cm. The height of the cylinder is 10 cm.
(a) Work out the volume of the cylinder. Give your answer in terms of π.
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Answer ................................................. cm3
(3)
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(b) Twenty of the cylinders are packed in a box of height 10 cm. The diagram shows how the cylinders are arranged inside the box. The shaded area is the space between the cylinders.
Work out the volume inside the box that is not filled by the cylinders. Give your answer in terms of π.
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Answer ................................................. cm3
(4) (Total 7 marks)
Q46. In the diagram OACD, OADB and ODEB are parallelograms.
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(a) Express, in terms of a and b, the following vectors. Give your answers in their simplest form.
(i)
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Answer ................................................. (1)
(ii)
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Answer ................................................. (1)
(iii)
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Answer ................................................. (1)
(b) The point F is such that OCFE is a parallelogram.
Write the vector in terms of a and b.
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Answer ................................................. (2)
(c) What geometrical relationship is there between the points O, D and F? Justify your answer.
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(Total 7 marks)
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Q47. A square-based pyramid with a base of side 2 cm has a volume of 2.75 cm3.
What is the volume of a similar square-based pyramid with a base of side 6 cm?
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Answer ................................................. cm3
(Total 2 marks)
Q48. A ruined tower is fenced off for safety reasons. To find the height of the tower Rashid stands at a point A and measures the angle of elevation as 18°. He then walks 20 metres directly towards the base of the tower to point B where the angle of elevation is 31°.
Not to scale
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Calculate the height, h, of the tower.
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Answer ................................................. m (Total 6 marks)
Q49. The sketch below is of the graph of y = x2
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On the axes provided, sketch the following graphs.
The graph of y = x2 is shown dotted on each set of axes to act as a guide.
(a) y = x2 + 2
(1)
(b) y = (x – 2)2
(1)
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(c)
(1)
(Total 3 marks)
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Q50. The graph of y = x2 – 4x + 8 is shown below.
(a) (i) By drawing the graph of an appropriate straight line, solve the equation
x2 – 4x + 8 = 3x – 2
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Answer ................................................. (3)
(ii) Hence, or otherwise, solve x2 – 7x + 10 = 0
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Answer ................................................. (1)
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(b) The graph of y = x2 – 4x + 8 is to be used to solve the equation x2 – 5x + 4 = 0
What straight line graph would need to be drawn? (You do not need to draw it, just state its equation.)
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Answer y = ................................................. (2)
(Total 6 marks)
Q51. In the diagram, the sides of triangle ABC are tangents to the circle. D, E and F are the points of contact. AE = 5 cm and EC = 4 cm
(a) Write down the length of CD.
Answer .................................................cm (1)
(b) The perimeter of the triangle is 32 cm. Calculate the length of DB.
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Answer .................................................cm (2)
(Total 3 marks)
Not to scale
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Q52. Tom is investigating the equation y = x2 – x + 5
He starts to complete a table of values of y for some integer values of x.
Tom says, "When x is an integer, y is always a prime number". Find a counter-example to show that Tom is wrong. Explain your answer.
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Answer ................................................. (Total 2 marks)
x –2 –1 0 1 2 3
y 11 7 5 5 7 11
Q53. A water tank is 50 cm long, 34 cm wide and 24 cm high. It contains water to a depth of 18 cm.
Four identical spheres are placed in the tank and are fully submerged. The water level rises by 4.5 cm.
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Calculate the radius of the spheres.
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Answer ................................................. cm (Total 5 marks)
Q54. ABCD is a quadrilateral. AB = 7 cm, AD = 6 cm and BC = 9 cm. Angle ABC = 75° and angle ADC = 90°
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Calculate the perimeter of ABCD.
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Answer .................................................. cm (Total 5 marks)
Q55. AB is a chord of a circle, centre O, radius 6 cm. AB = 7 cm
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Calculate the area of the shaded segment.
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Answer ................................................. cm2
(Total 6 marks)
Q56. (a) Complete the table of values for y = (0.8)x
(1)
x 0 1 2 3 4
y 1 0.8 0.64 0.41
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(b) On the grid below, draw the graph of y = (0.8)x for values of x from 0 to 4.
(2)
(c) Use your graph to solve the equation (0.8)x = 0.76
Answer ................................................. (1)
(Total 4 marks)
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Q57.
Enlarge the shaded shape by scale factor with centre of enlargement (–1, 0). (Total 2 marks)
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Q58. A tin of diameter 7 cm and height 12 cm has a label around it. The label is glued together using a 1 cm overlap. There is a 1 cm gap between the label and the top and the bottom of the tin.
Find the length and the height of the label.
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Answer Length = ......................................... cm
Height = ......................................... cm (Total 4 marks)
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Q59. Dario is using trial and improvement to find a solution to the equation
x + = 5
The table shows his first trial.
Continue the table to find a solution to the equation. Give your answer to 1 decimal place.
Answer x = ................................................. (Total 4 marks)
x x + Comment
4 4.25 Too low
Q60. (a) Points P, Q, R and S lie on a circle.
PQ = QR
Angle PQR = 116°
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Explain why angle QSR = 32°.
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(b) The diagram shows a circle, centre O. TA is a tangent to the circle at A. Angle BAC = 58° and angle BAT = 74°.
(i) Calculate angle BOC.
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Answer Angle BOC = ............................... degrees (1)
(ii) Calculate angle OCA.
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Answer Angle OCA = ............................... degrees (3)
(Total 6 marks)
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Q61. Which one of the following kites is a cyclic quadrilateral? Give a reason for your answer.
Answer ......................................................
Reason .....................................................................................................................
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Q62. A square-based pyramid has a base of edge 5 cm. The vertex of the pyramid is directly over the midpoint of the base.
The volume of the pyramid is 100cm3.
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Find the length of the slant edge of the pyramid (marked x in the diagram).
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Answer ..................................... cm (Total 5 marks)
Q63. A solid cube has a square hole cut through horizontally and a circular hole cut through vertically.
Both holes are cut centrally in the appropriate faces.
The dimensions of the cube and the holes are as shown in the diagram.
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Calculate the volume remaining after the holes have been cut.
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Answer ............................................ (Total 5 marks)
Q64. In triangle ABC, AB = 11 cm, BC = 9 cm and CA = 10 cm.
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Find the area of triangle ABC.
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Answer ..................................... cm2
(Total 5 marks)
Q65. ABCD is a rectangle with length 25 cm and width 10 cm.
The length of the rectangle is increased by 10%. The width of the rectangle is increased by 20%. Find the percentage increase in the area of the rectangle.
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Answer ................................. % (Total 3 marks)
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Q66. Solve the equation
x2 – 10x – 5 = 0
Give your answers to 2 decimal places.
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Answer ............................................ (Total 3 marks)
Q67. (a) ABC is a right-angled triangle. AB = 5.1 cm
CAB = 48°
Find the length of BC (marked x in the diagram). Give your answer to a suitable degree of accuracy.
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Answer ..................................... cm (4)
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(b) PQRS is a parallelogram. PQ = 5.1 cm PS = 6.8 cm
QPS = 48°
Calculate the area of PQRS.
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Answer .................................... cm2
(2) (Total 6 marks)
Q68. The diagram shows a circle with centre O and radius 2.5 cm. TA is a tangent to the circle, of length 6 cm. The line from A to the centre O of the circle cuts the circumference at B.
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Calculate the length of AB.
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Answer .................................... cm (Total 4 marks)
Q69. (a) Liquid is poured at a steady rate into the bottle shown in the diagram.
As the bottle is filled, the height, h, of the liquid in the bottle changes.
Which of the five graphs below shows this change?
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Give a reason for your choice.
Graph ..............................................................................................................
Reason ............................................................................................................
......................................................................................................................... (2)
(b) Liquid is poured at a steady rate into another container. The graph shows how the height, h, of the liquid in this container changes.
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Sketch a picture of this container.
(1)
(Total 3 marks)
Q70. The diagrams, which are not drawn to scale, show the graph of y = x2 and four
other graphs A, B, C and D.
A, B, C and D represent four different transformations of y = x2.
Find the equation of each of the graphs A, B, C and D.
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(Total 4 marks)
Q71. The diagram shows two right-angled triangles. AD = 15 cm. CD = 6 cm.
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(a) Given that cos x° = , calculate the length BD.
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Answer BD = ............................................ (2)
(b) Find the value of sin y°.
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Answer sin y° = ............................................ (3)
(Total 5 marks)
Q72. ABC is a triangle. ACD is a straight line.
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Using a ruler and compasses only, make an accurate construction of this diagram. You must show clearly all your construction arcs. The line AD has been drawn for you.
(Total 6 marks)
Q73. (a) The diagram shows a circle with centre O.
Work out the size of the angle marked x.
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Answer .................................. degrees (1)
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(b) The diagram shows a different circle with centre O.
Work out the size of the angle marked y.
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Answer ................................ degrees (1)
(c) A, B and C are points on the circumference of a circle with centre O. BOC is a straight line. Angle ABC = 20°
Work out the size of the angle marked z. Explain your answer.
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Answer ................................... degrees (2)
(Total 4 marks)
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Q74. Two similar bottles are shown below. The smaller bottle is 20 cm tall and holds 480 ml of water. The larger bottle is 30 cm tall.
How much water does the larger bottle hold?
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Answer ................................................. (Total 3 marks)
Q75. A thin-walled glass paperweight consists of a hollow cylinder with a hollow cone on top as shown. The paperweight contains just enough sand to fill the cylinder.
The paperweight is now turned upside down.
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Calculate the depth of the sand, (marked x in the diagram).
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Answer ................................................. cm (Total 5 marks)
Q76. Two ships, A and B, leave port at 13 00 hours. Ship A travels at a constant speed of 18 km per hour on a bearing of 070°. Ship B travels at a constant speed of 25 km per hour on a bearing of 152°.
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Calculate the distance between A and B at 14 00 hours.
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Answer ................................................. km (Total 4 marks)
Q77. ABCDEFGH is a cuboid with sides of 5 cm, 5 cm and l2 cm as shown.
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Calculate angle DFH.
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Answer ................................................. degrees (Total 5 marks)
Q78. In the diagram, the lines AC and BD intersect at E.
AB and DC are parallel and AB = DC.
Prove that triangles ABE and CDE are congruent.
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Q79. ABC is a right-angled triangle. BC = 125 m. Angle CAB = 33°.
Find the length of AC (marked x in the diagram). Give your answer to an appropriate degree of accuracy.
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Answer ................................................. m (Total 4 marks)
Q80. Triangles ADE and ABC are similar. DE is parallel to BC. AD = 4 cm, DE = 6 cm and BC = 9 cm.
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Calculate the length of BD.
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Answer ................................................. cm (Total 3 marks)
Q81. The graph of y = sin x for 0° ≤ x ≤ 360° is shown on the grid below. The point P(90, 1) lies on the curve.
On both of the grids that follow, sketch the graph of the transformed function. In both cases write down the coordinates of the transformed point P.
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(a) y = sin (x – 45)
P (......................., ......................) (2)
(b) y = 2sinx
P (......................., ......................) (2)
(Total 4 marks)
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Q82. A firm makes cone shaped containers out of card. The card is in the shape of a sector of a circle of radius 12 cm. The angle of the sector is 270°. The straight edges are brought together to make the cone.
(a) Find the arc length of the card used to make the cone. Give your answer in terms of π.
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Answer ................................................. cm (2)
(b) Calculate the radius of the base of the cone.
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Answer ................................................. cm (2)
(Total 4 marks)
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Q83.
OAB is a triangle where M is the mid-point of OB.
P and Q are points on AB such that AP = PQ = QB.
= a and = 2b
(a) Find, in terms of a and b, expressions for
(i)
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Answer ................................................. (1)
(ii)
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Answer ................................................. (2)
(iii)
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Answer ................................................. (2)
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(b) What can you deduce about quadrilateral OMQP? Give a reason for your answer.
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(Total 7 marks)
Q84. (a)
(i) Write down the value of x.
Answer ................................................. degrees (1)
(ii) Calculate the value of y.
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Answer ................................................. degrees (1)
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(b) A and C are points on the circumference of a circle centre B. AD and CD are tangents. Angle ADB = 40°.
Explain why angle ABC is 100°.
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(c) ABCD is a cyclic quadrilateral. PAQ is a tangent to the circle at A. BC = CD. AD is parallel to BC. Angle BAQ = 32°.
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Find the size of angle BAD.
You must show all your working.
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Answer Angle BAD = ........................... degrees (5)
(Total 9 marks)
Q85. The map below shows three boats, A, B and C, on a lake. Along one edge of the lake there is a straight path.
Treasure lies at the bottom of the lake.
The treasure is: between 150 m and 250 m from B, nearer to A than C, more than 100 m from the path.
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Using a ruler and compasses only, shade the region in which the treasure lies.
You must show clearly all your construction arcs. (Total 5 marks)
Q86. (a)
(i) Write down the value of x.
Answer ................................................. degrees (1)
(ii) Calculate the value of y.
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Answer ................................................. degrees (1)
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(b) A and C are points on the circumference of a circle centre B. AD and CD are tangents. Angle ADB = 40°.
Explain why angle ABC is 100°.
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(c) P is a point on the circumference of a circle with centre O. PQ is a tangent of length 8 cm.
The area of triangle OPQ is 24 cm2.
Calculate the area of the circle. Give your answer in terms of π.
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Answer ................................................. cm2
(3) (Total 7 marks)
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Q87. Three circles fit inside a rectangle as shown.
Two of the circles are identical and the third is larger. The circles have radii 9 cm, 9 cm and 25 cm.
Calculate the length, l, of the rectangle.
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Answer ............................................ cm (Total 6 marks)
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Q88. This diagram is made from 25 small squares and 16 large squares.
What percentage of the diagram is shaded?
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Answer .............................................. % (Total 6 marks)
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Q89. The diagram shows a regular pentagon and a regular decagon joined at side XY.
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Show that the points A, B and C lie on a straight line.
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Q90. In the figure, AC = 9 cm, AE = 6 cm, BD = 8.5 cm, BE = 4.5 cm and DF = 5 cm, BEDF and AEC are straight lines.
Not drawn accurately
(a) Show that triangles BEC and AED are similar.
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(b) By considering triangles BEC and FEA show that AF is parallel to BC.
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(Total 8 marks)
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Q91. Four points, A(– 4, –1), B(5, 11), C(20, 11) and D(11, –1) are joined to form a quadrilateral.
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Prove by calculation that ABCD is a rhombus.
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Q92. Find the reflection of the point A(1, 4) in the line y = 2x – 3
Use the grid below to help you.
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Answer ( ...................... , ….................... )
(Total 3 marks)
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Q93. The diagram shows four points A, B, C and D on the circumference of a circle, centre O. PAQ is the tangent to the circle at A. PBD is a straight line.
Angle QAD = 75° Angle APB = 30°
Work out angle BCD. You must show your working.
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Answer .................................. degrees (Total 3 marks)
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Q94. A manufacturer designs a set of three similar containers to fit inside each other. The diagram shows a sketch of the containers and their oval cross-sections.
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Some information about the containers is shown in this table.
(a) Work out the height of the large container.
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Answer …................................... cm (3)
Base length Height Area of card used in
manufacture
Large 30 cm
Medium 20 cm 12 cm 1080 cm
Small 10 cm
2
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(b) 1080 cm of card is used to manufacture the medium container.
Work out the area of card used to manufacture the small container.
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Answer …................................... cm
(2) (Total 5 marks)
2
2
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Q95. The shaded shape is made from two different right-angled triangles.
The area of each of the triangles is 6cm .
Each of lengths a, b, c and d are a whole number of centimetres. a + c = 10 cm a > b and c > d
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Work out the perimeter of the shaded shape.
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Answer ........................................... cm (Total 5 marks)
2
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Q96. The sector AOB of a circle is shown below. The length of its arc AB is 10π cm.
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The sector is folded so that the straight edges meet and form a cone as shown.
(a) Calculate the radius of the base of the cone.
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Answer .......................................... cm (3)
(b) The volume of the cone is 80π cm .
Work out the perpendicular height of the cone.
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Answer ......................................... cm (3)
(Total 6 marks)
3
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Q97. OABC is a quadrilateral. D, E, F and G are midpoints of OA, AB, BC and OC respectively.
= 2a, = 2b and = 2c
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Find the following vectors in terms of a, b and c.
For example = c – a
(a)
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Answer .................................................. (1)
(b)
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Answer .................................................. (1)
(c) Use your answers to parts (a) and (b) to show that = c – a
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(d) Explain how you can tell that DEFG is a parallelogram.
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(Total 4 marks)
Q98. The diagrams show a trapezium and a parallelogram.
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(a) Use the trapezium to explain why 2x + y = 180
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(b) The parallelogram can be used to form another equation connecting x and y.
Tick a box to show the correct equation.
(1)
3x + y = 130 3x + y = 230
3x = y – 50 3x + y = 410
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(c) Hence, or otherwise, work out the values of x and y.
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Answer x = ................ , y = …............... (3)
(Total 5 marks)
Q99. Airport runways have a two-digit number painted on them. These numbers are used to work out the direction of the runway.
To work out the three-figure bearing, multiply the runway number by 10.
Here is a diagram of a runway on a three-figure bearing of 280° and a runway on a three-figure bearing of 040°.
(a) (i) Write down the three-figure bearing for a runway pointing due South.
Answer .............................................. ° (1)
(ii) Write down the runway number for a runway pointing due South.
Answer ................................................. (1)
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(iii) A runway has a three-figure bearing of 060°.
Write down the runway number.
Answer ................................................. (1)
(b) A runway is being painted.
By measuring the three-figure bearing, work out the runway number.
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Answer .................................................. (2)
(c) Runways are used in both directions. This means that they have two different runway numbers, one at each end. A runway has the number 30 at one end.
What runway number is at the other end?
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Answer .................................................. (3)
(Total 8 marks)
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Q100. This is the graph of y = sin x for values of x from 0° to 360°
On each of the following grids the solid line shows a transformation of the graph of y = sin x.
Write down the equation of each of the transformed graphs.
On each grid, the graph y = sin x is shown dotted to help you.
(a)
Answer y = ...................................... (1)
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(b)
Answer y = ...................................... (1)
(c)
Answer y = ...................................... (1)
(Total 3 marks)
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Q101. In triangle ABC the length of AB is 13.2 cm. Angle BAC = 40° Angle BCA = 114°
Work out the length of BC. Give your answer to an appropriate degree of accuracy.
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Answer ....................................... cm (Total 4 marks)
Q102. Advice about wheelchair ramps is that for a ramp that rises more than 7.5 cm the
maximum
For example
Not drawn accuratelly
gradient should be
Not drawn accuratelly
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(a) Access to a Village Hall is by two steps each 16 cm high. It is proposed to build a ramp alongside the steps as shown in the diagram.
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Does the proposed ramp follow the advice given? You must show your working.
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(b) For a wheelchair ramp that rises less than 7.5 cm the maximum gradient should be
For example
A ramp is designed for a step that rises 6 cm.
Work out the length of a ramp that uses the maximum gradient.
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Answer ..................................... cm (4)
(Total 6 marks)
Not drawn accurately
Not drawn accurately
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Q103. The diagram shows a kite ABCD. AB = x + 1 and CD = 3x – 2
(a) Show that, in terms of x, the perimeter of the kite is 8x – 2
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(b) The perimeter of the kite is 16 cm.
Write down and solve an equation to work out the value of x.
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Answer x = ............................... cm (3)
(Total 4 marks)
Q104. A, B and C are three points such that
= 5a – 3b and = 7.5a – 4.5b
(a) Write down a fact about the points A, B and C.
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(b) Write down the ratio of the lengths AB : BC in its simplest form.
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(Total 2 marks)
Q105. The diagram shows the graph of x + y = 25
(a) By drawing a linear graph, write down one solution to the simultaneous equations
x + y = 25 and y = x
Answer x = ................ , y = ................ (2)
2 2
2 2
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(b) Explain why the x-coordinate in your answer to part (a) is an approximation of
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(Total 4 marks)
Q106. (a) The diagram shows a circle with centre O. TS is a tangent.
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Work out the value of x.
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Answer ............................... degrees (1)
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(b) The diagram shows another circle.
Write down the value of y.
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Answer ............................... degrees (1)
(Total 2 marks)
Q107. A baker is weighing out amounts of bread dough using a machine. The machine is set to weigh 400 grams of dough. The amounts of dough are each within 10% of the weight set. 300 amounts of dough are produced by the machine.
What is the maximum total amount of bread dough that could be produced? Give your answer in kilograms.
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Answer ......................................... kg (Total 3 marks)
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Q108. The diagram shows an accurate scale drawing of the plan view of a house and garage. The drawing is on a centimetre grid. The front of the house is 12.5 metres long.
(a) Explain why the scale is 1 centimetre represents 2.5 metres.
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(b) The owner of the house is designing an extension to be joined to the back wall of the house.
The extension is rectangular and is 7.5 metres long and 5 metres wide. It must be at least 2.5 metres from the garage.
Make an accurate sketch of the possible extension on the grid. (3)
(Total 4 marks)
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Q109. The diagram shows a rhombus made of two triangles X and Y. M is the midpoint of diagonal AC.
(a) Describe fully a single transformation that maps triangle X onto triangle Y.
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(b) Describe fully a different single transformation that maps triangle X onto triangle Y.
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(Total 5 marks)
Q110. AB is a diameter of the circle, centre O. ABC is a straight line. DTC is a tangent to the circle at T. Angle BCT = 32° and angle TAB = x°
Not drawn accuratelly
Not drawn accurately
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Find the value of x. Give reasons for all angles you write down or calculate.
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Answer .................................... degrees (Total 4 marks)
Q111. This right-angled triangle has sides of lengths (x – 2) cm, (x + 5) cm and 10 cm.
Not drawn accurately
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Calculate the value of x. Give your answer to an appropriate degree of accuracy.
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Answer ........................................... cm (Total 5 marks)
Q112. An old windmill is the shape of a truncated cone. The mill is 12 metres high and has 4 floors, equally spaced. The diameter of the ground floor is 8 metres and the diameter of the roof is 6 metres.
The mill is for sale. This is the advert.
DEVELOPMENT OPPORTUNITY OLD MILL FOR SALE
Over 150 square metres of floor space
Not drawn accurately
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Is the claim about floor space justified? You must show your working.
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Q113. 251 is a prime number.
(a) (i) Write down √251 Give your answer to 1 decimal place.
Answer .............................................. (1)
(ii) Explain how to test that 251 is a prime number.
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(b) (i) Express 2008 as a product of its prime factors.
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Answer .............................................. (2)
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(ii) Write down all the factors of 2008
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Answer .............................................. (1)
(c) (i) Show that (x + y)(x – y) = x – y
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................................................................................................................ (1)
(ii) Hence find both pairs of integers x and y such that
x – y = 2008
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Answer x = ..................., y = ...................
x = ..................., y = ................... (4)
(Total 11 marks)
2 2
2 2
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Q114. The diagram shows two right-angled triangles. AD = 40 cm CD = 7 cm
cos x =
Find the value of sin y.
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Answer .............................................. (Total 6 marks)
Not drawn accurately
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Q115. OABC is a quadrilateral.
P, Q, R and S are the mid-points of OA, AB, BC and CO respectively.
= 2a, = 2b and = 2c
Not drawn accurately
(a) Write down, in terms of a and b, the vector .
Answer .............................................. (1)
(b) Write down, in terms of c and b, the vector .
Answer .............................................. (1)
(c) Show that = = b
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(d) Using your answer to part (c) write down a geometrical fact about the line joining the mid-points of two sides of a triangle.
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(e) What type of quadrilateral is formed by joining the mid-points of the four sides of a quadrilateral?
Give a reason for your answer.
Type of quadrilateral ..................................................
Reason............................................................................................................
......................................................................................................................... (2)
(Total 7 marks)
Q116. The rule for this sequence is that each term is the mean of the two previous terms.
(a) Find an expression for a in terms of x and y.
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Answer .............................................. (2)
a
b
(b) Find an expression for b in terms of x and y. Simplify your answer.
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Answer .............................................. (2)
(Total 4 marks)
Q117. Four identical circular discs fit into a rectangle 10 cm long.
x y
Not drawn accurately
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Ten of the same discs fit into a rectangle 22 cm long.
24 discs are placed together in the same way.
How long is the rectangle?
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Answer ..............................................cm (Total 3 marks)
Q118. Joe uses a ruler and compasses to find the centre of the circle drawn below. He starts by drawing a chord on the circle.
Complete Joe’s construction to find the centre of the circle.
(Total 3 marks)
Not drawn accurately
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Q119. The rule for continuing a Fibonacci sequence is to add the last two terms to make the next term.
For example, the sequence that starts 1, 1, … continues as 1, 1, 2, 3, 5, 8, …
Two other Fibonacci sequences start a, 2a, …… and b, 4b, …
The fifth terms of these two sequences are equal.
Given that a + b = 11, work out the values of a and b.
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Answer a = ...................... b = ....................... (Total 4 marks)
Q120. The diagram shows a cone.
The diameter of the base of the cone is x cm.
The height of the cone is also x cm.
The volume of the cone is V cm .
3
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Find a formula for x in terms of V and π.
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Answer ....................................................................... (Total 4 marks)
Q121. ABCDEF is a regular hexagon.
AFGH and AJKB are squares.
Not drawn accurately
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Show that triangle AHJ is equilateral.
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Q122. Solve the equation 3x – 5x – 7 = 0
Give your answers to 2 decimal places.
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Answer ....................................................................... (Total 3 marks)
2
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Q123. Triangle ABC has a right angle at B.
Angle BAC = 38°
AB = 7.21 cm
Calculate the length of BC.
Give your answer to an appropriate degree of accuracy.
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Answer ................................................................. cm (Total 4 marks)
Q124. The diagram shows the graph of y = sin x° for 0 ≤ x ≤ 360
Not drawn accurately
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(a) Write down a possible equation of the following graph.
Answer ....................................................................... (1)
(b) Write down a possible equation of the following graph.
Answer ....................................................................... (1)
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(c) Write down a possible equation of the following graph.
Answer ....................................................................... (1)
(Total 3 marks)
Q125. The diagram shows a hollow cylinder and a solid sphere.
The radius of the cylinder = 3 cm
The radius of the sphere = 3 cm
The height of the cylinder = 6 cm
Not drawn accurately
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The sphere just fits inside the cylinder as shown.
Work out the volume of the space left inside the cylinder.
Give your answer in terms of π as simply as possible.
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Answer ................................................................ cm
(Total 5 marks)
Not drawn accurately
3
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Q126. (a) In the diagram O is the centre of the circle.
What is the value of x?
Answer .......................................................... degrees (1)
(b)
What is the value of y?
Answer .......................................................... degrees (1)
(Total 2 marks)
Not drawn accurately
Not drawn accurately
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Q127. The diagram shows two identical shapes A and B.
Describe fully the single transformation which takes shape A to shape B.
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Q128. In the diagram AB and CD are parallel.
Not drawn accurately
(a) Write down the value of x.
Answer .......................................................... degrees (1)
(b) Work out the value of y.
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Answer .......................................................... degrees (2)
(Total 3 marks)
Q129. This is the graph of y = x – 4x + 1
2
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By drawing an appropriate linear graph, solve the equation x – 5x + 3 = 0
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Answer ................................................ (Total 4 marks)
Q130. (a) A test tube is formed from a cylinder and a hemisphere as shown.
Work out the total volume of the test tube.
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Answer .......................................... cm
(4)
2
3
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(b) The test tube is filled with water to a depth of d cm, as shown in the next diagram.
The water occupies exactly half the full capacity of the test tube.
Work out the value of d.
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Answer ............................................. cm (4)
(Total 8 marks)
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Q131. O is the centre of the circle.
Angle PRS = 134°
Not drawn accurately
Work out the size of the reflex angle POQ.
You must show your working.
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Answer .................................. degrees (Total 3 marks)
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Q132. Triangle ABC has AB = 6 cm, AC = 10 cm, BC = 14 cm
Not drawn accurately
Calculate the largest angle in the triangle.
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Answer .................................. degrees (Total 3 marks)
Q133. Katy is using the quadratic formula to solve a quadratic equation.
After correctly substituting the values, she writes
(a) What is the quadratic equation Katy is trying to solve?
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Answer ............................................... (3)
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(b) Explain why Katy will not be able to find any solutions to the equation.
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(Total 4 marks)
Q134. (a) The right-angled triangle has sides shown.
Not drawn accurately
Show that x = 9 cm
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(b) This right-angled triangle has sides n, m and n + 1.
m and n are integers.
Prove that m must be an odd number.
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(Total 7 marks)
Q135. Here are four equations of graphs.
A y = 3x + 2 B 2x + 3y = 6 C y = 3x D y = x
(a) Here are three sketch graphs.
Match each graph to its equation.
Equation ................... Equation ................... Equation ................... (3)
2 3
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(b) On the axes below, sketch the graph of the other equation.
(1)
(Total 4 marks)
Q136. A gold bar has a trapezium cross-sectional area.
The dimensions are shown in the diagram.
Not drawn accurately
(a) Calculate the cross-sectional area of the gold bar.
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Answer .......................................... cm
(2)
2
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(b) Gold has a density of 19.3 grams per cm .
Work out the mass of the gold bar.
Give your answer in kilograms.
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Answer ........................................... kg (4)
(Total 6 marks)
Q137. (a) Factorise x + 10x
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Answer ............................................... (1)
(b) Factorise y – 36
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Answer ............................................... (1)
(c) Solve the equation 5w + 6 = 9 – w
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Answer w = ......................................... (3)
3
2
2
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(d) Solve the equation
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Answer x = ......................................... (4)
(Total 9 marks)
Q138. A restaurant serves garlic bread.
All the garlic breads are circular and the same thickness.
They can be made with different diameters as shown.
Robert is going to order a 14-inch garlic bread.
The restaurant has a special offer.
Special Offer
Get one 7-inch garlic bread and one 10-inch garlic bread for the same price as a 14-inch garlic bread.
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Robert says that if he has the special offer he will get less garlic bread.
Is Robert correct?
You must show your working.
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Q139. You have a square piece of paper which is folded in half to form a rectangle as shown.
The perimeter of the rectangle is 39 centimetres.
What is the area of the square you started with?
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Answer ……................................... cm
(Total 4 marks)
2
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Q140. Is the statement below always true, sometimes true or never true?
Tick the correct box.
The circumference of a circle of diameter 10 cm is greater than the perimeter of a triangle with a base 10 cm.
Explain your answer
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Q141. Triangle T is drawn on the grid.
(a) Draw the image of T after a rotation of 90° anticlockwise about O. (3)
Always true Sometimes true Never true
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(b) The triangle T is reflected to form a new triangle S.
The coordinates of S are (–4, 4), (–3, 3), and (–4, 1).
Work out the equation of the mirror line.
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Answer .............................................. (2)
(Total 5 marks)
Q142. The radius of the Earth and the radius of Jupiter are in the approximate ratio 1 : 11.
The mass of the Earth and the mass of Jupiter are in the approximate ratio 1 : 320.
You will need the following information.
• The Earth and Jupiter are spherical
• The volume of a sphere of radius r is
(a) Show that the approximate ratio of the volume of the Earth to the volume of Jupiter is 1 : 1331.
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(b) You are given density =
Work out the approximate ratio of the average density of the Earth to the average density of Jupiter in the form 1 : n
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Answer 1:................................................ (2)
(Total 3 marks)
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M1. (a) M1
2880 implies M2 M1
or k = 2880 and equation seen using k
A1
(b) M1
(W = their =) their 9.79.... A1 ft
so need 10 waiters. ft answer rounded up if M1 awarded.
A1 ft [6]
M2. (a) 65 B1
angle at centre B1
(b) 115 ft 180 – their 65 provided reason given is not contradictory
B1 ft
Opposite angles (of cyclic quad) or other valid explanation eg x + y = 180
B1 [4]
M3. (a) Correct enlargement
B1 for enlargement any scale factor (not 1) Accept any orientation
B2
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(b) 36 ÷ 4 or 3 × 3 or 54 ÷ 6
M1
9
SC1 for their (SF in (a))2
Accept ratio 1 : 9 or 9 : 1 A1
[4]
M4. AC2 = 72 + 92 – 2 × 9 × 7 × cos 75
M1
AC2 = 97.3888…
Accept 97.4 A1
DC2 = (their AC)2 – 62
M1
DC = 7.8(35…) or 7.84
A1 ft
29.8(35…) or 29.84 or 30 with correct working
A1 ft [5]
M5. (a) (i) – b + a or a – b B1
(ii) b – a
oe B1
(b) oe
M1
= a + b – a A1
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(c) = 2or OBN a straight line or BN = OB or B is midpoint of ON
B1 [5]
M6. (a) 360 ÷ 8
or 45 seen or 6 × 180 or 1080 or (2 × 8 – 4) right angles M1
180 – (their 45) (their 1080) ÷ 8
M1 dep
135 135
A1
(b) 360 – (their 135 + 135) or 2 × 45 M1
90° in X A1
Sides of X are equal or (regular) octagons so sides are equal 4 lines of symmetry or rotational symmetry of order 4 scores 3 marks Other symmetry scores B1
B1 [6]
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M7. (3x + 2)(x + 1)
Rectangle M1
3x2 + 5x + 2
Rectangle A1
x × 3x + 5(x + y) or x × 3x + x × 5 + y × 5 or x(3x + 5) + y × 5 or (3x + 5)(x + y) – 3x × y
M1
3x2 + 5x + 5y
L shape A1
5y = 2 oe dependent on a previous M1 and a term in y
M1 dep
0.4 oe
A1 [6]
M8. (a) 6 B1
(b) Plot points B1
Draw curve B1
(c) x = 1.4 and –1.4 B1
(d) (3x2 – 6) – (3x2 – 5x – 6)
Sight of (±) 5x (+ k) M1
= 5x
Draw y = 5x B1 ft
x = 2.5, –0.8 Accept 2.4 to 2.55 and –0.75 to –0.85
A1 [7]
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M9. (a) A ∞ d 2 or A = kd 2
When d = 15, A = 90 M1
90 = 225k
k = 0.4 or M1
A = 0.4d 2
Accept A = oe A1
(b) d = 20 =>
A = 0.4 × 202
A = 160 160 unsupported SC1
A1
(c) A = 250
250 = 0.4d 2
d 2 =
Dep on M2 in (a) M1
= 625
Accept M1 dep
d = 25 A1
[7]
M10. (a) (0, 1)
Generally marked B1
(b) Matching any (non zero for x) Values eg, a1 = 3, a2 = 9, etc
Must show as power M1
a = 3 A1
[3]
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M11. Volume of one (or two) spheres
Allow 10 for r for M1 M1
= 523.6 (1047.2) {524} {1048, 1050}
oe 500π/3 or 1000π/3 (523.3 to 523.7) or (1046.6 to 1047.4)
A1
Volume of cylinder
Allow 10 for h or 10 for r for M1 (not both) M1
= 1570.8 {1570, 1571}
oe 500π (1570 to 1571) A1
Volume remaining (1570.8 – 1047.2 =) 523.6, 524
oe
Due to different values of π an answer between 523.3 and 523.7 gets full marks ft If both Ms awarded and one value is correct.
A1 ft [5]
M12. Breaks down into areas of rectangles and areas of (quarter) circles
Any combination of rectangles and circles
or 12.56... or is enough evidence for area of circles NB 12.56 from 2 × π × 2, if seen is M0 NB 3.14 on its own is not evidence of the area of a quarter circle as it is π
M1
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Uses an ‘addition’ method (method 1) and finds
Area of one (or 5) ‘external’ quadrants
or
or Uses a ‘subtraction’ method (methods 2 and 3) and finds 5 × area one quadrant
or M1 dep
= 0.8584..., {0.9, 0.86, 0.858}
or = (× 5) 4.292... {4.3, 4.29} 15.71, 15.7
A1
52.3 or 52.29... 68 – 5π Allow 52 if 52.3 or 52.29... or a full method seen
A1 [4]
M13. (a) or or M1
k = 3200 or M1
P = 3200/Q or PQ = 3200 or Q = 3200/P A1
(b) Correct sketch graph B1
(c) (Their 3200) or 2Q = (Their 3200) ÷ Q or Q = (Their 3200) ÷ 2Q
M1
(Q =) 40 ft Their value of k
A1 ft [6]
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M14. (a) B1
(b) B1
(c) oe
B1 [3]
M15. (a)
Or M1
l = 2r Clearly shown since answer given
A1
(b) h2 = 4r2 – r2
Attempt to use Pythagoras’ theorem correctly M1
h = √3r
h2 = 3r2 is sufficient or h = √(3r2)
A1
(c) √3r :
ft with Their h if 1st M1 earned
M1 dep
√3 : 2 A1
[6]
M16. (a)
Multiply through by x, 4 = x(9 – 2x) is enough M1
Expanding and rearranging must be seen (answer given) A1
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(b) Attempt at
identified as being required line M1
Points worked out eg, table of values (2 points minimum)
M1 dep
Correct line plotted With ruler, must intersect the curve twice
A1
(c) Solutions can come from factorising
ie, B1
x = 4 No ft from incorrect factors
B1 [7]
M17. or Condone use of π = 3.(14...)
M1
and M1
(Their 50π) – (Their 25π) M1 dep
25 A1
[4]
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M18. (a) y = cosx + 1
y = 1 + cosx B1
(b) y = 2cosx B1
(c) y = cos2x B1
(d) y = cos(90 – x), y = cos (x +270) y = cos (x – 90) or y = sin x
B1 [4]
M19. ABD = 66 (Alt segment) or angles in triangle if ADB found first
B1
DCB = 104 (opposite in cyclic) In all alternatives, for first 3 B marks do not award B1 the first time no reason or wrong reason given, otherwise accept angles identified in answer or on diagram. NB Mark ‘positively’ ie, ignore wrong values or reasons unless totally contradictory.
B1
DBC = 38 (isosceles)
CBA = 104 B1
CBA + BAD = 180 (interior) In all alternatives, reason must be given for final B1 Accept ‘allied’ or ‘angles between parallel lines’. Dependent on correct angles.
B1
Alt. 1
ADB = 38 (Alt segment) B1
DCB = 104 (opposite in cyclic) B1
CBD = 38 (isosceles) B1
CBD = ADB (alternate)
Use of ‘Z angles’ is not acceptable Dependent on correct angles
B1
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Alt. 2
ADB = 38 (Alt segment) B1
DCB = 104 (opposite in cyclic) B1
BDC = 38 (isosceles)
ADC = 76 B1
BDC + BCD = 180 (interior) Dependent on correct angles
B1
Alt. 3
ADB = 38 and ABD = 66 (Alt segment) B1
DCB = 104 (opposite in cyclic) B1
CBD = CBD = 38 (isosceles) B1
DCB = CBA and CDA and BAD = (isosceles trapezium) B1
[4]
M20. (a) 0.007 B1
(b) (i) 0.9119215(052) B1
(ii) 0.9, 0.91, 0.912, 9 or 9.1 or 9.12 × 10-1
ft their answer for (b)(i) to 1, 2 or 3sf eg Gradians (b)(i) 0.02221673729 B0
(b)(ii) 0.02, 2 × 10–2, etc B1 ft
B1 ft
(c) 0.00805 or 8.05 × 10-3
B1 [4]
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M21. (a) π r 3 = 2 × π r 2 x
Must include the factor of 2 Allow use of h instead of x
M1
Simplified to give x = 2r Alternatively Allow substitution of 2r for height of cone and verification of result
ie 2 × Vol cone = 2 × × π × r 2 × 2r M1
= π r 3 (must be seen) A1
A1
(b) (l)2 = r 2 + 4r 2
(l)2 = r 2 + (2r)2 is M1
(l)2 = r 2 + 2r 2 is M0 M1
(l) = √5 r A1
Surface area cone = π × r × √5 r Using their l if from an attempt at Pythagoras
M1
4 : √5 Allow √5 : 4 SC2 for a complete numerical solution
A1 [6]
M22. (a) Correct Pythagoras in two appropriate right-angled triangles
or simply BH ² = 12² + 3² + 4² M1
13 A1
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(b) HB = 13, HC = 5 or DB = √153 with attempt at trig. Ratio
Explanations may not involve any calculations eg BC < BD or HC > HD together with some comparison such as BH is common (diagrams drawn, to illustrate, are appropriate)
M1
Two correct, comparable trig. ratios
eg sin x = and sin y =
For example: BH is common and triangles BHD and BHC are right-angled, so y must be bigger because the height is greater
A1
y Good explanation and correct conclusion … this earns all 3 marks
A1 [5]
M23. YZ = ZY B1
Angle MZY = angle NYZ base angles of (Isosceles) ∆ XYZ
Note Reason necessary eg you might see If XZ = XY then angle XZY = angle XYZ
B1
Angle MYZ = angle NZY B1
Triangles congruent, ASA Note Dependent on earning first 3 marks Must give correct reason for congruence (ASA) Only allow AAS if complete argument stating ‘third angles equal’
B1 dep [4]
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M24. Sight of correct ratio or scale factor ie 20 : 30, 2 : 3, , 1
oe sight of or earns this mark M1
× 40
oe eg might work out then subtract M1
24 Note 2 : 3 ratio might be scaled up to give ratio of 16 : 24 (M1, M1) Must state h = 24 for A1 Alternatively h/30 = (40 – h)/20 M1 20h = 30(40 – h) 20h = 1200 – 30h M1 50h = 1200 h = 24 A1
A1 [3]
M25. (a) M1
A1
80.2(1...) r= 12 giving 40.1 is Ml, Al, A0 r = 3 giving 160.4 is Ml, Al, A0
A1
(b) M1
h = 9 (cm) h = 12 gives Ml, A0
A1
Ml for difference of two cone volumes Al if all correct
M1, A1
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(V) = 327 or 326.7 ............................(cm3)
Accept 330 if working seen, ft their h if both M's awarded.
A1 ft [8]
ALTERNATIVE
linear scale factor 1:3 Must be used. Just writing it down does not qualify as a method unless progress is made.
M1
Volume scale factor 1:27 A1
Volume small cone M1
Volume large cone 27 × (their 12.566) 339.292...
DM1
(V) = 327 or 326.7 ............................(cm3)
Accept 330 if working seen. A1
Scs
12.566 only B1
339.29 only M1, A1, M1
M1,A1
[8]
M26. (a) y = x+ 1, y = – x – 2
x + 1= –x – 2 B1
–x – 2 = x2 + 2x – 3
M1
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(b) x2 + 3x – 1 = 0
Simplified to 3 terms in x2, x and constant
e.g. x2 = 1 – 3x
A1
x2 + 2x – 3 = × + 1
M1
(c) 1.6, –2.6 Accept 1.5 to 1.6, – 2.5 to – 2.6
A1 [5]
M27. M1
A1
oe e.g. A1
[3]
M28. (a)
Accept M1
Sin B = 0.9679(1...) A1
B = 75.4(...) A1
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(b) x2 = 222 + 232 – 2 × 22 × 23 × cos 48
M1
x2 = 335.8(...)
A1
x = 18.32(....)
ft only if an error made in calculation of x2
but not on (222 + 232 – 2 × 22 × 23 ( = 1)) cos 48
(=√ 0.669 = 0.818) A1 ft
18 or 18.3 Independent mark. Award if value > 3sf seen or calculation seen.
B1 ft [7]
M29. (a) 4a + 3 = 2b + 5 M1
(b) 4a – 2b = 2 (-2) Must indicate division by 2
A1
4a+3+2b+5+2a+b=32
6a + 3b = 24
2a + b = 8 Bl for any version
M1
(1) × 3: 6a – 3b = 3 M1
12a = 27 For attempt to eliminate AB or 4a + 3 =12 and BC or 2b + 5 = 12
M1
a = 2.25 A1
[6]
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M30. (a) Sight of tan unless alternative method used M1
Tan–1(5.59/1.5)
90° – tan–1(l.5/5.59), 1.5tan70 and 1.5tan80
DM1
74.(98) or 75° so safe 4.1(2) and 8.5(1)
A1
(b) Sight of cos M1
4 × cos80 DM1
0.69 0.7 with working
A1 [6]
M31. Vol Hemisphere =
Ml, M1 M1
144π
(3 × 4 × 63 × π) ÷ (2 × 3 × 27 × π) = r2 or...
A1
their 144π
...(4 × 63) ÷ (2 × 27) = r2 A1
M1
(r =) 4 (r =) 4
A1 [4]
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M32. angle OSQ = angle OQS = 50°
Isosceles triangle OQS Penalise ‘no reason’
B1
angle OSR = 90° → angle QSR = 40° Tangent-radius property first time only
B1
angle QSR = angle QRS (Isosceles) B1
[3]
M33. (Graph 1) D B1
(Graph 2) A B1
(Graph 3) E B1
(Graph 4) C B1
[4]
M34. (a) Line y= 1 drawn or points on curve
Accept y = 1 written in body of script. M1
0.8, –3.8 (±0.1) A1
(b) Attempt to split equation into
x2 + 3x – 2 = ax + b
Or x2 + 3x – 2 -(x2 + 2x – 1)
Or x2 + 3x – 2 + ax + b = x2 + 2x – 1 M1
Line (y = x – 1) drawn A1
0.4, –2.4 (±0.1) f.t. on their line if Ml awarded, e.g. y = x + 1(1, –3), y = 1 – x(0.6 (0.7), –4.6 (–4.7)),y = –1 –x(0.2, –4.2)
A1 ft [5]
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M35. (a) Linear scale factor is 2
Allow numerical examples but must be complete 23 = 8, 43 = 64, 64 = 8 × 8, but the increase by a factor of 8 must be shown and not assumed B2
B1
Volume scale factor is lsf3
Allow algebra (2x)3 = 8x3
B1
(b) (14.5 ÷ 8)3 or 1.81253
3√6 = 1.817 83 × 6, 14.53 ÷ 6
M1
=5.95(4)
8 × 1.817 ≈ 14.5 ≈ 14.53 ≈ 83
A1
Volume increases by about 6 × so claim justified. Allow ‘Almost but not quite’
A1 [5]
M36. Use of
Must use 4 or 8 as radius. M1
(Volume hemisphere =) 133.9 to 134.1 (inclusive)
133.97 if π = 3.14 used. A1
(Volume paperweight =) 500+(their 134) (=634)
If Ml awarded. A1 ft
cm3
This mark is independent B1
[4]
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M37. Area semicircle – area circle
Accept π × 102 ÷ 2 and/or π × 52 for M1
M1
π52 ÷ 2 – π(2.5)2 (= 12.5π – 6.25π = 39.27 – 19.63)
Accept fractions, decimals or in terms of π A1
19.6(....)(= 6.25π) ft on one error only, e.g. Accept fractions, decimals or in terms of π. Use of π as 3.14 gives 19.625 Al Common errors e.g.
π × 102 ÷ 2 – π × 52 = 157.1 – 78.5 = 78.6
= Ml,A0,A0ft. π × l02 ÷ 2 – π × 2.52 = 157.1 – 19.6 = 137.5 = M1,AO,A1 ft
π × 52 ÷ 2 – π × 52 = 39.3 – 78.5 = –39.3
Ml, AO, AO (non-sensible answer) A1 ft
[3]
M38. y = kx2 or y a x2
oe 5 = k × 16 M1
k = 0.3125 oe
A1
20 A1
[3]
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M39. Breaks problem down into sum of lines and (semi-)circles M1
Length of lines 4.1 + 5.9 + 4.7 + 2.9 (= 17.6) Sc 17.6 only B1
A1
Use of 2 πr ÷ 2 or πd ÷ 2 but must use with numbers.
DM1
Length of semi-circles 0.9π + 0.6π + 0.7π (= 6.9(11..))
2.8, 1.9,2.2 A1
Total = 24.5(...) ft on 1 arithmetical or ‘reading from scale’ error and both M's awarded. 4.1 = 2.9 + 0.6 + 0.6, 5.9 = 0.6 + 0.6 + 2.9 + 1.8, 4.7 = 2.9 + 1.8, 2.9 = 2.9
A1 ft [5]
M40. (a) BC2 = 192 – 92(= 280)
x2 + 92 = 192
M1
BC = √280 For squaring, subtracting and evidence of square rooting
DM1
BC = 17 or 16.7(....) 17 with no working gets 3
A1
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(b) Sight of tangent M1
or
Angle = tan–l (1 ÷ 24)
tan –1 (0.458)
M2 for any complete correct method Sin = 11/√697 or 11/26.4 Cos = 24/√697 or 24/26.4
DM1
25 or 24.6(....) 25 with no working gets 3 Radians 0.43 gradians 27.35 Penalise on first occurrence only.
A1 [6]
M41. 100 ÷ 60 or 80 – 50
1.66 or 1.6 M1
100/60 × 60 or 80/50 × 60 100 min or 96 min
DM1
Their(100 – 96) or reversed DM1
4 A1
[4]
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M42. (a) (2x – 3)2 = 4x2 – 6x – 6x + 9
condone one error 4x2+ 9 is two errors
M1
4x2 – 12x + 9 = 8x – 16
or 4x2 – 20x +25 (= 0)
for equating expressions and/or simplifying this must lead to a quadratic equation
M1
(2x – 5)(2x – 5) (– 0) ft from their quadratic equation (if 'formula' used, substitution must be completely correct)
M1
x = 2.5 A1
y = 2 A1
(b) Only one solution so straight line must be a tangent to the curve Hence sketch 2
ft from their solution(s) to (a)
clear solution(s) in (a) ‘correct’ sketch in (b) can earn B1 (no explanation) or B2 (with explanation)
B2 ft
ALTERNATIVE
(a) oe for setting up attempt to eliminate x
M1
y2 – 4y + 4 (= 0)
oe condone one error this must lead to a quadratic equation
M1
(y – 2)(y – 2)(–0) ft from their quadratic equation (if ‘formula’ used, substitution must be completely correct)
M1
y = 2 A1
x = 2.5 A1
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(b) Only one solution so straight line must be a tangent to the curve Hence sketch 2
ft from their solution(s) to (a)
clear solution(s) in (a) ‘correct’ sketch in (b) can earn B1 (no explanation) or B2 (with explanation)
B2 ft [7]
M43. (a) x = 110
430° scores B1 B1
(b) x = 250, x = 290 B1 for each
B2
(c) correct sketch of ‘double’ cycle B1 for sketch only as far as 180 or slight inaccuracy
B2
(d) x = 35, x = 55 must have 35 and ½ of their (a)
B1
x = 215 B1 ft
x = 235 B1 for each of (their 35 + 180), (their 55 + 180)
B1 ft [8]
M44. (a) 0 = 4 + 2p + q –5 = 9 – 3p + q
for substitution of both sets of coordinates allow one error
M1
5 = –5 + 5p oe for correct attempt at elimination of p or q
DM1
p = 2 A1
q = –8 p = 2 and q = –8 from no obvious working scores 4
A1
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(b) Solving their x2 + px + q = 0
if ‘formula’ used substitution must be completely correct M1
(–4, 0) A1
[6]
M45. (a) π(×)52
condone 3.1... × 52
M1
π (×) 52 × 10 or (their area) × 10π
(×) 52 × 10 or (their area) × 10
condone 3.1... × 10 × 52
their area must contain π M1
250π or 250 × π or π × 250 775 to 790 scores M2 A0 do not accept π250 ignore fw 250π can be recovered in (b)
A1
(b) 40 × 50 10 × 10 × 10 40 × 50
M1
their 2000 × 10
their 1000 – their 250π 20 × their (π × 52)
M1
20 × their 250π 20 × their (1000 – 250π) their 2000 – their 500π
M1
20000 – 5000π
20(1000 – 250π) 10(2000 – 500π)
4290 to 4500 scores M3 A0 ignore fw except 15000π
A1 [7]
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M46. (a) (i) a + b
b + a B1
(ii) 2a + b b + 2a
B1
(iii) b – a – a + b
B1
(b) CF = OE M1
a + 2b a + b + b oe
A1
(c) Straight line because OD = a + b
B1
3 times bigger because OF = 3a + 3b
B1 [7]
M47. 2.75 × 27
M1 for 33
M1
74.25..... Accept 74 or 74.3 Height first pyramid is 2.0625 Height second is 6.1875
Volume is × 62 × 6.1875 = 74.25
This line M1, A1. A1
[2]
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M48. Angle ATB = 13° B1
M1 for use of sine rule, A1 for correct substitution.
= M1, A1
BT = 27.47 (41539..) AT = 45.79112344
A1
H = BT × sin31 M1
H=14.2 or 14.15(........) Ft only if both Ms awarded. NB 14.2 can come from BT = 27.5 or AT = 46 Deduct 1 for pa if seen.
A1 ft [6]
M49. (a) Parallel curve translated up y axis
‘2’ need not be marked, needs to look symmetrical
B1
(b) Parallel curve translated in positive direction along x axis Must 'sit on' x axis and look symmetrical
B1
(c) Curve through (0,0) nearer to x axis than original Must look symmetrical
B1 [3]
M50. (a) (i) y = 3x – 2 plotted
must draw correct line M1
x = 2, x = 5 A1 for each, must be correct answers...no ft. coordinates given ... lose 1 mark
A2
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(ii) x = 2, x = 5 must have both solutions (ft answers from part (a) earns 1 mark)
B1
(b) x2 – 4x +8 = x + 4
allow one slip in manipulation M1
y = x + 4 Straight line to be clearly stated
A1 [6]
M51. (a) 4 B1
(b) (32 – 4 – 4 – 5 – 5) (÷ 2) or 14 or 16 – 4 – 5
or equivalent M1
7 A1
[3]
M52. Any correctly evaluated counter example with non-prime conclusion.
Examples –4 and 5 => 25 and not prime –5 and 6 => 35 and not prime –8 and 9 => 77 and not prime accept any indication of "not prime"
B1 any correctly evaluated trial with no conclusion Examples – 3 and 4 => 17 –4 and 5 => 25 –5 and 6 => 35 –6 and 7 => 47 –7 and 8 => 61 –8 and 9 => 77 or incorrectly evaluated trial that gives a counter example with non-prime conclusion
B2 [2]
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M53. Extra volume = 50 × 34 × 4.5
= 7650 M1
1912.5 A1
r3 = their 1912.5
Dependent on ‘their 1912.5’ coming from a volume calculation. M1
r3 = (3 × their 1912.5) ÷ 4π
Allow (3 × 7650) ÷ 4π DM1
R = 7.7, 7.70, 7.700… A1
[5]
M54. AC2 = 72 + 92 – 2 × 7 × 9 × cos75
M1
AC2 = 97……. , AC= 9.9, 9.86……
A1
Their AC2 – 62
AC2 must be > 36 = 61.38888 if correct
M1
DC = 7.8(3…) Answer must be accurate to 2 sf or better
A1 ft
Perimeter = 29.8(.…) ft their DC + 22 but both Ms must be awarded.
A1 ft [5]
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M55. Angle at centre = 2 × sin–1( )
Half angle sin–1( ) gets M1
M1
= 71.(...) A1
Area sector = ‘their 71’ ÷ 360 × π × 62
M for use of area sector formula not for πr2 ÷ 4 for example.
M1
Area sector = 22.4(….)_ A1
Their sector – their triangle area Must make a valid attempt at calculating the area of the triangle. (17.06…) and at least one of the previous M marks must be awarded.
DM1
Area segment = 5.3….. A1
[6]
M56. (a) 0.51(2) B1
(b) Correct plots B1 ft
Smooth curve ±0.5 square Use of ruler or ”double‘ lines or discontinuities B0
B1 ft
(c) 1.2 ft their graph. If ”double‘ line at y = 0.76 then B0. Within tolerance of their graph
B1 ft [4]
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M57.
B2 fully correct B1 for any translation of correct answer. Alternative scheme. M1 for ”rays‘ from at least 3 corners through (–1, 0) and attempt at drawing a reduced shape in 3rd quadrant. A1 if correct shape
B2 [2]
M58. C = π × 7
C = 2π × 3.5 Must substitute numbers. C = πd or 2πr is M0 until used. NB π × 3.5 is M0 as wrong method (πr)
M1
= 21.98 – 22
3.14 × 7 = 21.98, × = 22 A1
Length = 22.98 to 23 ft their 21.99 + 1 if M1 awarded.
A1 ft
Height = 10 cm Allow answers transposed.
B1 [4]
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M59. Trial for x > 4
All trials correctly evaluated to at least 1 d.p., rounded or truncated. NB Condone odd error as this may be ”recovered‘ later.
B1
Trial for 4 < x ≤ 5 5 → 5.2, 4.5 → 4.72, 4.6 → 4.81, 4.7 → 4.91
B1
Trials for 4.7 = ≤ x ≤ 4.85 and answer 4.8 4.75 → 4.96, 4.76 → 4.97, 4.77 → 4.979…, 4.78 → 4.989…, 4.79 → 4.998…, 4.8 → 5.008..or 5 4.85 → 5.056
B1
Trial for 4.75 ≤ x < 4.8 and answer 4.8 NB. Minimum for full marks. e.g. test 4.75, test 4.8, state 4.8 as answer.
B1 [4]
M60. (a) angle QPR = 32°
Base angle of isosceles triangle B1
angle QSR = 32° equal to angle QPR, angles in same segment oe precise explanations for 2 marks
B1
(b) (i) 116° B1
(ii) Line from O to A, creating 90° Join OA using alt seg thm
angle OAB = 16° angle CA(X) = 48° M1 angle BCA = 74° M1
M1
angle OAC = 42° angle OAC = 42° M1 angle BCO = 32° M1
M1
angle OCA = 42° angle OCA = 42° A1 angle OCA = 42° A1
A1 [6]
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M61. Kite 2 and reason e.g. 110 + 70 = 180 sum of (opposite) angles = 180
B1 Kite 2 B2
[2]
M62. Height = 100 × 3 ÷ 25
Alternative: B1 for length of median of side M1
= 12 (= 12.25)
A1
Diagonal base = √50 = 7.07 (or half base = 3.535...) M1 for pythagoras on half of side triangle.
B1
x = √(122 + 3.5352)
x = √(12.252 + 2.52)
M1
= 12.5 A1
[5]
M63. Volume cube = 203 = 8000
B1
Square hole = 102 × 20 = 2000
B1
Circular holes = 2π × 42 × 5 = 502.7
B1
Volume left = 5500 ft 8000 – 2000 – their cylinders. Only ft if attempt made to find volume circular holes not using 20 cm as height (e.g. 5 or 4 or 10)
B1 ft
cm3
Units mark independent B1
[5]
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M64. cos A =
If other angles found answers are (to 1d.p) C = 70.5, B = 59.0
M1
cos A = 0.6363.. Mark as scheme with different values
A1
A = 50.5° A1
Area = 0.5 × 10 × 11 × sin 50.5 DM1
= 42.43 cm2
A1 ft [5]
M65. 27.5 × 12 – 250 (=80)
330/250 or 330/2.5 get M1 M1
cv/250 (×100) For completion of method
DM1
32% increase 32% must be stated. Special cases all get M1, DM1, A0 Misreads both as 10% => 21% Misreads both as 20% => 44% Misread both as decreases => 28%
A1
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ALTERNATIVE
1.10 × 1.20 M1 for 110% × 120%
M1
= (their 1.32) – 1 A1 for 132% or equivalent
DM1
32% increase A1 stating answer Special cases above with equivalent values.
A1 [3]
M66. (x – 5)2 – 30 = 0
For attempt at (x – 5)2,
M1
x = ±√30 + 5 For –5 and –30
A1
x = 10.48, –0.48 Both answers (Accept 10.5, –0.477)
A1
ALTERNATIVE
x =
For substitution into formula (allow one error) M1
x = Correct substitution
A1
x = 10.48, –0.48 Both answers (accept 10.5, –0.477)
A1 [3]
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M67. (a) Sight of sine 48 or cos 42 M1
x = 5.1 × sin 48
5.1 × cos 42. √(5.12 – (5.1 cos48)2) M2
DM1
x = 3.79003861 cm A1
x = 3.8, 3.79 Rounding mark is independent for a value or calculation that is 4sf or greater
B1
(b) Area = 6.8 × (their x) 5.1 × 6.8 × sin 48
M1
= 25.77 cm2
f.t. their value for x. A1 ft
[6]
M68. 62 + 2.52 ( = 42.25)
Squaring and adding M1 sight of tan M1
√42.25 Square root DM1 For 6 ÷ cos22.6 or 2.5 ÷ sin22.6
DM1
6.5 A1
AB = 4 Ft their 6.5 – 2.5 s.c. 2.95 from incorrect pythagoras B1
A1 ft [4]
M69. (a) Graph D B1
Slow, steady, faster steady Steady rate and an indication why quicker at top.
DB1
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(b) Any container with uniform vertical cross-section Allow 2-D, e.g. Rectangle, and/or substantially uniform X-section
B1 [3]
M70. Graph A is y = (x – 3)2
B1
Graph B is y (x + 3)2
B1
Graph C is y = –x2
B1
Graph D is y = 3 – x2
B1 [4]
M71. (a) BD/15 = 2/3 M1
BD =10 Or equivalent eg. BD = 2/3 × 15
A1
(b) Correct use of Pythagoras M1
BC = 8 A1 ft
sin y = 8/10 f.t. with their BD iff BD > 6 for this M mark
A1 ft [5]
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M72. 1st arc AD, capable of being used in 60° construction
M1
2nd arc from ‘correct’ point crossing 1st arc
M1 dep
Angle of 60° (± 2°) A1
‘B’ correctly marked Need not be labelled ... only award if first M1 earned
B1
Complete construction for perp. from ‘B’ to AD (i.e. all arcs) M1
‘C’ correct (±2°) Need not be labelled
A1 [6]
M73. (a) 100 B1
(b) 130 B1
(c) 70 seen B1
Full explanation. E.g 180 – (90 + 20) Minimum requirement 90 – 20
B1 [4]
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M74. 1.53 (× 480)
M1 for s.f.3 e.g. (1.5)3 or (0.666)3
M1
= 1620 A1
ml or cm3
NB 1.62 1 gets M1, A1, B1. This is an independent units mark. So 720 ml gets M0, A0, B1 cl and I are acceptable iff an attempt made to convert answer.
B1 [3]
M75. volume cylinder = 113.(...) cm3
Accept 36π B1
volume cone = 18.8(...)cm3
Accept 6π B1
Volume (their cylinder – their cone) ÷ 9π Accept 30π ÷ 9π
M1
3.3(3) A1
= 5.3(3...) Accept fraction. (5 1/3) f.t. iff M1 awarded. Consistent use of diameter for radius gives 144π (= 452.39) for cylinder and 24π (=75.40) for cone. Volume = 120π (376.99). Volume ÷ 36π = 3.333.. + 2 = 5.333 Give B0, B1, M1, A1, A1 f.t. Hence do not give full marks if answer seen on answer line. Check working before awarding full marks. Do not accept 5 as a answer.
A1 ft [5]
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M76. Angle APB = 82° B1
x2 = 182 + 252 – 2 × 18 × 25 × cos 82
Use of cosine rule 324 + 625 – 900 cos (their angle) ( = 949 – 125.25 = 823.7....) f.t their angle for M1.
M1
x2 = 823.7(..........)
x = 49 cos 82 or 6.81948.....gets M1, A0, A0 A1 ft
x = 28.7(.....) Follow through on an incorrect angle only
A1 ft [4]
M77. HF2 = 52 + 122
DF2 = 52 + 52 + 122 = 194
DF = 13.9(2....) gets M1,A1. M1
HF = 13 B2 if HF = 13 stated
A1
Correct Right angled triangle DFH Follow through their HF if first M1 awarded Do not accept lines on diagram joining DF and FH as evidence unless right angled marked or something done with lengths
M1
tan-1(5/13)
DM1 if both previous M’s awarded. DM1 for appropriate ratio if other lengths used.
DM1
=21(.0...)° Ft on HF or DF only.
A1 ft [5]
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M78. AB = DC B1
Angle EAB = Angle ECD Alternate angles or Z angles must be stated
Angle EBA = angle EDC Alternate angles or Z angles must be stated
Angle AEB = Angle DEC Opposite must be stated NB only maximum of two can be scored here.
B1, B1
Therefore congruent because ASA, AAS etc Dependent on first B1(AB=DC) but can be awarded if angles stated as equal but not justified. This final step is needed for full marks. Can be given in words e.g. Angle, Side, Angle
DB1
This is a proof and the explanation must be rigid. Other versions of the proof are (for example)
Stating that ABCD a parallelogram B1
AE = EC
DE = EB
AB = CD
Any angles as above (with justification) NB B marks for length are dependent on the first B1 but angles are not. NB 3 statements need to be made here to get both marks. Sides equal do not need justifying but angles do. NB if only 2 length statements are made and no other statements then give DB1
B2
Conclusion SAS or SSS. Dependent on first B1
DB1 [4]
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M79. Sight of sine M1
125 ÷ sin 33 Accept 125 ÷ 33sin
DM1
229(.5..........) A1
230 or 229 Follow through any value ≥ 4 s.f. or calculation seen, e.g. 125 × sin 33 = 68 or 68.1
B1 [4]
M80. Scale factor = 9 ÷6 = 1.5
Accept 9/6 or 6/9 or 4/6 or 6/4 or ratios
or M1
AB = 4 × 1.5 (= 6) Correct use of their scale factor 36 = 24 + 6x or equiv
M1
BD = their AB – 4 = 2 A1 cao
[3]
M81. (a) Attempt at translation of 45° to the right M1
P = (135,1) A1
(b) Attempt at sine curve of twice the amplitude of the original M1
P = (90, 2) A1
[4]
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M82. (a) × (2 × π × 12) M1
18π Not π18, unless notation previously penalised π × 18 is acceptable
A1
(b) 2 × π × r = their 18π Or their 18π ÷ 2π
M1
r = 9
r = of 12 = 9 scores 2 marks A1 ft
[4]
M83. (a) (i) BA = a – 2b
or equivalent B1
(ii) MQ = MB + BA = b + (a – 2b) Attempt to set up a route, must include substitution of a and b, condone one error
M1
MQ = a + b or (a + b) Need not be simplified
A1
(iii) OP = OA + AB = a + (2b – a) or OP = OB + BA = 2b + (a – 2b)
M1
OP = a + b M1 as above, need not be simplified for A1
A1
(b) OP = 2 × MQ B1
Trapezium only if accompanied by a sound reason
B1 [7]
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M84. (a) (i) 40 B1
(ii) 140 or 180 – their x Do not ft if answer = 140 in (a)(i)
B1 ft
(b) Logical and precise explanation (Either written or as a calculation)
One angle labelled or stated correctly, no reason, B1 B2
(c) angle ADB = 32° Reasons not needed in any part Alt seg.thm. In all parts accept angles
B1
angle DBC = 32° Alternate angles marked on the diagram
B1 dep
angle BDC = 32° Base angles Isos. Triangle as ‘evidence’
B1 ft
angle BCD = 116° Angle sum of triangle
B1 ft
angle BAD = 64° Opp. angles cyclic quad.
B1 ft [9]
##
All loci must be within lake
Arcs centre B radius 3 cm (±2mm) and 5cm (±2mm)
Consistent use of scale 1 cm to 100m B3 max (-1 MR)
B1
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Construction arcs for perp. Bisector
Two sets of intersecting arcs centres A and C B1
Perp. Bisector of AC (±2mm) B1
Line 2cm (±2mm) from path B1
Correct region shaded B1
[5]
M86. (a) (i) 40 B1
(ii) 140 or 180 – (their x) Do not ft if answer = 140 in (a)(i)
B1 ft
(b) Logical and precise explanation (either written or as calculation)
B1 for 1 angle labelled or stated correctly, no reason B2
(c) 24 ÷ 8 × 2 or OP = 6
M1
π(or 3.14) × (their 6)2
M1
36π or 36 × π or π × 36 allow π36 SC2 108 to 114 or π × 9 oe SC1 27 to 28.5
A1 [7]
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M87. Side of 34 B1
50 – 18 or 25 – 9 M1
Side of 32 or 16 A1
162 + x2 = 342
M1
x = 30 2 × ‘8, 15, 17’ triangle get M1, A1
A1
64 A1
[6]
M88. Outer of total
Outer shaded area = of total B1
Inner squares of total B1
Outer shaded area = 16% B1
Inner shaded area = M1
11.52% A1
Total = 27.52% A1
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Alt
Some indication that 3L = 5W B1
L side 15 and S side 9 B1
Diagram is 75 × 75 (= 5 625) B1
M1 For using either part of numerator A1 For both and adding
M1, A1
Total = 27.25% A1
[6]
M89. 360 ÷ 10 or 360 ÷ 5
or 36 or 72 or 144 or 108 NB Angles may be marked on diagram
M1
144 and 108
or 36 and 72 A1
∠BXC = 360 – (144 + 108) or ∠BXD = 36 + 72 or 108 (X is point where decagon and pentagon meet between B and C)
M1
∠XBC = ∠XCB = (180 – 108) ÷ 2 or ∠XBC = 36
M1
∠ABX + ∠XBC = 144 + 36 ( = 180) oe eg, ∠CBX calculated from ΔBXC equals exterior angle of decagon
A1 [5]
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M90. (a) Lengths CE = 3, ED = 4
These will probably be marked on diagram B1
Angle BEC = angle AED B1
3 ÷ 4.5 = 4 ÷ 6 ( or vice versa) oe B1 For any comparison of value from both triangles eg, ratio 3 ÷ 4 = 4.5 ÷ 6 ( or vice versa)
B1
(b) EF = 9 B1
9 ÷ 6 = 4.5 ÷ 3 or 9 ÷ 4.5 = 6 ÷ 3 (or vice versa) BCE and AEF have two sides in same ratio.
Angle BEC = angle AED so similar Angle AFE = angle EBC (alternate) Hence AF parallel to BC
E4 Full explanation E3 Partial explanation with 3 relevant facts E2 Partial explanation with 2 relevant facts E1 Partial explanation with 1 relevant fact
E4 [8]
M91. BC = AD = 15 B1
92 + 122 (= 225)
Reference to 3, 4, 5 triangle M1
Both AB and CD (√225) = 15 Scaling up to 9, 12, 15
A1
ABCD is a rhombus as all sides same length A1
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Alt 1
Gradient BD –2 and
gradient AC = M1
Hence BD perpendicular to AC A1
Showing any two sides = 12 M1
ABCD is a rhombus as diagonal perpendicular and two sides same A1
Alt 2
Gradient BD = –2 and
gradient AC = M1
Hence BD perpendicular to AC A1
Midpoint AC and midpoint BD = (8, 5)
Method for finding midpoint (calculation or drawing) must be shown
M1
ABCD is a rhombus as diagonal perpendicular and intersect at midpoint A1
Alt 3
Vector BD = 12i and vector AD = 12i
Vector AB = 9i + 12j and vector DC = 9i + 12j
M1, A1
Showing either of AB or DC = 12 A1
ABCD is a rhombus as opposite sides of equal lengths parallel and two adjacent sides same
A1 [4]
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M92. y = 2x – 3 drawn B1
(1, 4) marked and perpendicular to line drawn Perpendicular can be implied.
B1
(5, 2) Accept coordinates on graph
B1 [3]
M93. ∠ABD = 75
or ∠PDA (= 75 – 30) = 45 and ∠PAB = 45 Allow on diagram
B1
∠BAD = 60 Allow on diagram
B1
120 ft (their) 60 not ∠BAD = 90 Allow on diagram
B1 ft [3]
M94. (a) or or 1.5
or or or 0.666(6 …) M1
(their) 1.5 × 12 or 12 ÷ (their) 0.66
M1
18 A1
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(b) 1080 ÷ or 1080 ×
or
1080 ÷ or 1080 ×
or 1080 ÷ 4 or 1080 × 0.25 M1
270 A1
[5]
M95. a, b, c and d must be integers (including ft)
(a =) 6 and (b =) 2 or (c =) 4 and (d =) 3
or (a or c =) 12 and (b or d =) 1 Any indication eg, on diagram
B1
(their) 62 + (their) 22
or
(their) 42 + (their) 32
ft (their) a and b or (their) c and d M1
√((their) 40 or 6.3(...) or √((their) 25 or 5
M1
6.3( …) or √40 or 2√10 and 5
A1
22.3( …) or 16 + √40 or 16 + 2√10 Allow 22 with method
A1 [5]
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M96. (a) 10π = 2πr or 10π = πd
M1
10π ÷ 2π d = 10 or 10 ÷ 2
M1 dep
5 A1
(b) 80π = π52h
oe M1
oe eg, 240 = 25h (no fraction and no π)
M1 dep
9.6 or or or or
oe ft 240 ÷ their radius2
Ignore fw A1 ft
[6]
M97. (a) 2b – 2a
oe b + b – a – a B1
(b) 2c – 2b
oe B1
(c) (2b – 2a) + (2c – 2b) or b – a + c – b B1
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(d) DG and EF are parallel and equal in length
or DG and EF are equal (vectors)
or GF and DE are equal (vectors) oe Both c – a Both are same vectors Do not need vector arrows Do not accept properties of parallelogram Not enough to say they are parallel
B1 [4]
M98. (a) Valid explanation
eg allied angles (add up to 180) inside parallel lines (add up to 180) y + y + 2x + 2x = 360 so y + 2x = 180 2y + 4x = 360 In a C add up to 180 Condone interior angles (add up to 180)
B1
(b) 3x + y = 230 oe
B1
(c) Attempt to eliminate a variable (with 2x + y = 180)
eg 6x + 2y = 460 and 6x + 3y = 540 and subtraction Note: Full marks can be awarded for this part on follow through
M1
x = 50 A1 ft
y = 80 A1 ft
3x + y = 130 → x = –50, y = 280 3x = y –50 → x = 26, y = 128 3x + y = 410 → x = 230, y = –280
[4]
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M99. (a) (i) 180 B1
(ii) 18 Their (a)(i) ÷ 10 Not ft 280 or 040 or > 360
B1 ft
(iii) 06 Do not allow 6
B1
(b) 210 Allow 208 – 212
B1
21 SC1 15 Not 21.0
B1 ft
(c) 300 – 180 180 ÷ 10
M1
120 30 – 18
A1
12 A1 ft
[8]
M100. (a) y = sin x + 2 B1
(b) y = 3 sin x B1
(c) y = sin (x – 90) B1
[3]
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M101. oe
M1
13.2 × M1 dep
9.287( ...) or 9.29 A1
9.29 or 9.3 ft Their BC to 1 dp or 3 sf
B1ft [4]
M102. (a) No and showing
oe No and … eg 0.125 > 0.083 ...
or
144 ÷ 12 < 16 16 × 12 or 192 > 144 The length should be 192
B1 or or 0.111 …seen
192 seen oe Deduct 1 for Yes
B2
(b) 48 B1
62 + (their 48)2
or 2340 M1
√(their) 2340 M1dep
48.3(7 ...) or 48.4 A1
[6]
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M103. (a) x + 1 + x + 1 + 3x – 2 + 3x – 2
oe eg, (4x – 1) × 2 B1
(b) 8x – 2 = 16 or 8x = 18
B1
8x = 18 oe
M1
(x =) A1
[4]
M104. (a) Collinear
oe B1
(b) 2 : 1 B1
[2]
M105. (a) y = x drawn M1
(x =) 3.5 and (y =) 3.5 ± 0.1 Readings from their y = x
A1
(b) 2x2 = 25
x2 + x2 = 25
M1
x2 = 12.5
Must see method A1
[4]
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M106. (a) 23 B1
(b) 44 B1
[2]
M107. 300 × 440
Note 300 × 440 or 300 × 400 scores M0 M1
132 000 A1
132 kg B1ft
[3]
M108. (a) eg, 5 squares = 12.5 or 5 cm = 12.5
12.5 ÷ 5 (= 2.5) 12.5 ÷ 2.5 = 5
B1
(b) Rectangle drawn 3 cm by 2 cm B1
Distance of rectangle from garage ≥1 cm B1
Attached to back wall B1
[4]
M109. (a) Reflection
Note parts (a) and (b) are marked as a pair with either answer in either part
B1
(In the line) AC B1
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(b) Rotation B1
180° B1
About M B1
[5]
M110. Angle ATB = 90° (angle on diameter)
B1
Angle BTC = x° (alternate segment)
B1
32 + x + x + 90 = 180 oe
M1
29 A1
Alt
Angle OTC = 90° (radius perp to tangent)
B1
Angle TOC = 58° B1
2x = 58 oe
M1
29 A1
[4]
M111. (x + 5)2 + (x – 2)2 = 102
x2 + 25 + x2 ± 4 = 100 implies M1
M1
x2 + 10x + 25 + x2 – 4x + 4 = 100
SC1 x2 + 10 x + 25 + x2 – 4 x + 4 = 10
A1
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2x2 + 6x – 71 = 0
For rearranging into a 3 term quadratic = 0 or going to cts straight away.
M1 dep
Use of quadratic formula or completing the square to solve Any evidence of formula or cts gets M1 Allow use of graphical calculator
M1 dep
4.6, 4.64, 4.65 4.64(4..) for T & I
A1 [5]
M112. Diameters = , 7,
M1 for attempt to find radii or diameters eg, 2 ÷ 4 or 1 ÷ 4
M1, A1
π × 42 + π × 3 + π × 3 + π × 3
M1 for Σ π r2 for at least 3 floors with any different radii
50.26, 44.18, 38.48, 33.18, (28.27) values rounded or truncated
M1, A1
Yes as 165 or 166... Must reach conclusion
A1 [5]
M113. (a) (i) (±) 15.8 B1
(ii) Divide by numbers and reference to answers being whole numbers and/or decimals
E1
Reference to dividing minimum 3 primes E1
Full explanation E2, partial explanation E1
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(b) (i) Evidence of dividing 2008 by primes or splitting into a ‘tree’ Must do at least 2 divisions or 2 ‘branches’
M1
(1) × 2 × 2 × 2 × 251
23 × 251
A1
(ii) 1, 2, 4, 8, 251, 502, 1004, 2008 B1
(c) (i) x2 + xy – xy (yx) – y2
B1
(ii) (x + y)(x – y) = 2008 × 1 or 1004 × 2 or 502 × 4 or 251 × 8
M1
x + y = 2008 or 1004 or 502 or 251 x – y = 1 or 2 or 4 or 8
M1
x = 503, y = 501 x = 253, y = 249
A1 for any correct solution including x = 1004.5, (y = 1003.5) x = 129.5, (y = 121.5)
A2 [11]
M114. BD = × 40 M1
BD = 25 A1
BC = √(their 252 – 72)
cos y = 7/their 25 M1 M1
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√ their 576 y = 73.7 A1
M1
24 Sin their 73.7 M1
A1
or 0.96 Ignore y = 73.7° after answer seen
A1 [6]
M115. (a) 2b – 2a
oe B1
(b) 2b – 2c
ft 2b – 4c if 2b – 4a given in (a) B1 ft
(c) oe ft their vectors
M1
c + (2b –2c) or a + (2b – 2a)
NB Answer given so could be fiddled A1
(d) Parallel to third side (other side, base) or half length oe
B1
(e) Parallelogram B1
Opposite sides same vector oe
B1 [7]
M116. (a)
oe M1
(a = ) 2y – x A1
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(b) M1
(b = ) oe
A1 [4]
M117. Evidence of searching for a pattern or r = 2 or d = 4 or 6 extra discs gives extra 12 cm
eg, 4 → 10, 10 → 22 or 5 → 10, 11 → 22 or markings on diagram or diagram of 24 discs (2 rows) 6 extra discs gives extra 12 cm
M1
2n + 2 or 2(n + )
or 14 extra discs gives 28 cm or 20 extra discs gives 40 cm
12d + r or 25r 22 + 28 10 + 40
M1
50 A1
[3]
M118. Attempt at perpendicular bisector of given chord (intersecting arcs centred on end points both sides)
Two 90° angles constructed at each end of chord and extended to meet circle
M1
A perpendicular bisector within tolerance (± 1 mm) Must cut both sides of circle unless another perpendicular bisector drawn Two diagonals of rectangle drawn
A1
Centre within ± 1 mm A1
[3]
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M119. 8a
8a = 14b or 4a = 7b M1 B1
14b 4a + 4b = 44 or 7a + 7b = 77 M1
B1
a = 7 answers only with no working is zero marks 11b = 44 or 11a = 77 A1
B1
b = 4 allow answers reversed a = 7 and b = 4 A1
B1 [4]
M120. (V) = πCondone missing bracket
M1
A correct step moving towards (their) formula for x3
eg, 3V ∕ π = x3 ∕ (their 4)
or V = π × x3 oe
M1dep
x3 = (their 12)V ∕π
Rearranging to get (their) formula for x3
M1dep
x = 3√(12V ∕ π)
SC3 x = 3√(6V ∕ π) oe
SC2 x = 3√(3V ∕ π) oe
A1 [4]
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M121. BAF = 120°
This can just be stated or exterior angle of hexagon = 60°
or reflex FAB = 240° B1
360 – (120 + 90 + 90) = 60° oe
HAJ must be shown to be 60° by calculation B1
AH = AJ This can just be stated or shown on diagram
B1
AJH = AHJ = (180 – 60) ÷ 2 Dep on first B2
B1dep [4]
M122. Allow one error or [5 ± √59] ÷ 6
M1
(5 ± √ 109) ÷ 6 A1
2.57 and –0.91 SC1 2.3 and –0.45
A1 [3]
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M123. tan 38 = BC/7.21
or BC ∕ sin 38 = 7.21 ∕ sin 52 M1
7.21 × tan 38 or 7.21 × sin 38 ÷ sin 52
M1 dep
5.6 (3 …) A1
5.6 or 5.63 B1 ft
[4]
M124. (a) y = 2 sin x
oe y = 2 f( x ) ‘y =’ or ‘x =’ missing penalize first time only
B1
(b) y = 1 + sin x oe y = 1 + f( x )
B1
(c) y = – sin x oe y = – f( x ) y = sin (x + 180) y = sin (x – 180)
B1 [3]
M125. π × 3 × 3 × 6
Cylinder M1
54π 169-170
A1
4 × π ×3 × 3 × 3 ÷ 3
Sphere × π ×3 × 3 × 3 M1
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36π 113 - 114
A1
18π Their 54π – their 36π Follow through dependent on award of at least one M1
A1 ft [5]
M126. (a) 53 B1
(b) 85 B1
[2]
M127. Reflection B1
y = –x oe
B1 [2]
M128. (a) 41 B1
(b) 180 – 67 71 + their 41 oe 360 – 41 – 67 – 139
M1
113 A1
[3]
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M129. Attempt to rewrite
x2 – 5x + 3 = 0 as x2 – 4x + 1 = x – 2 M1
Identify (y =) x – 2 A1
Accurately draws line y = x – 2 M1 dep
(x =) 0.7 and 4.3 ft their line if both M1s earned
A1 [4]
M130. (a) (height of cylinder =) 9 B1
π × 32 × their 9 or 81π
M1
× π × 33 or 18π
oe M1
99π A1
(b) Their 49.5π – their 18π or 31.5π
Must see use of ‘ of their 99π’ M1
Their 31.5π = π × 32 × h
dep on previous M1 M1 dep
h = their 31.5π ÷ (π × 32) or 3.5
M1
d = 6.5 Strand (ii) For correct answer supported by logical working showing key steps
Q1 [8]
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M131. Angle PRQ = 180° – 134° or 46° M1
Angle POQ = 2 × their 46° or 92° M1
Reflex angle POQ = 268° A1
[3]
M132. Cos A = (102 + 62 – 142) ÷ (2 × 6 × 10)
M1
–A1
120° A1
[3]
M133. (a) 2x2 – 7x + 9 = 0
oe –1 eeoo B2 fully correct expression
B3
(b) Reference to square root of negative number B1
[4]
M134. (a) x2 = 412 – 402
M1
x2 = 81 or x = (= 9)
A1
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(b) (n + 1)2 – n2 = m2
M1
n2 + 2n + 1 – n2 = m2
M1 dep
m2 = 2n + 1
A1
m2 is odd since 2n + 1 is odd
A1
m is odd since odd × odd = odd A1
[7]
M135. (a) D B1
A B1
C B1
(b) Negative gradient and through point on positive y-axis ‘2’ need not be marked
B1 [4]
M136. (a) × (7 + 11) × 5 M1
45 A1
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(b) Their 45 × 16 or 720
M1
19.3 × their 720 M1
13896 A1
13.896 ft if both Ms awarded
A1 ft [6]
M137. (a) x(x + 10) B1
(b) (y + 6)(y – 6) B1
(c) 5w + w = 9 – 6 Allow one sign error
M1
6w = 3 For collecting like terms ft their first line
M1
oe Accept A1
(d) LCM of 12 used correctly or attempt at LHS multiplied by 12 M1
6x + 9 + 4x – 20 Allow one error
M1
10x – 11 = 18 10x – 11 = 3 scores A0
A1
2.9 ft from one arithmetic error but not from 10x – 11 = 3
A1 ft [9]
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M138. π × 3.5 × 3.5 or π × 5 × 5 or π × 7 × 7
12.25π or 25π or 49π M1
π × 3.5 × 3.5 + π × 5 × 5 M1
47.25π and 49π A1
He is correct ft if both Ms awarded
A1 ft [4]
M139. 39 ÷ 3 or 39 ÷ 6 or 19.5 ÷ 3 or 19.5 ÷ 6
oe M1
13 or 6.5 seen A1
13 × 13 M1
169 A1
[4]
M140. Sometimes true B1
Valid explanation eg, height of triangle can vary
B1 [2]
M141. (a) Fully correct rotation
B1 180° rotation with centre 0 B1 90° clockwise rotation with wrong centre B2 90° clockwise rotation with centre 0 B2 90° anticlockwise rotation with wrong centre
B3
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(b) x = –1 B1 for coordinates plotted or line shown on graph
B2 [5]
M142. (a) Ratio is 1:113 and 113 = 1331
B1
(b) : oe
M1
1 : 0.24 A1
[3]
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