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)

   

Page 3 of 227

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

Page 4 of 227

(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)

   

Page 6 of 227

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)

Page 8 of 227

(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)

Page 9 of 227

(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)

Page 10 of 227

(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)

   

Page 11 of 227

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

Page 12 of 227

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)

   

Page 13 of 227

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)

Page 15 of 227

(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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......................................................................................................................... (2)

(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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......................................................................................................................... (2)

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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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......................................................................................................................... (3)

(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

Page 20 of 227

          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)

Page 21 of 227

(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

Page 22 of 227

          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)

Page 23 of 227

(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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......................................................................................................................... (2)

Page 24 of 227

(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

Page 25 of 227

(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.

Page 26 of 227

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

Page 27 of 227

          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

Page 28 of 227

(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

Page 29 of 227

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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Page 30 of 227

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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

Page 31 of 227

          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

Page 32 of 227

(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

Page 33 of 227

(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

Page 34 of 227

          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

Page 35 of 227

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)

   

Page 36 of 227

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

Page 37 of 227

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)

   

Page 38 of 227

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)

Page 39 of 227

(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

Page 41 of 227

          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

Page 42 of 227

          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)

   

Page 43 of 227

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.

Page 44 of 227

          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.

Page 45 of 227

          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)

Page 46 of 227

(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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......................................................................................................................... (2)

(Total 7 marks)

   

Page 47 of 227

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)

Page 48 of 227

(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)

   

Page 49 of 227

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)

   

Page 50 of 227

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)

Page 51 of 227

(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.

Page 52 of 227

(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)

   

Page 53 of 227

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

Page 54 of 227

          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

Page 55 of 227

          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)

Page 56 of 227

(c)    

(1)

(Total 3 marks)

   

Page 57 of 227

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)

Page 58 of 227

(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

Page 59 of 227

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.

Page 60 of 227

          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°

Page 61 of 227

          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

Page 62 of 227

          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

Page 63 of 227

(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)

   

Page 64 of 227

Q57.         

          Enlarge the shaded shape by scale factor  with centre of enlargement (–1, 0). (Total 2 marks)

   

Page 65 of 227

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)

   

Page 66 of 227

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°

Page 67 of 227

          Explain why angle QSR = 32°.

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........................................................................................................................... (2)

(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)

   

Page 68 of 227

Q61.          Which one of the following kites is a cyclic quadrilateral? Give a reason for your answer.

Answer ......................................................

Reason .....................................................................................................................

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.................................................................................................................................. (Total 2 marks)

   

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.

Page 69 of 227

          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.

Page 70 of 227

          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.

Page 71 of 227

          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)

   

Page 72 of 227

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)

Page 73 of 227

(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.

Page 74 of 227

          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?

Page 75 of 227

          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.

Page 76 of 227

          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.

Page 77 of 227

(Total 4 marks)

   

Q71.          The diagram shows two right-angled triangles. AD = 15 cm. CD = 6 cm.

Page 78 of 227

(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.

Page 79 of 227

          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)

Page 80 of 227

(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)

   

Page 81 of 227

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.

Page 82 of 227

          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°.

Page 83 of 227

          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.

Page 84 of 227

          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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................................................................................................................................... (Total 4 marks)

   

Page 85 of 227

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.

Page 86 of 227

          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.

Page 87 of 227

(a)     y = sin (x – 45)

P (......................., ......................) (2)

(b)     y = 2sinx

P (......................., ......................) (2)

(Total 4 marks)

   

Page 88 of 227

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)

   

Page 89 of 227

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)

Page 90 of 227

(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)

Page 91 of 227

(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°.

Page 92 of 227

          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.

Page 93 of 227

          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)

Page 94 of 227

(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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.......................................................................................................................... (2)

Page 95 of 227

(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)

   

Page 96 of 227

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)

 

 

Not drawn accurately

Page 97 of 227

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)

   

Page 98 of 227

Q89.          The diagram shows a regular pentagon and a regular decagon joined at side XY.

Not drawn accurately

          Show that the points A, B and C lie on a straight line.

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Page 99 of 227

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)

   

Page 100 of 227

Q91.          Four points, A(– 4, –1), B(5, 11), C(20, 11) and D(11, –1) are joined to form a quadrilateral.

Not drawn accurately

          Prove by calculation that ABCD is a rhombus.

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Page 101 of 227

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)

   

Page 102 of 227

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)

   

Not drawn accurately

Page 103 of 227

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.

Not drawn accurately

          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

Page 104 of 227

(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

Page 105 of 227

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

Not drawn accurately

          Work out the perimeter of the shaded shape.

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Answer ........................................... cm (Total 5 marks)

   

2

Page 106 of 227

Q96.          The sector AOB of a circle is shown below.           The length of its arc AB is 10π cm.

Not drawn accurately

          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

Page 107 of 227

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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......................................................................................................................... (1)

(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

Page 109 of 227

(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)

Page 110 of 227

(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)

   

Page 111 of 227

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)

Page 112 of 227

(b)    

Answer y = ...................................... (1)

(c)    

Answer y = ...................................... (1)

(Total 3 marks)

   

Page 113 of 227

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

         

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gradient should be

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Page 114 of 227

(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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........................................................................................................................ (2)

Page 115 of 227

(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)

   

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Not drawn accurately

Page 116 of 227

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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Page 117 of 227

(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

Page 118 of 227

(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)

Page 119 of 227

(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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Page 120 of 227

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)

   

Page 121 of 227

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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......................................................................................................................... (2)

(b)     Describe fully a different single transformation that maps triangle X onto triangle Y.

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......................................................................................................................... (3)

(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°

     

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Not drawn accurately

Page 122 of 227

          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.

      

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Page 123 of 227

          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

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Page 124 of 227

          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)

Page 125 of 227

(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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(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

Page 126 of 227

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)

   

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Page 127 of 227

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

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(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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......................................................................................................................... (2)

(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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Page 128 of 227

(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            

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Page 129 of 227

          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)

   

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Page 130 of 227

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

Page 131 of 227

          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.

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Page 132 of 227

          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

Page 133 of 227

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

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Page 134 of 227

(a)          Write down a possible equation of the following graph.

Answer ....................................................................... (1)

(b)     Write down a possible equation of the following graph.

Answer ....................................................................... (1)

Page 135 of 227

(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

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Page 136 of 227

          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)

   

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3

Page 137 of 227

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

Page 138 of 227

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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.................................................................................................................................. (Total 2 marks)

   

Page 139 of 227

Q128.          In the diagram AB and CD are parallel.

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(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

Page 140 of 227

          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

Page 141 of 227

(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)

   

Page 142 of 227

Q131.          O is the centre of the circle.

          Angle PRS = 134°

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          Work out the size of the reflex angle POQ.

          You must show your working.

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Answer .................................. degrees (Total 3 marks)

   

Page 143 of 227

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)

Page 144 of 227

(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.

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          Show that x = 9 cm

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Page 145 of 227

(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

Page 146 of 227

(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

Page 147 of 227

(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

Page 148 of 227

(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.

Page 149 of 227

          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

Page 150 of 227

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

Page 151 of 227

(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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......................................................................................................................... (1)

Page 152 of 227

(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)

   

Page 153 of 227

 

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

Page 154 of 227

(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

Page 155 of 227

(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]

   

Page 156 of 227

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]

   

Page 157 of 227

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]

   

Page 158 of 227

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

Page 159 of 227

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]

   

Page 160 of 227

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

Page 161 of 227

(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]

   

Page 162 of 227

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

Page 163 of 227

          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]

   

Page 164 of 227

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

Page 165 of 227

(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]

   

Page 166 of 227

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

Page 167 of 227

(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

Page 168 of 227

(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

Page 169 of 227

(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]

   

Page 170 of 227

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]

   

Page 171 of 227

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]

   

Page 172 of 227

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]

   

Page 173 of 227

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]

   

Page 174 of 227

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

Page 175 of 227

(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]

   

Page 176 of 227

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

Page 177 of 227

(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

Page 178 of 227

(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]

   

Page 179 of 227

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]

   

Page 180 of 227

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

Page 181 of 227

(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]

   

Page 182 of 227

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]

   

Page 183 of 227

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]

   

Page 184 of 227

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]

   

Page 185 of 227

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]

   

Page 186 of 227

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]

   

Page 187 of 227

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

Page 188 of 227

          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]

   

Page 189 of 227

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

Page 190 of 227

(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]

   

Page 191 of 227

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]

   

Page 192 of 227

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]

   

Page 193 of 227

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]

   

Page 194 of 227

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]

   

Page 195 of 227

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]

   

Page 196 of 227

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]

   

Page 197 of 227

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

Page 198 of 227

          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]

   

Page 199 of 227

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

Page 200 of 227

          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]

   

Page 201 of 227

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

Page 202 of 227

          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]

   

Page 203 of 227

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

Page 204 of 227

(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]

   

Page 205 of 227

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

Page 206 of 227

(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]

   

Page 207 of 227

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]

   

Page 208 of 227

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]

   

        

Page 209 of 227

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]

   

Page 210 of 227

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

Page 211 of 227

(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

Page 212 of 227

          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

Page 213 of 227

(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

Page 214 of 227

 √ 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

Page 215 of 227

(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]

   

Page 216 of 227

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]

   

Page 217 of 227

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]

   

Page 218 of 227

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

Page 219 of 227

          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]

   

Page 220 of 227

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]

   

Page 221 of 227

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

Page 222 of 227

(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

Page 223 of 227

(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]

   

Page 224 of 227

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

Page 225 of 227

(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]

   

Page 226 of 227

  

Page 227 of 227