HigherN2c/C Number

1 Calculate, giving your answer in standard form correct to 3 significant figures.

1.52 10 5 - 4.6 10 4 4.56 10-2

152 000 - 46 000 = 106 000106 000 0.0456 = 2.32 106

4

N2c/C2 The distance of the Earth from the

Sun at a particular moment is 93.5 million miles.a Write 93.5 million in standard form.

The distance of the Earth from the moon at a particular moment is 2.5 105 miles.b Write 2.5 105 as an ordinary number.

c How many times further is the Earth from the Sun than it is from the moon? Give your answer in standard form.

a) 9.35 107

b) 250 000c) 93 500 000 250 000 = 3.74 102

5

N2c/C/B3 a Write down the value of

i 50

ii 4-2

b Simplify 1634

−×8

13

a) i) 1ii) 1/16

b) 1/8 2=1/4

4

N2c/B4 The area of the Earth covered by sea

is 362 000 000 km2.a Write 362 000 000 in standard form.

The surface area, A km2, of the Earth may be found using the formulaA = 4r2

where r km is the radius of the Earth.R = 6.38 103.b Calculate the surface area of the Earth. Give your answer in standard

a) 3.62 108

b) From 5.11 108 to 5.12 108

c) (a)/(b) 100 = between 70 and 71 inclusive

6

form, correct to 3 significant figures.

c Calculate the percentage of the Earth's surface which is covered by sea. Give your answer correct to 2 significant figures

N2c/B5 Express 0.327 105 in standard

form.3.27 104 2

N2c/B6 My computer can carry out 2.7

108 calculations in one hour.Work out how many of these calculations my computer can carry out in one second. Give your answer in standard form.

2.7 108 3600 (= 75000) = 7.5 104 3

N2c/B7 A US Centillion is the number

A UK Centillion is the number

a How many US Centillions are there in a UK Centillion? Give your answer in standard form.

b Write the number 40 US Centillions in standard form.

a) 10 600 ÷ 10 303 = 10 600-303 = 1×10 297 or 10 297

b) 40×10 303 = 4×10 304

4

N2c/B8 The diameter of an atom is

0.000 000 03 m.a Write 0.000 000 03 in standard form.

Using the most powerful microscope, the smallest objects which can be seen have diameters which are one hundredth of the diameter of an atom.b Calculate the diameter, in meters, of the smallest objects which can be seen using this microscope.Give your answer in standard form.

a) 3×10 -8

b) (a) ÷ 100 = 3×10 -104

N3d/C9 Marcus sees a motorbike advertised

for £750.

This is the price after a reduction of 15%. Work out the original price of the motorbike.

750 is 85%1% = 750 ÷ 85100% = 750 × 85 ÷ 100 = £882.35

3

N3e/C10 Use your calculator to evaluate

560.3 20.3 (0.2 + 4.5)2

Write down all the figures on your calculator.

_11374.09_ = 11374.09 (0.2 + 4.5)2 4.72

= 514.89769...

3

N3e/C11 In this question you MUST use

your calculator and you MAY write down any stage in your calculation.Evaluate

( . . ) .. ( . . )

23 4 35 6 5 7200 3 16 2 815

__59 5.7_ = _336.3_200.3 8.05 1612.415

= 0.208569 = 0.208 or 0.209

2

N4a/C12 Shreena put £484 in a new savings

account.At the end of every year, interest of 4.3% was added to the amount in her savings account at the start of that year.Calculate the total amount in Shreena's savings account at the end of 2 years.

484 × 4.3/100

484 + 20.81 (= 504.81)504.81 × 4.3/100 = 21.7069504.81 + 21.71OR484 × 1.043 = 504.81"504.81" × 1.043 = 526.52OR484 × 1.0432 = 526.52Accept £526.51 or £526.52

5

N4a/C13 Astrid bought a motor car for £10

000 on the First of January 1996.1st Jan 1997 £85001st Jan 1998 £76501st Jan 1999 £6885

5

It lost 15% of its value during 1996 and then 10% during every year from the First of January 1997.

Work out the value of the car on the First of January 1999.

N4d/C14 Natalie measured the distance

between two points on a map.The distance she measured was 5 cm correct to the nearest centimetrea Write down the i least upper bound of the measurement, ii greatest lower bound of the measurement,

The scale of the map is 1 to 20 000b Work out the actual distance in real life between the upper and lower bounds. Give your answer in kilometres.

a) i) 5.5 cmii) 4.5 cm

b) 0.2 km

4

NFMa/A15 The temperature from a factory

furnace varies inversely as the square of the distance from the furnace.The temperature 2 metres from the furnace is 50ºC.Calculate the temperature 3.5 metres from the furnace. Give your answer to 2 decimal places.

T 1/d2

T = k × 1/d2

50 = k × ¼k = 200T = 200 × 1/3.52

T = 16.33ºC

5

NFMb/A 16 Draw a circle around each irrational

number in the list below.

2.3

circles 2/3 and 4/3 2

NFMb/A17 Change 0 45. into a fraction in its

lowest terms.x = 0.4545454545454545...100x = 45.454545454...99x = 45

3

so x = 5/9NFMb/A

18 Right angled triangles can have sides with lengths which are a rational of irrational number of units.Give an example of a right angled triangle to fit each description below.i All sides are rational

ii The hypotenuse is rational and the other two sides are irrational

iii The hypotenuse is irrational and the other two sides are rational.

iv The hypotenuse and one of the other sides are rational and the remaining side is irrational.

e.g.i) 3, 4, 5ii) 7, 2, 3iii) 2, 3, 13iv) 2, 5, 3

4

NFMb/A19 Put a tick in the box underneath

those numbers that are rational.

2/3 1.6 5 4/17

2/3, 1.6, 4/17 ticked 2

NFMb/A20 Put a tick in the box underneath the

rational numbers.

12 3

10 - 3 9.5 136

12

tick 12 and 1 3 36

2

NFMb/A21 a Give an example of two different

irrational numbers q and r such that q/r is a rational number.bWrite down a number which is greater than 15 and less than 16 and which has a rational cube root.

a) e.g. 8 2 (= 4 = 2)b) x3 such that 315 < x < 316 x3such that 2.446 < x < 2.519

4

NFMb/A22 a Write down a number which is

greater than 17 and less than 18 that has a rational square root.

b Give an example of two different irrational numbers c and d such that c d is a rational number.

a) x2: 17 < x < 18or x2: 4.123 x 4.242

b) e.g. 3 12 = 36 = 6

4

NFMc/A23 Cleo uses a pair of scales to

measure, in kilograms, the weight of a brick.

Since the scales were only accurate to 0.1kg then 1.4 should be the answer.

2

The scales were accurate to the nearest 100g.She read the scales as accurately as she could and wrote down the weight as 1.437 kg. Anthony said that this was not a sensible answer to write down. Explain why Anthony was correct.

NFMc/A24 The speed of light is 186 000 miles

per second correct to the nearest thousand miles per second.The distance of the earth from the sun is 93.5 million miles correct to the nearest one hundred thousand miles.A ray of light leaves the sun and travels to earth. It takes time T.Calculate the range within which the time T taken by this ray lies.

93 550 000 ÷ 185 500 = 50493 450 000 ÷ 186 500 = 501

5

NFMc/A25 Correct to 3 decimal places, a =

2.236.a For this value of a, write down i the upper bound, ii the lower bound

Correct to 3 decimal places, b = 1.414.b Calculate i the upper bound for the value of a + b ii the lower bound for the value of a + b.

Write down all the figures on your calculator display for parts c and d of this question.c Calculate the lower bound for the value ab.

d Calculate the upper bound for

the value of

a) i) 2.2365ii) 2.2355

b) i) 3.651ii) 3.649

c) 2.2355 × 1.4135 = 3.15987925d) 2.2365 ÷ 1.4135 = 1.58224266

8

NFMc/A/A*26 PQR is a right angled triangle.

RQ= 6.0cm and PR=8.3cm, both correct to 1 decimal place

a) i) 6.05ii) 5.95

b) (6.05×8.35)/2 = 25.25875

9

a Write down i the upper bound of the length

of RQii the lower bound of the length

of RQ.

b Calculate the upper bound of the area of the triangle PQR.

c Calculate the upper bound of the angle PQR. Give your value correct to 1 decimal place.

c) Tan PQR = 8.35/5.95 so PQR = 54.5°

NFMd/A27 Evaluate:

i

ii

i) 9ii) 4/9

2

NFMc/A*28 x = 40, correct to the nearest 10.

y = 60, correct to the nearest 10.a i Write down the lower bound of x.

ii Write down the upper bound of y.

b Calculate the greatest possible value of xy.

c Calculate the least possible value

of .

Give your answer correct to 3 significant figures.

d Calculate the greatest possible

a) i) 35ii) 65

b) 45 65 = 2925c) 35 65 (= 0.53846...) = 0.538d) = 1 + 10/x = 1 + 10/35 (= 1.2857...) = 1.29

8

value of

.Give your answer correct to 3 significant figures.

NFMc/A*29 Kim is doing an experiment using a

pendulum. She uses the formula

where g is a constant acceleration, L the length of the pendulum, and T is the time for one full swing of the pendulum.In Kim's experiment the length L is 1 metre, correct to the nearest centimetre.She measured the value of T to be 2 seconds, correct to the nearest 0.2 of a second.Calculate the upper bound and the lower bound of Kim's values for g.Give your answer in metres per second per second correct to two decimal places.

40 1.005 = 11.141.92

40 0.995 = 9.022.12

6

N3a/N2c/C30 The number 1998 can be written as

2 3n p, where n is a whole number a p is a prime number.i Work out the values of n and p.ii Using your answers to part i, or otherwise, work out the factor of 1998 which is between 100 and 200.

i) 2 999 = 2 9 111 = 2 27 37 = 2 33 37 ; n = 3, p = 37ii) 3 37 = 111

3

A2b/C Algebra31 Work out an algebraic expression

for the nth term of this sequence of numbers.2, 8, 18, 32, 50, .......

2n × n, 2(n × n) = 2n2 2

A2b/C32 The expression n ( n +1)

2

is the nth term of the sequence of

10n ( n +1) = 5n(n +1) 2

1

triangular numbers1, 3, 6, 10, ...

Write down an expression, in terms of n, for the nth term of the sequence

10, 30, 60, 100, ...

A2b/C33 Here are the first terms of a number

sequence.5, 8, 11, 14, 17,Write down an expression for the nth term of the sequence.

3n + 2 2

A2b/C34 The numbers of lines joining points

form a number sequence.

Complete the table to find the number of lines joiningi 6 pointsii 10 pointsiii n points

Points Lines2 13 34 65 10610n

i) 6 15ii) 10 45iii) n n ( n -1)

2

6

A2d/C35 a Make a table of values for y = 6 -

2x

xy

b Draw the graph of y = 6 - 2x on the grid below.

a) -2, -1, 0, 1, 2, 3, 410, 8, 6, 4, 2, 0, -2Ignore outside -2 to 4

b) graph drawnc) i) x = -1.5, y = 9

ii) y = 3.4, x = 1.3

6

c Use your graph to findi the value of y when x = -1.5ii the value of x when y = 3.4

A2c/B36 The diagram shows part of a

distance/time graph for a bus after if had left a bus stop.

a Use the graph to find the distance the bus travelled in the first 20 seconds after it had left the bus stop.

b Describe fully the journey of the bus represented by the parts AB,BC and CD of the graph.

a) 88 mb) AB constant speed BC gradually slowing down CD stationary.

4

A2b/C37 i On the grid below draw the

graph ofy = x2 - 3x - 5for values of x between -2 and

+5

i) x -2, -1, 0, 1, 2, 3, 4, 5y 5, -1, -5, -7, -7, -5, -1, 5

ii) x = -1.2 x = 4.2

6

x -2 -1 0 1 2 3 4 5

y

ii Use your graph to solve the equation

x2 - 3x - 5 = 0

A3c/C38 x is an integer, such that -3 < x 2.

List all the possible values of x.-2, -1, 0, 1, 2 2

A3c/C39 x is an integer. Write down the

greatest value of x for which 2x < 7.3 1

A3d/C40 Use the method of trial and

improvement to find the positive solution of x3 + x = 37Give your answer correct to 1 decimal place.

3.2 4

A3d/C41 Use a trial and improvement

method to solve the equation x3 - x = 15

x x3 - xx = 2 23 - 2 = 6 Too smallx = 3 33 - 3 = 24 Too big

Complete the working below and find a solution correct to one decimal place.

2.53 - 2.5 = 13.1252.63 - 2.6 = 14.9762.73 - 2.7 = 16.983so x = 2.6

3

A3b/B/C42 The velocity of a particle is given

by the formula v 2 = u 2 + 2as

a Calculate the value of the

velocity v when u = -5, a = and s = 5.67.

b Rearrange the formula to make u the subject.

a) 25 + 2 2/3 5.6732.56 (C)

b) u2 = v2 -2asu = (v 2 - 2as) (B)

5

A3b/B43 y = ab + c

Calculate the value of y when

a = , b =

and c =

Give your answer in the form

where p and q are integers.

(21 - 16)/32 = 5/32 3

A3b/B44 The volume, V, of the barrel is

given by the formula

= 3.14, H = 60, R = 25 and r = 20.

( 60)/3000 (2 252 + 202) = ( 60)/3000 1650 = 103.67...3.14 103.623.142 103.68..22/7 103.71..so V = 104

3

Calculate the value of V.Give your answer correct to 3 significant figures.

A3b/c/B45 In the diagram, each side of the

square ABCD is (3 + x) cm.

a Write down an expression in terms of x for the area, in cm2, of the square ABCD.

The actual area of the square ABCD is 10cm2.b Show that x 2 + 6x = 1

a) (3 + x)(3 + x) or (3 + x)2 = (x + 3)2

b) (3 + x)(3 + x) = 10 9 + 3x + 3x + x2 = 10 x2 + 6x + 9 = 10 and x2 + 6x = 1

4

A3b/c/B46 v 2 = u 2 + 2as

a Calculate the value of v when u = -6, a = 5 and s = 0.8.Give your answer to one significant figure.b Make u the subject of the formula v 2 = u 2 + 2as.

a) (-6)2 + 2 5 0.8 = 7or -62 + 2 5 0.8 = 7

b) u 2 = v 2 -2asu = (v 2 - 2as)

5

A3c/B47 a Factorise completely

I 4st + 8tu - 2tvii 15x + 3x2

b Expand 2a(4 - a)

c Expand and simplify (2c + 3)(c - 4)

a) i) 2t(2s + 4u - v)a) ii) 3x(5 + x)b) 8a - 2a2

c) 2c2 - 8c + 3c - 12 = 2c2 - 5c - 12

7

A3c/B48 a Simplify

i 12 x 5 3 y 3 9x2y

ii (4p2q3)2

b Factorise completelyi 9a2b3 + 15a3b2

ii x2 + 7x - 60

c On the number line below show the solution to these inequalities.

-7 2x - 3 < 3

a) i) 4x3y2

ii) 16p4q6

b) i) 3a2b2(3b + 5a)ii) (x + 12)(x - 5)

c)

11

A3c/B49 Solve the inequality

7y > 2y - 37y - 2y > - 3 5y > - 3; y > -3/5

2

A3d/B50 Solve the simultaneous equations

3x + 2y = 11 x - y = 7

3x + 2y = 112x - 2y = 14Adding gives5x = 25 so x = 5Substituting gives15 + 2y = 11so 2y = -4therefore y = -2

3

A3d/B51 Solve the simultaneous equations

3p + 2q = 62p + 5q = -7

6p + 4q = 126p + 15q = -21Subtracting gives11q = -33 q = -3; p = 4

4

A3c/A

52 Factorise completely 2p3q2 - 4p2q3 2p2q2(p - 2q) 2

A3d/A53 Solve the simultaneous equations

3x + y = 132x - 3y = 16

Eqn (1) × 3 9x + 3y = 39Adding 11x = 55÷ by 11 x = 5Substitute 15 + y = 13

y =-2ORy = 3 - 3x2x - 3(13 - 3x) = 1611x = 55, x = 5Substitute 15 + y = 13

4

AFMa/A54

Work out the value of x.

p1.5 p0.5 p-5 = p1.5+0.5-5 = p-3

so x = -33

AFMa/A55 a On the grid below, draw the

graph of y = 5 + 2x - x2 for -2 < x < 4.

b By drawing a suitable straight line on your graph, find the approximate solutions to x + 4 = 5 + 2x - x2

a) graph of y = 5 + 2x - x2 drawnb) line y = x + 4 drawn x values at intersection = -0.6 and 1.6.

6

AFMa/A56 a Complete the table of values for

the graphs ofy = x3 - 2 andy = 3x2 + 3x -6.

x -2 -1 0 1 2 3 4y = x3 - 2 -3 -2 -1 25 62y = 3x2 + 3x -6 0 -6 0 12 30

b i On the graph paper below

a) -10 6-6 54

b) i) Plot point and draw graphsii) x3 - 2 = 3x2 + 3x -6 x3 - 3x2 - 3x +

4 = 0 Read off x at intersectionsx = -1.35, 0.85 and 3.5 (all 0.1)

3

draw the graphs ofy = x3 - 2 and y = 3x2 + 3x - 6.

ii Use your graph to solve the equation x3 - 3x2 - 3x + 4 = 0.

AMFa/A57 The graph of the equation

y = ax + bintersects the graph of the equation

y = cx + dat point P.Show that the x-coordinate of P is

xd b

a c

ax + b = cx + d(a - c)x = d - bso x = d - b a - c

3

AFMa/A58 Solve the equations

i 4y2 - 81 = 0

ii

i) (2y + 9)(2y - 9) = 0 so y = 4.5 or y = -4.5ii) (3 + (x +2))/(3(x + 2)) = -1

x + 5 = -3(x + 2)x + 5 = -3x - 6

4x = -11 so x = -2.75

6

AFMa/A59 Solve the equation

2x2 - 5x - 4 = 0x = 5 (25 + 32) ; x = 3.14 and x = -0.637

4

4

AFMd/A60 The graph shows how Narinder's

height increased in the first 16 years of his life.

a) Tangent drawn at age 14; gradient = 13.b) rate of growth

5

a Calculate an estimate for the gradient of the graph when Narinder was 14 years old.

b What does the gradient represent?

AFMe/A61 The table shows the distance, s

metres, travelled by an object from the point P in t seconds.

t (seconds) 0.5 1.0 1.5 2.0 2.5s (metres) 1.125 1.2 1.325 1.5 1.725

It is thought the relationship between s and t has the form s = at2 + b, where a and b are constants.a Confirm the relationship by plotting a suitable graph on the grid below.

b Use the graph to estimate the values of a and b.

a) Use of t2 values: 0.25, 1, 2.25, 4, 6.25b) a = 0.1, b = 1.1

5

AFMa/A*62 a Factorise

x2 - 4x - 21a) (x - 7)(x + 3)b) ( x - 1) - 3( x + 2)

(x -1)(x + 2)

7

b Simplify

c Make f the subject of the formula

= _-2 x - 7_ x2 + x -2

c) (u + v)/uv = 1/fso f = __uv__

u + v

AFMa/A*63 Factorise completely

ax - by - bx - aya(x + y) - b(x + y) = (a - b)(x + y) 2

AFMa/A*64 The diagram represents the graph of

a function of x

Draw and label on the same axes the graphs of i y = f(-x)ii y = f(x + 2)

4

AFMa/A*65 a i Factorise x2 - 4x - 12.

ii Solve x2 - 4x - 12 = 0.a) i) (x + 2)(x - 6) ii) (a)i = 0, (x + 2)(x - 6) = 0; x = -2 and x = 6b) A = (6,0) B = (-2,0)c) Sketch graph, the given one moved +2 parallel to x-axis

6

The diagram shows a sketch of the graph of y = x2 - 4x - 12.The curve cuts the x-axis at the points A and B.b Write down the coordinates of A and B.

f(x) = x2 - 4x - 12.c Sketch on the axes above the graph of y = f(x - 2).

AFMc/A*66 Here is the velocity time graph for

an oscillating particle.

In part a you must write down the units with your answer.a Calculate an estimate for the acceleration of the particle at 10 seconds.

b Calculate an estimate for the distance travelled by the particle in the first 6 seconds.

a) tangent drawn at t = 10, or chord drawn about t = 10 vert/horiz; 2.3 - 2.8 ms-2

b) attempts to find area, splits area or count squares; 3 marks for answers in the range: 30 - 35 or 41 - 46; 4 marks for answers in the range 35 - 41.

8

AFMe/A*67 Tarquin measures the length of a a = 5.4 b = 90 5

pendulum and the time it takes for the pendulum to swing forwards and backwards.

He records the results in a table.

Length (l) cm Time (t) secs10 11.720 13.930 15.940 17.750 19.160 20.570 21.780 22.8

Tarquin has been told that the formula connecting the time (t) and the length (l) is of the type

t2 = al + b

By plotting a suitable graph on the grid below find the value of a and b.

A2d/A3c/C/B68 a Make y the subject of the

equation x + 2y = 6b On the grid, draw the line with equation x + 2y = 6

a) 2y = 6 - x or x/2 + y = 3; or

(B)

6

c On the grid, shade the region for which

x + 2y 6, 0 x 4 and y 0.

b) e.g.(0,3 ), (6, 0), (2, 2), (4, 1) (C)c)

(B)

A2d/A3d/C69 a On the grid below, draw the

graphs of i x + y = 4 ii y = x + 2

b Use the graphs to solve the simultaneous equations x + y = 4 y = x + 2

a) i) graph of x + y = 4 or y = -x + 4a) ii) graph of y = x + 2b) x = 1, y = 3

4

A2c/B/A70 A car travels between two sets of

traffic lights.

The diagram represents the

a) Velocity = 11.4 m/s. Do not accept responses which merely read off values without adding interpretation (B)b) The car accelerates (the speed increases). Then it travels at a constant speed (or velocity), then there is a constant deceleration (or the car brake steadily). (B)c) Draw tangent at t = 10. Calculate the gradient as 0.57 - 0.73. Alternative: allow a chord about the required point e.g (11.4 - 5)/10 = 0.64 (B)d) Realises area needed. Splits area up

11

velocity/time graph of the car.The car leaves the first set of traffic lights.

a Use the graph to find the velocity of the car after 15 seconds.

b Describe fully the journey of the car between the two sets of traffic lights.

The car leaves the first set of traffic lights.c Calculate an estimate for the acceleration of the car, in m/s-2, after 10 seconds.

The car leaves the first set of traffic lights. It travels for 20 seconds.d Use the graph to estimate the distance, in metres, travelled by the car in the first 20seconds.

e.g. use of trapezium rule, or dividing area up into smaller parts to count squares. (A)

A3b/AFMa/B/A*71 A cylinder of radius R cm and

height h cm has a cylindrical hole of radius r cm drilled through it. Its volume V cm3 is given by the formulaV = (R2 - r2)h

a Find the value of V when

, R = 4, r = 0.5, h =

The surface area, A cm2, of the shape can be written asA = 2 (R + r)(R - r + h)For a particular solid of this type the numerical value of the surface area is twice the numerical value of the

a) V = 22/7 (42 - 0.52) 2¾ = 136 1/8 or 136.125b) 2(R + r)(R - r + h) = 2(R2 - r2)h

(R + r)(R - r + h) = (R2 - r2)h(R + r)(R - r + h) = (R + r)(R - r)hR - r + h = (R - r)hR - r = (R - r -1)h

8

volume.b For this particular shape express h in terms of R and r.

S2e/C Shape72 Calculate the length of AB.

Give your answer correct to 1 decimal place.

(182 - 122) = 13.4 3

S2e/f/C73 The diagram is part of a map

showing the positions of three Nigerian towns.Kaduna is due North of Aba.

a Calculate the direct distance between Lagos and Kaduna.Give your answer to the nearest kilometre.

b Calculate the distance between Kaduna and Aba.Give your answer to the nearest kilometre.

a) ( )440 4402 2 = 622 kmb) Tan 20º = x/440

x = 440 Tan 20º = 160Distance = 440 + 160 = 600 km

5

S2e/f/C74 Sidney places the foot of his ladder

on horizontal ground and the top against a vertical wall.The ladder is 16 feet long.The foot of the ladder is 4 feet from the base of the wall.

a Work out how high up the wall the ladder reaches. Give your answer to 3 significant figures.

b Work out the angle the base of the ladder makes with the ground. Give your answer to 3 significant figures.

a) ( )16 42 2 = 15.5b) cos angle = 4/16; angle = 75.5º

6

S2e/f/C/B75 Here is a side view of a swimming

pool.ABCD is a horizontal straight line. AH, BG, CF and DE are vertical lines.

a Calculate the length of the line FG.Give your answer correct to 3 significant figures.

a) 5.32 + "0.9"2 = 5.38 (C)b) tan angle = "0.9"/5.3; angle = 9.6º

(B)

6

b Calculate the angle that the line GF makes with the horizontal. Give your answer correct to 1 decimal place.

S2f/C76 In the diagram

AB = 17.9 m, BD = 8.2 m, angle CBD = 37º and angle BDC = 90º.

ADC is a straight line.a Calculate the length of DC.Give your answer, in metres, correct to 3 significant figures.

b Calculate the size of angle DAB.Give your answer correct to 1 decimal place.

a) DC/8.2 = tan 37ºDC = 8.2 × tan 37º = 6.179...

b) sin DAB = 8.2/17.9 = 0.4581... so DAB = 27.26... º = 27.3º

6

S2f/C77 The diagram shows a house and a

garage on level ground.

A ladder is placed with one end at the bottom of the house wall.The top of the ladder touches the top of the garage wall.The distance between the garage

a) x/1.4 = tan 62ºx = 1.4 × tan 62º (= 1.4 × 1.8807...) =

2.63(30..)b) cos x = 1.4 / 3.5

x = 66.4218… x = 66.4º

6

wall and the house is 1.4 m.The angle the ladder makes with the ground is 62º.a Calculate the height of the garage wall. Give your answer correct to 3 significant figures.

A ladder of length 3.5 m is then placed against the house wall.The bottom of this ladder rests against the bottom of the garage wall.b Calculate the angle that this ladder makes with the ground.Give your answer correct to 1 decimal place.

S2e/B78 Calculate the length of a diagonal of

this rectangle.Give your answer in centimetres correct to one decimal place.

(152 + 122) = 19.2 cm 3

S2f/A79 Triangle ABC is isosceles. ½ × 12sin55 × 12cos55 × 2 = 67.7 cm2 6

AB = AC = 12 cmAngle ABC is 55ºCalculate the area of the triangle.Give your answer to 3 significant figures.

S3d/C80 The Fast Foto company uses two

letter F's as part of their logo.

The letter F's are similar in shape.The lengths of the large F and the lengths of the small F are in the ratio 3:2.The height of the large F is 0.9m.a Work out the height of the small F.

The width of the small F is 28 cm.b Work out the width of the large F.

a) 0.9 ÷ 3 × 2 = 0.6 mb) 28 ÷ 2 × 3 = 42 cm

4

S3d/C81 Seamus and Mick set off on a

journey from a point AThey travelled 30 km on a bearing of 060º to a point B.From B they travelled 48 km on a bearing of 210º to a point C.a Using a scale of 1 cm to represent 4 km draw a scale drawing of their journey.

a) accurate journey drawnb) i) 6.7 km

ii) 356º

9

b i Work out how far C is from A.

ii Write down the bearing of A from C.

S3d/C82 Shape A is shown in the diagram.

Shape A is enlarged to obtain the shape B.

a Write down the scale factor of the enlargement.b Complete the drawing of shape B on the diagram.

a) 1/3b)

3

S3e/C83 P and Q are two points marked on

the grid.Constructs the perpendicular bisector of P and Q.

2

Construct accurately the locus of all points which are equidistant from P and Q.

S3b/C/B84 a Reflect shape S in the line x = 0.

Label the new shape T.

b Rotate the new shape T through an angle of 90º anticlockwise using (0, 0) as the centre of rotation. Label the new shape U.

c Describe fully the single transformation that will move shape U back onto shape S.

a) and b)

(C)c) Reflection in the line y = -x (B)

6

S3d/B85 Triangle ABC is similar to triangle

PQR.i) 3 × 5/4 = 3.75 cmii) 6.5 ÷ (5/4) or 6.5 × (4/5) = 5.2 cm

5

Angle ABC = angle PQR.Angle ACB = angle PRQ.Calculate the length ofi PQii AC

S3d/B86 Here is a diagram of a company

logo.

The diagram is enlarged so that the length BC becomes 7.5 cm.a Work out the length of the enlarged side AD.

The enlarged side AB is 6 cm.b Work out the length AB on the original diagram.

c What is the size of angle A in the enlarged diagram?

a) 9 × 1.5 = 13.5 cmb) 6 ÷ 1.5 = 4 cmc) 60º

5

S4a/C87 There are 12 inches in 1 foot.

There are 3 feet in 1 yard.There are 2.54 centimetres in 1 inch.Express 1 metre in yards. Give your answer correct to 3 decimal places.

100 cm = 100 ÷ 2.54 = 39.3700 ÷ 36 = 1.0936

= 1.094

3

S4a/C88 There are 2.54 centimetres in 1

inch.There are 12 inches in 1 foot.There are 3 feet in 1 yard.i Calculate the number of yards in 10 metres.ii Calculate the number of metres

i) 1000/(2.54 3 12) = 10.9(36..) ydsii) 100/"10.9" = 9.1743... = 9.1 - 9.2 m

4

in 10 yards.

S4a/C89 There are 14 pounds in a stone.

There are 2.2 pounds in a kilogram.A man weighs 13 stone 6 pounds.Work out his weight in kilograms.Give your answer to the nearest kilogram.

(13 × 14) + 6 = 188188/2.2 = 85 kg

3

S4d/C90 The diagram shows a triangular

prism.

BC = 4 cm, CF = 12 cm and angle ABC = 90º.The volume of the triangular prism is 84 cm3.Work out the length of the side AB of the prism.

Area of base × 12 = 84½ × 4 × h = 7h = 3.5

3

S4d/C91 The shape below is the cross section

of a prism 10 cm long.(2 + 4)/2 × 2 × 10 = 60 cm2 3

Calculate the volume of the prism.

S4d/C92 The diagram shows a cylinder.

The height of the cylinder is 26.3 cm.The diameter of the base of the cylinder is 8.6 cm.

Calculate the volume of the cylinder.Give your answer correct to 3 significant figures.

3.14 × 4.3 × 4.3 × 26.3 = 1526.9 ( 1527.7) 3.142 1527.9 = 1530

3

S4a/C/B93 There are 6 groats in a florin and 5

florins in 10 shillings.Work out how manyi groats there are in 5 florins,ii florins there are in 24 shillings,iii groats there are in 8 shillings.

i) 5 × 6 = 30 groats (C)ii) 10 shillings = 5 florins

1 shilling = 0.5 florins24 shillings = 24 × 0.5 = 12 florins (C)

iii)1 shilling = 0.5 florins3 groats = 1 shilling

8 shillings = 24 groats (B)

6

S4d/B94 The expressions shown below can

be used to calculate lengths, areas or volumes of various shapes.The letters r and h represent lengths. , 2, 3, 4, 5 and 10 are numbers which have no dimensions.

circle around: r(r + 4h), rh4

, 3 3rh

3

r( ) 2 4 2rh r(r + 4h)

rh4

45

3r

10 3r ( )r h 2 3 3rh

r h r2 ( )

Draw a circle around each of the expressions which can be used to calculate an area.

S4d/B95 The diagram represents a solid

shape.

From the expressions below, choose the one that represents the volume of the solid shape.

and 13

are numbers which have no

dimensions.a, b and h are lengths.

13(b2 - ab + a2),

13h(b2 + ab +

a2), 13h2(b2 - a2),

13(a2 + b2),

13h2(b2 - ab +

a2).

Write down the correct expression.

13h(b2 + ab + a2) 1

SFMa/A96 Two similar boxes have volumes of

2000 cm3 and 16 000 cm3.The area of the base of the larger box is 60 cm3.

scale factor (vol) = 16 000/2000 = 8scale factor (length) = 81/3 = 2req'd area = 60 ÷ 4 = 15 cm2

4

Calculate the area of the base, in cm2, of the smaller box.

SFMa/A97 A child's toy is made out of plastic.

The toy is solid.The top of the toy is a cone of height 10 cm and base radius 4 cm.The bottom of the toy is a hemisphere of radius 4 cm.

Calculate the volume of plastic needed to make the toy.

1/3 × 42 × 10hemi: = 134.04167.55 + 134.04 = 301.59

6

SFMc/A98 The depth of water in harbour varies

according to the formulay = 10 + 5 sin(30t)º(y is the depth of the water in feet; t is the time in hours)Here is a sketch of the graph of this formula.

a Complete the labelling on the t and y axes for this graph.

A ship wishes to leave the harbour, but needs a depth of water of 13 feet

a)

b) 13 = 10 + 5sin (30t) sin (30t) = 0.6 t = 1.23

1.23 60 = 73.8 = 74 mins so ship leaves at 13:14

6

to do so safely.When the time is 1200 hours the value of t is zero.b At what time can the ship first leave the harbour safely?

SFMb/A/A*99 In triangle ABC, M is the midpoint

of AC.

N is a point on AB so that AN = 2NB.L is the midpoint of CN.p is the vector AB, q is the vector AC.a Express in terms of p and q the vectors

i AM,ii AN,iii NL.

b Write down two different facts about the lines AB and LM.

a) i) ½ qii) 2/3 qiii) ½ q - 1/3 q (A)

b) AB // LMAB = 3LM (A*)

6

SFMc/A/A*100 On the scales in Ali's book shop the

weight of a book correct to 2 decimal places is 0.62 kg.a Write down

i the lower bound of the weight of the book,

ii the upper bound of the weight of the book.

Ali needs to work out the weight of 50 copies of the book. He uses his value for the weight of one book.b Calculate

i the lower bound of the weight of 50 books,

a) i) 0.615ii) 0.625 (A)

b) i) 0.615 × 50 = 30.75ii) 0.625 × 50 = 31.25 (A)

c) (b)ii - 31 or 31 - (b)i or ½((b)ii - (b)i) = 0.25 (A*)d) (c) × 10 = 2.5 (A*)

8

ii the upper bound of the weight of 50 books.

c Calculate the greatest possible error that could occur in calculating the weight of 50 copies of the book.

d Write down the greatest possible error that could occur in calculating the weight of 500 copies of the book.

SFMd/A/A*101 Two tangents are drawn from a

point T to a circle centre O. They meet the circle at points A and B.Angle AOB is equal to 128º.

In this question you MUST give reasons for your answers.Work out the size of the anglesi APB,ii BAO,iii ABT.

i) APB = 64 (Angle subtended at the circumference is half that subtended at the centre) (A)ii) BAO = 26 Triangle OAB is Isosceles so BAO = (180 - 128) 2(A*)iii) ABT = 64 ( OBT is 90 because BT is a tangent and OB is a radius;

ABT = 90 - OBA = 90 -26(A*)

5

SFMa/A*102 Q is the midpoint of the side PR and

T is the midpoint of the side PS of triangle PRS.

a Write down, in terms of a and b,

a) i)

ii)

iii)

b) Either QT and RS parallel or RS = 2QT

4

the vectors

i

ii

iii

b Write down one geometrical fact about QT and RS which could be deduced from your answers to part a.

SFMb/A*103 ABCD is a parallelogram.

The diagonals of the parallelogram intersect at O.

, a Write an expression, in terms of a and b, for

i

ii

iii

X is the point such that =2a - b

b i Write down an expression, in

terms of a and b, for ii Explain why B, A and X lie on the same straight line.

a) i) 2a ii) a - b iii) -a - b

b) i) a - b ii)

6

SFMc/A*104 In the diagram, XY represents a

vertical tower on level ground.

A and B are points due West of Y. The distance AB is 30 metres.The angle of elevation of X from A is 30º. The angle of elevation of X from B is 50º.Calculate the height, in metres, of the tower XY.Give your answer correct to 2 decimal places.

XB/sin30º = 30/sinAXBXB = 30 × 0.5/sin 20º = 43.86XY = XB sin50º = 33.596 m = 33.60 m

6

SFMc/A*105 Find two different values of x

between 0 and 180 for whichsin (2x)º = sin 30º

2x = 30, 150; x = 15º, 75º 2

SFMc/A*106 A straight road UW has been

constructed to by-pass a village V.

The original straight roads UV and VW are 4 km and 5 km in length respectively.V lies on a bearing of 052º from U.W lies on a bearing of 078º from V.The average speed on the route

UVW = 102 + 52 = 154ºUW2 = 42 + 52 - 2×4×5×cos154 = 76.95UW = 8.7722Time for UVW = 9/30 × 60 = 18Time for UW = 8.7722/65 ×60 = 8.097Time saved = 10 mins

8

UVW, through the village is 30 kilometres per hour.The average speed on the by-pass route UW is 65 kilometres per hour.Calculate the time saved by using the by-pass route UV.Give your answer to the nearest minute.

SFMd/A*107 The diagram shows a circle centre

O.PQ and QR are tangents to the circle at P and Q respectively.S is a point on the circle.Angle PSR = 70º.PS = SR.

a i Calculate the size of angle PQR.

ii State the reason for your answer.

b i Calculate the size of angle SPO. ii Explain why PQRS cannot be a cyclic quadrilateral.

a) i) 40ºa) ii) OPQ = ORQ = 90º

between tan and radPOR = 140º at circumPQR = 360 - (90 + 90 + 140)

b) i) 35ºb) ii) opp angles of cyclic quad add to 180º

PSR + PQR = 110SPQ + SRQ = 250

7

SFMd/A*108 A, B, C and T are points on the

circumference of a circle.a) ACT = 25º alt angle

ATP = 25º alt segAPT = 25º isos triangle

b) TAP = 130º 's in triangleCAT = 50º 's in straight lineBTS = 75º alt seg

6

Angle BAC = 25º.The line PTS is the tangent at T to the circle.AT = AP.AB is parallel to TC.a Calculate the size of angle APT. Give reasons for your answer.

b Calculate the size of angle BTS. Give reasons for your answer.

S2d/S3a/S4d/C109 The scale diagram shows the

position of a radio mast, M.1 cm on the diagram represents 20 km.

M

Signals from the radio mast can be received up to a distance of 100 km.a Shade the region on the scale diagram in which signals from the radio mast can be received.

The distance of a helicopter from the radio mast is 70 km correct to the nearest kilometre.b Write down

i the maximum distance the helicopter could be from the radio mast,

ii the minimum distance the helicopter could be from the radio

a) circle drawn at 5 cm radius circle shadedb) i) 70.5

ii) 69.5

4

mast.

D2d/C Handling Data110 Andrew did a survey at the seaside

for his science coursework.He measured the lengths of 55 pieces of seaweed.The results of the survey are shown in the table.

Length ofseaweed(L cm)

Frequency

0 < L 20 220 < L 40 2240 < L 60 1360 < L 80 10

80 < L 100 5100 < L 120 2120 < L 140 1

Andrew needs to calculate an estimate for the mean length of the pieces of seaweed.a Work out an estimate for the mean length of the piece of seaweed.Give your answer correct to 1 decimal place.

b Write down the interval which contains the median length of a piece of seaweed.

a) 2 × 10 + 22 × 30 + 13 × 50 + 10 × 70 + 5 × 90 + 2 × 110 + 1 × 130 = 20 + 660 + 650 + 700 + 450 + 220 + 130 = 2830 (= 51.4545...) = 51.5b) 40 < L 60

6

D2d/C111 A survey was carried out to find

how much time was needed by a group of pupils to complete homework set on a particular Monday evening.The results are shown in the table below.

(3 × 0) + (14 × 0.5) + (17 × 1.5) + (5 × 2.5) + (1 × 3.5)= 48.548.5 ÷ 40 = 1.21

4

Time, thours,spent onhomework

Numberof pupils

0 30 < t 1 141 < t 2 172 < t 3 53 < t 4 1

Calculate an estimate for the mean time spent on homework by the pupils in the group.

D2d/C112 Bronwen owns a pet shop.

The table gives information about the weights of hamsters in Bronwen's shop.

Weight w ofhamsters in g

Number ofhamsters

28 < w 30 930 < w 32 532 < w 34 434 < w 36 2

Calculate an estimate for the mean weight of the hamsters in Bronwen's shop.

29 × 9 = 26131 × 5 = 15533 × 4 = 13235 × 2 = 70

618618 ÷ 20 = 30.9

4

D2c/f/C113 Information about oil was recorded

each year for 12 years.The scatter graph shows the amount of oil produced (in billions of barrels) and the average price of oil (in £ per barrel).

a) Line of best fitb) Draw line "amount = 10.4" on graph or state use of amount = 10.4; price about £16.50

4

a Draw a line of best fit on the scatter graph.

In another year the amount of oil produced was 10.4 billion barrels.

b Use your line of best fit to estimate the average price of oil per barrel in that year.

D2c/f/C114 The table list the weights of twelve

books and the number of pages in each one.

Number of pages Weight (g)80 160

155 330100 200125 260145 32090 180

140 290160 330135 260100 180115 230165 350

This information is presented below as a scatter graph.

a Draw a line of best fit on your scatter graph.

b Use your line of best fit to estimate

i the number of pages in a book of weight 280 g,

a) Line of best fit (only str. line)b) i) accept 136 - 140 pages ii) accept 216 - 220 g

4

ii the weight, in grams, of a book with 110 pages.

D2b/B115 Martin, the local Youth Centre

leader, wishes to know why attendance at the Youth Centre is less than at the same time last year.He thinks that it could be due to a number of changes that occurred during the course of the year.These changes were:

the opening hours changeda new sports centre opened

nearbysome of the older members

started bullying the younger members.Design a suitable question, that is easily answered, to find out why people do not attend the Youth Centre.

Look for a question that links to the original problem via the perceived causes and a way of completing the questionnaire easily e.g. boxes

2

D2c/B116 Pippa collected data for the heights

(h) of the students in her class.Here is the grouped frequency table of her results.

Height (h) in cm Frequency160 h < 165 7165 h < 170 6170 h < 175 2175 h < 180 10180 h < 185 5

a Use the table to calculate an estimate of the mean for her results.

b Complete the cumulative frequency table for Pippa's data and hence draw a cumulative frequency graph for the data.

a) Mid point Total162.5 1137.5167.5 1005172.5 345177.5 1775182.5 912.5Total 5175Mean = 5175 ÷ 30 = 172.5 cm

b) 7, 13, 15, 25, 30c) Median = 175

Interquartile range = 17

12

Group Frequency Cumulative frequency160 h < 165 7 7160 h < 170 6160 h < 175 2160 h < 180 10160 h < 185 5

c Use the cumulative frequency graph to calculate an estimate of

I median of the dataii interquartile range of the

data.

D2a/B117 The table gives information about

the weights of 100 new born babies.

Weight (w) in kg Frequency1.0 w < 1.5 41.5 w < 2.0 92.0 w < 2.5 112.5 w < 3.0 213.0 w < 3.5 263.5 w < 4.0 184.0 w < 4.5 94.5 w < 5.0 2

a Complete the cumulative frequency table below.

a) 4, 13, 24, 45, 71, 89, 98, 100b) Points joined on a cumulative frequency graph; ignore graph below x = 1.5.

c) 3.15 kg

6

Weight (w) in kg CumulativeFrequency

1.0 w < 1.51.0 w < 2.01.0 w < 2.51.0 w < 3.01.0 w < 3.51.0 w < 4.01.0 w < 4.51.0 w < 5.0

b Draw a cumulative frequency graph for your table.

c Use your cumulative frequency diagram to estimate the median weight, in kilograms, of the new born babies.Show your method clearly.

D2c/B118 A survey is made of all 120 houses

on an estate.The floor are, in m2, of each house is recorded.The results are shown in the cumulative frequency table.

a) plot points, joined with smooth curve/lineb) segments 240 - 170 = 70c) 10% is 12 houses. Read off where cf = 120 - 12 = 108; floor area = 275 m2

6

Floor area (x) in m2 CumulativeFrequency

0 < x 100 40 < x 150 200 < x 200 490 < x 250 970 < x 300 1140 < x 350 1180 < x 400 120

a On the grid draw a cumulative frequency graph for the table.

b Use your cumulative frequency graph to estimate the interquartile range of the floor areas of the houses.

The houses on the estate with the greatest floor areas are called luxury houses.10% of the houses are luxury houses.c Use your graph to estimate the minimum floor area for a luxury house.

H2c/d/B119 150 year 11 pupils took a

mathematics examination.The table shows information about their marks.

a Complete the cumulative frequency table below.

a) 6, 23, 45, 90, 116, 135, 144, 150b) cum freq diag drawn: ignore line between first point and the originc) 60% of 150 = 90

150 - 90 = 60 pass mark about 42 - 45 marks

7

Marks (x) CumulativeFrequency

0 x < 200 x < 300 x < 400 x < 500 x < 600 x < 700 x < 800 x < 100

b On the grid below, draw a cumulative frequency diagram to show these marks.

60% of the pupils passed the examination.c Use your diagram to find an estimate for the pass mark for the examination.

D3b/C120 A dice has six faces numbered 1, 2,

3, 4, 5 and 6.The dice, which is biased, is thrown 200 times and the number on the upper face is recorded.The frequencies of the numbers obtained are shown in the table.

46/200 or 23/100 or 0.23 or 23% 2

Numbershown on dice

1 2 3 4 5 6

Frequency 38 22 46 25 53 16

Estimate the probability that the next time the dice is thrown it will show the number 3.

D3a/B/C121 The probability of a machine being

able to manufacture a component within a tolerance of one tenth of a millimetre is 0.995.a Work out the probability of the machine not being able to manufacture a component to within a tolerance of one tenth of a millimetre.

Ten thousand components are manufactured in one day.b Work out an estimate of how many components will be outside the tolerance of one tenth of a millimetre.

a) 1 - 0.995 = 0.005 (C)b) 10 000 × (a) - 50 (B)

4

D3e/B122 Two dice with faces numbered 1 to

6 are rolled and the sum of the scores on upward facing faces noted.

The score on these dice is 1 + 4 = 5a Complete the table of probabilities for the chance of scoring all the totals available.

i) 0.0625 × 0.032 × 0.044 = 0.0000915ii) 0.065 × 0.968 × 0.954 = 0.06iii) 0.935 × 0.968 × 0.954 = 0.863

7

Score Possibilities Probability2 1 + 13 1 + 2, 2 + 14 1 + 3, 3 + 1, 2 + 25 1 + 4, 4 + 1, 2 + 3, 3 + 26789101112

b Work out the probability that the score will be i more than 9, ii less than 5, iii more than 6 and less than 10.

D3e/B123 The probability of a car chosen at

random having defectivetyres is 0.065brakes is 0.032steering is 0.044Work out the probability that a vehicle chosen at random will havei defective tyres, brakes and steering,ii has defective tyres but no other defects,iii has no defects.

i) 0.0625 × 0.032 × 0.044 = 0.0000915ii) 0.065 × 0.968 × 0.954 = 0.06iii) 0.935 × 0.968 × 0.954 = 0.863

7

D3f/B124 Nikki and Ramana both try to score

a goal in netball.The probability that Nikki will score a goal on her first try is 0.65.The probability that Ramana will score a goal on her first try is 0.8.i Work out the probability that Nikki and Ramana will both score a goal on their first tries.ii Work out the probability that neither Nikki nor Ramana will score a goal on their first tries.

i) 0.65 × 0.8 = 0.52ii) 1 - 0.65 (= 0.35)

1 - 0.8 (= 0.2)0.35 × 0.2 = 0.07

5

D3e/A/A*125 Susan has 8 eggs in her fridge. a) 9

Two of the eggs have passed their sell by date and are "bad".She selects 3 eggs at random to bake a cake.a Complete the probability tree diagram.

b Work out the probability that Susan will select 3 "good" eggs.

c Work out the probability that Susan will select at least one "bad" egg.

(A)

b) ¾ × 5/7 × 2/3 = 5/14 (A*)c) 1- (b) = 9/14 (A*)

DFMb/A126 Kim sowed some seeds in her

greenhouse.10 weeks later she measured the heights of the plants.Some of the results are shown in the table and the histogram.

Height(h) in cm

Number ofplants

0 < h 5 05 < h 20 30

20 < h 30 12030 < h 3535 < h 4040 < h 50 96

Over 50 0

a) 3rd interval: 120 plants need 12 cm2 so 10 plants per cm2. 4th interval: Area 11.5cm2 so 11.5 × 10 = 115 plants. 5th interval: Area 10.5 cm2 so 105 plants. 2nd interval: 30 plants need 3 cm2 so height = 3 ÷ 3 = 1 cm. 6th interval: 96 plants need 9.6 cm2 so height = 9.6 ÷ 2 = 4.8 cm.ORf.d. for 3rd interval = 120 ÷ 10 = 12, so scale on f.d. axis is 1 cm = 12 ÷ 6 = 2. 4th

interval: 2 × 11.5 × 5 = 115. 5th interval: 2 × 10.5 × 5 = 105. 2nd interval: f.d. = 30 ÷ 15 = 2 (2 ÷ 2 = 1 cm). 6th interval: f.d. = 96 ÷ 10 = 9.6 (9.6 ÷ 2 is 4.8 cm).

b) 500 - (30 + 120 + 115 + 105 + 96) = 500 - 466 = 34

6

a Use the information to complete the table and the histogram.

Kim had sown 500 seeds.b Calculate the number of seeds that had not produced plants.

DFMb/A127 John measured the time, in seconds,

that birds spent on each individual visit to his bird table. The birds made a total of 113 individual visits.The histogram shows some of the results.a Use the information in the histogram to complete the frequency table below.

Time(x seconds)

Frequency

0 < x 1010 < x 20 2820 < x 2525 < x 3030 < x 50 12 x > 50 0Total 113

b Use the information in the frequency table to complete the histogram.

a) 2nd interval 28 ÷ 14 = 2 birds per cm2 . 10 × 2 = 20, 15.5 × 2 = 31, 113 - (20 + 28 + 31 + 12) = 22orf.d. for 2nd interval = 28 ÷ 10 = 2.8, so scale on f.d. axis is 1 cm = 2.8 ÷ 7 = 0.40.4 × 5 × 10 = 20, 0.4 × 15.5 × 5 = 31 113 - (20 + 28 + 31 + 12) = 22

b) 4th interval, 22 ÷ 2 = 11 cm2. 5th interval, 12 ÷ 2 = 6 cm2, 6 ÷ 4 = 1.5 cmorf.d. = 22 ÷ 5 = 4.4, (4.4 ÷ 0.4 = 11 cm) f.d. = 12 ÷ 20 = 0.6, (0.6 ÷ 0.4 = 1.5 cm)

6

DFMb/A128 The histogram gives information

about the ages of the teacher at a school on 1st September last year.

a Use the information in the histogram to complete the frequency table below.

Age (A) years Frequency22 A < 2424 A < 2727 A < 3030 A < 35 1635 A < 40 1940 A < 5050 A < 65 27

b Use the information in the frequency table to complete the histogram

a) e.g. 16 ÷ 5 = 3.2; 64 ÷ 3.2 = 20 so 76 ÷ 20 × 5 = 19 etc. Frequencies are: 7, 6, 9, 36b) 27 ÷ 15 × 20 = 36

6

DFMc/A

129 Pippa's friend Wayne was also measuring the heights of people in his class.Here are his results for the 10 boys in the class.

162 165 178 182 175185 172 178 180 175

All the measurements are in centimetres correct to the nearest centimetre.a Calculate the standard deviation of the data.

Wayne made a mistake when he was measuring the heights.He started at the 2.5 centimetre mark on the tape.b i Work out the actual mean of the data.

ii Work out the actual standard deviation of the data.

a) mean 175.2sum of squares 307420standard deviation = 6.85

b) i) 177.7ii) 6.85

7

DFMc/A130 Calculate the standard deviation of

the numbers3, 3, 4, 5, 7, 9, 11.

mean = 6(3 - 6)2 + (3 - 6)2 + (4 - 6)2 + (5 - 6)2 + (7 - 6)2 + (9 - 6)2 + (11 - 6)2 (= 58) (58 7) = 2.88OR use formula

2

DFMd/A131 The diagram shows two boxes A

and B.

Box A contains 5 white beads and 3 black beads.Box B contains 4 white beads and 4 black beads.A bead is to be taken at random from box A and placed in box B.A bead is then to be taken at random from box B and placed in box A.a Calculate the probability that

a) 5/8 × 5/9 = 25/72b) BB = 3/8 × 5/9 = 15/72 + (a) = 40/72 or 5/9

6

both beads taken will be white.

b Calculate the probability that after both beads have been taken, there will be exactly 5 white beads in box A.

DFMd/A/A*132 There are two sets of traffic lights

on Paul's route to school.The probability that the first set of lights will be green is 3/5.If he finds the first set of lights green, the probability that the second set of lights will be green, when he gets to them, is 2/7.If he finds the first set of lights are not green, the probability that the second set of lights will be green, when he gets to them, is 4/7.Part of the tree diagram showing these probabilities is shown.a Complete the probability tree diagram.

b Calculate the probability that Paul will find the first set of lights is not green and the second set of lights is green.

c Calculate the probability that Paul will find the second set of lights is green.

a) 2/5; 2/7; 5/7; 4/7; 3/7; b) 2/5 × 4/7 =

8/35

c) 3/5 × 2/7 + 2/5 × 4/7 = 14/35 = 2/5

8

DFMd/A/A*133 There are 4 red balls, 5 blue balls

and 3 green balls in a bag.A ball is to be taken at random and not replaced.A second ball is then to be taken at random.

a) 4/12; 3/11; 5/11; 3/115/12; 4/11; 4/11; 3/113/12; 4/11; 5/11; 2/11 (A)

b) i) 4/12 × 3/11 = 1/11ii) (5/12 × 4/11) + (3/12 × 2/11) +

"1/11" = 19/66 (A)

11

a Complete the tree diagram below.

b Use the tree diagram to calculate the probability that both balls taken will be

i red,ii the same colour.

c Calculate the probability that exactly one of the balls taken will be red.

c) (4/12 × 5/11) + (4/12 × 3/11) + (5/12 × 4/11) + (3/12 × 4/11) = 16/33 (A*)

DFMc/A*134 a Show that the mean of five

consecutive numbers with median n is also n.

b Show that the mean of the squares of these 5 numbers exceeds the median of the squares by the number 2.

a) If median is n the numbers must be n-2, n-1, n+1, n +2

Sum = 5n mean = 5n 5 = nb) (n - 2)2 = n2 - 4n + 4

(n - 1)2 = n2 - 2n + 1n2 = n2

(n + 1)2 = n2 + 2n + 1(n + 2)2 = n2 + 4n + 4Sum = 5n2 + 10Mean = n2 + 2Median = n2

7

DFMc/A*135 In an experiment, Nazia measured

the time it took for a ball bearing to sink down to the bottom of a tube of oil. She made 5 measurements. Her results are given below.

2.4s, 2.5s, 2.4s, 2.6s, 2.7s

a) x = 12.6x2 = 31.82

standard deviation = 0.01166 = 0.0117

b) i) 2.52 + 0.2 = 2.72ii) (a) unchanged = “0.0117”

6

a Calculate the standard deviation of these times.Give your answer correct to 3 decimal places.

Nazia found that her timing had been 0.2 seconds too short each time.b For these corrected times, find

i the mean time,ii the standard deviation.

DFMc/A*136 The table shows the income per sale

from the sale of CD players in a shop over one week.

CD player Income per sale (£) Number ofsales

Good Value 75 4Special 130 3Super 190 1

De-Luxe 220 2

The mean income per sale is £132.

Calculate the standard deviation.

Using fx

fx

22∑

∑−

fx2 = 4 752 + 3 1302 + 1 1902 + 2 2202 =2250 + 50700 + 36100 + 96800 = 206100206100 ÷ 10 = 20610var = 20610 - 1322 = 3186s.d. = 3186 = 56.444.. = £56.44or

usingf x x

f

( )−

× × × ×

∑∑

2

2 2 2 24 57 3 2 1 58 2 8812996 12 3364 15488 31860

31860 ÷ 10 = 3186s.d. = 3186 = £56.44

4

D3c/e/DFMd/A*/B/C137 Peter and Asif are both taking their

driving test for a motor cycle for the first time.The table below gives the probabilities that they will pass the test at the first attempt or, if they fail the first time, the probability that they will pass at the next attempt.

Probabilityof passing atfirst attempt

Probability ofpassing at nextattempt if they failthe first attempt

Peter 0.6 0.8Asif 0.7 0.7

On a particular day 1000 people will take the test for the first time.For each person the probability that

a) 1000 0.7 = 700 (C)b) 0.6 0.7 = 0.42 (B)c) 0.6 0.3 = 0.18 (B)d) P(pass) + P(fail) P(pass)

0.7 + 0.3 0.7 = 0.91 (A*)

9

they will pass the test at the first attempt is the same as the probability that Asif will pass the test at the first attempt.a Work out an estimate for how many of these 1000 people are likely to pass the test at the first attempt.

b Calculate the probability that both Peter and Asif will pass the test at the first attempt.

c Calculate the probability that Peter will pass the test at the first attempt and Asif will fail the test at the first attempt.

d Calculate the probability that Asif will pass the test within the first two attempts.

N4d/S4b/C Integrated questions138 Jomo is going to design a circular

roundabout. The roundabout will have a circumference of 7 metres.Jomo is given three estimates for the length of the diameter of the roundabout.The estimates are: 2.2278803 metres 2 metres 2.23 metresa Give a reason why 2.23 metres is the most reasonable estimate to use.

b Explain why 2.2278803 metres and 2 metres are not appropriate to use.

a) e.g. 2.23 m can be measured: normally give to 2 d.p.b) 2nd not accurate; not near enough to 7m; 1st too accurate: cannot be measured

3

S4d/N4d/C139 This container is made from a

cylinder and a cube.a) r2h = 82 20 = 4021.. cm3

b) 16 16 16 = 4096 cm3 (a) + "4096" = 8117 cm3

c) i) 7.5 cmii) 8.5 cm

7

The cylinder has a height of 20 cm. It has a base radius of 8 cm.The cube has sides of edges 16 cm.a Calculate the total volume, in cm3, of the cylinder. Give your answer to the nearest cm3.

b Calculate the total volume, in cm3, of the container. Give your answer to the nearest cm3.

When the radius of 8 cm was measured, this measurement was rounded to the nearest centimetre.c i Write down the minimum length, in cm, it could be.

ii Write down the maximum length, in cm, it could be.

A3b/N4c/B140 Matthew uses this formula to

calculate the value of D.

D a ca c

−−32

a Calculate the value of D when a = 19.9 and c = 4.05.Write down all the figures on your calculator display.

Matthew estimates the value of D without using a calculator.b i Write down an approximate value for each of a and c that Matthew could use to estimate the value of D. ii Work out the estimate that these approximations give for the

a)

19 9 3 4 0519 9 4 05

7 753 4975

2 215868477

2

. .. ..

..

− ×−

b) i) a = 20, c = 4

ii)20 1220 16

−−

= 84

= 2

6

value of D. Show all your working.

SFMa/NFMc/A/A*141 A cone has a height of 18 cm and

the radius of its base is 3 cm.

a Calculate the volume of the cone.

The measurements of the cone are correct to the nearest millimetre.b Write down the lower bound of the radius of the cone.

c Calculate the difference between the upper and lower bounds of the volume of the cone expressed as a percentage of the volume of the cone found in part (a).

a) 1/3 3.14 32 18 ( = 169.56) = 169.6 (A)b) 2.95 (A)c) Least volume = 1/3 3.14 2.952 17.95 = 163.4997 (163.58)

Greatest volume = 1/3 3.14 3.052 18.05 = 175.7459 (175.84)

Difference = 12.246 (12.25)12.246/169.56 100% = 7.2(2..)%

(A*)

9

SFMc/AFMa/A*142 In triangle ABC, AB = 5 cm, AC = x

cm, BC = 2x cm and angle BAC = 60.

a Show that 3x2 + 5x - 25 = 0.

b Solve the equation 3x 2 + 5x - 25

a) (2x)2 = 52 + x2 - 25xcos604x2 = 25 + x2 - 10x 0.5

b) 3x2 + 5x - 25 = 0x = (-5(52 + 300)) 6= (-518.027..) 6

= 2.17 or -3.84

c) BD

sin sin60

5

104

BD

BD

5 60

1044.4626

sin

sin. . .

9

= 0.Give your answers correct to 3 significant figures.

D is the point on AC such that angle ADB = 104.c Calculate the length of BD.