Higher.doc - mathsteachers



|Higher | | | |

| |N2c/C |Number | |

|1 |Calculate, giving your answer in standard form |152 000 - 46 000 = 106 000 |4 |

| |correct to 3 significant figures. |106 000 ( 0.0456 = 2.32 ( 106 | |

| | | | |

| |1.52 ( 105 - 4.6 ( 104 | | |

| |4.56 ( 10-2 | | |

| | | | |

| |N2c/C | | |

|2 |The distance of the Earth from the Sun at a |a) 9.35 ( 107 |5 |

| |particular moment is 93.5 million miles. |b) 250 000 | |

| |a Write 93.5 million in standard form. |c) 93 500 000 ( 250 000 = 3.74 ( 102 | |

| | | | |

| |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. | | |

| | | | |

| |N2c/C/B | | |

|3 |a Write down the value of |a) i) 1 |4 |

| |i 50 |ii) 1/16 | |

| |ii 4-2 |b) 1/8 ( 2=1/4 | |

| | | | |

| |b Simplify [pic][pic] | | |

| | | | |

| |N2c/B | | |

|4 |The area of the Earth covered by sea is 362 000 |a) 3.62 ( 108 |6 |

| |000 km2. |b) From 5.11 ( 108 to 5.12 ( 108 | |

| |a Write 362 000 000 in standard form. |c) (a)/(b) ( 100 = between 70 and 71 inclusive | |

| | | | |

| |The surface area, A km2, of the Earth may be found| | |

| |using the formula | | |

| |A = 4(r2 | | |

| |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 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/B | | |

|5 |Express 0.327 ( 105 in standard form. |3.27 ( 104 |2 |

| | | | |

| |N2c/B | | |

|6 |My computer can carry out 2.7 ( 108 calculations |2.7 ( 108 ( 3600 (= 75000) = 7.5 ( 104 |3 |

| |in one hour. | | |

| |Work out how many of these calculations my | | |

| |computer can carry out in one second. Give your | | |

| |answer in standard form. | | |

| | | | |

| |N2c/B | | |

|7 |A US Centillion is the number [pic] |a) 10 600 ÷ 10 303 = 10 600-303 = 1×10 297 or 10 297 |4 |

| |A UK Centillion is the number [pic] |b) 40×10 303 = 4×10 304 | |

| |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. | | |

| | | | |

| |N2c/B | | |

|8 |The diameter of an atom is |a) 3×10 -8 |4 |

| |0.000 000 03 m. |b) (a) ÷ 100 = 3×10 -10 | |

| |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. | | |

| | | | |

| |N3d/C | | |

|9 |Marcus sees a motorbike advertised for £750. |750 is 85% |3 |

| |[pic] |1% = 750 ÷ 85 | |

| | |100% = 750 × 85 ÷ 100 = £882.35 | |

| |This is the price after a reduction of 15%. Work | | |

| |out the original price of the motorbike. | | |

| | | | |

| |N3e/C | | |

|10 |Use your calculator to evaluate |_11374.09_ = 11374.09 |3 |

| | |(0.2 + 4.5)2 4.72 | |

| |560.3 ( 20.3 | | |

| |(0.2 + 4.5)2 |= 514.89769... | |

| | | | |

| |Write down all the figures on your calculator. | | |

| | | | |

| |N3e/C | | |

|11 |In this question you MUST use your calculator and |__59 ( 5.7_ = _336.3_ |2 |

| |you MAY write down any stage in your calculation. |200.3 ( 8.05 1612.415 | |

| |Evaluate | | |

| |[pic] |= 0.208569 = 0.208 or 0.209 | |

| | | | |

| |N4a/C | | |

|12 |Shreena put £484 in a new savings account. |484 × 4.3/100 |5 |

| |At the end of every year, interest of 4.3% was |484 + 20.81 (= 504.81) | |

| |added to the amount in her savings account at the |504.81 × 4.3/100 = 21.7069 | |

| |start of that year. |504.81 + 21.71 | |

| |Calculate the total amount in Shreena's savings |OR | |

| |account at the end of 2 years. |484 × 1.043 = 504.81 | |

| | |"504.81" × 1.043 = 526.52 | |

| | |OR | |

| | |484 × 1.0432 = 526.52 | |

| | |Accept £526.51 or £526.52 | |

| |N4a/C | | |

|13 |Astrid bought a motor car for £10 000 on the First|1st Jan 1997 £8500 |5 |

| |of January 1996. |1st Jan 1998 £7650 | |

| | |1st Jan 1999 £6885 | |

| |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/C | | |

|14 |Natalie measured the distance between two points |a) i) 5.5 cm |4 |

| |on a map. |ii) 4.5 cm | |

| |The distance she measured was 5 cm correct to the |b) 0.2 km | |

| |nearest centimetre | | |

| |a 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 20 000 | | |

| |b Work out the actual distance in real life | | |

| |between the upper and lower bounds. Give your | | |

| |answer in kilometres. | | |

| | | | |

| |NFMa/A | | |

|15 |The temperature from a factory furnace varies |T ( 1/d2 |5 |

| |inversely as the square of the distance from the |T = k × 1/d2 | |

| |furnace. |50 = k × ¼ | |

| |The temperature 2 metres from the furnace is 50ºC.|k = 200 | |

| |Calculate the temperature 3.5 metres from the |T = 200 × 1/3.52 | |

| |furnace. Give your answer to 2 decimal places. |T = 16.33ºC | |

| | | | |

| |NFMb/A | | |

|16 |Draw a circle around each irrational number in the|circles (2/3 and 4/3( |2 |

| |list below. | | |

| | | | |

| |[pic] 2.3 [pic][pic] [pic] [pic] [pic] | | |

| | | | |

| |NFMb/A | | |

|17 |Change [pic] into a fraction in its lowest terms. |x = 0.4545454545454545... |3 |

| | |100x = 45.454545454... | |

| | |99x = 45 | |

| | |so x = 5/9 | |

| |NFMb/A | | |

|18 |Right angled triangles can have sides with lengths| e.g. |4 |

| |which are a rational of irrational number of |i) 3, 4, 5 | |

| |units. |ii) (7, (2, 3 | |

| |Give an example of a right angled triangle to fit |iii) 2, 3, (13 | |

| |each description below. |iv) 2, (5, 3 | |

| |i All sides are rational | | |

| |[pic] | | |

| |ii The hypotenuse is rational and the other two | | |

| |sides are irrational | | |

| |[pic] | | |

| |iii The hypotenuse is irrational and the other two| | |

| |sides are rational. | | |

| |[pic] | | |

| |iv The hypotenuse and one of the other sides are | | |

| |rational and the remaining side is irrational. | | |

| |[pic] | | |

| | | | |

| |NFMb/A | | |

|19 |Put a tick in the box underneath those numbers |2/3, 1.6, 4/17 ticked |2 |

| |that are rational. | | |

| | | | |

| |[pic] | | |

| |NFMb/A | | |

|20 |Put a tick in the box underneath the rational |tick (12 and (1 |2 |

| |numbers. |(3 (36 | |

| | | | |

| |[pic] | | |

| |NFMb/A | | |

|21 |a Give an example of two different irrational |a) e.g. (8 ( (2 (= (4 = 2) |4 |

| |numbers q and r such that q/r is a rational |b) x3 such that 3(15 < x < 3(16 | |

| |number. |x3such that 2.446 < x < 2.519 | |

| |b Write down a number which is greater than 15 and| | |

| |less than 16 and which has a rational cube root. | | |

| | | | |

| |NFMb/A | | |

|22 |a Write down a number which is greater than 17 and|a) x2: (17 < x < (18 |4 |

| |less than 18 that has a rational square root. |or x2: 4.123 ( x ( 4.242 | |

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

| |b Give an example of two different irrational | | |

| |numbers c and d such that c ( d is a rational | | |

| |number. | | |

| | | | |

| |NFMc/A | | |

|23 |Cleo uses a pair of scales to measure, in |Since the scales were only accurate to 0.1kg then 1.4 |2 |

| |kilograms, the weight of a brick. |should be the answer. | |

| |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/A | | |

|24 |The speed of light is 186 000 miles per second |93 550 000 ÷ 185 500 = 504 |5 |

| |correct to the nearest thousand miles per second. |93 450 000 ÷ 186 500 = 501 | |

| |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. | | |

| | | | |

| |NFMc/A | | |

|25 |Correct to 3 decimal places, a = 2.236. |a) i) 2.2365 |8 |

| |a For this value of a, write down |ii) 2.2355 | |

| |i the upper bound, |b) i) 3.651 | |

| |ii the lower bound |ii) 3.649 | |

| | |c) 2.2355 × 1.4135 = 3.15987925 | |

| |Correct to 3 decimal places, b = 1.414. |d) 2.2365 ÷ 1.4135 = 1.58224266 | |

| |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 | | |

| |[pic] | | |

| | | | |

| |NFMc/A/A* | | |

|26 |PQR is a right angled triangle. |a) i) 6.05 |9 |

| |RQ= 6.0cm and PR=8.3cm, both correct to 1 decimal |ii) 5.95 | |

| |place |b) (6.05×8.35)/2 = 25.25875 | |

| |[pic] |c) Tan PQR = 8.35/5.95 so PQR = 54.5° | |

| |a Write down | | |

| |i the upper bound of the length of RQ | | |

| |ii 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. | | |

| | | | |

| |NFMd/A | | |

|27 |Evaluate: |i) 9 |2 |

| |i [pic] |ii) 4/9 | |

| | | | |

| |ii[pic][pic][pic] | | |

| | | | |

| |NFMc/A* | | |

|28 |x = 40, correct to the nearest 10. |a) i) 35 |8 |

| |y = 60, correct to the nearest 10. |ii) 65 | |

| |a i Write down the lower bound of x. |b) 45 ( 65 = 2925 | |

| |ii Write down the upper bound of y. |c) 35 ( 65 (= 0.53846...) = 0.538 | |

| | |d) = 1 + 10/x = 1 + 10/35 (= 1.2857...) = 1.29 | |

| |b Calculate the greatest possible value of xy. | | |

| | | | |

| |c Calculate the least possible value of [pic]. | | |

| |Give your answer correct to 3 significant figures.| | |

| | | | |

| |d Calculate the greatest possible value of | | |

| |[pic][pic]. | | |

| |Give your answer correct to 3 significant figures.| | |

| | | | |

| |NFMc/A* | | |

|29 |Kim is doing an experiment using a pendulum. She |40 ( 1.005 = 11.14 |6 |

| |uses the formula |1.92 | |

| |[pic] |40 ( 0.995 = 9.02 | |

| | |2.12 | |

| |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. | | |

| | | | |

| |N3a/N2c/C | | |

|30 |The number 1998 can be written as 2 ( 3n ( p, |i) 2 ( 999 = 2 ( 9 ( 111 = 2 ( 27 ( 37 = 2 ( 33 ( 37 ; n|3 |

| |where n is a whole number a p is a prime number. |= 3, p = 37 | |

| |i Work out the values of n and p. |ii) 3 ( 37 = 111 | |

| |ii Using your answers to part i, or otherwise, | | |

| |work out the factor of 1998 which is between 100 | | |

| |and 200. | | |

| | | | |

| |A2b/C |Algebra | |

|31 |Work out an algebraic expression for the nth term |2n × n, 2(n × n) = 2n2 |2 |

| |of this sequence of numbers. | | |

| |2, 8, 18, 32, 50, ....... | | |

| | | | |

| |A2b/C | | |

|32 |The expression n(n +1) |10n(n +1) = 5n(n +1) |1 |

| |2 |2 | |

| | | | |

| |is the nth term of the sequence of triangular | | |

| |numbers | | |

| |1, 3, 6, 10, ... | | |

| |Write down an expression, in terms of n, for the | | |

| |nth term of the sequence | | |

| |10, 30, 60, 100, ... | | |

| | | | |

| |A2b/C | | |

|33 |Here are the first terms of a number sequence. |3n + 2 |2 |

| |5, 8, 11, 14, 17, | | |

| |Write down an expression for the nth term of the | | |

| |sequence. | | |

| | | | |

| |A2b/C | | |

|34 |The numbers of lines joining points form a number |i) 6 ( 15 |6 |

| |sequence. |ii) 10 ( 45 | |

| |[pic] |iii) n( n(n -1) | |

| |Complete the table to find the number of lines |2 | |

| |joining | | |

| |i 6 points | | |

| |ii 10 points | | |

| |iii n points | | |

| | | | |

| |[pic] | | |

| |A2d/C | | |

|35 |a Make a table of values for y = 6 - 2x |a) -2, -1, 0, 1, 2, 3, 4 |6 |

| | |10, 8, 6, 4, 2, 0, -2 | |

| |[pic] |Ignore outside -2 to 4 | |

| | |b) graph drawn | |

| |b Draw the graph of y = 6 - 2x on the grid below. |c) i) x = -1.5, y = 9 | |

| |[pic] |ii) y = 3.4, x = 1.3 | |

| | | | |

| |c Use your graph to find | | |

| |i the value of y when x = -1.5 | | |

| |ii the value of x when y = 3.4 | | |

| | | | |

| |A2c/B | | |

|36 |The diagram shows part of a distance/time graph |a) 88 m |4 |

| |for a bus after if had left a bus stop. |b) AB constant speed | |

| | |BC gradually slowing down | |

| |[pic] |CD stationary. | |

| | | | |

| |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. | | |

| | | | |

| |A2b/C | | |

|37 |i On the grid below draw the graph of |i) x -2, -1, 0, 1, 2, 3, 4, 5 |6 |

| |y = x2 - 3x - 5 |y 5, -1, -5, -7, -7, -5, -1, 5 | |

| |for values of x between -2 and +5 |ii) x = -1.2 | |

| | |x = 4.2 | |

| |[pic] | | |

| |[pic] | | |

| |ii Use your graph to solve the equation | | |

| |x2 - 3x - 5 = 0 | | |

| | | | |

| |A3c/C | | |

|38 |x is an integer, such that -3 < x [pic] 2. |-2, -1, 0, 1, 2 |2 |

| |List all the possible values of x. | | |

| | | | |

| |A3c/C | | |

|39 |x is an integer. Write down the greatest value of |3 |1 |

| |x for which 2x < 7. | | |

| | | | |

| |A3d/C | | |

|40 |Use the method of trial and improvement to find |3.2 |4 |

| |the positive solution of | | |

| |x3 + x = 37 | | |

| |Give your answer correct to 1 decimal place. | | |

| | | | |

| |A3d/C | | |

|41 |Use a trial and improvement method to solve the |2.53 - 2.5 = 13.125 |3 |

| |equation |2.63 - 2.6 = 14.976 | |

| |x3 - x = 15 |2.73 - 2.7 = 16.983 | |

| | |so x = 2.6 | |

| |[pic] | | |

| |Complete the working below and find a solution | | |

| |correct to one decimal place. | | |

| | | | |

| |A3b/B/C | | |

|42 |The velocity of a particle is given by the formula|a) 25 + 2 ( 2/3 ( 5.67 |5 |

| |v 2 = u 2 + 2as |32.56 (C) | |

| | |b) u2 = v2 -2as | |

| |a Calculate the value of the velocity v when u = |u = ((v 2 - 2as) (B) | |

| |-5, a = [pic]and s = 5.67. | | |

| | | | |

| |b Rearrange the formula to make u the subject. | | |

| | | | |

| |A3b/B | | |

|43 | y = ab + c |(21 - 16)/32 = 5/32 |3 |

| | | | |

| |Calculate the value of y when | | |

| |a =[pic], b = [pic][pic]and c =[pic] | | |

| | | | |

| |Give your answer in the form [pic] where p and q | | |

| |are integers. | | |

| | | | |

| |A3b/B | | |

|44 |The volume, V, of the barrel is given by the |(( ( 60)/3000 ( (2 ( 252 + 202) = (( ( 60)/3000 ( 1650 = |3 |

| |formula |103.67... | |

| | |3.14 ( 103.62 | |

| |[pic][pic] |3.142 ( 103.68.. | |

| | |22/7 ( 103.71.. | |

| |[pic]= 3.14, H = 60, R = 25 and r = 20. |so V = 104 | |

| | | | |

| |[pic] | | |

| | | | |

| |Calculate the value of V. | | |

| |Give your answer correct to 3 significant figures.| | |

| | | | |

| |A3b/c/B | | |

|45 |In the diagram, each side of the square ABCD is |a) (3 + x)(3 + x) or (3 + x)2 = (x + 3)2 |4 |

| |(3 + x) cm. |b) (3 + x)(3 + x) = 10 | |

| | |9 + 3x + 3x + x2 = 10 | |

| |[pic] |x2 + 6x + 9 = 10 and | |

| | |x2 + 6x = 1 | |

| |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 | | |

| | | | |

| |A3b/c/B | | |

|46 |v 2 = u 2 + 2as |a) (-6)2 + 2 ( 5 (0.8 = 7 |5 |

| |a Calculate the value of v when u = -6, a = 5 and |or -62 + 2 ( 5 (0.8 = 7 | |

| |s = 0.8. |b) u 2 = v 2 -2as | |

| |Give your answer to one significant figure. |u = ((v 2 - 2as) | |

| |b Make u the subject of the formula v 2 = u 2 + | | |

| |2as. | | |

| | | | |

| |A3c/B | | |

|47 |a Factorise completely |a) i) 2t(2s + 4u - v) |7 |

| |I 4st + 8tu - 2tv |a) ii) 3x(5 + x) | |

| |ii 15x + 3x2 |b) 8a - 2a2 | |

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

| |b Expand 2a(4 - a) | | |

| | | | |

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

| | | | |

| |A3c/B | | |

|48 |a Simplify |a) i) 4x3y2 |11 |

| |i 12x5 ( 3y3 |ii) 16p4q6 | |

| |9x2y |b) i) 3a2b2(3b + 5a) | |

| | |ii) (x + 12)(x - 5) | |

| |ii (4p2q3)2 |c) | |

| | |[pic] | |

| |b Factorise completely | | |

| |i 9a2b3 + 15a3b2 | | |

| |ii x2 + 7x - 60 | | |

| | | | |

| |c On the number line below show the solution to | | |

| |these inequalities. | | |

| |-7 ( 2x - 3 < 3 | | |

| | | | |

| |[pic] | | |

| | | | |

| |A3c/B | | |

|49 |Solve the inequality |7y - 2y > - 3 |2 |

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

| | | | |

| |A3d/B | | |

|50 |Solve the simultaneous equations |3x + 2y = 11 |3 |

| |3x + 2y = 11 |2x - 2y = 14 | |

| |x - y = 7 |Adding gives | |

| | |5x = 25 so x = 5 | |

| | |Substituting gives | |

| | |15 + 2y = 11 | |

| | |so 2y = -4 | |

| | |therefore y = -2 | |

| |A3d/B | | |

|51 |Solve the simultaneous equations |6p + 4q = 12 |4 |

| |3p + 2q = 6 |6p + 15q = -21 | |

| |2p + 5q = -7 |Subtracting gives | |

| | |11q = -33 ( q = -3; p = 4 | |

| |A3c/A | | |

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

| | | | |

| |A3d/A | | |

|53 |Solve the simultaneous equations |Eqn (1) × 3 ( 9x + 3y = 39 |4 |

| |3x + y = 13 |Adding ( 11x = 55 | |

| |2x - 3y = 16 |÷ by 11 ( x = 5 | |

| | |Substitute ( 15 + y = 13 | |

| | |y =-2 | |

| | |OR | |

| | |y = 3 - 3x | |

| | |2x - 3(13 - 3x) = 16 | |

| | |11x = 55, x = 5 | |

| | |Substitute ( 15 + y = 13 | |

| |AFMa/A | | |

|54 |[pic] |p1.5 ( p0.5 ( p-5 = p1.5+0.5-5 = p-3 |3 |

| | |so x = -3 | |

| |Work out the value of x. | | |

| | | | |

| |AFMa/A | | |

|55 |a On the grid below, draw the graph of y = 5 + 2x |a) graph of y = 5 + 2x - x2 drawn |6 |

| |- x2 for -2 < x < 4. |b) line y = x + 4 drawn | |

| |[pic] |x values at intersection = -0.6 and 1.6. | |

| | | | |

| |b By drawing a suitable straight line on your | | |

| |graph, find the approximate solutions to | | |

| |x + 4 = 5 + 2x - x2 | | |

| | | | |

| |AFMa/A | | |

|56 |a Complete the table of values for the graphs of |a) -10 6 |3 |

| |y = x3 - 2 and |-6 54 | |

| |y = 3x2 + 3x -6. |b) i) Plot point and draw graphs | |

| | |ii) x3 - 2 = 3x2 + 3x -6 ( x3 - 3x2 - 3x + 4 | |

| |[pic] |= 0 Read off x at intersections | |

| |b i On the graph paper below draw the graphs of |x = -1.35, 0.85 and 3.5 (all ( 0.1) | |

| |y = x3 - 2 and y = 3x2 + 3x - 6. | | |

| | | | |

| |ii Use your graph to solve the equation x3 - 3x2| | |

| |- 3x + 4 = 0. | | |

| |[pic] | | |

| | | | |

| |AMFa/A | | |

|57 |The graph of the equation y = ax + b |ax + b = cx + d |3 |

| |intersects the graph of the equation y = cx + d |(a - c)x = d - b | |

| |at point P. |so x = d - b | |

| |Show that the x-coordinate of P is [pic] |a - c | |

| | | | |

| |AFMa/A | | |

|58 |Solve the equations |i) (2y + 9)(2y - 9) = 0 so y = 4.5 or y = -4.5 |6 |

| |i 4y2 - 81 = 0 |ii) (3 + (x +2))/(3(x + 2)) = -1 | |

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

| |ii [pic] |x + 5 = -3x - 6 | |

| | |4x = -11 so x = -2.75 | |

| |AFMa/A | | |

|59 |Solve the equation |x = 5 (((25 + 32) ; x = 3.14 and x = -0.637 |4 |

| |2x2 - 5x - 4 = 0 |4 | |

| | | | |

| |AFMd/A | | |

|60 |The graph shows how Narinder's height increased in|a) Tangent drawn at age 14; gradient = 13. |5 |

| |the first 16 years of his life. |b) rate of growth | |

| | | | |

| |[pic] | | |

| | | | |

| |a Calculate an estimate for the gradient of the | | |

| |graph when Narinder was 14 years old. | | |

| | | | |

| |b What does the gradient represent? | | |

| | | | |

| |AFMe/A | | |

|61 |The table shows the distance, s metres, travelled |a) Use of t2 values: 0.25, 1, 2.25, 4, 6.25 |5 |

| |by an object from the point P in t seconds. |b) a = 0.1, b = 1.1 | |

| | | | |

| |[pic] | | |

| | | | |

| |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. | | |

| |[pic] | | |

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

| | | | |

| |AFMa/A* | | |

|62 |a Factorise |a) (x - 7)(x + 3) |7 |

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

| | |(x -1)(x + 2) | |

| |b Simplify |= _-2x - 7_ | |

| |[pic] |x2 + x -2 | |

| | |c) (u + v)/uv = 1/f | |

| |c Make f the subject of the formula |so f = __uv__ | |

| |[pic] |u + v | |

| | | | |

| |AFMa/A* | | |

|63 |Factorise completely |a(x + y) - b(x + y) = (a - b)(x + y) |2 |

| |ax - by - bx - ay | | |

| | | | |

| |AFMa/A* | | |

|64 |The diagram represents the graph of a function of | |4 |

| |x |[pic] | |

| |[pic] | | |

| |Draw and label on the same axes the graphs of | | |

| |i y = f(-x) | | |

| |ii y = f(x + 2) | | |

| | | | |

| |AFMa/A* | | |

|65 |a i Factorise x2 - 4x - 12. |a) i) (x + 2)(x - 6) |6 |

| |ii Solve x2 - 4x - 12 = 0. |ii) (a)i = 0, (x + 2)(x - 6) = 0; x = -2 and | |

| | |x = 6 | |

| |[pic] |b) A = (6,0) B = (-2,0) | |

| |The diagram shows a sketch of the graph of y = x2 |c) Sketch graph, the given one moved +2 parallel to | |

| |- 4x - 12. |x-axis | |

| |The curve cuts the x-axis at the points A and B. | | |

| |b Write down the coordinates of A and B. | | |

| |[pic] | | |

| |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|a) tangent drawn at t = 10, or chord drawn about t = 10 |8 |

| |particle. |vert/horiz; 2.3 - 2.8 ms-2 | |

| |[pic] |b) attempts to find area, splits area or count squares; 3 | |

| |In part a you must write down the units with your |marks for answers in the range: 30 - 35 or 41 - 46; 4 | |

| |answer. |marks for answers in the range 35 - 41. | |

| |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. | | |

| | | | |

| |AFMe/A* | | |

|67 |Tarquin measures the length of a pendulum and the |a = 5.4 b = 90 |5 |

| |time it takes for the pendulum to swing forwards | | |

| |and backwards. | | |

| |[pic] | | |

| | | | |

| |He records the results in a table. | | |

| | | | |

| |[pic] | | |

| |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. | | |

| |[pic] | | |

| | | | |

| |A2d/A3c/C/B | | |

|68 |a Make y the subject of the equation x + 2y = 6 |a) 2y = 6 - x or x/2 + y = 3; [pic][pic] or [pic] |6 |

| |b On the grid, draw the line with equation x + 2y |(B) | |

| |= 6 | | |

| |[pic] |b) e.g.(0,3 ), (6, 0), (2, 2), (4, 1) | |

| |c On the grid, shade the region for which |(C) | |

| |x + 2y ( 6, 0 ( x ( 4 and y ( 0. |c) | |

| | |[pic] | |

| | |(B) | |

| |A2d/A3d/C | | |

|69 |a On the grid below, draw the graphs of |a) i) graph of x + y = 4 or y = -x + 4 |4 |

| |i x + y = 4 |a) ii) graph of y = x + 2 | |

| |ii y = x + 2 |b) x = 1, y = 3 | |

| |[pic] | | |

| |b Use the graphs to solve the simultaneous | | |

| |equations | | |

| |x + y = 4 | | |

| |y = x + 2 | | |

| | | | |

| |A2c/B/A | | |

|70 |A car travels between two sets of traffic lights. |a) Velocity = 11.4 m/s. Do not accept responses which |11 |

| | |merely read off values without adding interpretation (B) | |

| |[pic] |b) The car accelerates (the speed increases). Then it | |

| | |travels at a constant speed (or velocity), then there is | |

| |The diagram represents the velocity/time graph of |a constant deceleration (or the car brake steadily). (B) | |

| |the car. |c) Draw tangent at t = 10. Calculate the gradient as 0.57 | |

| |The car leaves the first set of traffic lights. |- 0.73. | |

| | |Alternative: allow a chord about the required point e.g | |

| |a Use the graph to find the velocity of the car |(11.4 - 5)/10 = 0.64 (B) | |

| |after 15 seconds. |d) Realises area needed. Splits area up e.g. use of | |

| | |trapezium rule, or dividing area up into smaller parts to | |

| |b Describe fully the journey of the car between |count squares. (A) | |

| |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. | | |

| | | | |

| |A3b/AFMa/B/A* | | |

|71 |A cylinder of radius R cm and height h cm has a |a) V = 22/7 ( (42 - 0.52) ( 2¾ = 136 1/8 or 136.125 |8 |

| |cylindrical hole of radius r cm drilled through |b) 2((R + r)(R - r + h) = 2((R2 - r2)h | |

| |it. Its volume V cm3 is given by the formula |(R + r)(R - r + h) = (R2 - r2)h | |

| |V = [pic](R2 - r2)h |(R + r)(R - r + h) = (R + r)(R - r)h | |

| | |R - r + h = (R - r)h | |

| |[pic] |R - r = (R - r -1)h | |

| | |[pic] | |

| |a Find the value of V when[pic], R = 4, r = 0.5, h| | |

| |=[pic] | | |

| | | | |

| |The surface area, A cm2, of the shape can be | | |

| |written as | | |

| |A = 2[pic](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 volume. | | |

| |b For this particular shape express h in terms | | |

| |of R and r. | | |

| | | | |

| |S2e/C |Shape | |

|72 |Calculate the length of AB. |((182 - 122) = 13.4 |3 |

| | | | |

| |[pic] | | |

| | | | |

| |Give your answer correct to 1 decimal place. | | |

| | | | |

| |S2e/f/C | | |

|73 |The diagram is part of a map showing the positions|a) [pic] = 622 km |5 |

| |of three Nigerian towns. |b) Tan 20º = x/440 | |

| |Kaduna is due North of Aba. |x = 440 Tan 20º = 160 | |

| | |Distance = 440 + 160 = 600 km | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |S2e/f/C | | |

|74 |Sidney places the foot of his ladder on horizontal|a) [pic]= 15.5 |6 |

| |ground and the top against a vertical wall. |b) cos angle = 4/16; angle = 75.5º | |

| |The ladder is 16 feet long. | | |

| |The foot of the ladder is 4 feet from the base of | | |

| |the wall. | | |

| | | | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |S2e/f/C/B | | |

|75 |Here is a side view of a swimming pool. |a) 5.32 + "0.9"2 = 5.38 (C) |6 |

| |ABCD is a horizontal straight line. AH, BG, CF and|b) tan angle = "0.9"/5.3; angle = 9.6º (B) | |

| |DE are vertical lines. | | |

| | | | |

| |[pic] | | |

| | | | |

| |a Calculate the length of the line FG. | | |

| |Give your answer correct to 3 significant figures.| | |

| | | | |

| |b Calculate the angle that the line GF makes with | | |

| |the horizontal. Give your answer correct to 1 | | |

| |decimal place. | | |

| | | | |

| |S2f/C | | |

|76 |In the diagram |a) DC/8.2 = tan 37º |6 |

| |AB = 17.9 m, BD = 8.2 m, angle CBD = 37º and angle|DC = 8.2 × tan 37º = 6.179... | |

| |BDC = 90º. |b) sin DAB = 8.2/17.9 = 0.4581... | |

| | |so DAB = 27.26... º = 27.3º | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |S2f/C | | |

|77 |The diagram shows a house and a garage on level |a) x/1.4 = tan 62º |6 |

| |ground. |x = 1.4 × tan 62º (= 1.4 × 1.8807...) = 2.63(30..) | |

| | |cos x = 1.4 / 3.5 | |

| |[pic] |x = 66.4218… | |

| | |x = 66.4º | |

| |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 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. | | |

| | | | |

| |[pic] | | |

| | | | |

| |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/B | | |

|78 |Calculate the length of a diagonal of this |((152 + 122) = 19.2 cm |3 |

| |rectangle. | | |

| |Give your answer in centimetres correct to one | | |

| |decimal place. | | |

| | | | |

| |[pic] | | |

| | | | |

| |S2f/A | | |

|79 |Triangle ABC is isosceles. |½ × 12sin55 × 12cos55 × 2 = 67.7 cm2 |6 |

| | | | |

| |[pic] | | |

| | | | |

| |AB = AC = 12 cm | | |

| |Angle ABC is 55º | | |

| |Calculate the area of the triangle. | | |

| |Give your answer to 3 significant figures. | | |

| | | | |

| |S3d/C | | |

|80 |The Fast Foto company uses two letter F's as part |a) 0.9 ÷ 3 × 2 = 0.6 m |4 |

| |of their logo. |b) 28 ÷ 2 × 3 = 42 cm | |

| | | | |

| |[pic] | | |

| |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. | | |

| | | | |

| |S3d/C | | |

|81 |Seamus and Mick set off on a journey from a point |a) accurate journey drawn |9 |

| |A |b) i) 6.7 km | |

| |They travelled 30 km on a bearing of 060º to a |ii) 356º | |

| |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. | | |

| | | | |

| |[pic] | | |

| | | | |

| |b i Work out how far C is from A. | | |

| |ii Write down the bearing of A from C. | | |

| | | | |

| |S3d/C | | |

|82 |Shape A is shown in the diagram. |a) 1/3 |3 |

| |Shape A is enlarged to obtain the shape B. |b) | |

| | |[pic] | |

| |[pic] | | |

| | | | |

| |a Write down the scale factor of the enlargement. | | |

| |b Complete the drawing of shape B on the diagram. | | |

| | | | |

| |S3e/C | | |

|83 |P and Q are two points marked on the grid. |Constructs the perpendicular bisector of P and Q. |2 |

| | | | |

| |[pic] | | |

| | | | |

| |Construct accurately the locus of all points which| | |

| |are equidistant from P and Q. | | |

| | | | |

| |S3b/C/B | | |

|84 |a Reflect shape S in the line x = 0. Label the |a) and b) |6 |

| |new shape T. |[pic] (C) | |

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

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |S3d/B | | |

|85 |Triangle ABC is similar to triangle PQR. |i) 3 × 5/4 = 3.75 cm |5 |

| | |ii) 6.5 ÷ (5/4) or 6.5 × (4/5) = 5.2 cm | |

| |[pic] | | |

| | | | |

| |Angle ABC = angle PQR. | | |

| |Angle ACB = angle PRQ. | | |

| |Calculate the length of | | |

| |i PQ | | |

| |ii AC | | |

| | | | |

| |S3d/B | | |

|86 |Here is a diagram of a company logo. |a) 9 × 1.5 = 13.5 cm |5 |

| | |b) 6 ÷ 1.5 = 4 cm | |

| |[pic] |c) 60º | |

| | | | |

| |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? | | |

| | | | |

| |S4a/C | | |

|87 |There are 12 inches in 1 foot. |100 cm = 100 ÷ 2.54 |3 |

| |There are 3 feet in 1 yard. |= 39.3700 ÷ 36 | |

| |There are 2.54 centimetres in 1 inch. |= 1.0936 | |

| |Express 1 metre in yards. Give your answer |= 1.094 | |

| |correct to 3 decimal places. | | |

| | | | |

| |S4a/C | | |

|88 |There are 2.54 centimetres in 1 inch. |i) 1000/(2.54 ( 3 ( 12) = 10.9(36..) yds |4 |

| |There are 12 inches in 1 foot. |ii) 100/"10.9" = 9.1743... = 9.1 - 9.2 m | |

| |There are 3 feet in 1 yard. | | |

| |i Calculate the number of yards in 10 metres. | | |

| |ii Calculate the number of metres in 10 yards. | | |

| | | | |

| |S4a/C | | |

|89 |There are 14 pounds in a stone. |(13 × 14) + 6 = 188 |3 |

| |There are 2.2 pounds in a kilogram. |188/2.2 = 85 kg | |

| |A man weighs 13 stone 6 pounds. | | |

| |Work out his weight in kilograms. | | |

| |Give your answer to the nearest kilogram. | | |

| | | | |

| |S4d/C | | |

|90 |The diagram shows a triangular prism. |Area of base × 12 = 84 |3 |

| | |½ × 4 × h = 7 | |

| |[pic] |h = 3.5 | |

| | | | |

| |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. | | |

| | | | |

| |S4d/C | | |

|91 |The shape below is the cross section of a prism 10|(2 + 4)/2 × 2 × 10 = 60 cm2 |3 |

| |cm long. | | |

| | | | |

| |[pic] | | |

| | | | |

| |Calculate the volume of the prism. | | |

| | | | |

| |S4d/C | | |

|92 |The diagram shows a cylinder. |3.14 × 4.3 × 4.3 × 26.3 = 1526.9 (( ( 1527.7) 3.142 ( |3 |

| | |1527.9 = 1530 | |

| |[pic] | | |

| | | | |

| |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.| | |

| | | | |

| |S4a/C/B | | |

|93 |There are 6 groats in a florin and 5 florins in 10|i) 5 × 6 = 30 groats (C) |6 |

| |shillings. |ii) 10 shillings = 5 florins | |

| |Work out how many |1 shilling = 0.5 florins | |

| |i groats there are in 5 florins, |24 shillings = 24 × 0.5 = 12 florins (C) | |

| |ii florins there are in 24 shillings, |iii) 1 shilling = 0.5 florins | |

| |iii groats there are in 8 shillings. |3 groats = 1 shilling | |

| | |8 shillings = 24 groats (B) | |

| |S4d/B | | |

|94 |The expressions shown below can be used to |circle around: r(r + 4h), [pic], [pic] |3 |

| |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. | | |

| |[pic] [pic] r(r + 4h) [pic] [pic] | | |

| | | | |

| |[pic] [pic] [pic] [pic] | | |

| | | | |

| |Draw a circle around each of the expressions which| | |

| |can be used to calculate an area. | | |

| | | | |

| |S4d/B | | |

|95 |The diagram represents a solid shape. |[pic](h(b2 + ab + a2) |1 |

| | | | |

| |[pic] | | |

| | | | |

| |From the expressions below, choose the one that | | |

| |represents the volume of the solid shape. | | |

| |( and [pic]are numbers which have no dimensions. | | |

| |a, b and h are lengths. | | |

| | | | |

| |[pic]((b2 - ab + a2), [pic](h(b2 + ab + a2), | | |

| |[pic](h2(b2 - a2), | | |

| | | | |

| |[pic]((a2 + b2), [pic](h2(b2 - ab + a2). | | |

| | | | |

| |Write down the correct expression. | | |

| | | | |

| |SFMa/A | | |

|96 |Two similar boxes have volumes of 2000 cm3 and 16 |scale factor (vol) = 16 000/2000 = 8 |4 |

| |000 cm3. |scale factor (length) = 81/3 = 2 | |

| |The area of the base of the larger box is 60 cm3. |req'd area = 60 ÷ 4 = 15 cm2 | |

| |Calculate the area of the base, in cm2, of the | | |

| |smaller box. | | |

| | | | |

| |SFMa/A | | |

|97 |A child's toy is made out of plastic. |1/3( × 42 × 10 |6 |

| |The toy is solid. |hemi: | |

| |The top of the toy is a cone of height 10 cm and |= 134.04 | |

| |base radius 4 cm. |167.55 + 134.04 = 301.59 | |

| |The bottom of the toy is a hemisphere of radius 4 | | |

| |cm. | | |

| | | | |

| |[pic] | | |

| | | | |

| |Calculate the volume of plastic needed to make the| | |

| |toy. | | |

| | | | |

| |SFMc/A | | |

|98 |The depth of water in harbour varies according to |a) |6 |

| |the formula |[pic] | |

| |y = 10 + 5 sin(30t)º |b) 13 = 10 + 5sin (30t) | |

| |(y is the depth of the water in feet; t is the |sin (30t) = 0.6 | |

| |time in hours) |t = 1.23 | |

| |Here is a sketch of the graph of this formula. |1.23 [pic]60 = 73.8 = 74 mins | |

| | |so ship leaves at 13:14 | |

| |[pic] | | |

| | | | |

| |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 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. |a) i) ½ q |6 |

| | |ii) 2/3 q | |

| |[pic] |iii) ½ q - 1/3 q (A) | |

| | |b) AB // LM | |

| |N is a point on AB so that AN = 2NB. |AB = 3LM (A*) | |

| |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. | | |

| | | | |

| |SFMc/A/A* | | |

|100 |On the scales in Ali's book shop the weight of a |a) i) 0.615 |8 |

| |book correct to 2 decimal places is 0.62 kg. |ii) 0.625 (A) | |

| |a Write down |b) i) 0.615 × 50 = 30.75 | |

| |i the lower bound of the weight of the book, |ii) 0.625 × 50 = 31.25 (A) | |

| |ii the upper bound of the weight of the book. |c) (b)ii - 31 or 31 - (b)i or ½((b)ii - (b)i) | |

| | |= 0.25 (A*) | |

| |Ali needs to work out the weight of 50 copies of |d) (c) × 10 = 2.5 (A*) | |

| |the book. He uses his value for the weight of one| | |

| |book. | | |

| |b Calculate | | |

| |i the lower bound of the weight of 50 books, | | |

| |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 |i) ( APB = 64( (Angle subtended at the circumference is |5 |

| |centre O. They meet the circle at points A and B. |half that subtended at the centre) (A) | |

| |Angle AOB is equal to 128º. |ii) ( BAO = 26( Triangle OAB is Isosceles so | |

| | |( BAO = (180 - 128) ( 2 (A*) | |

| |[pic] |iii) ( ABT = 64( (( OBT is 90( because BT is a tangent and| |

| | |OB is a radius; | |

| |In this question you MUST give reasons for your |( ABT = 90 - (OBA = 90 -26 (A*) | |

| |answers. | | |

| |Work out the size of the angles | | |

| |i APB, | | |

| |ii BAO, | | |

| |iii ABT. | | |

| | | | |

| |SFMa/A* | | |

|102 |Q is the midpoint of the side PR and T is the |a) i) |4 |

| |midpoint of the side PS of triangle PRS. |[pic] | |

| | | | |

| |[pic] |ii) | |

| | |[pic] | |

| |[pic][pic] |iii) | |

| |a Write down, in terms of a and b, the vectors |[pic] | |

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

| |i [pic] | | |

| | | | |

| |ii [pic] | | |

| | | | |

| |iii [pic] | | |

| | | | |

| |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. |a) i) 2a |6 |

| | |ii) a - b | |

| |[pic] |iii) -a - b | |

| | |i) a - b | |

| |The diagonals of the parallelogram intersect at O.|ii) | |

| | |[pic] | |

| |[pic], [pic] | | |

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

| | | | |

| |i [pic] | | |

| | | | |

| |ii [pic] | | |

| | | | |

| |iii [pic] | | |

| | | | |

| |X is the point such that[pic]=2a - b | | |

| | | | |

| |b i Write down an expression, in terms of a and | | |

| |b, for [pic] | | |

| |ii Explain why B, A and X lie on the same | | |

| |straight line. | | |

| | | | |

| |SFMc/A* | | |

|104 |In the diagram, XY represents a vertical tower on |XB/sin30º = 30/sinAXB |6 |

| |level ground. |XB = 30 × 0.5/sin 20º = 43.86 | |

| | |XY = XB sin50º = 33.596 m = 33.60 m | |

| |[pic] | | |

| |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. | | |

| | | | |

| |SFMc/A* | | |

|105 |Find two different values of x between 0 and 180 |2x = 30, 150; x = 15º, 75º |2 |

| |for which | | |

| |sin (2x)º = sin 30º | | |

| | | | |

| |SFMc/A* | | |

|106 |A straight road UW has been constructed to by-pass|( UVW = 102 + 52 = 154º |8 |

| |a village V. |UW2 = 42 + 52 - 2×4×5×cos154 = 76.95 | |

| | |UW = 8.7722 | |

| |[pic] |Time for UVW = 9/30 × 60 = 18 | |

| | |Time for UW = 8.7722/65 ×60 = 8.097 | |

| |The original straight roads UV and VW are 4 km and|Time saved = 10 mins | |

| |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, 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. |a) i) 40º |7 |

| |PQ and QR are tangents to the circle at P and Q |a) ii) OPQ = ORQ = 90º | |

| |respectively. |( between tan and rad | |

| |S is a point on the circle. |POR = 140º ( at circum | |

| |Angle PSR = 70º. |PQR = 360 - (90 + 90 + 140) | |

| |PS = SR. |b) i) 35º | |

| | |b) ii) opp angles of cyclic quad add to 180º | |

| |[pic] |PSR + PQR = 110 | |

| | |SPQ + SRQ = 250 | |

| |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. | | |

| | | | |

| |SFMd/A* | | |

|108 |A, B, C and T are points on the circumference of a|a) ACT = 25º alt angle |6 |

| |circle. |ATP = 25º alt seg | |

| | |APT = 25º isos triangle | |

| |[pic] |b) TAP = 130º ('s in triangle | |

| | |CAT = 50º ('s in straight line | |

| |Angle BAC = 25º. |BTS = 75º alt seg | |

| |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/C | | |

|109 |The scale diagram shows the position of a radio |a) circle drawn at 5 cm radius circle shaded |4 |

| |mast, M. |b) i) 70.5 | |

| |1 cm on the diagram represents 20 km. |ii) 69.5 | |

| | | | |

| | | | |

| | | | |

| | | | |

| | | | |

| |( | | |

| |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 mast. | | |

| | | | |

| |D2d/C |Handling Data | |

|110 |Andrew did a survey at the seaside for his science|a) 2 × 10 + 22 × 30 + 13 × 50 + 10 × 70 + 5 × 90 + 2 × |6 |

| |coursework. |110 + 1 × 130 = 20 + 660 + 650 + 700 + 450 + 220 + 130 = | |

| |He measured the lengths of 55 pieces of seaweed. |2830 (= 51.4545...) = 51.5 | |

| |The results of the survey are shown in the table. |b) 40 < L ( 60 | |

| | | | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |D2d/C | | |

|111 |A survey was carried out to find how much time was|(3 × 0) + (14 × 0.5) + (17 × 1.5) + (5 × 2.5) + (1 × 3.5)=|4 |

| |needed by a group of pupils to complete homework |48.5 | |

| |set on a particular Monday evening. |48.5 ÷ 40 = 1.21 | |

| |The results are shown in the table below. | | |

| | | | |

| |[pic] | | |

| |Calculate an estimate for the mean time spent on | | |

| |homework by the pupils in the group. | | |

| | | | |

| |D2d/C | | |

|112 |Bronwen owns a pet shop. |29 × 9 = 261 |4 |

| |The table gives information about the weights of |31 × 5 = 155 | |

| |hamsters in Bronwen's shop. |33 × 4 = 132 | |

| | |35 × 2 = 70 | |

| |[pic] |618 | |

| |Calculate an estimate for the mean weight of the |618 ÷ 20 = 30.9 | |

| |hamsters in Bronwen's shop. | | |

| | | | |

| |D2c/f/C | | |

|113 |Information about oil was recorded each year for |a) Line of best fit |4 |

| |12 years. |b) Draw line "amount = 10.4" on graph or state use of | |

| |The scatter graph shows the amount of oil produced|amount = 10.4; price about £16.50 | |

| |(in billions of barrels) and the average price of | | |

| |oil (in £ per barrel). | | |

| | | | |

| |[pic] | | |

| |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/C | | |

|114 |The table list the weights of twelve books and the|a) Line of best fit (only str. line) |4 |

| |number of pages in each one. |b) i) accept 136 - 140 pages | |

| | |ii) accept 216 - 220 g | |

| |[pic] | | |

| |This information is presented below as a scatter | | |

| |graph. | | |

| | | | |

| |[pic] | | |

| | | | |

| |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, | | |

| |ii the weight, in grams, of a book with 110 | | |

| |pages. | | |

| | | | |

| |D2b/B | | |

|115 |Martin, the local Youth Centre leader, wishes to |Look for a question that links to the original problem via|2 |

| |know why attendance at the Youth Centre is less |the perceived causes and a way of completing the | |

| |than at the same time last year. |questionnaire easily e.g. boxes | |

| |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 changed | | |

| |a new sports centre opened nearby | | |

| |some 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. | | |

| | | | |

| |D2c/B | | |

|116 |Pippa collected data for the heights (h) of the |a) Mid point Total |12 |

| |students in her class. |162.5 1137.5 | |

| |Here is the grouped frequency table of her |167.5 1005 | |

| |results. |172.5 345 | |

| | |177.5 1775 | |

| |[pic] |182.5 912.5 | |

| |a Use the table to calculate an estimate of the |Total 5175 | |

| |mean for her results. |Mean = 5175 ÷ 30 = 172.5 cm | |

| | |b) 7, 13, 15, 25, 30 | |

| |b Complete the cumulative frequency table for |c) Median = 175 | |

| |Pippa's data and hence draw a cumulative frequency|Interquartile range = 17 | |

| |graph for the data. | | |

| | | | |

| |[pic] | | |

| |[pic] | | |

| |c Use the cumulative frequency graph to calculate | | |

| |an estimate of | | |

| |I median of the data | | |

| |ii interquartile range of the data. | | |

| | | | |

| |D2a/B | | |

|117 |The table gives information about the weights of |a) 4, 13, 24, 45, 71, 89, 98, 100 |6 |

| |100 new born babies. |b) Points joined on a cumulative frequency graph; ignore | |

| | |graph below x = 1.5. | |

| |[pic] |c) 3.15 kg | |

| |a Complete the cumulative frequency table below. | | |

| | | | |

| |[pic] | | |

| |b Draw a cumulative frequency graph for your | | |

| |table. | | |

| | | | |

| |[pic] | | |

| | | | |

| |c Use your cumulative frequency diagram to | | |

| |estimate the median weight, in kilograms, of the | | |

| |new born babies. | | |

| |Show your method clearly. | | |

| | | | |

| |D2c/B | | |

|118 |A survey is made of all 120 houses on an estate. |a) plot points, joined with smooth curve/line |6 |

| |The floor are, in m2, of each house is recorded. |b) segments 240 - 170 = 70 | |

| |The results are shown in the cumulative frequency |c) 10% is 12 houses. Read off where cf = 120 - 12 = 108; | |

| |table. |floor area = 275 m2 | |

| | | | |

| |[pic] | | |

| |a On the grid draw a cumulative frequency graph | | |

| |for the table. | | |

| | | | |

| |[pic] | | |

| |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/B | | |

|119 |150 year 11 pupils took a mathematics examination.|a) 6, 23, 45, 90, 116, 135, 144, 150 |7 |

| |The table shows information about their marks. |b) cum freq diag drawn: ignore line between first point | |

| | |and the origin | |

| |a Complete the cumulative frequency table below. |c) 60% of 150 = 90 | |

| | |150 - 90 = 60 | |

| |[pic] |pass mark about 42 - 45 marks | |

| |b On the grid below, draw a cumulative frequency | | |

| |diagram to show these marks. | | |

| | | | |

| |[pic] | | |

| | | | |

| |60% of the pupils passed the examination. | | |

| |c Use your diagram to find an estimate for the | | |

| |pass mark for the examination. | | |

| | | | |

| |D3b/C | | |

|120 |A dice has six faces numbered 1, 2, 3, 4, 5 and 6.|46/200 or 23/100 or 0.23 or 23% |2 |

| |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. | | |

| | | | |

| |[pic] | | |

| |Estimate the probability that the next time the | | |

| |dice is thrown it will show the number 3. | | |

| | | | |

| |D3a/B/C | | |

|121 |The probability of a machine being able to |a) 1 - 0.995 = 0.005 (C) |4 |

| |manufacture a component within a tolerance of one |b) 10 000 × (a) - 50 (B) | |

| |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. | | |

| | | | |

| |D3e/B | | |

|122 |Two dice with faces numbered 1 to 6 are rolled and|i) 0.0625 × 0.032 × 0.044 = 0.0000915 |7 |

| |the sum of the scores on upward facing faces |ii) 0.065 × 0.968 × 0.954 = 0.06 | |

| |noted. |iii) 0.935 × 0.968 × 0.954 = 0.863 | |

| | | | |

| |[pic] | | |

| | | | |

| |The score on these dice is 1 + 4 = 5 | | |

| |a Complete the table of probabilities for the | | |

| |chance of scoring all the totals available. | | |

| | | | |

| |[pic] | | |

| |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/B | | |

|123 |The probability of a car chosen at random having |i) 0.0625 × 0.032 × 0.044 = 0.0000915 |7 |

| |defective |ii) 0.065 × 0.968 × 0.954 = 0.06 | |

| |tyres is 0.065 |iii) 0.935 × 0.968 × 0.954 = 0.863 | |

| |brakes is 0.032 | | |

| |steering is 0.044 | | |

| |Work out the probability that a vehicle chosen at | | |

| |random will have | | |

| |i defective tyres, brakes and steering, | | |

| |ii has defective tyres but no other defects, | | |

| |iii has no defects. | | |

| | | | |

| |D3f/B | | |

|124 |Nikki and Ramana both try to score a goal in |i) 0.65 × 0.8 = 0.52 |5 |

| |netball. |ii) 1 - 0.65 (= 0.35) | |

| |The probability that Nikki will score a goal on |1 - 0.8 (= 0.2) | |

| |her first try is 0.65. |0.35 × 0.2 = 0.07 | |

| |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. | | |

| | | | |

| |D3e/A/A* | | |

|125 |Susan has 8 eggs in her fridge. |a) |9 |

| |Two of the eggs have passed their sell by date and|[pic] | |

| |are "bad". |(A) | |

| |She selects 3 eggs at random to bake a cake. | | |

| |a Complete the probability tree diagram. |b) ¾ × 5/7 × 2/3 = 5/14 (A*) | |

| | |c) 1- (b) = 9/14 (A*) | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |DFMb/A | | |

|126 |Kim sowed some seeds in her greenhouse. |a) 3rd interval: 120 plants need 12 cm2 so 10 plants per |6 |

| |10 weeks later she measured the heights of the |cm2. 4th interval: Area 11.5cm2 so 11.5 × 10 = 115 | |

| |plants. |plants. 5th interval: Area 10.5 cm2 so 105 plants. 2nd | |

| |Some of the results are shown in the table and the|interval: 30 plants need 3 cm2 so height = 3 ÷ 3 = 1 cm. | |

| |histogram. |6th interval: 96 plants need 9.6 cm2 so height = 9.6 ÷ 2 =| |

| | |4.8 cm. | |

| |[pic] |OR | |

| |[pic] |f.d. for 3rd interval = 120 ÷ 10 = 12, so scale on f.d. | |

| |a Use the information to complete the table and |axis is 1 cm = 12 ÷ 6 = 2. 4th interval: 2 × 11.5 × 5 = | |

| |the histogram. |115. 5th interval: 2 × 10.5 × 5 = 105. 2nd interval: f.d.| |

| | |= 30 ÷ 15 = 2 (2 ÷ 2 = 1 cm). 6th interval: f.d. = 96 ÷ | |

| |Kim had sown 500 seeds. |10 = 9.6 (9.6 ÷ 2 is 4.8 cm). | |

| |b Calculate the number of seeds that had not | | |

| |produced plants. |b) 500 - (30 + 120 + 115 + 105 + 96) = 500 - 466 = 34 | |

| | | | |

| |DFMb/A | | |

|127 |John measured the time, in seconds, that birds |a) 2nd interval 28 ÷ 14 = 2 birds per cm2 . 10 × 2 = 20, |6 |

| |spent on each individual visit to his bird table. |15.5 × 2 = 31, 113 - (20 + 28 + 31 + 12) = 22 | |

| |The birds made a total of 113 individual visits. |or | |

| |The histogram shows some of the results. |f.d. for 2nd interval = 28 ÷ 10 = 2.8, so scale on f.d. | |

| |a Use the information in the histogram to complete|axis is 1 cm = 2.8 ÷ 7 = 0.4 | |

| |the frequency table below. |0.4 × 5 × 10 = 20, 0.4 × 15.5 × 5 = 31 113 - (20 + 28 + 31| |

| | |+ 12) = 22 | |

| |[pic] | | |

| |b Use the information in the frequency table to |b) 4th interval, 22 ÷ 2 = 11 cm2. 5th interval, 12 ÷ 2 =| |

| |complete the histogram. |6 cm2, 6 ÷ 4 = 1.5 cm | |

| | |or | |

| |[pic] |f.d. = 22 ÷ 5 = 4.4, (4.4 ÷ 0.4 = 11 cm) f.d. = 12 ÷ 20 = | |

| | |0.6, (0.6 ÷ 0.4 = 1.5 cm) | |

| |DFMb/A | | |

|128 |The histogram gives information about the ages of |a) e.g. 16 ÷ 5 = 3.2; 64 ÷ 3.2 = 20 so 76 ÷ 20 × 5 = 19 |6 |

| |the teacher at a school on 1st September last |etc. Frequencies are: 7, 6, 9, 36 | |

| |year. |b) 27 ÷ 15 × 20 = 36 | |

| | | | |

| |a Use the information in the histogram to complete| | |

| |the frequency table below. | | |

| | | | |

| |[pic] | | |

| |b Use the information in the frequency table to| | |

| |complete the histogram | | |

| |[pic] | | |

| | | | |

| |DFMc/A | | |

|129 |Pippa's friend Wayne was also measuring the |a) mean 175.2 |7 |

| |heights of people in his class. |sum of squares 307420 | |

| |Here are his results for the 10 boys in the class.|standard deviation = 6.85 | |

| |162 165 178 182 175 185 172 178 180 175 |b) i) 177.7 | |

| |All the measurements are in centimetres correct to|ii) 6.85 | |

| |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. | | |

| | | | |

| |DFMc/A | | |

|130 |Calculate the standard deviation of the numbers |mean = 6 |2 |

| |3, 3, 4, 5, 7, 9, 11. |(3 - 6)2 + (3 - 6)2 + (4 - 6)2 + (5 - 6)2 + (7 - 6)2 + (9 | |

| | |- 6)2 + (11 - 6)2 (= 58) ((58 ( 7) = 2.88 | |

| | |OR use formula | |

| |DFMd/A | | |

|131 |The diagram shows two boxes A and B. |a) 5/8 × 5/9 = 25/72 |6 |

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

| |[pic] | | |

| | | | |

| |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 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 |a) 2/5; 2/7; 5/7; 4/7; 3/7; |8 |

| |route to school. |b) 2/5 × 4/7 = 8/35 | |

| |The probability that the first set of lights will |c) 3/5 × 2/7 + 2/5 × 4/7 = 14/35 = 2/5 | |

| |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. | | |

| | | | |

| |[pic] | | |

| | | | |

| |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. | | |

| | | | |

| |DFMd/A/A* | | |

|133 |There are 4 red balls, 5 blue balls and 3 green |a) 4/12; 3/11; 5/11; 3/11 |11 |

| |balls in a bag. |5/12; 4/11; 4/11; 3/11 | |

| |A ball is to be taken at random and not replaced. |3/12; 4/11; 5/11; 2/11 (A) | |

| |A second ball is then to be taken at random. |b) i) 4/12 × 3/11 = 1/11 | |

| |a Complete the tree diagram below. |ii) (5/12 × 4/11) + (3/12 × 2/11) + "1/11" = 19/66 (A) | |

| | |c) (4/12 × 5/11) + (4/12 × 3/11) + (5/12 × 4/11) + (3/12 ×| |

| |[pic] |4/11) = 16/33 (A*) | |

| | | | |

| |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. | | |

| | | | |

| |DFMc/A* | | |

|134 |a Show that the mean of five consecutive numbers |a) If median is n the numbers must be n-2, n-1, n+1, n +2 |7 |

| |with median n is also n. |Sum = 5n (mean = 5n ( 5 = n | |

| | |b) (n - 2)2 = n2 - 4n + 4 | |

| |b Show that the mean of the squares of these 5 |(n - 1)2 = n2 - 2n + 1 | |

| |numbers exceeds the median of the squares by the |n2 = n2 | |

| |number 2. |(n + 1)2 = n2 + 2n + 1 | |

| | |(n + 2)2 = n2 + 4n + 4 | |

| | |Sum = 5n2 + 10 | |

| | |Mean = n2 + 2 | |

| | |Median = n2 | |

| |DFMc/A* | | |

|135 |In an experiment, Nazia measured the time it took |a) (x = 12.6 |6 |

| |for a ball bearing to sink down to the bottom of a|(x2 = 31.82 | |

| |tube of oil. She made 5 measurements. Her |standard deviation = 0.01166 = 0.0117 | |

| |results are given below. | | |

| |2.4s, 2.5s, 2.4s, 2.6s, 2.7s |b) i) 2.52 + 0.2 = 2.72 | |

| |a Calculate the standard deviation of these times.|ii) (a) unchanged = “0.0117” | |

| |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 |Using [pic] |4 |

| |of CD players in a shop over one week. |fx2 = 4 ( 752 + 3 ( 1302 + 1 ( 1902 + 2 ( 2202 =2250 + | |

| | |50700 + 36100 + 96800 = 206100 | |

| |[pic] |206100[pic]10 = 20610 | |

| |The mean income per sale is £132. |var = 20610 - 1322 = 3186 | |

| | |s.d. = [pic] = 56.444.. = £56.44 | |

| |Calculate the standard deviation. |or | |

| | |using | |

| | |[pic] | |

| | |31860[pic]10 = 3186 | |

| | |s.d. = [pic]= £56.44 | |

| |D3c/e/DFMd/A*/B/C | | |

|137 |Peter and Asif are both taking their driving test |a) 1000 ( 0.7 = 700 (C) |9 |

| |for a motor cycle for the first time. |b) 0.6 ( 0.7 = 0.42 (B) | |

| |The table below gives the probabilities that they |c) 0.6 (0.3 = 0.18 (B) | |

| |will pass the test at the first attempt or, if |d) P(pass) + P(fail) ( P(pass) | |

| |they fail the first time, the probability that |0.7 + 0.3 ( 0.7 = 0.91 (A*) | |

| |they will pass at the next attempt. | | |

| | | | |

| |[pic] | | |

| |On a particular day 1000 people will take the test| | |

| |for the first time. | | |

| |For each person the probability that 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 questions | |

|138 |Jomo is going to design a circular roundabout. |a) e.g. 2.23 m can be measured: ( normally give to 2 |3 |

| |The roundabout will have a circumference of 7 |d.p. | |

| |metres. |b) 2nd not accurate; not near enough to 7m; 1st too | |

| |Jomo is given three estimates for the length of |accurate: cannot be measured | |

| |the diameter of the roundabout. | | |

| |The estimates are: | | |

| |2.2278803 metres | | |

| |2 metres | | |

| |2.23 metres | | |

| |a 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. | | |

| | | | |

| |S4d/N4d/C | | |

|139 |This container is made from a cylinder and a cube.|a) (r2h = ( ( 82 ( 20 = 4021.. cm3 |7 |

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

| |[pic] |c) i) 7.5 cm | |

| | |ii) 8.5 cm | |

| |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/B | | |

|140 |Matthew uses this formula to calculate the value |a) |6 |

| |of D. |[pic] | |

| |[pic] |b) i) a = 20, c = 4 | |

| | |ii)[pic] = [pic] = 2 | |

| |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 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|a) 1/3 ( 3.14 ( 32 ( 18 ( = 169.56) = 169.6 (A) |9 |

| |base is 3 cm. |b) 2.95 (A) | |

| | |c) Least volume = 1/3 ( 3.14 ( 2.952 ( 17.95 = 163.4997 | |

| |[pic] |(((163.58) | |

| | |Greatest volume = 1/3 ( 3.14 ( 3.052 ( 18.05 = 175.7459 | |

| |a Calculate the volume of the cone. |(((175.84) | |

| | |Difference = 12.246 (12.25) | |

| |The measurements of the cone are correct to the |12.246/169.56 ( 100% = 7.2(2..)% (A*) | |

| |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). | | |

| | | | |

| |SFMc/AFMa/A* | | |

|142 |In triangle ABC, AB = 5 cm, AC = x cm, BC = 2x cm |a) (2x)2 = 52 + x2 - 2(5(x(cos60 |9 |

| |and angle BAC = 60(. |4x2 = 25 + x2 - 10x ( 0.5 | |

| | |b) 3x2 + 5x - 25 = 0 | |

| |[pic] |x = (-5(((52 + 300)) ( 6 | |

| | |= (-5(18.027..) (6 | |

| |a Show that 3x2 + 5x - 25 = 0. |= 2.17 or -3.84 | |

| | |c) [pic] | |

| |b Solve the equation 3x 2 + 5x - 25 = 0. | | |

| |Give your answers correct to 3 significant |[pic] | |

| |figures. | | |

| | | | |

| |D is the point on AC such that angle ADB = 104(. | | |

| |c Calculate the length of BD. | | |

| | | | |

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