By Mr Kiapene 2008. Be Fully Prepared for Your SAT Exam

THIS BOOKLET IS FOR YOU.

I SACRIFICED MY TIME TO COMPILE THIS FOR YOU BECAUSE YOU DESERVE IT.

GOOD LUCK DEAR BY MR KIAPENE (April 2008)

By Mr Kiapene 2008. Be fully prepared for your SAT exam Number Number Adding and subtracting fractions Example To add and subtract fractions, write them with the 6 3 Work out  (common denominator is 35) common denominator 7 5 Example 6 3 30 21 30  21 9 2 4      Work out  (common denominator is 15) 7 5 35 35 35 35 3 5 1 2 2 4 10 12 10 12 22 7 Work out 4 1 (subtract whole number first)       1 6 4 3 5 15 15 15 15 15 1 2 1 2 2 6 2  6 1 2 4 1 =(4 – 1)   3   3 (each one of Work out 3 1 (add the whole numbers first) 6 4 6 4 12 12 12 4 3 the 3 whole number is equal to 12) 3 2 3 2 9 8 17 5 2  6 12  2  6 14  6 8 2 3 1  (3 1)   4   4  5 3  2  2  2  2 4 3 4 3 12 12 12 12 12 12 12 12 3

Multiplying Fractions Dividing fractions In multiplying fractions, we multiply the numerators We change division to multiplication by flipping the and the denominators separately after cancelling the fraction after the division sign. common factors Example 2 3 3 9 Work out  Work out  (change division to multiplication first) 3 8 5 10 2 3 2 3 6 1 3 9 1 3 210 1 2 2          3 8 38 24 4 5 10 1 5 3 9 1 3 3 It is easier to cancel out the common factors first before multiplying 5 7 1 1 2  2 3 2 3 11 1 Work out (change mixed fractions to improper fractions)      8 16 3 8 1 3 4 8 1 4 4 5 7 21 7 3 21 216 2 3 6 2         6 1 3 1 1 Work out 5 1 (change fractions to improper fractions) 8 16 8 16 8 7 11 1 4 7 1 3 21 10 3 21 510 3 5 15 1 5 1        7 4 7 4 7 2 4 1 7 21 2 2 BIDMAS Example B=Bracket Work out 9 ÷ (7-4) - 5 +2 x (12÷3)² Brackets first I = Indices Follow this order in 9 ÷ 3 – 5 + 2 x 42 (indices next) D= Division all calculations 9 ÷ 3 – 5 + 2 x 16 (division next) M = Multiplication involving more than 3 – 5 + 2 x 16 (multiplication next) A = Addition one operation 3 – 5 + 32 (add the positive numbers) 3 +32 – 5 S = Subtraction 35 – 5 30 Percentage increase and decrease Example Remember 100% is divide by 1 The price of a laptop is £320.00 with 18% price reduction. 50% is divide by 2 Find the new price of the laptop. 25% is divide by 4 Solution 10% is divide by 10 10% 0f 320 = 32 1% is divide by 100 5% of 320 = 16 (half of 10%) % increase = Original amount + % 1% 0f 320 = 3.2 % decrease = Original amount - % 1% of 320 = 3.2 1% of 320 = 3.2 18% of 320 =57.6 New price = 320 – 57.6 =£262.40 Reverse of percentage increase and decrease Example This is the process of finding the original amount After a 10% price decrease, a hi-fi system now costs £288. How after a % increase or decrease. much was it before the decrease? Solution Example Original price = 100% After 20% pay rise, John now gets £1500 per month. New price = Original price - % decrease By Mr Kiapene 2008. Be fully prepared for your SAT exam How much per month did he gets before the pay rise = 100% - 10% = 90% Solution Original Salary = 100% 90% = £288 Original salary + increase = 100% + 20% = 120% Divide both sides by 100 90% 288 120% = 1500  Divide both sides by 120 90 90 120% 1500 1% = 3.2  120 120 Multiply both sides by 100 1% = 12.5 100% = 320 Multiply both sides by 100 Original price = £320. 100% = 1250 Original salary = £1 250.00 Compound interest Alternative method (use of multiplier) Compound interest is the concept of adding accumulated Original Amount = 100% interest back to the principal, so that interest is earned on New amount = 100% + 3%= 103% = 1.03 interest as well. (1.03 is called a multiplier) Example After first year: 450 x 1.03 = 463.5 How much would you have in the bank if you invest as £450 at After second year: 463.5 x 1.03 = 477.41 3% interest per annum for 4 years After third year: 477.41 x 1.03 = 491.73 After fourth year: 491.73 x 1.03 = 506.48

Method 4 After 1st yr 3% of 450 = 13.5 at the end of 1st yr amt is The quickest way Amt = 450 x 1.03 = 506.48 450 + 13.5 = 463.5 From the foregoing we can develop a formula for After 2nd yr 3% 0f 463.5 = 13.91 at the end of 2nd yr amt is compound interest as follow. 463.5 + 13.91 = 477.41 rd rd %rate After 3 yr 3% of 477.42 = 14.32 at the end of 3 yr amt is Amount = initial amount x (1  )time 477.42 + 14.32 = 491.74 100 After 4th yr 3% of 491.74 = 14.75 at the end of 4th yr amt is We use + when investment is earning interest 491.74 + 14.75 = 506.49 We use – when investment is making loss Ratio and proportion Example Ratio compares one part to another. We simplify Divide £480 in the ratio 3 : 5 ratio by cancelling out common factors. Solution Example Sum of ratios = 3 + 5 = 8 parts Each part = 480 ÷ 8 = 60 simplify the following ratio 3 : 5 a) 14: 16 b) 0.4:3 c) 2.4 : 3.2 7 : 8 4:30 24:32 3 x 60 : 5 x 60 2:15 3:4 180 : 300 Example Write each of the following in the form (i) 1:m (ii) n:1 Divide £240 in the ratio 1 : 1.5 Solution a) 5:4 b) 6:10 Sum of ratios = 1+1.5 = 2.5 a) i) 5:4 = 1: 0.8 (divide both parts by 5) Each part = 240 ÷ 2.5 = 96 ii) 5:4 = 1.25:1 (divide both parts by 4) 1:1.5 b) i) 8:20 = 1: 2.5 (divide both parts by 8) ii) 8:20 = 0.4:1 (divide both parts by 20) 1 x 96 : 1.5 x 96 96 : 144 Significant figures and estimation Example Significant figure means the most important Round the following numbers to 1 significant figure and estimate Example the answer Round each of the following to 1 significant figures a) (3124 x 476) ÷ 283 a) 582  600 (1 sf) a)582  580 ( 2 sf) solution (3124 x 476) ÷ 283  (3000 x 500) ÷300 b) 0.0893  0.09 (1 sf) b) 0.0893  0.089 ( 2 sf)  1 500 000 ÷ 300 c) 7351  7000 (1 sf) c) 7351  7400 ( 2 sf)  5 000 d) 23.82  20 (1 sf) d) 23.82  24 ( 2 sf) Numbers between 0 and 1 Example 1. When a number is multiplied by a number 2 560 x 0.2 = 112 ( the same as 560 x ) between 0 and 1 it gets smaller. 10 2. When a number is divided by a number 9 between 0 and 1 it gets bigger. 450 x 0.9 = 405 ( the same as 450 x ) 10 By Mr Kiapene 2008. Be fully prepared for your SAT exam example 8 10 320 ÷ 0.8 = 400 ( the same as 320 ÷ = 320 x ) a. 41.6 x 0.1 = 4.16 10 8 12 100 b. 26 ÷ 0.1 = 260 720 ÷ 0.12 = 6000 ( the same as 720 ÷  720 ) 100 12 c. 370 x 0.001 = 370 000 d. 8 ÷ 0.01 = 0.08 Powers of 10 Example Examples 1 34.5 x 10-2 = 34.5 ÷ 100= 0.345 ( i.e. 10-2 = 0.01= ) 23.4 x 102 =23.4 x 100 = 2340 100 0.345 x 103 = 0.345 x 1000 = 345 -2 -2 1 2 132.7 ÷ 10 = 132.7 x 100 = 13270 ( i.e. 10 = 0.01= ) 2.78 ÷ 10 = 2.78 ÷ 100 = 0.0278 100 2456 ÷ 103 = 2456 ÷ 1000 = 2.456 34.78 x 0.001 = 0.03478 5.238 ÷ 0.0001 = 52380 Recurring decimals Example Example Write 0. . . as an exact fraction Write the following as a recurring decimal 27 . . 5 7 4 4 Let F = 0. 27 a) b) c) d) 3 9 11 7 9 F = 0.27272727272727272 ……………….(1) 5 Multiply equation (1) by 100 a) = 0.555555555555 = 0. . 9 5 100F = 27.2727272727272727 ……………...(2) Subtract equation (1) from (2) 7 . . b) = 0.6363636363636363 = 0. 99F = 27 11 63 Divide both side by 99 4 c) = 0.571428571428571428 = 0. . . 27 3 7 571428 F =  99 11 4 . When you have 3 dp places in your recurring decimal d) 3 =3. 4 9 divide it by 999, 1 dp divide by 9, 4 d.p divide by 9999 etc. Efficient use of calculator Example The best use of a calculator is to insert brackets 153.7(16  2.912 ) To work out correct to 2dp separate between two or more operations 36.7 18.32 2 x  squared Key (153.7(16 – 2.91 x 2 )) ÷ (36.7 x 18.32) = 1.72 x y = y to power x Example = square root 13 11 To work out  3 y = cube root of y 15 18 23  y = x root of y Key 13 a b 15 - 11 a b 18 = c c 90 a b  fraction 3 c To work out 81 4 x a b Key 81 y 3 c 4 = 27 Index Notation Examples Rules 1. 3b3 x 2b4 =(3x2)b3+4=6b7 1. an x am = an + m 2. 24c3 ÷ 3c2=(24 ÷3)c3-2 = 8c 2. an ÷ am = an – m 1 1 1 3. 2-3 = = = 1 3 3. i) a-1 = 2 2x2x2 8 a1 1 4. (c3)4 = c3 x 4 = c12 ii) a-2 = a 2 4. (an)m = anm Standard Form Example 1. standard form is always in the form Work out the following and leave your answer in standard A x 10n where 1≤A<10 form 2. (a x 10n) x (b x 10m) = (a x b) x 10n+m a) (3.3 x 103) x (5.0 x 102) 3. (a x 10n) ÷ (b x 10m) = (a ÷ b) x 10n-m b) (3.5 x 105) ÷ (7.0 x 103)

By Mr Kiapene 2008. Be fully prepared for your SAT exam a) (3.3 x 5) x (103 x 102) Example 16.5 x 103 + 2 1. express the following in standard form 16. 5 x 105 a) 234000 b) 0.0000067 1.65 x 106 b) (3.5 ÷ 7.0) x (105 x 102) a) 234000 = 2.34 x 105 0.5 x 105-2 b) 0.0000067 = 6.7 x 10-6 0.5 x 103 5.0 x 102 Squares, square roots, cubes and cube roots of Example number 1 1 Simplify a) 2 b) 4 c) 1 64 81 16000000 1. n 2   n 1 a) 64 2 =  64  8 1 1 b) 4 4 2. n 3  3 n 81   81  3 16000000   16  1000000 c) 3. n3 = n x n x n =  4x1000 =  4000

Upper and lower bound Example Rules A rectangle has a length of 20cm and a width of 12cm, both 1. Lower bound = Value - Half the degree of accuracy measured to the nearest cm. What is the upper and lower 2. Lower bound = Value + Half the degree of accuracy bounds of the area the rectangle. 3. For two numbers a and b with upper and lower bounds Length amin  a  amax and bmin  b  bmax Operation Maximum Minimum Degree of accuracy = 1cm (half = 0.5) Lower bound = 20 – 0.5 = 19.5 a + b a  b a  b max max min min Upper bound = 20 + 0.5 = 20.5 a – b amax  bmax amin  bmin Width a x b amax  bmax amin  bmin Degree of accuracy = 1cm (half = 0.5 a  b a ÷ b max max amin  bmin Lower bound = 12 – 0.5 = 11.5 Example Upper bound = 12 + 0.5 = 12.5 A football crowd which is 3600 to the nearest hundred. Solution Area The degree of accuracy is 100 (half = 50) Lower bound area = 11.5 x 19.5 = 224.25 Lower bound = 3600 – 50 = 3550 Upper bound area = 12.5 x 20.5 = 256.25 Upper bound = 3600 + 50 = 3650 224.25  Area  256.25 3550  C  3650 (C = crowd)

Past Questions on Number (Level 7 and 8)

1. A book gives this information: 5. People who live to be 100 years old are called centenarians. In 1998 there were 135 000 A baby giraffe was born that was 1.58 metres high. centenarians. The ratio of male to female was 1 : 4 How many female centenarians were there in 1998? It grew at a rate of 1.3 centimetres every hour. Show your working.

Suppose the baby giraffe continued to grow at this 8 4 13 rate. 6. (a) Show that (4 x 10 ) x (8 x 10 ) = 3.2 x 10 8 4 About how many days old would it be when it was (b) What is (4 x 10 ) ÷ (8 x 10 )? 6 metres high? Write your answer in standard form. 2. The diagram shows a square and a circle. 7. (a)Find a and b from the equations below. The circle touches the edges of the square. 48 = 3 × 2a 56 = 7 × 2b

(b) 48 × 56 = 3 × 7 × 2c By Mr Kiapene 2008. Be fully prepared for your SAT exam 6 cm What is the value of c?

8. I start with any two consecutive integers.

I square each of them, then

I add the two squares together.

Prove that the total must be an odd number. What percentage of the diagram is shaded? 1 9. Writing Numbers, is equal to 0.0004 3. A groundsman marks out a football pitch. 2500

93 metres (a) Write 0.0004 in standard form.

1 ,(b) Write in standard form. 25000 50 metres 1 1 (c) Work out  2500 25000

Show your working, and write your answer in standard (a) He makes the pitch 93 metres long, to the form. nearest metre. What is the shortest possible length of the pitch? 10. A shop had a sale. All prices were reduced by 15%

(b) He makes the pitch 50 metres wide, to the nearest metre. What is the shortest possible Sale width of the pitch? 15% off (c) Des wants to know how many times he should run around the outside of this pitch to be sure of running at least 3km. A pair of shoes cost £38.25 in the sale.

Use your answer to parts (a) and (b) to find What price were the shoes before the sale? how many times Des should run around the pitch.

4. (a)Write the values of k and m.

64 = 82 = 4k = 2m

(b) if 215 = 32 768 what is 214 ? Shape, Space and Measure

Geometrical Facts 9. The exterior angle of a triangle is equal to the sum of the 1. Supplementary angles add up to 1800 interior opposite angles 2. Complementary angles add up to 900 10. Two pairs of adjacent sides of a kite are equal, and one 3. Angles in a straight line add up to 180 pair of opposite angles are equal. Diagonals intersect at 4. Angles at a point add up to 360 right angles. One diagonal is bisected by the other 5. Vertically opposite angles are equal 11. The diagonals of a square and rhombus bisect one 6. Corresponding angles of parallel lines are equal another at right angle. 7. Alternate angles of parallel lines are equal 12. The opposite angles of a rhombus and parallelograms 8. Allied or co-interior angles of parallel lines are are equal. supplementary 13. Congruent figures are exact duplicates of each other.

By Mr Kiapene 2008. Be fully prepared for your SAT exam Sum of interior angles of polygon Name of Number Number of Sum of interior Example polygon of sides triangles angles Find the missing angle below. Triangle 3 1 1 x 1800 = 1800 Quadrilateral 4 2 2 x 1800 = 3600 130 Pentagon 5 3 3 x 1800 = 5400 67 Hexagon 6 4 4 x 1800 = 7200 83 Heptagon 7 5 5 x 1800 = 9000 Octagon 8 6 6 x 1800 = 10800 250 Nonagon 9 7 7 x 1800 = 12600 Decagon 10 8 8 x 1800 = 14400 110 n sides n n - 2 Sum =(n – 2)1800 n Find the sum of interior angles of 19 sided polygon Sum = (n – 2) 180 Since it is a six sided polygon 0 Sum of 20 sided polygon = (19 – 2) 180 67 + 130 + 83 + 110 + 250 + n = 720 = 17 x 180 640 + n = 720 = 30600 n = 720 – 640 n = 800 Angles of regular polygon (all sides and angles Examples are equal) Find x and y

1. Sum of ext. angle of a polygon = 3600 Since it is an hexagon 2. an ext. angle + int. angle = 1800 x = 360 ÷ 6 = 60 0 360 x  y  180 3. Number of sides = ext.angle 60  y  180 y = 180 – 60 y = 1200 Tessellation and regular polygons Example

A tessellation is created when a shape is repeated over and over again covering a plane without any gaps or overlaps. Better still we can call it tiling.

This shows that a regular hexagon tessellates.

Circle and its parts Radius: The distance from the centre to any point on the circumference of the circle is called the radius. A diameter is twice the distance of a radius.

Circumference: The distance around a circle is called its circumference. It is also the perimeter of the circle

Arc: An arc is a part of the circumference of a circle. The longer arc is called the major arc while the shorter one is called the minor arc.

Chord: A chord is a straight line joining two points on the circumference. The longest chord in a circle is the diameter. The diameter passed through the centre.

Sector: A sector is a region enclosed by two radii and an arc. The larger segment is the major sector whiles the smaller one, the minor sector.

Segment: A segment of a circle is the region enclosed by a chord and an arc of the circle. The larger segment is the major segment whiles the smaller one, the minor segment.

By Mr Kiapene 2008. Be fully prepared for your SAT exam Tangent: This is a straight line that slightly touches a circle on its circumference. Minor sector Major sector

Chord Radius P

Circumference

R Tangent

Minor arc Minor segment Major segment major arc

Area and circumference of circles d= 12cm r = 12 ÷ 2 = 6cm Circumference =  d or 2 r 12cm Circumference =  d Area =  r2 = 3.14 x 12 = 37.68 Where d = diameter  37.7 cm r = radius Area =  r2 = 3.14 x 62 Example = 3.14 x 36 Find the circumference and area of the circle on = 113.04 the right correct to 1 d.p.. Take  = 3.14  113.0cm2 Volume and surface area of prisms Three steps to find the volume of prisms are: a) identify the cross-section of the prism (the 6cm 7.8cm face that is the same all the way through) b) Find the area of the cross-section c) Multiply the area of the cross section by the 5cm 10cm length of the prism Area of cross-section = (6 x 5) ÷ 2 = 15cm2 Volume = 15 x 10 = 150cm3 Volume = area of cross-section x length The prism has 5 faces (two triangles + 3 rectangles) Total Surface area = area of all the faces of the Area of triangle = (6 x 5) ÷ 2 = 15cm2 prism. Area of triangle 2 = (6 x 5) ÷ 2 = 15cm2 Area of Front rectangle = 10 x 7.8 = 78cm2 Area of back rectangle = 6 x 10 = 60cm2 Area of bottom rectangle = 5 x 10 = 50cm2 +

Total surface area = 218cm2 Length of arc and area of sectors Example Find the length of minor arc and the area of the minor sector below correct to 1 d.p. take  = 3.14

Radius = 8cm Diameter = 16cm

 Length of arc = x  d 3600  40 The length of arc is a fraction ( 0 ) of the =  3.14 16 360 3600 By Mr Kiapene 2008. Be fully prepared for your SAT exam circumference of the circle. 2009.6 = 3600  Length of Arc AB = x  d = 5.5822222 3600  5.6cm  Area of sector = x  r2 The area of the sector is a fraction of the area of 3600 the whole circle. 40 =  3.14 82 3600  Area of sector AOB = x  r2 40 3600 =  3.14  64 3600 8038.4 = = 22.3288888  22.3cm2 3600 Pythagoras Theorem Example 1 This theorem states that in every right-angled triangle, Find the missing length (sides) of the following right- the square of the hypotenuse (the longest side) equals angled triangles. the sum of the squares of the other two sides To find a shorter side

a2 + b2 = c2

152 = 225

92 = 81 minus

.b2 = 144

32 + 42 = 52 b = 144

b = 12

8cm y cm To find the longest side

82 = 64 So, the square of a (a²) plus the square of b (b²) is 62 = 36 + equal to the square of c (c²): 6cm y2 = 100 a2 + b2 = c2 .y = 100 .y = 10 cm Similar triangles Example Two triangles are called similar triangles if one is an enlargement Explain why the two triangles below are f the other. Thus satisfy the following conditions similar. And hence fine the length of side QR

From the above diagram Angle C = 180 – 88 – 35 = 570 1. AB:A’B’ =AC:A’C’=BC:B’C’ Angle Q = 180 – 88 – 57 = 350 AB AC BC Since all the included angles are the same, this means   A' B ' A'C ' B 'C ' the two triangles are similar. RQ AB  2.

By Mr Kiapene 2008. Be fully prepared for your SAT exam 12 6 RQ = = 8cm 9 Trigonometry Example Trigonometry is derived from two Greek trigōnon Calculate the length w correct to the nearest whole means "triangle" and metron which means number. "measure"). It is a branch of mathematics that deals with triangles. At this level we shall focus Use CAH because A and H on right-angled triangles. are what we need w The sine, cosine, and tangent of right-angled Cos 520 = triangles can be remembered by representing 28 them as strings of letters, as in Cross multiply w = 28 Cos 520 SOH-CAH-TOA. w = 28 x 0.6157 Opposite w = 17.2396 Sine = SOH w  17cm Hypothenuse Example Adjacent Cosine = CAH Calculate the size of angle x correct to the nearest degree. Hypothenuse

Opposite Tangent = TOA Adjacent

H O Opposite

0 b A O Sin b0 = We will use TOA because O and A are known H 42 Tan x0 = A 60 Cos b0 = 0 H Tan x = 0.7 Divide both sides by tan O tan x 0.7 Tan b0 =  A tan tan Cancel tan on the left hand side and remember 1  tan 1 tan x0 = tan-1 0.7 x0 = 34.9920 x0  350 Volume of cylinder Example A cylinder is a prism. Find the volume of the cylinder below. Take  = 3.14 Volume = area of cross-section x length Area of cross section = r2 Length = height Volume =  r2h Volume =  r2 x h =  r2h Where  = 3.14 r = 4cm h = 7cm Volume = 3.14 x 42 x 7 = 3.14 x 16 x 7 = 351.68cm3 By Mr Kiapene 2008. Be fully prepared for your SAT exam Metric units for area and volume Unit converters of Area 100mm² = 1 cm² 1000000 mm² = 1m² 10 000cm² = 1m² 10000m² = 1 hectares(ha)

Unit Converters of Volume 1cm³ = 1000 mm³ 1m³ = 1000 000 000mm³

Unit Converters of Capacity 1m³ = 1000 litres (l) 1 litre = 1000cm³ 1cm³ = 1 millilitres (ml) 1 centilitre (cl) = 10 ml 1litre = 100cl = 1000m

Example The cuboidical container measured 50cm, 40cm and 20cm. What is the maximum number of litres that can be contained in the container? Volume = 50 x 40 x 20 = 40 000cm3 (but 1000cm3 = 1 litre) Capacity of the container = 40 000 ÷ 1000 = 40 litres Rate of change Example A coach travels 180km on a motorway at an average speed Distance = speed x time of 80km/h. Find the time take for the journey. Solution dis tan ce dis tan ce Speed = Time = (D = 180km, S = 80km/h) time speed 180 dis tan ce D Time = = 2.25 Time = S T 80 speed Time = 2hours 15 minutes.

Mass = Density x Volume Example Find the mass of a stone which has a volume of 40cm³ ad a Mass density of 3.25g/cm³ Density = Volume Mass = Density x Volume (D = 3.25g/cm3 and V = 40cm3) M Mass Mass = 3.25 x 40 Volume = D V Mass = 130g Density Enlargement Example Enlargement is the act or increasing or reducing the size of a shape. Two conditions are required in enlargement: 1. Scale factor: the number of times the corresponding sides of the object (original shape) has been enlarge or reduce to achieve the image (new shape) 2. Centre of enlargement. The point that determines the position of the image.

By Mr Kiapene 2008. Be fully prepared for your SAT exam A enlarge to B Scale factor is -2 A enlarge to B (upside down) Scale factor of 3 Centre (-1, -2) And centre 0,0 (origin)

Past Questions on Shape, Space and Measure (Level 7 and 8)

1. Calculate the length of the unknown side of B this right-angled triangle correct to 1 d.p. 3cm

12cm Complete the table: Prism A Prism B 17cm

(b) Calculate the length of the unknown side height 5cm 3cm of the right-angled triangle below correct to 3 3 1 d.p. volume 200cm .....……..... cm

4. In the diagram of the wooden frame shown below, PQ is parallel to BC. 5cm A

200 cm 11cm P Q NOT TO SCALE 2. (a) Look at this triangle: 400 cm

B C 600 cm

Calculate length PQ using similar triangles.

10cm 8cm 5. Look at the diagram: A

xº NOT TO SCALE 6cm D Show working to explain why angle x must be a right angle. xº 3xº (b) What is the volume of this prism? B C Side AB is the same length as side AC. Side BD is the same length as side BC. Calculate the value of x

By Mr Kiapene 2008. Be fully prepared for your SAT exam 6. The diagram shows a square inside a triangle. DEF is a 10cm straight line.

8cm 15 cm D 7cm 6cm Not drawn accurately E C

You must show each step in your working. F A B 3. Prisms A and B have the same cross- 12 cm sectional area. The side length of square ABCE is 12 cm. A The length of DE is 15 cm.

Show that the length of EF is 20 cm. 5cm

Handling Data

Probability of single events Example P(A) = The Number Of Ways Event A Can Occur A fair 6 sided dice is tossed. What is the probability of The Total Number Of Possible Outcomes obtaining: a) prime number b) not a square number Solution P(Not A) = 1 p(A) Possibilities: 1, 2, 3, 4, 5, 6 3 1 P(Prime number) = = (prime Nos. are: 2, 3 and 5) 0 ≤ p(A) ≤ 1 (means the probability of an event 6 2 must be within 0 and 1) 2 1 P(square) = = (square Nos are: 1 and 4) 6 3 If P(A) = 0 event A is an impossible event. 1 2 If P (A) = 1, event A is certain or must happen. P (Not square) = 1- = 3 3 Mutually events are events that cannot happen Example at the same time. The occurrence of one prevents A bag contains a large number of discs, each labelled either the other from occurring. Probabilities of all A, B, C or D. mutually event add up to 1. The probabilities that a disc picked at random will have a given letter are shown below. Exhaustive events: A set of events are said to P(A) = 0.25 P(B) = 0.4 P(C) = 0.3 P(D) = ? be exhaustive when they are such that at least a) What is the probability of choosing a disc with a letter D one of the events must occurs. For example, In on it? the experiment of tossing a coin, the probability b) What is the probability of choosing a disc with a letter A of throwing a tail or head is an exhaustive event. or B on it? c) What is the probability of choosing a disc which does not P (A or B) = P(A) + P(B). (“OR” and have the letter C on it? “EITHER” means Plus in probability) d) If there are 60 disc in the bag, how many are labelled A?

P(A and B) = P(A) x P(B) (“BOTH” and a) P(D) = 1 – (0.25 + 0.4 + 0.3) = 1 – 95 = 0.05 “AND” means times in probability) b) P(A or B) = P(A) + P(B) = 0.25 + 0.4 = 0.65 c) P(Not C) = 1 – P(C ) = 1 – 0.3 = 0.7 d) Number of A = P(A) x total number of disc = 0.25 x 60 = 15 disc

Combining probability

By Mr Kiapene 2008. Be fully prepared for your SAT exam The probability tree diagram is the easiest method to work out combined probabilities. Example A bag contains 3 black balls and 5 white balls. Paul picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag. 9 a) Draw a probability diagram to represent b( i) P(BB) = the information 64 15 15 30 15 b) Use the probability diagram to find the ii) (BW or WB) =    probability that: 64 64 64 32 i) both are black 9 25 34 17 iii) (BB or WW) =    ii) one black and one white 64 64 64 32 iii) the same colour Estimates of probability and relative frequency Example Researcher can keep a record of an event e.g. the A spinner was spun several times and the numbers of times it timely, late and early arrival of trains in Birmingham land on red are shown in the table below. train station over a period and then use the results of a) Find the relative frequency correct to 2 these trials to predict the probability that a train will d.p. and complete the table. b) Which of relative frequency is the best be late in the future. This is called an estimate of estimate? probability. c) If the spinner is spun 720 times, how Example many times do you expect it to land on Arrival of train Early On time Late red? Number of trains 25 60 15 Number of spins 10 20 50 75 85 100 a) estimate the probability the next train will be: Number of red 4 7 15 20 30 35 i) Early ii) On time iii) Late Relative 0.40 0.35 b) If 500 trains arrive at the train station over a frequency week, how many do you expect to arrive on time? Solution a) Relative frequency = Number of red Number of spins a) Estimate = Number of successful trials Probability Total Number Of Trials RF1 = 0.40 RF2 = o.35 RF3 = 0.3 RF4 = 0.27 RF6 = 0.35 Total trial = 25 + 60 + 15 = 100 25 1 60 3 i) P(Early) = = ii) P(On time) =  b) RF 4 = 0.27 is the best estimate because it is the closest to 100 4 100 5 theoretical probability.

15 3 c) Number of times it will land on red = RF4 x 720 iii) P(Late) =  100 20 = 0.27 x 720 b) No. that will arrive on time = P(on time) x Total No. of trains = 194 3 =  500  300 5 Mean Solution Mean = Sum of all values Score Frequency Score x frequency Number of values 4 5 20 In a frequency table we find the mean as follows. 5 10 50 Mean = Sum of frequency x value 6 20 120 7 10 70 Sum of frequencies 8 3 24 Example 9 2 18 The table below shows the scores of 50 students. In a Total 50 302 Maths test. Calculate the mean score. 302 Mean =  6.04 Score 4 5 6 7 8 9 50 Frequency 5 10 20 10 3 2

By Mr Kiapene 2008. Be fully prepared for your SAT exam Cumulative frequency curve Example This is the running total of the frequencies. It is drawn to estimate the median and 120 interquratile range 1 1st Quartile (Q1) = th value 100 4 y

rd n 80

3 Quartile (Q3) = th value e

4 u q e

1 r f 60

Median = th value e v

2 i t a Interquartile range = Q3 – Q1 l

Example m u

Complete the cumulative frequencies of the C 20 table below Age Frequency Cumulative (t years) frequency 0 0 < t  5 41 41 0 5 10 15 20 25 30 26 41+26 = 67 5 < t  10 Ages 10 < t  15 20 67+20 = 87 15 < t  20 10 87 + 10 = 97 st rd 20 < t  25 3 97 +3 = 100 1 Quartile = 2.5 Median = 6.5 3 Quartile = 11.5 Plot the bigger number in age against the cumulative frequency. Interquartile range = 11.5 – 2.5 = 9.0 Comparing sets of data Example At this level, using the mean and the range alone is The scores of top sets of two schools are represented in the not enough to compare sets of data. The best way to Box plot below. Find the interqurtile range of both schools compare data is to plot a box plot. We need the and which school has better top set results. Explain your following to plot a box plot. Median, 1st quartile, 3rd answer. quartile, lowest value and the highest value of the sets of data.

Year 7 girls lower upper shortest quartile median quartile tallest

Q1 of A = 24 Q1 of B = 27 Q3 of A = 29 Q3 of B = 31

Interquartile range of A = 29 – 24 = 5 Interquartile range of B = 31 – 27 = 4 136 140 144 148 152 156 160 Height (cm) Group B is better than group A because it is more consistent and had higher scores than school A.

Past Questions on Handling Data (Level 7 and 8)

1. I have two fair 4-sided dice. 4. 40 students worked on a farm one weekend. The cumulative frequency graph shows the One dice is numbered 2 , 4 , 6 and 8 distribution of the amount of money they earned. No one earned less than £15. The other is numbered 2 , 3 , 4 and 5

I throw both dice and add the scores. What is the probability 30 that the total is even? Cumulative 20 Frequency 2. Owls eat small mammals. They regurgitate the bones and fur in balls called pellets. The table shows the contents of 62 10 pellets from long-eared owls. 0 0 5 10 15 20 25 30 35 40 45 50 Amount of money earned (£)

By Mr Kiapene 2008. Be fully prepared for your SAT exam Number of mammals per pellet 1 2 3 4 5 6 (a) Read the graph to estimate the median amount of money earned. Frequency 9 17 24 6 5 1 (b)Estimate the percentage of students who (a) Show that the total number of mammals found is 170 earned less than £40.

(b) Calculate the mean number of mammals found in each pellet. c)Show on the graph how to work out the Correct to 2 d.p. interquartile range of the amount of money earned and write down the value of the (c) There are about 10 000 long-eared owls in Britain. On average, interquartile range. a long-eared owl regurgitates 1.4 pellets per day. 5. A company makes computer disks. Altogether, how many mammals do the 10 000 long-eared owls It tested a random sample of disks from a eat in one day? large batch. The company calculated the probability of any disk being defective as 3. The goldcrest is Britain’s smallest species of bird. On winter 0.25. days, a goldcrest must eat enough food to keep it warm at night. During the day, the mass of the bird increases. Glenda buys two disks. (a) Calculate the The scatter diagram shows the mass of goldcrests at different probability that both disks are defective. times during winter days. It also shows the line of best fit. (b) Calculate the probability that only one of

6.0 the disks is defective.

5.5 (c) The company found 3 defective disks in the sample they tested. 5.0 Mass (g) How many disks were likely to have been 4.5 tested?

4.0 6. A bag contains 3 green, 5 blue and 2 red marbles. Two marbles are picked from the 3.5 bag one after the other with replacement.

0 7am 8am 9am 10am 11am 12pm 1pm 2pm 3pm a) Draw a tree diagram to represent the Time (GMT) information. a) What kind of correlation does the scatter graph shows b) Calculate the probability that they are: and describe the correlation i) Both red. (ii) both blue (iii) both green b) Estimate the mass of the goldcrest at 11.30am (iv) the same colours. Algebra

Sequences Write down the first five terms of the following sequences a) 3n – 5 Write the nth term of the following T1 = 3 x 1 – 5 = -2 sequences T2 = 3 x 2 – 5 = 1 a) 4, 8, 12, 16, 20 ..... 4n T3 = 3 x 3 – 5 = 4 The sequence is the 4 times table. T4 = 3 x 4 – 5 = 7 b) 2, 5, 8, 11,...... 3n - 1 T5 = 3 x 5 – 5 = 10 the sequence goes up in 3s but we need to The sequence is: -2, 1, 4, 7 , 10 take 1 away from the 3 times table to get the b) 2n2 + 3 sequence. T1 = 2 x 12 + 3 = 5 d) 5, 9, 13, 17, ...4n + 1 T2 = 2 x 22 + 3 = 11 the sequence goes up in 4s but we need to T3 = 2 x 32 + 3 = 21 add 1 to the 4 times table to get the T4 = 2 x 42 + 3 = 35 sequence. T5 = 2 x 52 + 3 = 53 The sequence is: 5, 11, 21, 35, 35

By Mr Kiapene 2008. Be fully prepared for your SAT exam Spotting patterns The fastest way to solve a problem with a pattern is to write the nth term of the sequence and use it to solve the questions that follow. Example A series of patterns is drawn using dots.

a) How many dots are in (i) Pattern 4 (ii) Pattern 5 The table shows the number of dots needed for different patterns. Pattern 1 2 3 4 5 6 7 Number of dots 5 8 11

(b) Complete the table. (c) Explain how you would work out the number of dots needed for Pattern 12. Solution d) the nth term of the number of dots is 3n + 2 i) 4th pattern = 3 x 4 + 2 = 14 dots ii) 5th pattern = 3 x 5 + 2 = 17 dots e) the 6th and 7th pattern has 20 and 23 dots respectively (because the sequence goes up in 3s) f) 12th Pattern = 3 x 12 + 2 = 38 Inverse functions Example Example Input ( x ) Output (function is) Find the inverse of Find the inverse of Function 5 36 7 x + 1 x '  2 x  2 This can be written as y = 7x + 1 Multiply both sides by 2 Lets rewrite the function as Inverse of the function is the same thing as writing follow 2x  x '  2 x in terms f y. (6  y) x  Example 1 2 ' 2 4 Find inverse of y = 3 x - 5 2x  x  2 ' Multiply both sides by 4 +5 5 x  2x  2 y = 3x  5 4x 6  66  y y – 5 = 3x (divide both sides by 3) y = 4x - 6 y  5  x We have written functions in three different ways in the 3 above examples. y  5 x  3 Limits of sequences Example 1. Finite sequence: this is a sequence that has a last Use the term-to-term rule, divide by 5 and add 4, find the first 10 term. terms of the sequence, starting from 1 2. Infinite sequence:- This is a sequence that has no What is the limit of the sequence? final term. It goes on forever indicated by … Solution 1, 4.2, 4.84, 4.968, 4.9936, 4.99872, 4.999744, 4.9999488, 3. Limit of a sequence:- the value that a sequence gets 4.99998976, 4.999997952 closer and closer to but will never be equal to is called limit of that sequence Limit of the sequence is 5. Equations and formula Examples Examples Solve 3(3m + 4) = 2(2m - 8) Solve 7(2w - 3) -5(3w -1) =0 a) Solve 3d - 5 = 37 Solve 5s - 2 = 7s – 12 Expand brackets Expand brackets (mind the signs) +5 +5 +2 +2 3d – 5 = 37 5s – 2 = 7s – 12 9m + 12 = 4m – 16 14w – 21 – 15w + 5 = 0 3d÷ 3 = 42÷ 3 5s-7s = 7s-7s – 10 9m + 12-12 = 4m – 16-12 Collect like terms ÷ -2 ÷ - 2 d = 14 -2s = -10 9m-4m = 4m-4m - 28 14w – 15w -21 + 5 = 0 s = 5 5m = -28 -w – 16 = 0 Solve 2(w + 4) = 24 Solve 3(2n – 4) = 30 5m÷5= -28÷5 -w – 16+16 = 0+16 Expand bracket Divide both sides by 3 m = -5.6 2w + 8-8 = 24-8 2n – 4+4 = 10+4 - w = 16 By Mr Kiapene 2008. Be fully prepared for your SAT exam 2w÷ 2 = 16÷2 2n÷ 2 = 14÷2 -w ÷ -1 = 16÷ - 1 w = 8 n = 7 w = - 16 Equations involving squares Example When the unknown term is squared it is quadratic Solve (n - 6)² = 169 equation. Every quadratic equation has two answers To remove the square sign, we square root the RHS (remember Solve n²+ 3 = 19 Solve 150 - c² = 101  must following whenever you introduce square root n² + 3-3 = 19-3 150-150- c² = 101-150 (n - 6)² = 169 Solve (e + 5)² = 81 n² = 16 - c² = -49 n – 6 =  169 Remove the square on the RHS n =  4 Divide both sides by -1 n – 6 =  13 e + 5 =  81 Solve g²- 12 = 888 c² = 49 (add 6 to both sides, do e + 5 =  9 +12 +12 g²- 12 = 888 c =  7 this to RHS before the  ) minus 5 from both sides, do g² = 900 n = 6  13 this to RHS before the  g =  30 When n+ n = 6 + 13 = 19 e = -5  9 + When n- n = 6 - 13 = -5 When e e = -5 + 9 = 4 -  n = -5 or 19 When e e = -5 - 9 = -14  e = - 14 or 4 Equations involving fractions Examples To solve equations with fractions, we need to non e  2 2e  8 4 2 Solve  Solve (2a 1)  (a  3) fractional equation by multiplying by the LCM of 3 4 5 3 the denominators. LCM of 3 and 4 = 12 LCM of 5 and 3 = 15 Example Times both sides by 12 Times both sides by 15 3b 4s  5 412(e  2) 312(2e  8) Solve  6 Solve  7  5 3 1 3 1 4 315 4 515 2 Times both sides by 5 Multiply both sides by 3 (2a 1)  (a  3) 4(e – 2) = 3(2e – 8) 1 5 1 3 3b = 5 x 6 4s + 5 = 7 x 3 +8 +8 ÷ 3 ÷ 3 -5 -5 e – 8 = 6e – 24 3 x 4 (2a + 1) = 5 x 2(a - 3) 3b = 30 4s + 5 = 21 -6e -6e ÷ 4 ÷4 4e = 6e – 16 12(2a + 1 ) = 10(a – 3) b = 10 4s = 16 ÷ -2 ÷ -2 s = 4 -2e = - 16 Expand brackets e = 8 24a + 12-12 = 10a – 30-12 24a-10a = 10a-10a – 42 14a÷ 14 = -42÷ 14 a = -3 Direct proportion and graph Example Direct proportion involves situations where two 1. A baby squid is weighed from birth at Midday for its first five values vary, but the ratio between the values stays days. The results are shown in the table below the same. The best way to solve direct proportion questions is to write the formula connecting the variables. a) Find the weight of the baby squid in the 25th day Example b) At what day will the squid first weighed over 15kg If 10 calculators cost £120, how much will 8 a) W= 0.3 + 1.4D b) W = 0.3 + 1.4D calculators cost? (W = weight and D = Day) (when W = 15kg) C = 12N (C = cost and N = number of calculators) W = 0.3 + 1.4 x 25 (when D =25) 15-0.3= 0.3-0.3+ 1.4D C = 12 x 8 ( N = 8) W = 0.3 + 35 14.7÷1.4 = 1.4D÷1.4 C = £96 W = 35.5 kg 10.5 = D The baby squid weighs over 15kg on the 11th day. Solving problems using equation Example This involves write out a mathematical equation A length of a rectangle is 4cm bigger than the width of the and using algebra to work out the solution. rectangle. If the perimeter of the rectangle is 28, find the Example dimentions of the rectangle Three friends have 200 game cards between them. Joe Solution has 36 more cards than Chris. Kay has twice as many Let the width of the rectangle be n cm. cards as Chris. How many cards does each have? The length of the rectangle = (n + 4)cm Solution (n + 4)cm 4n÷4 = 20÷ 4 Let y = number of cards Chris has. n = 5 2y = number of cards Kay has n cm Width = 5 cm y + 36 = number of cards Joe has. Length = 5 + 4 = 9cm y + 2y + y + 36 = 200 4y + 36-36 = 200-36 4y÷ 4 = 164÷ 4 n + n + (n +4) + (n+4)=28 By Mr Kiapene 2008. Be fully prepared for your SAT exam y = 41 (Chris has 41 cards) 4n + 8 = 28 Kay has 2 x 41 = 82 cards and 4n + 8-8 = 28-8 Joe has 41 + 36 = 77 cards 4n = 20 Trial and improvement Example The solution of some quadratic or cubic equations Use trial and error to work out b³- b = 43 may not give whole numbers answers. In such b b3 b3 - b comment cases, the best option is to use trial and error. 4 64 60 Too big Example 3 27 24 Too small Use trial and error to work out n²+ n = 37 3.5 42.875 39.375 Too small Solution 3.51 43.2436 39.7336 Too small 2 2 n n n + n Comment 3.6 46.656 43.056 correct 5 25 30 Too small 3.61 47.04588 43.43588 Too big 6 36 42 Too big

5.5 30.25 35.75 Too small 5.6 31.36 36.96 Correct b = 3.6 5.61 31.4721 37.0821 too big n = 5.6 Simultaneous equations Example Example This is the solving of two equations with two unknowns at Solve 4x + 3y = 34 ……(1) Solve 3x + 2y = 8 …….(1) the same time. 2x - 3y = 8 ……...(2) 2x - 3y = 1 ….….(2) Steps for elimination method The coefficients of y are the Times equ (1) by 3 and equ 1. Make the coefficient of y (the same the signs are different. (2) by to make the number before y) the same. Add equ (1) and equ (2) coefficients of y equal 2. When the signs are different add 6x = 42 9x + 6y = 24 ……(3) the two equations 6x÷ 6 = 42÷ 6 4x – 6y = 2 ……..(4) 3. When the signs are different x = 7 Add equ (3) and equ (4) subtract one equation from the other. Substitute x into equ (1) because the signs are 4. Solve the linear equation 4(7) + 3y = 34 different 5. Substitute the value of x into any of 28-28 + 3y = 34-28 13x = 26 the equation to find the value of y. 3y = 6 13x ÷ 13 = 26 ÷ 13 Example 3y ÷ 3 = 6÷ 3 x = 2 Solve 5x + y = 23 ……(1) Substitute value of x into equ (1) y = 2 Substitute x into equ (1) 2x + y = 11 …….(2) 5(4) + y = 23 (x, y ) = (7, 2) 3(2) + 2y = 8 The coefficient is the same (1) 20 + y = 23 6-6 + 2y = 8-6 and the signs are the same. 2y ÷ 2 = 2÷ 2 subtract equ (2) from equ (1) y = 23 – 20 y = 1 3x = 12 y = 3 (x, y) = (2, 1) x = 4 (x, y) = (4, 3) Inequalities Example Symbols of inequalities a) Solve 1 ≤ 4y - 3 < 13 1. > means "is greater than" b) List all the values of y 2. < means "is less than" c) Show the solution in a number line 3. ≥ means "is greater than or equal to" Solution 4. ≤ means "is less than or equal to" There are two symbols of inequalities in the expression. We 5. less than therefore need to split it into two, at the end of the first 6. Greater than For 7. Less than or equal to Number symbol and the beginning of the second symbol. + 3 + +3 +3 8. Greater than or equal to Line a) 4y – 3 < 13 3 and 1 ≤ 4y – 3 4y÷4 < 16÷ 4 4÷4 ≤ 4y÷ 4 Example y < 4 1 ≤ y Solve 3e - 4 < 11 and show the solution in a number line we can combined these solutions as 1 ≤ y < 4 3e – 4 + 4 < 11+4 a) y = 1,2,3 3e÷ 3 < 15÷ 3 c) e < 5 -2 -1 0 1 2 3 4 5 6

0 1 2 3 4 5 6 7 8 Quadratic graphs Expansion of brackets Example Expand is the act of removing the bracket. This is done Expand and simplify 3a(5a + 4) - 2a(3a - 5) by using the number outside the bracket to multiply Solution everything inside the bracket 3a(5a + 4) - 2a(3a - 5) = 15a2 + 12a – 6a2 + 10a (-2a x -5 = 10a) By Mr Kiapene 2008. Be fully prepared for your SAT exam Example = 15a2 – 6a2 + 12a + 10a 2(4a + 3) = 8a + 6 = 9a2 + 22a 4y(3y -2a) = 12y2 – 8ay Example Example Expand and simplify 4g(2f - 3y) - 3g(f +2y) Expand and simplify Solution 2(2a + 3b) + 3(2a + 5b) = 4a + 6b + 6a + 15b (collect like terms) 4g(2f - 3y) - 3g(f +2y) = 8gf – 12gy – 3gf – 6gy = 4a + 6a + 6b + 15b = 8gf -3gf – 12gy – 6gy = 10a + 21b = 5gf – 18gy Factorisation Example Factorisation is the opposite of expansion. It is the Factorise 5k²- 25k = 5k x k – 5k x 5 ( common factor is 5k) act of removing the common factor outside a = 5k (k – 5) bracket. Example Factorise 12an + 8am - 16ap = 4a x 3n + 4a x 2m – 4a x 4p Factorise 6m + 15 = 3 x 2m + 3 x 5 (common factor is 3) = 4a (3n + 2m – 4p) = 3 (2m + 5) Quadratic expansion Example Expand (k - 3)(k - 2) = k x k + k x – 2 – 3 x k – 3 x -2 = k2 – 2k – 3k + 6 (c + 4)(c + 6) = c x c + c x 6 + 4 x c + 4 x 6 = k2 – 5k + 6 = c2 + 6c + 4c + 24 Example 2 Expand (4w - 2)(3w - 5) = 4w x 3w + 4w x -5 – 2 x 3w – 2 x – 5 = c + 10c + 24 2 Example = 12w -20w – 6w + 10 = 12w2 – 26w + 10 Expand (g + 5)(g - 2) = g x g + g x -2 + 5 x g + 5 x -2 = g2 – 2g + 5g - 10 = g2 + 3g - 10 Quadratic factorisation Example The general form of quadratic expression is ax 2 + b x + c Factorise x 2  9x  20 Steps for quadratic factorisation x 2  20  20x 2 ax 2 6. multiply by c Now think of two parts when multiplied will give you 7. Think of two terms when multiplied will give you 2 20x and when added will give you  9x . answer to 1 and when added will give you bx . 2 8. Replace bx with the two terms  4x  5x  20x 9. factorise the terms  4x  5x  9x (now replace  9x with them) 2 Example x  4x  5x  20 Factorise x 2  6x  8 x(x  4)  5(x  4) x 2 8  8x 2 (x  4)(x  5) answer Now think of two parts when multiplied will give you 8x 2 and when added will give you 6x. Look at this quick way 2 4x  2x  8x 2 Factorise x  8x 12 4x  2x  6x (now replace 6x with the two) Write two brackets. ( x )( x ) x 2  4x  2x  8 2 Think of two numbers when multiplied will give you (x  4x)  (2x  8) 12 and when added will give you -8. Write one x(x  4)  2(x  4) number on each bracket. (x  4)(x  2) answer The numbers are -6 and – 2. Thus the answer is: x x ( - 6)( - 2) Change of subject Example Formulas are written so that a single variable, the subject of the formula, is p  2t W  make t the subject of the on the left hand side of the equation. Everything else goes on the right g hand side of the equation. V = U + at (V is the subject of the formula) Example formula V = U + at (make a the subject of the formula) Multiply both sides by g V u  U u  at Wg = p + 2t (subtract p from both sides) V – U = at (divide both side by t) Wg – p = 2t (divide both sides by 2) V U  a Wg  p t  t V U 2 a = t By Mr Kiapene 2008. Be fully prepared for your SAT exam Wg  p t = 2

Past Questions on Algebra (Level 7 and 8)

1. Solve these equations. Show your working. 4. a) The subject of the equation below is p

(a) 4 – 2y = 10 – 6y p = 2 ( e + f )

(b) 5y + 20 = 3(y – 4) Rearrange the equation to make e the subject. 2. (a) The nth term of a sequence is 3n + 4. What is the 8th term of this sequence? 1 (b) Rearrange the equation r = (c – d) 2 (b) The nth term of a different sequence to make d the subject. n – 2 is Write the first three terms 2 n . 5. Look at these expressions. of this sequence.

3. Multiply out the brackets in these expressions. 5y – 8 3y + 5

a) y (y – 6) = first second expression expression b) (k + 2)(k + 3) = …… What value of y makes the two expressions equal?

6. a) Solve these inequalities. 9. a) Explain how you know that (y + 3)2 is not Show your working. equal to y2 + 9 2(2y  7) < 2 (b) Multiply out and simplify these expressions. 3 i) (y + 2)(y + 5) 4(7 – 2 y) > 1 12 ii) (y – 6)(y – 6)

Remember: When you multiply or divide both sides of iii) (3y – 8)(2y + 5) the equality expression by minus you must inverse 10. Solve this equation. the sign. 5(2y – 3) 7. Solve these simultaneous equations using an  3 algebraic method. 3y

3x + 3y = 21 11. Each year a school has a concert of readings and songs. 2x + y = 8 In 1999 the concert had 3 readings and 9 8. (a)Find the values of a and b when p = 10 songs. It lasted 120 minutes.

3 p 3 In 2000 the concert had 5 readings and 5 a  i) 2 songs. It lasted 90 minutes. In 2001 the school plans to have 5 readings 2p 2 ( p – 3) b  and 7 songs. ii) 7 p Use simultaneous equations to estimate how (b) Simplify this expression as fully as long the concert will last. Call the time possible: estimated for a reading x minutes, and the time estimated for a song y minutes.

By Mr Kiapene 2008. Be fully prepared for your SAT exam 3 cd 2 5 cd

Answers

Number

1). 340 days 2). 78.5% 3 (a) 92.5m (b) 49.5m (c) 10 times 4(a) k = 3 and m = 6 (b) 16384 (5) 108000 7(a) a = 4 and b = 3 (b) c = 7 9(a) 4.0 x 10-4 (b) 4.0 x 10-5 c) 4.4 x 10-4 (10) £45

Space, Shape and Measure

1(a) 20.8cm (b) 9.8cm 2(b) 168cm3 (3) 120cm3 (4) 200cm (5) 22.50 (6) 20cm

Handling Data

1 1) 2(b) 2.74 (c) 38 360 3(a) positive correlation. As the time increases so does the mass (b) 4.85 2 1 1 9 19 4(a) 30.5 (b) 87.5% (c) 11 5(a) 0.0625 (b) 0.375 6b(i) b(ii) b(iii) b(v) 25 4 100 50

Algebra

1 p p  2 f 1 (a) y = 1.5 (b) y = -16 2(a) 28 (b) -1, 0, 3(a) y2 – 6y (b) k2 + 5k + 6 4(a) e =  f or e  9 2 2

3 4(b) d = c – 2r or d = -2r c (5) 6.5 6(a) y < - 2 (b) y < 2 7(a) x = 1 and y = 6 8a(i) 1500 (ii) 20 (b) d 5

9b(i) y2 + 7y + 10 (ii) y2 – 12y + 36 (iii) 6y2 – y – 40 (10) y = 15 (11) 112 minutes and x = 7 while y = 11

THE END

By Mr Kiapene 2008. Be fully prepared for your SAT exam