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Maths IGCSE.pdf

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Department
Biology
Course
BIOL-UA 6
Professor
davidjenkins
Semester
Spring

Description
Mathematics IGCSE notes Index 1. Decimals and standard form 2. Accuracy and Error 3. Powers and roots click on a topic to 4. Ratio & proportion visit the notes 5. Fractions, ratios 6. Percentages 7. Rational and irrational numbers 8. Algebra: simplifying and factorising 9. Equations: linear, quadratic, simultaneous 10. Rearranging formulae 11. Inequalities 12. Parallel lines, bearings, polygons 13. Areas and volumes, similarity 14. Trigonometry 15. Circles 16. Similar triangles, congruent triangles 17. Transformations 18. Loci and ruler and compass constructions 19. Vectors 20. Straight line graphs 21. More graphs 22. Distance, velocity graphs 23. Sequences; trial and improvement 24. Graphical transformations 25. Probability 26. Statistical calculations, diagrams, data collection 27. Functions 28. Calculus 29. Sets {also use the intranet revision course of question papers and answers by topic } 1. Decimals and standard form top (a) multiplying and dividing (i) 2.5×1.36 Move the decimal points to the right until each is a whole number, noting the total number of moves, perform the multiplication, then move the decimal point back by the previous total: →× 25 136 = 3400 , so the answer is 3.4 {Note in the previous example, that transferring a factor of 2, or even better, 4, from the 136 to the 25 makes it easier: 25×1=36 25×(4×34) = (25×4)×34 = 100×34 = 3400 } (ii)0.00175÷0.042 Move both decimal points together to the right until the divisor is a whole number, perform the calculation, and that is the answer. →1.75÷42, but simplify the calculation by cancelling down any factors first. In this case, both numbers share a 7, so divide this out:.25÷6 , and 0.0416▯ ▯ 60 .25 , so the answer i0.0416 (iii) decimal places To round a number to n d.p., count n digits to the right of the decimal point. If the digit following the n is ≥5 , then the n digit is raised by 1. e.g. round 3.012678 to 3 d.p. 3.012678 → 3.012|678 so 3.013 to 3 d.p. (iv) significant figures To round a number to n s.f., count digits from the left starting with the first non-zero digit, then proceed as for decimal places. e.g. round 3109.85 to 3 s.f.3109.85 → 310|9.85 so 3110 to 3 s.f. e.g. round 0.0030162 to 3 s.f.,0.0030162 → 0.00301|62 , so 0.00302 to 3 s.f. (b) standard form (iii) Convert the following to standard form: (a) 25 000 (b) 0.0000123 Move the decimal point until you have a number x where 1≤1 x < 0 , and the number of places you moved the point will indicate the numerical value of the power of 10. So 25000 =× 2.5 10 , and 0.0000123 =× 1.23 10 −5 (iv) multiplying in standard form: (4.4×1×0 ) (3.5×10 6) As all the elements are multiplied, rearrange them thus: =×(4.4 3.5) × 1()×10 6 = 15.4×10 11 = 1.54×10 12 2 3.2×10 12 (v) dividing in standard form: 3 Again, rearrange the calculation to 2.5×10 (3.2÷×2.5) (10 ÷10 ) 3 = 1.28×10 9 (vi) adding/subtracting in standard form:(2.5×1+0 ) (3.75×10 ) The hardest of the calculations. Convert both numbers into the same denomination, i.e. in this case 10 or 10 , then add. 7 7 7 =(×0.25 10 ) + (3.75×10 ) = 4×10 Questions (a) 2.54×1.5 (b) 2.55÷0.015 (c) Convert into standard form and multiply: 25000000×0.00000000024 3 2 − (d) (2.6×1÷0 ) (2×10 ) (e) (1.55× −0 ) −( )51× 0 Answers (a) →× 254 15 = 3810 , so2.54×1.5 = 3.81 (b) 2.55÷0=.015 2550÷15 . Notice a factor of 5, so let’s cancel it first: =÷510 3 = 170 7 1 − 0 −3 (c) =×(2.5 10 )×(2.4×10 ) = 6×10 (d) =÷(2.6 2)×(10 ÷10 ) = 1.3×10 5 (e) =×(1.55 10 )−(0.25×10 ) = 1.3×10 −3 3 2. Accuracy and Error top To see how error can accumulate when using rounded values in a calculation, take the worst case each way: e.g. this rectangular space is measured as 5m by 3m, each measurement being to the nearest 3m metre. What is the area of the rectangle? 5m To find how small the area could be, consider the lower bounds of the two measurements: the length could be as low as 4.5m and the width as low as 2.5m. So the smallest possible area i4.5×2.5 = 11.25 m . Now, the length could be anything up to 5.5m but not including the value 5.5m itself (which would be rounded up to 6m) So the best way to deal with this is to use the (unattainable) upper bounds and get a ceiling for the area as 5.5×=.5 19.25 m , which the area could get infinitely close to, but not equal 2 2 to. Then these two facts can be expressed as 11.25m ≤ area < 19.25m . Questions (a) A gold block in the shape of as cuboid measures 2.5cm by 5.0cm by 20.0cm, each to the nearest 0.1cm. What is the volume of the block? (b) A runner runs 100m, measured to the nearest metre, in 12s, measured to the nearest second. What is the speed of the runner? (c) a = 3.0, b = 2.5, both measured to 2 s.f. What are the possible value of a−b ? Answers 3 (a) lower bound volume = 2.45×4.95×1 =9.95 241.943625 cm . upper bound volume = 2.55×5.05×2 =0.05 258.193875 cm . 3 3 S o 241.943625cm ≤volume<258.193875cm distance (b) Since speed = , for the lower bound we need to take the smallest time value of distance with the biggest value of time, and vice-versa for the upper bound. 99.5 100.5 So < 4 , the answer would be x < 2 or x > 2. (c) 2 variable linear inequalities e.g.3x2− ≥ 6. Plot the boundary line32− = 6 , then take a trial point (e.g. the origin) to determine which side of the line to accept. y 2 1 –2 –1 1 2 3 x –2 –3 –4 The origin’s coordinates make 30 −× 2 0 which is not ≥ 6, so that side is rejected: y 2 1 –2 –1 1 2 3 x –2 –3 –4 22 Questions y 5 3 (a) Solve2(1− x) < 6 2 1 2 (b) Solve12−≤ x x –4–3–2–1 1 2 3 4 5 x –2 (c) Find the 3 inequalities which identify this region: –3 –4 Answers (a)2(1− x) < 6 [ ÷2] ∴1−3− 2 (b)12−≤ x x2 [rearrange] 2 xx+− 12 ≥ 0 ∴(x4+−)( 3) ≥ 0 giving critical values of –4 and +3. y 10 5 x –5–4–3–2––5 1 2 3 4 – 10 – 15 so x ≤− 4 or x ≥ 3 1 (c) The three line equations ayx 21 +=, y x 2, x + y= 4. 2 By considering a point (e.g. origin) in the shaded region, the inequalities are 1 y <21 ,y > x 2 , and x+ < 4. 2 23 12. Parallel lines, bearings, polygons top (a) Parallel lines correspondingangles equal allied or interior add up to alternate angles equal 180° N (b) bearings A Bearings are measured clockwise from North: bearing of B from A is 135º 45° B (c) polygons for a polygon with n sides, exterior angle sum of interior angles = (2− )180 º interior angle sum of exteriors = 360º 24 Questions a (a) In the diagram opposite, find A E the value ofθ in terms of a and b. D θ C (b) The bearing of B from A is 090º, B b and the bearing of C from B is 120º. Given also that AB = BC, find the bearing of C from A. (c) A pentagon has exactly one line of symmetry, and angles all of which are either 100º or 120º. Make a sketch of the pentagon, marking in the angles. Answers (a) DAE = a (opposite), DEC = b (corresponding), so ▯ ▯ AED =180−b (angles on a straight line). ADE =θ (opposite). We now have the three angles in triangle ADE, sab(180−+) θ =180 . A rearrangement gives θ = b−a . N N (b) Angle at point B means B 120° A ABC =−360 90−120 =150º ▯ C Triangle ABC is isosceles, so BAC =15º. The bearing of C from A is therefore 105º. (c) Sum of internal angles (n− 2)180 = 540 º for any pentagon. A line of symmetry means c the set up is like this: b b The only way of allocating 100º and 120º to a, b, c and make a total of 540º is to have three 100º’s a a and two 120º’s. So there are two possible pentagons: 100° 100° 100 ° 100° 120°12 0° 120°120° 100°100 ° 25 13. Areas and volumes, similarity top (a) Areas of plane figures CIRCLE TRIANGLE B a r h b C A b π r2 1 1 bh or absinC 2 2 TRAPEZIUM PARALLELOGRAM b h h b a 1 bh (a + b)h 2 (b) Surface area and volume Shape surface area volume PRISM Prism p×l Al× A p r l Cylinder 2πrh πr h h CYLINDER 1 2 CONE Cone πrl πr h h l 3 r 2 4 3 SPHERE r Sphere 4πr πr 3 1 Pyramid ×base area× h PYRAMID h 3 3 Pipe flow: number of m /s flowing through (or out of) a pipe = cross-sectional are× speed v 26 (b) Similarity Enlargement scale factor = k Area scale factor = 2 k Volume scale factor = k 3 Questions 3 (a) A cylinder has volume 100cm , and height 5cm. What is its diameter? (b) A cone of base radius 10cm and height 20cm is sliced parallel to the base half way up into two pieces. What is the volume of the base part? (frustum) 25m (c) The empty swimming pool shown opposite 10m is to be filled with water. The speed of flow of water in the pipe is 2m/s, and the radius of the 1m 3m pipe is 5cm. How long will the pool take to fill? (d) Two blocks are geometrically similar, and the big blocks weighs 20 times the small block. What is the ratio of surface areas of the two blocks? Answers (a)πr 51 = 00 ∴ r2 =100 , so r = 20 = 2.52 cm. Whoops! Diameter asked for! 5π π diameter = 5.05cm to 3sf {Note the pre-corrected value was doubled resulting in 5.05 when itself rounded, not 5.04} 27 (b) The upper small cone has base radius 5cm and height 10cm. The volume of 1 2 1 2 1 the base is thereforeπ 10 × 20 − π 5 × 10 which factorises to π1750 = 3 3 3 3 1830cm to 3sf (c) Pool is a prism with cross section the side, which is a trapezium. 1 3 So volume of pool = (1 3)25 10 = 500 m . 2 Rate of egress of water is c.s.a.peed = π5 ×2 =00 5000π cm , which 6 3 is5000π ÷10 m /s. (Units!!) So time taken = 105 500 (5000 π 10 ) = = 31831 s, i.e. approx 8hrs 51 mins. π (d) assuming same density material, weight is directly proportional to volume. The volume factor is 20, i.ek = 20 . ∴k = 20 , and so the ratio of surface areas,1 :k2, is 1 : 7.37 28 14. Trigonometry top O sinθ = H H O A θ cosθ = H A tanθ =O A a b c Sine rule: = = sin A sin B C sin 2 2 2 Cosine rule:ab=+ c − 2cc os A y x Two opposite pairs: use sine rule x y Three sides and one angle: use cosine rule z Angle between line and plane is the angle between the line and its projection on the plane: e.g. for the angle between this diagonal and the base, draw the projection, and the angle is shown here: Trigonometric functions for all angles: 1y 1 y 5y 90 180 270 360 x 90 180 270 360 x 90 180 270 360 x –1 –1 –5 tanx sinx cosx 29 C Questions 6cm (a) ADB is a straight line of length 20cm. .ind 40° θ A D B (b) In triangle ABC, AB = 5cm, BC = 8cm, andCA= 30º. Find ABC (c) A yacht sails 5 miles at 045º then 6 miles at 090º. How far and at what bearing is it from its original point? (d) Is an internal diagonal of a cube at 45º elevation from the base? (e) Find two values of x in the range 0º to 360º for which− 0.5 30 Answers (a) draw a diagram. No, a decent diagram! θ lies in the triangle on the right, and all the information we have is in the left triangle. To connect with the triangle BCD it would be helpful to calculate CD and AD. º CD º sin 40 = , ∴CD = 6 =sin40 3.85.... 6 º AD º cos40 = 6 , ∴AD = 6cos40 = 4.59....., so BD = 20− 4.59... = 15.4... Then tanθ = =.85... 0.250..., soθ =14.1 º to 3 s.f. 15.4... B (b) If we target angle A then we have 2 opposite side/angle pairs, so use the sine rule: 5 cm 8 cm º 30 ° sin A = sin30 , so A= 53.1.. , and A C 8 5 ˆ B =−80 53.1... −0 = 96.9º N (c) Using cosine rule, N A 6 B OB =+ 25 6 − 25×6 × ×cos135 º 5 so OB = 10.2 miles (to 3 s.f.) 45° Now using the sine rule, O sin AOB sin135 º º = , which gives AOB = 24.7 6 10.16... The bearing of B from O is therefore 069.7º (d) Assume the length of side of the cube is 1. (Enlargement won’t make any difference to the angles). Pythagoras gives the projection on the base as 2, and the opposite side is 1. 1 So tanθ = . an dθ = 35.3º to 3 s.f. 2 y 1 (e) First find the principal value from the calculator: -30º. Where are there -30 other angles in our window with the – 90 90 180 270 360 x same sine? -0.5 –1 y 1 -30 – 90 90 180 270 360 x -0.5 –1 Clearly at 30º beyond 180º and 30º back from 360º. So x = 210º, 330º 31 15. Circles top (a) arcs, sectors, segments arc Arc length = θ ×2πr θ sector 360 r Sector area = θ ×π r 2 360 segment Segment area = Sector – Triangle (b) circle theorems (i) Angle subtended at the centre =2× angle subtended at the circumference by the same arc (ii) Angles in the same segment are equal (iii) Angle in a semicircle = 90º (iv) Opposite angles of a cyclic quadrilateral add up to 180º (v) Exterior angle of a cyclic quadrilateral = interior opposite (vi) Alternate segment theorem: the angle between tangent and side = interior opposite 32 Questions (a) The arc of a sector of a circle of radius 20cm has length 10cm. Find the area of the sector. (b) A cylindrical tank, radius 50cm and length 2m with horizontal axis is partially filled with oil to a maximum depth of 25cm. How much oil is contained in the cylinder? (c) Findθ in the following diagrams: θ (a) 120° (b) (c) 35° θ θ 40 ° Answers θ (a) Arc length = 360 ×2π2 0 and this is given as 10cm. Rearranging gives θ = 90 . Therefore sector area =θ ×π 202 = 90 ×π 20 2 π 360 360π 2 which simplifies nicely to 100cm . 90 {Would you have reached for the calculator atθ = π , and missed the beautiful cancellation later?} (b) We need to find the area of the segment comprising the cross-section of the oil. Above the oil is an isosceles triangle, so 25 cm split it down the line of symmetry: OIL 25 cm 50 cm 25 cm −1 This gives an angle ofcos 0.5 = 60º, and a base of 43.3…cm by Pythagoras. So the angle at the centre of the sector is 120º. Therefore the area of the segment is120 ×− 502 43.3... 25 = 1535cm . The oil is in the 360 shape of a prism with volume 1535×200= 307092 cm 3 3 = 0.307m . 33 (c) The angle subtended at the centre is 360º – 120º = 240º, so =120 º by the angle at the centre theorem. θ C = 35º (angles in the same segment) ˆ C CAB = 90º (angle in a semicircle) so θ =−80 35−90= 55 º (angle sum of a triangle) 35° A θ B ABT = 40º (alternate segment theorem) C θ Isosceles triangle givesT = 70 º, and B A so C = 70 º (angles in same segment) T 40° 34 16. Similar triangles, congruent triangles top (a) Similar triangles same shape, different size; related by enlargement (may be different orientation) to prove similar: AAA (each pair of angles equal) to solve problems use either (a) scale factor or (b) ratio of sides equal (b) Congruent triangles same shape and size, i.e. identical though usually in different positions. SSS to prove congruent: SAS AAS ASA RHS but not ASS – there are sometimes two different triangles with the same ASS 35 Questions E D 6 (a) (i) Prove that triangles BCD and C ACE are similar. (ii) Hence find the lengths BD and 8 DE. (iii)If the B area of triangle BCD is 12 what is the A 4 area of the trapezium ABDE? (b) Use congruent triangles to prove that the diagonals of a parallelogram bisect each other. Answers (a) (i)AC = DBC and AEC = BDC (corresponding). The third angle is shared, so AAA is established and they are similar. 12 3 3 (ii) scale factor of enlargement =s . So BD =6 ÷ , or 82 2 2 3 6 × = 4. CE is6 ×= 9, so DE is 9 – 6 = 3 3 2 3 2 2 (iii) Area of triangle AC12 ( 2 {note area scale factor = k } = 27. So the trapezium has area 27 – 12 = 15 A B (b) The two pairs of marked angles are equal (alternate), and the top and bottom sides are X equal (parallelogram). So we have two congruent triangles ABX and DCX by D C ASA. (Note each would have to be rotated 180º about X to transform onto the other). So AX = XC and DX = XB, i.e. the diagonals bisect each other. 36 17. Transformations top a (i) translation by vector⎜ ⎟ shifts a to the right and b up. ⎝⎠ (ii) rotation about P throughθ . [Note e.g. +90º means 90º anticlockwise] perform a rotation using compasses, or if a multiple of 90º, use the L shape: To find the centre of a rotation already performed, perpendicularly bisect a line joining any point with its image. Repeat with
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