Mathematics IGCSE Higher Tier, November /4H (Paper 4H)

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1 Link to examining board: The question paper associated with these solutions is available to download for free from the Edexcel website. The navigation around the website sometimes changes. However one possible route is to follow the above link, then SUBJECTS Mathematics, QUALIFICATIONS (from the LH Panel), under INTERNATIONAL GCSE FROM 2003 choose MATHEMATICS. Otherwise you can order the paper from the Edexcel Publications by phoning them on +44 (0) These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or an organisation and would like to purchase these solutions please contact Chatterton Tuition for further details. Question 1 i) PU is the same as QR + ST PU = b + 3a = 3a + b ii) PQ is the same as UT RS PQ = (3a + 2b) (a b) = 3a + 2b a + b = 2a + 3b Question 2 a) Philip Nikos Total note 24 Note: whatever we did to Nikos to get from 4 to 24 we do the same to Philip 24 4 = 6 6 x 3 = 18 b) James Suki Total note 40 Note: whatever we did to the total to get from 8 to 40 we do the same to Suki 40 8 = 4 5 x 5 = Page 1

2 Question 3 a) area can be split into area of rectangle and area of triangle area rectangle 2 x 1.5 = 3 m 2 area triangle ½ x base x height ½ x 1.5 x 1.2 = 0.9 m 2 total area = = 3.9 m 2 b) the fraction we need of one litre is. = litres there are 1000 millilitres to one litre x 1000 = 195 ml Question 4 speed = so distance = speed x time and time = time = = 2.75 hours = 2 hours and 45 minutes (note 0.75 hours is not 75 minutes, it is 0.75 x 60 = 45 minutes) Question 5 a) probabilities always add up to 1 1 ( ) = = 0.7 b) 20 x 0.2 = Page 2

3 Question 6 a) 5x 10 b) subtract 3 from both sides = 7 multiply both sides by 4 x = 28 c) add 6 to both sides 5x 8 divide both sides by 5 x 1.6 Question 7 this is a right angled triangle so we can use Pythagoras Theorem h 2 = h 2 = = 52 square root both sides h = 7.21 cm (3 significant figures) Question 8 i) part A: this is a straight line which shows that as time increases the distance from home increases. This is a straight line so the speed is constant over this time. John is travelling at a steady speed ii) part B: this is flat showing that as time increases John doesn t get any further away. John is not moving iii) part C: the slope starts off fairly flattish but then gets steeper and steeper John s speed is increasing Page 3

4 Question 9 a) i) the members of A are the multiples of 3 that are also positive integers less than 11 positive integers less than 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 multiples of 3 that are also integers less than 11 are 3,6, 9 ii) A B means everything that is in A and is in B members of B are 2,4,6,8,10 A B = 2,3,4,6,8,9,10 b) i) students in class 12Y who study both mathematics and history ii) Students in class 12Y who do not study mathematics Question = 2 x 2 x 3 x Page 4

5 Question 11 a) First throw Second throw 2 3 Heads 2 3 Heads 1 3 Tails 2 3 Heads 1 3 Tails 1 3 Tails b) probability of heads and heads = x = c) this is the same as the probability that it doesn t show tails and tails 1 ( x ) = 1 - = alternatively we could have added together the probabilities that it was head and head, head and tails or tails and head ( x ) + ( x ) + ( x ) = + + = Page 5

6 Question 12 a) i) m will be the powers added together = 11 m = 11 ii) c is a x b c = ab b) 3.2 x 4.5 = p x 10 q = 10 p + q so we have 14.4 x 10 p + q but this is not standard form as 14.4 is 10 we need to make 14.4 smaller so to compensate we need to increase the power by x 10 p + q + 1 Question 13 a) we have to give our answers to 3 significant figures. This tells us that the equation won t factorise so we have to use the quadratic formula: 4 2 where a, b and c are the coefficients of the x 2 term, the x term and the units a = 1 b = 2 c = -1 substituting these values into the equation we have x = or = b) multiply both sides by (y + 4) 2 = 3(y + 4) expand the brackets 2 = 3y y + 12 = 2 subtract 12 from both sides 3y = -10 divide both sides by 3 y = = Page 6

7 Question 14 a) this is a right angled triangle so we can use basic trigonometry label the sides from the point of view of the angle h cm opp hyp 32⁰ 6 cm adj We don t have opp and don t want it. From SOHCAHTOA we can see that we need CAH as it uses adj and hyp. cos 32⁰ = multiply both sides by h hcos 32⁰ = 6 divide both sides by cos 32⁰ h = ⁰ = 7.08 cm (3 significant figures) b) area of triangle = ½absinC from formulae sheet where a and b are the sides that surround the angle C here we don t have the angle that we want but we can calculate it because angles in a triangle add up to 180⁰ (and we have the other two) C 3 cm 7 cm b a A 40⁰ c 40⁰ 32⁰ 25⁰ B angle C = 180 ( ) = = 115⁰ a = 3, b = 7 area = ½ x 3 x 7 x sin 115⁰ = 9.52 cm 2 (3 significant figures) Page 7

8 Question 15 we can solve simultaneous equations by using the elimination method or the substitution method elimination method Multiply one or both equations by a number (never the same number for both) so that we have the same number for x or the same number for y Here I will multiply the second equation by 5 so that I can eliminate the x terms 5x + 4y = 3 (unchanged) 5x 10y = 10 (everything multiplied by 5) we have 5x in both equations subtract the second equation from the first equation to make the x terms disappear 4y y = y = -7 divide both sides by 14 y = -½ substitute this value for y back into the first equation 5x + (4 x -½) = 3 5x 2 = 3 add 2 to both sides 5x = 5 divide both sides by 5 x = 1 we have x = 1 and y = -½ to check substitute these values back into the second equation 1 (2 x -½) = = = Page 8

9 substitution method rearrange the second equation to make x the subject x = 2y + 2 now substitute this value for x into the first equation (5 x (2y + 2)) + 4y = 3 10y y = 3 14y + 10 = 3 subtract 10 from both sides 14y = -7 divide both sides by 14 y = -½ substitute this value for y back into the rearranged second equation x = 2y + 2 x = (2 x -½) + 2 = x = 1 we have x = 1 and y = -½ to check substitute these values back into the first equation (5 x 1) + (4 x -½) = 5-2 = 3 Question 16 In histograms the area of the columns is directly proportional to the frequency for that category We are told that there are 120 stones with masses less than 30 g and there are 6 squares on the histogram representing this 6 squares = 120 stones 1 square = 20 stones We want the number of stones with masses between 35 g and 70 g, this is represented by (½ x 6) = 7 squares If 1 square = 20 stones 7 squares = 20 x 7 = 140 stones 140 stones Page 9

10 Question 17 a) this is a harder quadratic to factorise as there is a number in front of the x 2 term multiply the coefficient of x 2 by the unit term (2 x 3) to get 6 we must find two numbers that multiply to give 6 but add to give 5, this will be 2 and 3 rewrite the expression splitting the 5x into 2x and 3x 2x 2 + 2x + 3x + 3 factorise in pairs 2x(x + 1) + 3(x + 1) we should have the same thing in both brackets, which we do now factorise again (as (x + 1) goes into both terms) (x + 1)(2x + 3) b) this can be done in the same way as above by treating the coefficient of y as 0 but it is better to recognise this as the difference of two squares the square root of 4y 2 is 2y the square root of 9 is 3 so we have (2y 3)(2y + 3) Alternatively multiply the coefficient of y 2 by the unit term (4 x -9) to get -36 we must find two numbers that multiply to give -36 but add to give 0, this will be -6 and +6 rewrite the expression splitting the 0y into -6y and +6y 4y 2 6y + 6y 9 factorise in pairs 2y(2y 3) + 3(2y 3) we should have the same thing in both brackets, which we do now factorise again (as (2y 3) goes into both terms) (2y 3)(2y + 3) Question 18 a) when we raise powers to further powers we multiply the powers together (9 ½ ) 4 = 9 2 = 81 b) 5 20 = (25 ½ ) 20 = c) 8 = 8 ½ = (2 3 ) ½ = 2 3 x ½ = Page 10

11 Question 19 a) i) = + = x + 2 = x + 2y ii) = + + = -y + x + 2y = x + y b) = = x + y so lines AE and DC are both parallel to each other and the same length as each other therefore AD and EC must also be parallel to each other. AECD forms a parallelogram Question 20 a) to differentiate you multiply by the power and then drop the power by 1 i) = 6x 1 ii) first put as a negative power = x-1 = -1x-2 = -x -2 b) we first differentiate with respect to x, then set this equal to 12 and solve to find x once we have x we can put this back into the original equation to get the y co-ordinate = 3x2 = 12 divide both sides by 3 x 2 = 4 square root both sides x = +2 or -2 when x = +2 y = 2 3 = 8 when x = -2 y = (-2) 3 = -8 the two co-ordinates are (2, 8) and (-2, -8) Page 11

12 Question 21 a)f(2) = = b) we must exclude x = -3 from the domain this is because we can t have the denominator equal to 0 c) f(a) = = a + 3 = 10 subtract 3 from both sides a = 7 d) we replace the x in g(x) with f(x) gf(x) = f(x) + 2 = + 2 put 2 as so that we have a common denominator of x + 3 gf(x) = + = = = Page 12

13 Question 22 a) i) label all the sides from the point of view of the angle x A 1 hyp x B opp M SOHCAHTOA we have hyp and we want to find opp so we need sine (soh) sin x = = BM = sin x ii) M is the midpoint of BC so BC = 2 x BM BC = 2sin x b) using the whole triangle ABC the cosine rule is given in the formulae sheet a 2 = b 2 + c 2 2bc cos A BC 2 = AB 2 + AC 2 (2 x AB x AC x cos (2x)) BC 2 = cos (2x) BC 2 = 2 2cos(2x) c) from ii) we have BC = 2sin x and from b) we have BC 2 = 2 2cos(2x) so (2sin x) 2 = 2 2cos(2x) 4(sin x) 2 = 2 2cos(2x) divide both sides by 2 2(sin x) 2 = 1 cos(2x) add cos(2x) to both sides cos(2x) + 2(sinx) 2 = 1 subtract 2(sin x) 2 from both sides cos (2x) = 1 2(sin x) Page 13

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