Mathematics IGCSE Higher Tier, June 2010 …

1 Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 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)1623 467467 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.

2 Question 1 Apple Fool For 6 people For 1 person note 1 For 15 people note 2 For 5 people note 3 Cooking apples 900 g 150 g 2250 g 750 g Sugar 100 g g 250 g Double cream 300 ml 50 ml 750 ml Note 1: we have the ingredients for 6 people so divide each of these figures by 6 Note 2: we now have the ingredients for 1 person so multiply each of these figures by 15 Note 3: from the ingredients for 1 person, multiply by 5 a) for 15 people we need: 2250 g of cooking apples 250 g of sugar 750 ml of double cream b) for 5 people we need: 750 g of cooking apples Question 2 a) i) for this we need to ignore the line PR altogether x = 62 ii) alternate angles are equal b)i) here we need to ignore the line PQ altogether y = 71 ii) corresponding angles are equal Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 2 Question 3 a) the three numbers have a median of 4 and as there are only three numbers then the middle number must equal 4 if we put a + 2, b + 2 and c + 2 in order of size, they will have the same order as a, b and c when in size order the median will be the middle number.

3 This is 2 more than the previous median. Median = 4 + 2 = 6 b) the range of the numbers will be the largest take away the smallest this will be the same as the original range (7) as we have simply added 2 to the highest and to the lowest. The range is still 7. Question 4 a) 5(n + 6) = 5n + 30 b) y6 c) 4(x 2) = 3 4x 8 = 3 add 8 to both sides 4x = 11 divide both sides by 4 x = Question 5 a) we want to find of = x = x = x = b) if are women then this means that the total number must be a multiple of 12 (otherwise would give us a part of a person!) if wear glasses then this means that the total number must be a multiple of 8 (otherwise would give us a part of a person!) so our total number must be a multiple of both 12 and 8.

4 We want the least common multiple. Multiple of 12 are 12, 24, 36, .. Multiples of 8 are 8, 16, 24, 32 The least common multiple is 24 Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 3 Question 6 a) the modal class is the class that has the highest frequency. This is 400 V 500. b) Volume of water (V m3) Frequency Mid point Midpoint x frequency 0 V 100 2 50 100 100 V 200 4 150 600 200 V 300 6 250 1500 300 V 400 18 350 6300 400 V 500 44 450 19800 500 V 600 6 550 3300 Total 80 31600 Mean = 31600 80 = 395 c) Volume of water (V m3) Cumulative Frequency 0 V 100 2 0 V 200 6 0 V 300 12 0 V 400 30 0 V 500 74 0 V 600 80 To get the cumulative frequency you just keep adding the next category to the previous Eg cum freq for 2nd category is 2 + 4 = 6, cum freq for 3rd category is 6 + 6 = 12 The final cumulative frequency should be the same as the total frequency it is d) e) There are 80 families so the median will be the 40th family.

5 Draw a line across from 40 til it meets the curve then drop down to meet the x axis. (shown as a red dashed line). The median is 425 m3 Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 4 Question 7 We have a right angled triangle so we can use SOHCAHTOA Label all the sides from the point of view of the angle We have hyp (H) and we are trying to find Adj (A). We don t have Opp (O) and don t want it. SOHCAHTOA this leaves CAH (cosine) cos 41 = = . multiply both sides by 41 = x x = cos 41 = cm (3 significant figures) Question 8 It is really important that you read this question carefully. The $1786 represents her income after tax has been deducted.

6 Therefore $1786 is equivalent to 76%. We want to work out what income is equivalent to 100%. If we divide $1786 by 76 then we will know what 1% is worth. Then we scale this back up by multiplying by 100 to get what 100% is worth. 1786 76 x 100 = $2350 cm x cm 41 opp adj hyp Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 5 Question 9 a) this is a reflection in the mirror line y = -x b) c) this is a reflection in the mirror line y = 1 Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 6 Question 10 a) -4 x 3 (as there is a full circle above the -4 we use , and as there is an open circle about the 3 we use ) b) i) subtract 9 from both sides 2x -8 divide both sides by 2 x -4 ii) integer means whole number n could be -3, -2, or -1 Question 11 a)

7 Area of a circle is (given in formulae sheet) r2 if diameter is 16 then radius will be half of this. Radius (r) = 8 cm area = x 82 = 64 = 201 cm2 (3 significant figures) b) substituting the values for h, x and y we have V = x x x ((3x ( )) + ( 3 x 82) + ) x x x ( + 192 + ) x x x = = 5050 cm3 (3 significant figures) Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 7 Question 12 a) x -3 -2 -1 0 1 2 3 4 y 11 18 note 1 13 note 2 2 note 3 -9 note 4 -14 note 5 -7 18 Note 1: (-2)3 (12 x -2) + 2 = -8 - -24 + 2 = -8 + 24 + 2 = 18 Note 2: (-1)3 (12 x -1) + 2 = -1 - - 12 + 2 = -1 + 12 + 2 = 13 Note 3: (0)3 (12 x 0) + 2 = 0 0 + 2 = 2 Note 4: (1)3 (12 x 1) + 2 = 1 12 + 2 = -9 Note 5.

8 (2)3 (12 x 2) + 2 = 8 24 + 2 = -14 b) c) i) = 3x2 12x ii) to find the gradient at a particular point, we first differentiate to get , then substitute our particular value for x into the equation for gradient = (3 x 52) (12 x 5) = 75 60 = 15 Question 13 angle PQS = angle PRS = 36 because angles in the same segment are equal angle PQR = 90 because angles in semicircle make 90 (we know PQR is a semicircle since PR goes through the centre C it is a diameter of the circle) angle RQS = angle PQR angle PQS = 90 36 = 54 Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 8 Question 14 a) i) AB is 20 cm, correct to 1 significant figures.

9 It could be as low as 15 or as high as 25 (and still get rounded to 20) lower bound is 15 cm ii) BC is cm, correct to 2 significant figures. It could be as low as or as high as (and still get rounded to ) lower bound is cm b) area of a triangle is x base x height for this to be as low as possible we would want the base to be as small as possible and the height to be as small as possible. lower bound for area = x 15 x = cm2 c) tan x = for this to be as small as possible we need the top (opp) to be as small as possible but the bottom (adj) to be as big as possible the smallest the opp could be is , the largest the adj could be is 25 lower bound for tan x = 25 = Question 15 a) E 1/r2 can be written as E = k/r2 where k is a constant to be determined substituting in the values we have for E and r 4 = k 502 4 = k 2500 multiply both sides by 2500 10000 = k k = 10,000 E = 10,000/r2 b) substitute r = 20 E = 10,000 202 = 10,000 400 = 25 c)

10 Substitute E =1600 1600 = 10,000/r2 multiply both sides by r2 1600 x r2 = 10,000 divide both sides by 1600 r2 = 10,000 1600 = square root both sides r = = Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 9 Question 16 (3 - 5)(3 - 5) = 9 + 25 - 3 5 - 3 5 9 + 5 - 6 5 = 14 - 6 5 Question 17 The two prisms are mathematically similar which means that all the dimensions are in the same proportion as each other. If we double one length, then all the other lengths are also doubled. We are trying to find the surface area of the larger prism so we want a scale factor with the larger number on top. We have the two lengths in common so our linear scale factor is = we are trying to find an area though which is two dimensional so our area scale factor is the square of the linear scale factor area scale factor = total surface area of Q = 544 x = 1224 cm2 Question 18 factorise the top and the bottom x2 + 6x = x(x + 6) x2 36 = (x + 6)(x 6) difference of the square so we have = Mathematics IGCSE Higher tier , june 2010 4400/3H (Paper 3H) 0775 950 1629 Page 10 Question 19 a) probability of two 20 cent coins is x = x = b) of the nine outcomes on the right the 2nd, 3rd and 6th outcome all satisfy this.