Mathematics IGCSE Higher Tier, June /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 When you divide fractions you change to multiply and flip the second fraction over x = x = = 1 Question 2 i) Angelou put x 5 sweets in the bag Nina has 3 x (x 5) sweets and this equals 39 3(x 5) = 39 expand the brackets 3x 15 = 39 ii) add 15 to both sides 3x = 54 divide both sides by 3 x = 18 sweets Question 3 substitute the values into the expression = = = 3 x -1 = Page 1

2 Question 4 The median is the middle number when you line the data up in order (67 + 1) 2 = 34 the median here will be the 34 th piece of data By the end of shoe size 6 we have seen 20 people, by the end of shoe size 7 we have seen another 19 people (making 39 in total). The median falls in this shoe size. Median is size 7 Question 5 a) circumference of a circle is 2πr where r is the radius (given in formulae sheet) 2 x π x 40 = 251 m (3 significant figures) b)shaded area = area of rectangle area of circle area of rectangle = 10 x 8 = 80 cm 2 area of circle = πr 2 (given in formulae sheet) = π x 3 2 = cm 2 shaded area = = = 51.7 cm 2 (3 significant figures) Question 6 a) probabilities must add to = = 0.2 probability of 4 is 0.2 b) probability of an odd number = probability of 1 or 3 = = Page 2

3 Question 7 a) use tracing paper and put your pen in the centre o. Turn the tracing paper anticlockwise by 90⁰. Rotated figure shown in red. b) vector translation of 4 units to the left and 5 units up Page 3

4 Question 8 a) we have a right angled triangle so can use Pythagorus Theorem (given in formulae sheet) where c is the longest side x 2 = = = square root both sides x = 6.02 cm (3 significant figures) b) label the sides from the point of view of the angle A opp 32⁰ B adj 6.5 C we have adjacent (adj) and want to get opposite (opp) so do not want or have hypotenuse (hyp) SOHCAHTOA we use TOA (tan) tan 32⁰ = =. multiply both sides by 6.5 AB = 6.5 x tan 32⁰ = 4.06 cm (3 significant figures) Question 9 multiply both sides by x = 21 add x to both sides 12 = x + 21 rewrite x + 21 = 12 subtract 21 from both sides x = Page 4

5 Question = 2 x 2 x 3 x 11 = 2 2 x 3 x 11 Question 11 we should round 84.2 to 1 significant figure to 80 we should round 41.6 to 1 significant figure to 40 we should round 38.2 to 36 (as this is the closest square number and we are square rooting it) so we have = = = 12 Question 12 a) gradient is given by difference in y coordinates divided by the difference in the x coordinates = b)any line can be put in the form y = mx + c where m is the gradient and c is the y intercept the y intercept (c ) is where the line crosses the y axis. This happens when y = 2. y = ½x + 2 c) any line that is parallel to L will have the same gradient as L (ie ½). The intercept can be anything you like (apart from 2). y = ½x Page 5

6 Question 13 Two or more figures which are exactly the same shape but not necessarily the same size or the same way round are said to be mathematically similar a) the angles will be the same in both shapes x = 60⁰ b) find two sides that correspond to each other, this would be AB and PQ (5 and 4). y is on the smaller shape so we want the smaller number on the top. our scale factor will be y = x 7.5 = 6 cm c) find two sides that correspond to each other, this would be AB and PQ (5 and 4). z is on the bigger shape so we want the bigger number on the top. our scale factor will be z = x 3 = 3.75 cm Page 6

7 Question 14 a) First throw Second throw 3 4 Heads 3 4 Heads 1 4 Tails 3 4 Heads 1 4 Tails 1 4 Tails Probability of tails and tails = x = Question 15 a) 3c 5+2 d 1+4 = 3c 7 d 5 b)2 4 x 3x4 y4 = 16x 12 y 4 c) we need to factorise the top and the bottom and then something will cancel 2x 6 = 2(x 3) x 2 3x = x(x 3) = = Page 7

8 Question 16 a) we must first multiply 2 and -3 together to give -6 then we must find two numbers that multiply together to give -6 but add together to give -1 these two numbers are 2 and -3 rewrite the expression splitting the x term 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 factorise again bringing the bracket to the outside (x + 1)(2x -3) b) (x + 1)(2x -3) = 0 If two things multiply to give 0 then one of them must be 0 Either (x + 1) =0 or (2x -3) = 0 x + 1 = 0, x = -1 or 2x 3 = 0, 2x = 3, x = 1.5 x = -1 or 1.5 Question 17 a) = 2x + 3 b) put x = -4 into our expression for (2 x -4) + 3 = = -5 gradient is -5 c) at the minimum point the gradient will be equal to 0 so set = 0 and solve 2x + 3 = 0 subtract 3 from both sides 2x = -3 divide both sides by 2 x = -1.5 now substitute back into y = x 2 + 3x to get the y coordinate y = (-1.5) 2 + (3 x -1.5) = = minimum point is (-1.5, -2.25) Page 8

9 Question 18 a) MN is parallel to AB and is the same length so = = -x Alternatively: = + + = ½ ½ ½y x - ½y = -x b) = + = x + y c) = + = - - = -(y- ½x) (x + y) expanding the brackets -y + ½x x y grouping terms - ½x 2y Question 19 a) we know the number of leaves in the class width 3 w 4 is 15. There are 150 little squares in this class on the histogram so 150 squares = 15 leaves and 10 squares = 1 leaf 150 little squares = 15 leaves 10 little squares = 1 leaf In the class 0 w 2 there are 80 little squares = 8 8 x 1 = 8 leaves b) in the interval 4.5 w 5.5 there are = 90 little squares = 9 9 x 1 = 9 leaves Page 9

10 Question 20 a) as the triangle is equilateral all the angles will be 60⁰ if we just look at triangle ABM we have a right angled triangle with an angle of 60⁰ at B A hyp 2m opp 60⁰ B 1 m adj M i) SOHCAHTOA cos 60⁰ = = ii) sin 60⁰ = = we can calculate AM by using Pythagorus theorem (given in formulae sheet) where c is the longest side BM 2 + AM 2 = AB AM 2 = AM 2 = 4 subtract 1 from both sides AM 2 = 3 square root both sides AM = 3 Now put back into sin 60⁰ = = b) cos 60⁰ =, sin 60⁰ = so substituting these values in to the equation we have ( )2 + ( )2 = + = = 1 (cos 60⁰) 2 + (sin 60⁰) 2 = Page 10

11 Question 21 a) we have to give our answer to 3 significant figures so very unlikely to factorise we should use the quadratic formula (given in formulae sheet) where a = 2, b = 3 and c = -1 x = or (to 3 significant figures) b) first multiply by x 2 - = x now multiply by (x + 1) 2(x + 1) x = x(x + 1) expand the brackets 2x + 2 x = x 2 + x group terms x + 2 = x 2 + x subtract x from both sides 2 = x 2 x 2 = 2 square root both sides x = Page 11

12 Question 22 a) we must first adjust the x term or the y term so that we have the same power of 10 for each adjusting the y term y x 10 4 = y 10 x y 5 = 0.1y x 10 5 now we can add the two expressions (x x 10 5 ) + (0.1y x 10 5 ) = (x + 0.1y) x 10 5 z = x + 0.1y b) i) (3 x 10 n ) (4 x 10 m ) = x 10n-m = 0.75 x 10 n-m but this is not standard form as the number needs to be between 1 and 9.99 so we have 7.5 x 10 n-m-1 a = 7.5 ii) p = n m - 1 If you found this paper helpful then visit where you will find plenty more. It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by Edexcel. In addition these solutions may not necessarily constitute the only possible solutions Page 12