Mathematics IGCSE Higher Tier, November /3H (Paper 3H)

Transcription

1 Link to examining board: This question paper associated with this paper is not currently available to download for free from the Edexcel website. You can purchase your own copy of the question paper by phoning the Edexcel order line on or you may be able to get a copy from your school. The question paper should be available for free from the Edexcel website from about February or March 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 Number of Frequency (f) n x f children in the family (n) total Mean = = 2.64 (this seems a sensible answer given the data) A common mistake here is to divide by 5 as there were 5 categories. This would give a mean of 13.2 which should make you realise you have gone wrong. Question 2 a) i) 4c 12 ii) d 3 + 4d b) look for something that goes in to both terms. This would be x so put x on the outside of the brackets: x(3 2x) Page 1

2 Question 3 Triangle ABC is isosceles so angle BAC = angle ACB = 70⁰ PA is parallel to BC so angle PAC + angle ACB = 180⁰ (this uses a combination of corresponding angles and angles on a straight line add up to 180⁰) Angle PAC + 70 = 180 Angle PAC = 110⁰ x = angle PAC angle BAC x = = 40⁰ Question 4 a) area of a circle is πr 2 where r is the radius (in this case 8.9) area = π x = m 2 b) 250 m 2 (to 2 significant figures) Page 2

3 Question 5 a) when you divide fractions you change to a multiply and flip the second fraction over 4 is the same as x = = b) 3 = and 1 = in order to subtract (or add) fractions we need a common denominator (number on the bottom) the common denominator could be 15 = and = - = = 1 Alternative solution: in order to subtract (or add) fractions we need a common denominator (number on the bottom) the common denominator could be 15 3 = 3 and 1 = 1 3-1= 2-1= Page 3

4 Question 6 a) subtract 2x from both sides 5x + 3 = -4 subtract 3 from both sides 5x = -7 divide both sides by 5 x = - = -1 b) multiply both sides by y = 6 add 5y to both sides 16 = 5y + 6 swap sides 5y + 6 = 16 subtract 6 from both sides 5y = 10 divide both sides by 5 y = 2 Question 7 a) i) means intersection so A B means hats that are clothes of Mr Smith. Ie Mr Smith s hats ii) there are no members in A C because there are no clothes of Mr Smith that are also hats of Mrs Koshi No members b)i) means union B C = all hats and all hats belonging to Mrs Koshi. All hats belonging to Mrs Koshi are included in all hats anyway so this just equals B B C = B ii) Mr Smith s favourite shirt will be a member of ( A Mr Smith s favourite shirt A Page 4

5 Question 8 a) This is a right angled triangle involving angles so will be basic trigonometry (SOHCAHTOA) label everything from the point of view of the angle (opposite (opp), adjacent (adj) and hypotenuse (hyp)). We don t have hyp and don t need it so using SOHCAHTOA we see that we need TOA (tan) x cm opp hyp 36⁰ 9 cm adj tan 36⁰ = = = swap sides = tan 36⁰ multiply both sides by 9 x = 9 x tan 36⁰ = 6.54 (to 3 significant figures) b) this is a right angled triangle and has nothing to do with angles so will be Pythagorus s Theorem 10 cm y cm 4.5 cm y 2 = y 2 = 100 subtract from both sides y 2 = square root both sides y = 8.93 (to 3 significant figures) Page 5

6 Question 9 a) as there are an odd number of members (3) the middle number must be the same as the median the middle number will be 5 they have a mean of 4 which means that the three numbers must add up to 12 (12 3 = 4) we already have 5 as the middle number so the other two numbers must be either side of 5 and add up to 7. These must be 1 and 6 1, 5, 6 b) mode of 5 means that at least 2 of the numbers must be 5. If only two of the numbers are 5 then the other two numbers can t be the same. If three of the members were 5 then the median would be 5. Exactly four of the numbers can t be 5 because if they were then the median would also be 5. We can conclude that exactly two of the numbers must be 5 The median is 6 and as we have four numbers 6 must be the average of the 2 nd and 3 rd number. So the 3 rd number must be 7. We can have 5, 5, 7, 8 We could have anything we wanted for the 4 th number so long as it is bigger than 7. Question 10 a) we are going from the larger triangle to the smaller triangle so our scale factor will be (as opposed to ) the corresponding side to x is 14 cm (LN) x = 14 x = 10 cm b) we are going from the smaller triangle to the larger triangle so our scale factor will be (as opposed to ) the corresponding side to y is 18 cm (QR) y = 18 x = 25.2 cm Page 6

7 Question 11 a) to find the median we draw a horizontal line across from a cumulative frequency of 20 until it meets the curve then drop this down to meet the length (shown as a blue dashed line) the median is 15 cm b) to find the upper quartile we draw a horizontal line across from a cumulative frequency of 30 until it meets the curve then drop this down to meet the length (shown as a green dashed line) to find the lower quartile we draw a horizontal line across from a cumulative frequency of 10 until it meets the curve then drop this down to meet the length (shown as a green dashed line) upper quartile = 35 lower quartile = 6.5 interquartile range = = 28.5 c) we draw a vertical line up from 44 cm to meet the curve and then draw a horizontal line across to meet the cumulative frequency at 36 (shown as a red dashed line) there are 36 branches that are less than 44 cm = 4 There are 4 branches that are more than 44 cm Page 7

8 Question 12 We can solve by elimination or by substitution. Elimination method would be better here as we can t easily express x or y in terms of the other. Both methods are set out below. Elimination Label the equations eqn 1 and eqn 2 Multiply eqn 1 by 3 (so that we have the same number of xs in both equations) 6x 15y = 39 eqn 3 6x + 3y = 3 eqn 2 Start with eqn 2 and subtract eqn 3 (done this way around so that end up with a positive number of ys) 3y y = y = -36 divide both sides by 18 y = -2 now put back into eqn 1 2x (5 x -2) = 13 2x + 10 = 13 subtract 10 from both sides 2x = 3 divide both sides by 2 x = 1.5 x = 1.5, y = -2 Put both back into eqn 2 to see if it works: (6 x 1.5) + (3 x -2) = 9 6 = Page 8

9 Substitution Method Label the equations eqn 1 and eqn 2 Rearrange eqn 2 to make y the subject 3y = 3 6x divide by 3 y = 1 2x substitute this value for y into eqn 1 2x 5(1 2x) = 13 expand the brackets 2x x = 13 group terms 12x 5 = 13 add 5 to both sides 12x = 18 divide both sides by 12 x = 1.5 now put back into the rearranged equation for y y = 1 2x to give y = 1 (2 x 1.5) = 1 3 = -2 x = 1.5, y = -2 Put both back into eqn 1 to see if it works: (2 x 1.5) - (5 x -2) = 3-10 = 13 Question 13 a) we need to find two numbers that multiply to give 15 and add to give -8 These two numbers are -5 and -3 (x 5)(x 3) b) we can recognise this as the difference of two squares x 2 is a square number and 49 is a square number (x 7)(x + 7) Alternatively: Find two numbers that multiply to make -49 and add to give 0 (as there are no x terms) These two numbers are -7 and + 7 (x 7)(x + 7) Page 9

10 Question 14 a) draw the line y = 2 (shown as a red line) the solutions are where this line meets our curve (see dark blue dashed lines) x = 0.25 or 3.75 b) now draw the line y = x + 1 (shown as the green line) the solutions are where this line meets our curve (see purple dashed lines) x = 0.45 or Page 10

11 Question 15 We need to work out the volume in two parts and then add them together. Volume of cylinder Volume of a cylinder is πr 2 h (as given in formulae sheet) r = 1.5, h = 4 volume = π x x 4 = cm 3 Volume of hemisphere This is half the volume of a sphere (as given in formulae sheet) x πr3 r is 1.5 volume = x π x 1.53 = cm 3 total volume = = = 35.3 cm 3 (to 3 significant figures) Question 16 a) = 3x2 + 6x 24 b) at any turning point the gradient will be zero. So we set = 0 and solve to find the x coordinates. 3x 2 + 6x 24 = 0 divide by 3 x 2 + 2x 8 = 0 factorise (x + 4)(x 2) = 0 If two brackets multiply to give 0 then either of the brackets could be equal to 0 x + 4 = 0 or x 2 = 0 If x + 4 = 0 then x = -4 If x 2 = 0 then x = 2 We now need to find the corresponding y values to go with these x values We know y = x 3 + 3x 2 24x So when x = -4 we have y = (-4) 3 + (3 x (-4) 2 ) (24 x -4) = = 80 When x = 2 we have y = (3 x 2 2 ) (24 x 2) = = -28 (-4, 80) and (2, -28) Page 11

12 Question 17 a) the only way that Hari can have a total of 3 is if he rolls a 1 each time the probability of throwing a 1 (p(1)) is the probability of throwing three 1s is x x = b)there are 3 ways we can do this: (1, not 1, not 1) or (not 1, 1, not 1) or (not 1, not 1, 1) Each of these has the probability of x x = (but just put together in a different order) Probability of exactly one 1 is 3 x = Question 18 multiply both sides by x Px = 100(y x) expand the brackets Px = 100y 100x add 100x to both sides Px + 100x = 100y factorise x(p + 100) = 100y divide both sides by P x = Page 12

13 Question 19 Area of a triangle is ½absin C where C is the angle that is between the two sides a and b (given in formulae sheet) First we need to work out what this angle is. We can use Sine Rule to find the angle at A (given in formulae sheet) Start with the Sine Rule the following way up as it will save a step later: = ⁰ = multiply both sides by 5 ⁰ sin A = = take inverse sin (sin -1 ) of both sides A = ⁰ Angles in a triangle add up to 180⁰ Angle B = 180 ( ) = ⁰ Now we can apply ½absin C where C is ⁰, a is 6 and b is 5 (or other way around) Area = ½ x 5 x 6 x sin ⁰ = = 14.3 (to 3 significant figures) Question 20 a) = = 2-4 b) 2 3 = 8 so taking the cube root of both sides 2 = 8= 8 1/3 c) rationalise the denominator means to get rid of the from the bottom multiply the top and the bottom by = Factorise a out of the top The a on the top and the bottom will cancel out = Page 13

14 Question 21 a) let y = 2x + 1 swap the x and y around x = 2y + 1 now rearrange to make y the subject 2y + 1 = x subtract 1 from both sides 2y = x 1 divide both sides by 2 y = f -1 (x) = b) hg(x) = h(2 + x) = (2 + x) 2 h(x) = x 2 (2 + x) 2 = x 2 (2 + x)(2 + x) = x x + 2x + x 2 = x 2 subtract x 2 from both sides 4 + 4x = 0 subtract 4 from both sides 4x = -4 divide both sides by 4 x = -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 14