Edexcel Mathematics Higher Tier, May 2009 (1380/4H) (Paper 4, calculator)

Transcription

1 Link to examining board: You will be able to download this paper for free from the website. 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 a) First put the exchange rate into a table: see below Whatever we have done to 1 to get 325, we do the same to (325 1) x 1.68 = 546 b) We have a new exchange rate: see below 117 Whatever we have done to 1.50 to get 117, we do the same to 1. ( ) x 1 = Page 1

2 Question 2 a) The question has not told us to use any particular centre of enlargement so we can put the enlargement anywhere we like on the grid. b) The single transformation is a reflection along the mirror line x = 0 (we could also define the mirror line as being the y axis) Question 3 The first term is = = 2 Second term is = = 5 Third term is = = Page 2

3 Question a) two points plotted (65,100) and (80,110) b) positive correlation. As the height of the sheep increases so does the length of the sheep c) now draw the line of best fit. This is shown in red. The line should go through the points with as many points being above the line as below the line. See the dashed lines for a sheep of height 76 cm the length would be 109 cm Question 5 Each calculator costs = 7.56 So 31 calculators will cost 7.56 x 31 = Page 3

4 Question 6 a) Replacing C with -8 F = (1.8 x -8) + 32 = = 17.6 b) Replacing F with = 1.8C + 32 subtract 32 from both sides of the equation 36 = 1.8C now divide both sides by = C so C = 20 Question 7 When working out bearings you should always measure the angle in a clockwise direction Page 4

5 Question 8 a) 18:12 which simplifies to 3:2 b) oranges apples Total see below = 9 So 5 x 9 = 45 apples Question 9 x x x comment (20 x 2) = = 48 too low (20 x 2.5) = = too low (20 x 2.7) = = too high (20 x 2.6) = = too low 2.7 is too high and 2.6 is too low, therefore the answer is somewhere between the two. However the answer is only required to one decimal place so we must see which of 2.7 and 2.6 gives the closest answer to 71. This is 2.6 (which is away from 71 compared to 2.7 which gives an answer away). Answer is Page 5

6 Question 10 Question 11 Usually prime numbers are odd (3, 5, 7, 11, 13, 17 etc) but there is also one prime number which is even. This is 2. If we add 2 to any other prime number we would be adding an even to an odd and would get an odd number. For example = which is odd. Question 12 Time (t minutes) frequency midpoint Midpoint x frequency 0 t x 15 = 45 6 t x 25 = t x 20 = t x 12 = t x 8 = 216 total Estimated mean is = minutes Make sure that you divide by the total frequency (80) rather than the number of categories (5) this is a common mistake. The answer you get should be reasonable given the data Page 6

7 Question 13 Perimeter (circumference) of a full circle is πd or 2πr. In this case the diameter is 8 or the radius is 4. We have half a circle so we want πr. Curved length is π x 4 = cm We need to add to this the length of the diameter. Total perimeter = = 20.57cm Question 14 a) a 3 b) 5(3x 2) = 15x - 10 c) 3y(y + 4) = 3y y d) 2(x 4) + 3(x + 2) = 2x 8 + 3x + 6 = 5x 2 e) (x + 4)(x 3) = x 2 + 4x 3x 12 = x 2 + x - 12 Question 15 ( ) ( ) = = Question 16 a) t 6+2 = t 8 b) m 8-3 = m 5 c) (2x) 3 = 2 3 x x 3 = 8x 3 d) 3 x 4 x a 2+5 x h 1+4 = 12a 7 h Page 7

8 Question 17 The triangle is right angled so Pythagorus s theorem will apply a 2 + b 2 = c 2 we must have c as the longest side (hypotenuse) AB = 9 2 AB = 81 subtract 36 from both sides AB 2 = 45 square root both sides AB = 45 = 6.71 (3 sf) Question 18 a) the heaviest bag is 29kg (23 is where the upper quartile is and where the box ends) b) the median weight is 17kg (where the vertical line inside the box is) c) the upper quartile is 23, the lower quartile is 10. The interquartile range is = 13 d) 10kg is the lower quartile. This means that a quarter of the bags are less than 10kg. A quarter of 240 bags is 60 bags. 60 bags weigh less than 10kg. Question a) 4500 x (1.04) 2 = 4500 x = b) by trial and error we must solve 2400 x (1.075) n = when n = x (1.075) 3 = when n = x (1.075) 4 = when n = x (1.075) 5 = n = 5 years Page 8

9 Question 20 a) This is a right angled triangle and the question involves angles so we will use basic trigonometry (SOHCAHTOA) We have the adjacent (5) and the hypotenuse (8) so we will use cosine (cos = A/H) cos x = = = take the inverse cos (cos -1 ) of both sides x = cos = 51.3⁰ (1 dp) b) This is another basic trigonometry question. We have the angle and the adjacent (12.5) and need to find the opposite (y) so we will use tan (tan = O/A) tan 40 = =. multiply both sides by x tan 40 = y y = 10.5 cm (1 dp) Question 21 a) we want our sample size to be 50 so we need to scale everything down by dividing by 258 and multiplying by 50 ( ) Male Spanish is 26. So our sample needs 26 x = Now we can t have 5.04 boys so need so round this to 5 boys. b) in total we have 135 female students ( = 135) so our sample needs 135 x = females. Again we must round this figure to 26 females. Question 22 (3n + 1) 2 (3n 1) 2 = (3n + 1)(3n + 1) (3n 1)(3n 1) = (9n 2 + 3n + 3n + 1) (9n 2 3n 3n + 1) = (9n 2 + 6n + 1) (9n 2 6n + 1) = 9n 2 + 6n + 1 9n 2 + 6n 1 = 12n As 12 is a multiple of 4 so 12n will be a multiple of Page 9

10 Question 23 a) = + = -a + b b) = + = a + = a + (-a + b) = a - a + b = a + b = (2a + 3b) Question 24 Area of a triangle is given in formulae sheet and is absinc. This is an equilateral triangle so all angles are 60⁰. Area of triangle is x 6 x 6 x sin 60⁰ = cm2 Area of a sector of a circle is πr 2 x where θ is the angle of the sector. Area of this sector is π x 3 2 x = cm2 Shaded area = area of triangle area of sector = = = 10.9 cm 2 (3 sf) Page 10

11 Question 25 In all of these types of questions you can usually factorise the top and the bottom and then one of the factors will cancel. To factorise the top we need to find two numbers that multiply to give 15 and combine to give -8. These two numbers could be -3 and -5. Therefore the top factorises to give (x -3)(x -5). To factorise the bottom, it is a bit harder as we have a number greater than 1 in front of the x 2 term. Take the coefficient of the x 2 and the units figure (2 and -15) and multiply them together. This gives -30. Now we must find two numbers that multiply to give -30 but combine to give -7 (the coefficient of the x term). These two numbers could be +3 and -10. Rewrite the quadratic equation splitting the x term into these two components: 2x 2 7x 15 = 2x 2 + 3x 10x 15 factorise this into two pairs x(2x + 3) 5(2x + 3) if this is correct then it should be the same factor for both ((2x + 3) in this case) now factorise again (2x + 3)(x 5) So going back to our original expression we have and as expected one of the factors (x-5) will cancel on the top and the bottom leaving us with: Page 11

12 Question 26 4 orange 5 20 orange red yellow orange red red yellow orange red yellow 7 yellow The way we can have sweets not the same colour is orange, red orange, yellow red, orange red, yellow yellow, orange yellow, red We add up the probabilities of the above ( x ) + ( x ) + ( x ) + ( x ) + ( x ) + ( x ) = = = = Page 12

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