Edexcel Mathematics Higher Tier, June 2011 (1380/3H) (Paper 3, non-calculator)

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1 Link to examining board: This question paper is not currently available to download for free from the Edexcel website. You can purchase your own copy 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 August 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 We have the ingredients for 10 flapjacks but we want the ingredients for 15. We should halve all the ingredients to get what we need for 5 flapjacks and then scale this up by multiplying by 3 to get 15 flapjacks. Ingredients For 10 flapjacks For 5 flapjacks For 15 flapjacks Rolled Oats 80 g 40 g 120 g Butter 60 g 30 g 90 g Golden Syrup 30 ml 15 ml 45 ml Light Brown Sugar 36 g 18 g 54 g 120 g rolled oats 90 g butter 45 ml golden syrup 54 g light brown sugar Page 1

2 Question 2 a) this shows a positive correlation (that as the number of pages increases so the time taken to read it also increases) b)draw a line of best fit. Then draw a vertical line up from 150 pages on the x axis to meet the line of best fit. Then take this line horizontally to meet the y axis at 7.6 hours Estimated time 7.6 hours Question 3 i) x will be 55⁰ ii) x corresponds with the angle SRB. They form corresponding angles so will be equal Page 2

3 Question 4 First round each number to 1 significant figure 7.19 to to to 0.5 So we have. =. Multiply the top and bottom by 2 and we have = 280 Question 5 a) i) h = (5 x (-2) 2 ) + 2 h = (5 x 4) + 2 h = h = 22 ii) 47 = 5t t = 47 subtract 2 from both sides 5t 2 = 45 divide both sides by 5 t 2 = 9 square root both sides t = 3 b) n is an integer means that n is a whole number n could be -1, 0, 1, 2, or 3 n cannot equal 4 as we have rather than Question 6 The exterior angles of a polygon always add up to 360⁰. If we have a regular polygon then all the exterior angles will be the same = 12 Our regular polygon has 12 sides Page 3

4 Question 7 a) b) this is a vector translation as the shape P is simply picked and moved so many places horizontally and so many places vertically. It doesn t turn or change size. It has moved 6 places to the left (-6) and 1 place down (-1) Vector translation Page 4

5 Question 8 Make sure that questions are not ambiguous Make sure questions are not biased or leading Give some kind of time frame Give a response section making sure there are no gaps and no overlaps How many hours do you spend exercising on average each week? Please put a tick in one box only. None More than 0 but less than 1 hour 1 hour or more but less than 2 hours 2 hours or more but less than 3 hours 3 hours or more but less than 4 hours 4 hours or more but less than 5 hours 5 hours or more b) if Sophie only asks the people at her local swimming pool then she already knows that these people exercise so this would be a biased sample. She needs a random sample of people from different types of places. Question 9 a) substitute n = 1 and n = 2 into the expression 3n + 1 when n = 1, (3 x 1) + 1 = = 4 when n = 2, (3 x 2) + 1 = = 7 4,7 b) take the difference between each term to get 4 we know the sequence is something to do with the 4 times table write out the 4 times table see what we have to do to the 4 times table to get to our sequence 4n - 3 Sequence times table Page 5

6 Question 10 Area of cross section Divide into two rectangles by adding an internal vertical line Area of LHS rectangle is 2 x 7 = 14 cm 2 Area of RHS rectangle is 2 x (7 2) = 2 x 5 = 10 cm 2 Cross section area = = 24 cm 2 Volume of prism Volume of any prism is area of cross section x length of prism 2 m is the same as 200 cm Volume = 24 x 200 = 4800 cm 3 Density = Multiply by volume Mass = density x volume Mass = 8 x 4800 = 38,400 g Question 11 Peter has the smallest share. Let Peter s share be P. Tarish gets three times as much as Peter so Tarish gets 3P. Ben gets twice as much as Tarish so Ben gets 2 x 3P = 6P P + 3P + 6P = 54 10P = 54 Divide both sides by 10 P = 5.40 Ben gets 6P so Ben gets 6 x 5.40 = Page 6

7 Question 12 a) i) when multiplying with powers add the powers w = w 10 ii) when dividing with power subtract the powers h 8 3 = h 5 b) we can cancel the 12 with the 3 to get 4 on top (12 3 = 4) we can cancel the x to get x on the bottom (x x 2 = ) we can cancel the y 3 with each other (y 3 y 3 = 1) we are left with Question 13 We need to find the mid point of the weight. Then plot this point with the corresponding frequency. Join each point with a straight line Page 7

8 Question 14 Open your compass so that it is the exact same length as the line. You can do this by putting the point of the compass in at one end and then opening it until your pencil tip touches the other end. Keeping the compass at this setting, put the point in at both ends of the line and draw an arc above the line from each end. You should end up with two arcs (as shown below on the LHS). Then from this new point (where the arcs cross) draw a new arc. Also draw a new arc with your compass point in the end of our initial straight line (other end to the P). Where these two arcs meet (as shown below on RHS) join this point to the point P and you will have a perfect 30⁰ angle. Question 15 a) x(x + 2) = x 2 + 2x b) (x + 3)(x 4) = x x 4x group terms x 2 - x 12 c) 2y goes into both terms so bring that outside 2y(y 2) d) you need to be able to recognise this as the difference of two squares (x 3)(x + 3) Page 8

9 Question 16 a) to divide fractions you flip the second one over and multiply instead x = x Cancel down the 6 with the 3 to get 2 on top x = x = b) To be able to subtract fractions you must have a common denominator (number on the bottom) 15 is a multiple of both 3 and 5 so 15 can be our common denominator 2 = 2 1 = 1 So we have 2-1 but we still have a problem here in that we can t take 6 away from 5 so we need to convert one of the whole numbers to a fraction. One whole number is worth 2 = 1 + = 1 Now we can subtract the fractions 1-1 = Question 17 Angle APC = angle BPC = 90⁰ (since CP is perpendicular (at right angles) to AB) Angle PAC and angle ACP both add up to 90⁰ (since angles in a triangle add up to 180⁰ and we already know that the other angle is 90⁰). So we can say that angle ACP = 90 angle PAC Angle ACB is a right angle (90⁰) so angle BCP = 90 angle ACP Angle BCP = 90 (90 angle PAC) = angle PAC = angle PAC Angle BCP = angle PAC Now we have shown that two of the angles are the same, that means that the other angle also has to be the same. Angle ACP = angle CBP Page 9

Question 18 a) the modal class interval is the one with the highest frequency modal class interval is 90 m 100 b) Time (m minutes) Cumulative Frequency 70 m 80 4 70 m 90 16 70 m 100 50 70 m 110 82 70

10 Question 18 a) the modal class interval is the one with the highest frequency modal class interval is 90 m 100 b) Time (m minutes) Cumulative Frequency 70 m m m m m m Check that the final cumulative frequency is the same as the total frequency which it is. c) To find the median draw a horizontal line across from a cumulative frequency of 60 to meet the curve. Where this line meets the curve drop it down to meet the time axis. This will be the median. Median = 103 minutes Page 10

11 Question 19 4x + y = 10 equation 1 2x 3y = 19 equation 2 We can solve by elimination or by substitution Elimination method We need to have either the same number of x or the same number of y Multiply equation 1 by 3 so that we end up with 3y 12x + 3y = 30 equation 3 2x 3y = 19 equation 2 (unchanged) Add the two equations together to make the y terms disappear 14x = 49 Divide both sides by 14 x = = = 3.5 Now we have x we can put this back into equation 1 to get y (4 x 3.5) + y = y = 10 Subtract 14 from both sides y = -4 x = 3.5, y = -4 Now check the answers by putting both back into equation 2 (2 x 3.5) (3 x -4) = = = Page 11

12 Substitution Method We can easily rearrange equation 1 to get y as the subject y = 10 4x Now substitute this into the second equation replacing the y with 10 4x 2x (3 x (10 4x)) = 19 2x (30 12x) = 19 2x x = 19 Grouping terms 14x 30 = 19 Add 30 to both sides 14x = 49 Divide both sides by 14 x = = = 3.5 Now we have x we can put this back into the rearranged equation 1 to get y Y = 10 - (4 x 3.5) Y = y = -4 x = 3.5, y = -4 Now check the answers by putting both back into equation 2 (2 x 3.5) (3 x -4) = = = 19 Question 20 The main things to compare are the median and the interquartile range as that is where 50% of the data lies. 1. The median is 80 for both boys and girls 2. Interquartile range for boy is more than for girls IQR boys: = 30 IQR girls: = 22 This means that the girls are more consistent with each other with their spellings. We could also say that the boys had the lowest result at 38, the girls lowest result was much higher at 56. The boys had the highest result at 98, whilst the girls heighest result was 94 The boys had the biggest overall range with 60 (98 38) Girls range was 38 (94 56) Page 12

13 Question 21 a) to work out gradient we take two clear points and divide the difference in the y coordinates by the difference in the x coordinates two points could be (0, 2) and (4, 0) gradient = = = - (we have to be consistent if we take y 1 y 2 then we must also have x 1 x 2 otherwise we will get the sign of the gradient wrong) b) line M is parallel to line L so will have the same gradient of - line M will be of the form y = - x + c where c is to be determined we know line M goes through (6, 2) substitute y = 2 and x = 6 so we can work out what c is 2 = (- x 6) + c 2 = -3 + c Add 3 to both sides c = 5 y = - x Page 13

14 Question 22 a)anything to the power of a negative number means that the number goes to the bottom 27-2/3 = We can rewrite as x 2 = 27 1/3 = 27= 3 So we have = = b) multiply the top and bottom by 2 x = a= -3 and b = 4 = = Question 23 Multiply both sides by (k 2) t(k 2) = k expand brackets tk 2t = k subtract k from both sides tk k -2t = 0 add 2t to both sides tk k = 2t factorise out the k k(t 1) = 2t divide both sides by (t 1) k = Page 14

15 Question 24 First we need to calculate class width and frequency density. Frequency density = frequency class width Height (h cm) Frequency Class width (CW) Frequency Density (F CW) 100 h h note note h note note h h Note 1: We can use the top entry to work out the scale on the histogram Each 5 small squares equals 1. So we can see that the frequency density for the second and third categories are 4.2 and 6 Note 2: Now we have the frequency densities we can use these to calculate frequency. Frequency = class width x frequency density. Frequency = 20 x 4.2 = 84 Frequency = 10 x 6 = 60 b) Page 15

16 Question 25 The two shapes have the same sized circle on the bottom so if the two total surface areas are equal so too will the curved surface areas. Curved surface area of cone Curved surface area = πrl from formulae sheet where l is the sloping height and r is the radius We know the radius is x in this case. We don t have the sloping height but we can work it out by using Pythagoras s Theorem. l 2 = h 2 + x 2 Square root both sides l = Curved surface area = πx Curved surface area of hemisphere Surface area of sphere is given in formulae sheet as 4πr 2 so surface area of a hemi sphere will be half of this. Also we know that radius is x in this case Surface area = 2πx 2 Now put the two areas equal to each other 2πx 2 = πx Divide each side by πx 2x = Square both sides 4x 2 = h 2 + x 2 Subtract x 2 from both sides 3x 2 = h 2 h 2 = 3x 2 square root both sides h = 3 x Page 16

17 Question 26 a) = + = -2a + 3b b) = x = (-2a + 3b) = + = 2a + (-2a + 3b) = 2a - a + b = b + b = (a + b) as is a multiple of a + b this means that they are parallel to each other Question 27 multiply both sides by 2 x - = 2 multiply both sides by (x + 1) x(x + 1) 4 = 2(x + 1) expand the brackets x 2 + x 4 = 2x + 2 subtract 2x from both sides x 2 x - 4 = 2 subtract 2 from both sides x 2 x 6 = 0 factorise (find two numbers that multiply to get -6 but add to get -1) (x 3)(x + 2) = 0 if two things multiply to give 0 then either one or the other must equal 0 x 3 = 0 add 3 to both sides x = 3 x + 2 = 0 subtract 2 from both sides x = -2 x = 3 or Page 17

18 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. If you found this paper helpful then visit where you will find plenty more Page 18

2015 Non Calculator Paper 4 Name:

2015 Non Calculator Paper 4 Name: GCSE Mathematics (Linear) 1380 Formulae: Higher Tier You must not write on this formulae page. Anything you write on this formulae page will gain NO credit. Volume of