Edexcel Mathematics Higher Tier, November 2010 (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 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 milk dark Total note Note: Whatever we did to 2 to get to 24 we do the same to = 12 3 x 12 = 36 Total number of chocolates is 36 Question 2 a) p 4 b) 6cd Page 1

2 Question 3 a) note 1 15 note note 3 17 note 4 Note 1: = 13 Note 2: = 15 Note 3: = 15 Note 4: = 17 b) (4, 7), (6, 5), (8, 3) in each case the two numbers add to give 11 c) there are three scores of less than 6 in the table. This is out of a possible 20 scores. The probability of less than 6 is Question 4 a) we are adding 4 each time so the sequence is something to do with 4n (the 4 times table) write out the sequence with the 4 times table immediately below: n 2 note n Note: We can see that to get to the sequence from the 4 times table we simply subtract 2 each time. The nth term is 4n 2 b) i) when n = = 10 9 = 1 The third term is 1 ii) when n = = = -15 The fifth term is Page 2

3 Question 5 Area of a circle is πr 2 where r is the radius Area = 3.14 x 10 2 = 3.14 x 100 = 314 cm 2 Question 6 Round all the figures to 1 significant figure: 3870 rounds to rounds to rounds to 5 So we have = = 4 Question x 37 = 6475 pence (see table below for workings) x Total total pence = Page 3

4 Question 8 There are two ways to plot a straight line: We can either work out 3 coordinates using a table and plot these x y Or we can first rearrange to get y = -x + 4 so that it is in the form y = mx + c (m is the gradient and c is the y intercept) The gradient is the number in front of the x (-1) and the y intercept is 4 Question 9 a) and b) angles in a triangle always add up to 180⁰ this is an equilateral triangle so all the angles are the same and they are 60⁰ (180 3 = 60) angles on a straight line add to make 180⁰ = 120 x = 120⁰ Page 4

5 Question 10 a) first draw an unordered stem and leaf diagram make sure we have 15 pieces of data (we do ) now put each row in order Key: 8 3 means 83 b) the mode is the most often The mode is 77 Question 11 We first need to find the total price of the van (including VAT) 10% of 6000 = 600 5% of 6000 = 300 (half of 10%) 2.5% of 600 = 150 (half of 5%) 17.5% of 6000 = = 1050 Total price of van = = = 4050 Lizzie paid 3000 when she got the van, that leaves 4050 to pay in monthly instalments = equal monthly payments of Page 5

6 Question 12 a) the four possible transformations are: reflection enlargement vector translation rotation this is a rotation, the centre of rotation is (0, 0) (also called the origin), the direction is irrelevant here as the angle of rotation is 180⁰ If you put tracing paper over triangle A and put the point of your pen in at (0, 0). Turn the tracing paper 180⁰ then you form triangle B. b) we pick triangle A up and move it 3 places to the right and 0 places down Page 6

7 Question 13 subtract 2 from both sides t 2 = multiply both sides by 5 5(t 2) = v rewrite v = 5(t 2) expand the brackets v= 5t - 10 Question 14 To find the midpoint of two coordinates we find the average of the x coordinates and the average of the y coordinates (, ) = (7, 5) Question 15 The two nets that make a square based pyramid are B and E Question 16 a) 3x x 12 group terms 13x + 3 b) factorise 2x + 4 to give 2(x + 2) so we have cancel the 2 on the top and the bottom = x + 2 c) 5 goes into both 5x and into 10 so bring 5 out 5(x + 2) d) x goes into both terms, so does y so bring out xy xy(x + y) Page 7

8 Question 17 Open a compass to more than half the length of the line AB. Put the compass point in at A and draw an arc above and below the line. Put the compass point in at B and draw an arc above and below the line. Where the two arcs above the line meet, join this point to the other joining point below the line. This gives you the perpendicular bisector of line AB Page 8

9 Question 18 a) in order to subtract (or add) fractions we need a common denominator the common denominator could be 20 we can leave the whole numbers alone = 2-1 = 1 b) in order to multiply fractions we must have them as top heavy 2 = 1 = When we multiply fractions we multiply the top together and multiply the bottom together x = = = = Page 9

10 Question 19 We have two triangles that are mathematically similar. It is easier to do these questions if you separate out the two triangles: A A 10 cm 15 cm E 8 cm B D C a) the two sides that correspond to each other are AB and AC we are trying to find a length on the larger triangle so we want our scale factor to be = 1.5 length DC = 8 x 1.5 = 12 cm b) the area of a trapezium is ½(a + b)x h where a and b are the lengths of the parallel sides and h is the perpendicular height in this case a = 8, b = 12 and h = 5 ½(8 + 12) x 5 = ½ x 20 x 5 = 10 x 5 = 50 cm Page 10

11 Question 20 a) The median is represented by the vertical line in the box median is 13.2 cm b) the upper quartile is 13.8 the lower quartile is 12.6 the quartiles are represented by the box interquartile range = = 1.2 cm c) the range is the largest value less the smallest value. This will be effected by any one piece of data that may be particularly low or high. The interquartile range on the other hand shows how spread out the middle 50% of the data is and isn t effected by any extreme values Page 11

12 Question 21 We can solve simultaneous equations by elimination or by substitution Elimination method Label each equation eqn 1 and eqn 2 We need to have the same number for x or the same number for y Multiply eqn 1 by 2 and multiply eqn 2 by 3 (that way we will have 12x in both equations) 12x + 4y = -6 eqn 3 12x 9y = 33 eqn 4 subtract eqn 4 from eqn 3 4y - - 9y = y = -39 divide both sides by 13 y = -3 now put this value for y back into eqn 1 6x + (2 x -3) = -3 6x 6 = -3 add 6 to both sides 6x = 3 divide both sides by 6 x = 0.5 x = 0.5 and y = -3 Check by substituting both values into eqn 2: (4 x 0.5) (3 x -3) = = = 11 Substitution method Label each equation eqn 1 and eqn 2 rearrange eqn 1 to make y the subject 2y = -3 6x divide by 2 y = x eqn 3 now substitute eqn 3 into eqn 2 4x -3(-1.5 3x) = 11 4x x = 11 13x = 11 subtract 4.5 from both sides 13x = 6.5 divide both sides by 13 x = 0.5 now substitute this value for x back into eqn 3 y = -1.5 (3 x 0.5) = = -3 x = 0.5 and y = -3 Check by substituting both values into eqn 2: (4 x 0.5) (3 x -3) = = = Page 12

Question 22 a) tangents to a circle always make right angles with a radius at the point that they meet the circle b) as we have a right angled triangle we can apply Pythagorus s Theorem a 2 + b 2 = c

13 Question 22 a) tangents to a circle always make right angles with a radius at the point that they meet the circle b) as we have a right angled triangle we can apply Pythagorus s Theorem a 2 + b 2 = c 2 where c is the longest side (hypoteneuse) = OB = OB = OB 2 square root both sides 10 = OB We know that OC is 6 cm (since it is a radius) BC = OB OC = 10 6 = 4 cm Question 23 a) (x 3)(x + 5) x x + 5x x 2 + 2x 15 b) first see if we can factorise we want to find two numbers that multiply to give -9 but add to give +8 these two numbers are -1 and +9 (x 1)(x + 9) = 0 If two brackets multiply to give 0 then either one of the brackets must equal 0 If x 1 = 0 then x = 1 If x + 9 = 0 then x = -9 x = 1 or Page 13

14 Question 24 a) the area of a histogram is directly proportional to the frequency if 8 students watched between 10 and 15 hours then 8 students are represented by an area of 400 mm squares 8 students 400 squares divide both sides by 8 1 student 50 squares multiply by 6 6 students 300 squares The width for this group is 20 squares so the height needs to be 15 small squares (20 x 15 = 300) Alternatively frequency density = frequency class width For 10 to 15 hours: Frequency density = 8 5 = 1.6 This gives us the scale up the y axis For 0 to 10 hours: Frequency density = 6 10 = Page 14

15 b) 15 to 20 hours: frequency density = = frequency 5 multiply both sides by 5 Frequency = 5 x 1.2 = 6 20 to 30 hours: Frequency density = = frequency 10 multiply both sides by 10 Frequency = 10 x 0.5 = 5 Total frequency = = 25 students Question 25 If we want a sample of 50 this represents = = th of the actual population Each number must be divided by 20 (but don t round yet) Age (years) Total Number of teenagers Number in sample (unrounded) Number in sample note 50 Note: the 5.5 should have been rounded up to 6 but if this had been the case then the total of the sample would be 51 (and not 50 as we want). I have therefore changed the rounding on this to give 5. The number of 14 year olds in the sample should be 9 For this question you did not need to work out the number for each age, I have done that to show you what to do. It would have been sufficient to just calculate this for the 14 year olds Page 15

16 Question 26 a) P is inversely proportional to V We can write P Now replace with = so long as introduce a constant k P = when V = 8, P = 5, we can use this to calculate k 5 = multiply both sides by 8 40 = k P = b) P = = 20 Question 27 a) = ½ = ½( + ) = ½(a + b) = ½a + ½b b) = + = -a + ½a + ½b = -½a + ½b Page 16

Question 28 a) x 2 + y 2 = 9 is the equation of a circle with centre (0, 0) and radius 3 b) the line x + y = 1 can be rewritten as y = -x + 1 (so in the form y = mx + c) gradient is m = -1 and y

17 Question 28 a) x 2 + y 2 = 9 is the equation of a circle with centre (0, 0) and radius 3 b) the line x + y = 1 can be rewritten as y = -x + 1 (so in the form y = mx + c) gradient is m = -1 and y intercept is c = 1 the two graphs meet at two points: (2.55, -1.55) and (-1.55, 2.55) If you found this paper useful and would like to see some more then click here: 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 17

Mathematics IGCSE Higher Tier, June /4H (Paper 4H)

Mathematics IGCSE Higher Tier, June /4H (Paper 4H) Link to examining board: http://www.edexcel.com The question paper associated with these solutions is available to download for free from the Edexcel website. The navigation around the website sometimes