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

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1 Link to examining board: This question paper is not yet 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. 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) = b) first express as a fraction and then multiply by 100 to turn into a percentage x = x = = 30% c) we can either see this by inspection or first find half of 10 half of 10 is 5. The difference between the two prices is 1.50 so if we increase one by 75p and decrease the other by 75p they will still add up to 10 but the difference between them will be 1.50 Bus fare (more expensive) = p = 5.75 Just to check: food = 5 75p = = Page 1

2 Question 2 a) and b) Line Number x x x x x Across each row the three entries equal each other c) would be the entry for the 999 th line this would be the same as 2 x x 1,000, = 2,000, Page 2

3 Question 3 Internal angle of the square The internal angle of the square will be 90⁰ (right angle or 360/4) Internal angle of the hexagon The external angles of all polygons always add up to 360⁰ A regular hexagon as 6 equal external angles so each external angle is equal to 360/6 = 60⁰ The internal angle and the external angle add up to 180⁰ so the internal angle will be = 120⁰ angle a is on a point along with the internal angle of the square (90⁰) and the internal angle of the hexagon (120⁰). a = 360 a = 360 a = = 150⁰ Page 3

4 Question 4 the median is the middle number (when they are all in order as they are now). With 21 pieces of data the median will be the 11 th piece of data ( = 11). The 11 th piece of data is 3.1 Median = Page 4

5 Question 5 a) shape A moves 8 places to the left and 2 places down b) The red dashed line is the line y = x because everywhere along this line the x coordinate equals the y coordinate Page 5

6 Question 6 Izzy drives 25 miles to the meeting in Buckingham. Her speed is 50 mph. Time = distance/speed Time taken = 25/50 = ½ hour Izzy leaves home at 9 am, arrives at the meeting at 9.30 am (1/2 hour later) and leaves the meeting at pm (3 hours later) She then drives 45 miles to Reading and another 30 miles back to Oxford (total 75 miles). Speed is again 50 mph Time = distance/speed Time = 75/50 = 1 ½ hours Izzy arrives in Oxford at 2 pm Page 6

7 Question 7 a) expand the brackets 6t 12 = 2t + 12 Subtract 2t from both sides 4t 12 = 12 Add 12 to both sides 4t = 24 Divide both sides by 4 t = 6 b) expand (taking care with the negative before the second bracket) 2x 2y 3x + 6y Group terms -x + 4y c) (x 5)(x + 7) = x x + 7x group terms x 2 + 2x Page 7

8 Question 8 We need to replace each figure with it rounded to one significant figure 0.49 is rounded to is rounded to 0.6 So we have: (0.5 x 0.6) 2 = = Page 8

9 Question 9 10% of 180 = 18 5% of 180 = 9 2.5% of 180 = % = 10% + 5% + 2.5% = = % = 10% + 10% + 2.5% = = Difference in cost is = Page 9

10 Question 10 Triangle BCE is isosceles: Angle BCE = angle BEC = 48⁰ Angle CBE = 180 ( ) = = 84⁰ Angles on a straight line make 180⁰ Angle ABC = = 96⁰ Triangle ABC is also isosceles: Angle BAC = angle BCA = (180 96)/2 = 84/2 = 42⁰ DC is parallel to AB and there is a line going through them so angle BAC corresponds with angle DCA and so both equal 42⁰ Angle DCA = 42⁰ Page 10

11 Question 11 a) b) as the temperature increases the time taken decreases (this is a negative correlation) c) draw a vertical line up from 25 on the temperature until it meets a line of best fit. Then take this across horizontally to meet the time at 20 minutes. 20 minutes d) our line of best fit would cross the temperature axis at less than 100 implying it would take a negative amount of time which is clearly nonsense our data only goes up to 70⁰C, we can not safely make any assumptions about temperatures beyond this Page 11

12 Question 12 Multiply the first equation by 2 to get 6x + 8y = 400 Multiply the second equation by 3 to get 6x + 9y = 432 We now have the same number for x in both equations. If we take the first equation away from the second equation we will eliminate the x terms 9y 8y = y = 32 Substitute back into the first equation (before we multiplied by 2) 3x + 4(32) = 200 3x = 200 Subtract 128 from both sides 3x = 72 Divide both sides by 3 x = 24 Check by putting x = 24 and y = 32 into the second equation (2 x 24) + (3 x 32) = = 144 as expected x = 24, y = Page 12

13 Question 13 a) to multiply in standard form, multiply the ordinary numbers together and add the powers of x = 24 x Now we need to adjust this so we have 2.4 rather than 24 We have decreased 24 to get 2.4 so to compensate we need to increase the power of 10 from 15 to x b) in order to be able to add standard form we need to have the same power of 10 (a bit like adding fractions when you need the same denominator) adjust one of them (doesn t matter which) 6 x 10 8 = 60 x 10 7 We have decreased the power of 10 from 8 to 7 so to compensate I increase 6 to 60 Now we can add them: (60 x 10 7 ) + (4 x 10 7 ) = 64 x 10 7 Now adjust again to get back to proper standard form 6.4 x 10 8 We have decreased 6 to get 6.4 so to compensate we need to increase the power of 10 from 7 to Page 13

14 Question 14 a) i) the LHS of the equation is just y so look at the line y = 0 (the x axis) and see where the curve meets this line the curve meets the x axis at -0.5 and 5.5 x = -0.5, 5.5 ii) the LHS of the equation is just y so look at the line y = 6 see where the curve meets this line and read off the x values x = -1.4, Page 14

15 b) the LHS of the equation is just y so look at the line y = x - 4 see where the curve meets this line and read off the x values x = 0.2, 5.8 now we have the x values we can substitute back into y = x 4 to get the y values when x = 0.2, y = -3.8 when x = 5.8, y = Page 15

Question 15 a) the modal class interval is the one with the highest frequency modal class interval is 200 c 400 b) Cost ( C) Cumulative Frequency 0 C 200 7 0 C 400 18 0 C 600 27 0 C 800 37 0 C 1000

16 Question 15 a) the modal class interval is the one with the highest frequency modal class interval is 200 c 400 b) Cost ( C) Cumulative Frequency 0 C C C C C C Check that the final cumulative frequency is the same as the total frequency which it is. c) d)draw a vertical line up from a 700 to meet the curve. Where this line meets the curve take it horizontally to meet the cumulative frequency axis (at 32). This will be the number of repairs which cost less than 700. Subtract this from 50 to find (18) the number of repairs that cost more than 700. Answer = 18 repairs Page 16

17 Question 16 First we need to calculate the volume of the prism. Volume = area cross section x length (from formulae sheet) Volume = area trapezium x 20 Area of a trapezium = ½ (a + b) x h where a and b are the parallel sides and h is the height Area of trapezium = ½(8 + 12) x 6 = ½ x 20 x 6 = 10 x 6 = 60 cm 2 Volume = 60 x 20 = 1200 cm 3 We are given density = 5 g/cm 3 the units tell us that density = mass/volume We want mass so rearrange to get mass = density x volume Mass = 5 x 1200 = 6000 g = 6kg Page 17

18 Question 17 a) y = -10 (2 x 3 x (-5) 2 ) y = -10 (6 x 25) y = y = -160 b) add 2qx 2 to both sides (so the term containing x is positive) y + 2qx 2 = p subtract y from both sides 2qx 2 = p y Divide both sides by 2q x 2 = square root both sides x = Page 18

19 Question 18 a) anything to the power of 0 is always = 1 b) 2 y = = 2-2 y = -2 c) 9-3/2 = = = = Page 19

20 Question 19 AB is a diameter of the circle so that means that where it joins the edge of the circle at C we have a right angle (90⁰) We want angle CDB As BD is parallel to AC we can see that by alternate angles angle ACD equals angle CDB In addition triangle BCD is an isosceles triangle so that angle BCD also equals angle CDB We have that at point C the angles either side are the same as each other (angle ACD = angle BCD = angle CDB) so the line CD bisects angle ACB As the angle at C is 90⁰ we can see that angle CDB will be half of this = 45⁰ Angle CDB = 45⁰ Page 20

21 Question 20 a) this is a harder factorisation as there is a number in front of the x 2 take this number and multiply by the 4 to get 8 we must find two numbers that multiply to give 8 but combine to give -9 these two numbers are -1 and -8 rewrite the expression splitting the -9x term into x and -8x 2x 2 x 8x + 4 Factorise in pairs x(2x 1) - 4(2x 1) we should have the same thing in both brackets which we do factorise again (x 4)(2x 1) b) using what we have just worked out (x 4)(2x 1) = (2x 1) 2 Take the LHS over to the RHS and set the equation equal to 0 (2x 1) 2 - (x 4)(2x 1) = 0 Factorise (2x 1)(2x 1 (x 4)) = 0 (2x 1)(2x 1 x + 4) = 0 (2x 1)(x + 3) = 0 If two things multiply to give 0 then either one of them must be equal to 0 2x 1 = 0 2x = 1 x = ½ x + 3 = 0 x = -3 x = ½ or Page 21

22 Question 21 Area of a triangle is ½ x base x height A = ½ x 2 3 x height We don t have height but we can work it out using Pythagoras Theorem a 2 + b 2 = c 2 Let the height be h (2 3) 2 + h 2 = 6 2 (4 x 3) + h 2 = h 2 = 36 h 2 = 24 h = 24 = 4 x 6 = 4 x 6 = 2 6 now we can work out area (A) A = ½ x 2 3 x 2 6 = 2 3 x 6 = 2 3 x 2 x 3 = 2 x 3 x 2 = 6 2 A = 6 2 K = Page 22

Question 22 a) For box A the probabilities are out of 10. As Jan puts the first counter from box A into box B the probabilities for the box B are now out of 11.

23 Question 22 a) For box A the probabilities are out of 10. As Jan puts the first counter from box A into box B the probabilities for the box B are now out of 11. From A to B From B to A Outcome in box A 8 11 Black 6B and 4W 6 10 Black 3 11 White 5B and 5W 4 10 White Black White 7B and 3W 6B and 4W b) Outcomes Black then Black: started with 6 black, took one out so now 5 black. Black from box B put back into box A so back to 6 black (and 4 white). Black then White: started with 6 black, took one out so now 5 black. White from box B put back into box A so still 5 black (and 5 white). White then Black: started with 6 black, took a white out so still 6 black. Black from box B put into box A so now have 7 black (and 3 white). White then White: started with 6 black, took a white out so still 6 black. White from box B put nto box A so still have 6 black (and 4 white) Page 23

24 We want the probability of black and black or white then white Black then black = x = = White then white = x = = Adding these two probabilities together we have + = 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 24

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

Edexcel Mathematics Higher Tier, June 2011 (1380/3H) (Paper 3, non-calculator) Link to examining board: www.edexcel.com 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