Edexcel Mathematics Higher Tier, November 2009 (1380/3H) (Paper 3, non-calculator)

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Stuart Haynes
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Link to examining board: http://www.edexcel.com/migrationdocuments/qp%20current%20gcse/nov09-qp/1380_3h_que_20091105.

1 Link to examining board: As at the time of writing you can download this paper for free from the Edexcel website. Alternatively you can order directly from Edexcel by phoning on 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) 740 x 234 this is very similar to the original equation except that we are now multiplying by 740 instead of 74. So our answer will now be 173,160 (instead of 17,316) b) 74 x 2.34 this is very similar to the original equation except that we are now multiplying by 2.34 instead of 234. So our answer will now be (instead of 17,316) In both of the above we know that the answer will involve the digits and we could just use estimations to work out the answers. 740 x x 200 = (so actual answer will be 173,160). 74 x x 2 = 140 (so actual answer will be ) Question 2 We need to round each number to one significant figure To get rid of the decimal on the denominator multiply the denominator and the numerator by 10 to get = = 15 x 5 x 10 = 75 x 10 = Page 1

2 Question 3 a) x y -2 note note 2 6 note 3 8 note 4 note 1: when x = -2 then y = (2 x -2) + 2 = = -2 note 2: when x = 1 then y = (2 x 1) + 2 = = 4 note 3: when x = 2 then y = (2 x 2) + 2 = = 6 note 4: when x = 3 then y = (2 x 3) + 2 = = 8 b) c) i) when x = -1.5 then y = -1 (see red dashed line) ii) when y = 7, then x = 2.5 (see other red dashed line) Page 2

3 Question 4 a) b) Page 3

4 Question 5 a) first put all the data in an unordered stem and leaf diagram 3 3, 8, 7 4 6, 1, 4 5 4, 1, 5, 1, 5, 2 6 0, 2, 3 Now it easier to put them into an ordered stem and leaf diagram 3 3, 7, 8 4 1, 4, 6 5 1, 1, 2, 4, 5, 5 6 0, 2, 3 Key: 5 4 means 54 grams b) there are 10 eggs that have a weight of more than 45 grams out of a possible 15 eggs so the probability will be = Page 4

5 Question 6 a) first we need to calculate the mid point for each category. Then we plot mid point with frequency. Time (t Frequency Mid point minutes) 0 t t t t t b) the modal class interval is the class interval with the highest frequency. This is 20 t Page 5

6 Question 7 a) we need to have a common denominator in order to add or subtract fractions. Let the common denominator be 8. So we have + = b) we simply multiply the numerators together and multiply the denominators together. = c) 423 x 12 = (423 x 10) + (423 x 2) = = x Question 8 a) 1. There is no time unit. How much do you spend on your mobile phone per week/day/ year etc 2. What if you spend less than 1 or more than What box do you tick if you spend 5 or 10 (there are overlaps) b) How much money do you spend using your mobile phone per month? Please tick one box only. 0 to 5 More than 5 to 10 More than 10 to 15 More than Page 6

7 Question 9 a) area of one face would be 5 x 5 = 25 cm 2. A cube has 6 identical faces so the total surface area will be 6 x 25 = 150 cm 2 b) 1 cm = 10 mm so 1 cm 3 = 1000 mm cm 3 = 125,000 mm 3 c) i) minimum weight would be 86.5 grams ii) maximum weight would be (though 87.5 is perfectly acceptable here) Question 10 a) grouping terms 3a a + 4c + 3c = 2a + 7c b) y(2y 3) = 2y 2 3y c) x 2 4x = x(x 4) d) 2(x + 3) + 3(2x -1) = 2x x 3 = 8x + 3 e) 3(x + 2) = 8 expand brackets 3x + 6 = 8 subtract 6 from both sides 3x = 2 divide both sides by 3 x = Page 7

8 Question 11 a) the bearing of B from A (measure clockwise from the North at A) is 060⁰ b) measure 160⁰ clockwise from North at B. Then make this line 4cm long. Question 12 If he buys p packets of batteries with 4 in each packet he will have 4p batteries If he buys b boxes of batteries with 20 in each box he will have 20b batteries In total he will have 4p + 20b = N batteries 4p + 20b = N Question 13 a) 213,000 = 2.13 x 10 5 (we have moved the decimal point 5 places to the left) b) = 1.23 x 10-3 (we have moved the decimal point 3 places to the right) Page 8

9 Question 14 a) anything to the power of 0 is always 1 so 5 0 = 1 b) negative powers turn the number upside down (gives the reciprocal) 2-1 = Question 15 a) k is an integer which means it must be a whole number k could be -1, 0, 1, 2 (it can t be 3 as k 3) b) 6y y + 10 subtract y from both sides 5y 10 divide both sides by 5 y 2 Question 16 5(q + p) = 4 + 8p expand the brackets 5q + 5p = 4 + 8p subtract 5p from both sides 5q = 4 + 3p divide both sides by 5 q = Question 17 a) the highest mark in the English test is 50 b) The median for English is 38 compared to the median for Maths of 27. This means that on average the English marks are higher by 11 marks. The interquartile range for English is 17 (42 25). The interquartile range for Maths is 13 (35 22). This means that the Maths marks are more consistent and less spead out than the English marks Page 9

10 Question 18 a) where tangents meet a circle they make an angle of 90⁰ with the radius. Hence angle ABD will be = 55⁰ b) As BE is a diameter then angle BDE will be 90⁰ (angles in a semi-circle make 90⁰). Angles in a triangle always add up to 180⁰ so angle DEB will be 180 ( ) = 180 = 125 = 55⁰ Question 19 a) 5 7 Blue 5 7 Blue 2 7 Red 2 7 Red Blue Red As Emma puts the pen back into the box the probabilities for the second pen will be the same as for the first pen. b) we could have red then blue with probability x = or we could have blue then red with probability x = The probability that we have exactly one pen of each colour will be + = Page 10

11 Question 20 To solve simultaneous equations you can either solve by elimination or by substitution: Elimination We need the same number of x or the same number of y. We have 4x in both equations. If we subtract one equation from the other then the x terms will have been eliminated. (4x + y = -1) - (4x 3y = 7) y - -3y = y = -8 now divide both sides by 4 y = -2 substitute this value for y back into the first or the second equation. I am putting it back into the first equation. 4x + -2 = -1 4x 2 = -1 add 2 to both sides of the equation 4x = 1 divide both sides of the equation by 4 x = we have x = and y = -2 substitute both of these back into the other equation to make sure it works (I need to put back into the second equation) (4 x ) (3 x -2) = = = 7 Substitution First we need to rearrange one of the equations so that either x or y is the subject. The easiest one to rearrange here is the first equation. 4x + y = -1 subtract 4x from both sides y = -1 4x substitute this value for y into the second equation 4x (3 (-1 4x) = 7 4x (-3 12x) = 7 4x x = 7 16x + 3 = 7 subtract 3 from both sides Page 11

12 16x = 4 divide both sides by 16 x = = going back to the rearranged equation y = -1-4x and substituting in our value for x we get y = -1 (4 x ) = -1 1 = -2 we have x = and y = -2 substitute both of these back into the other equation to make sure it works (I need to put back into the second equation) (4 x ) (3 x -2) = = = 7 Question 21 (2 + 3)(2-3) = = 1 Question 22 a) = + = -a + b b) = + = a + = a + (-a + b) = a - a + b = a + b = (a + 2b) Question 23 Let R = = there are two numbers recurring so we multiply by R = R = subtracting these two equations we get 99R = 36 divide both sides by 99 R = = Page 12

13 Question 24 a) y = f(x 2) this is a translation. The change is inside the brackets so effects the x, and always effects the x in the opposite way to what you think. So the new graph is picked up and moved two places in the positive x direction. The new minimum point will be (5, -4) b) y = f(x + 5) + 6 this is made up of two translations. The first is a translation of 5 units in the negative x direction, followed by 6 units in the positive y direction. The new minimum point will be (-2, 2) Question 25 Let the first whole number be n, the next consecutive whole number will be n + 1. The sum of these two consecutive numbers will be n + n + 1 = 2n + 1. Since 2n will always be even, so 2n + 1 will always be odd. 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 13

Mathematics Edexcel Advanced Subsidiary GCE Core 1 (6663) January 2010

Mathematics Edexcel Advanced Subsidiary GCE Core 1 (6663) January 2010 Link to past paper on Edexcel website: http://www.edexcel.com/quals/gce/gce08/maths/pages/default.aspx These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you