Mathematics Higher Tier, November /2H (Paper 2, calculator)

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1 Link to past paper on AQA website: This question paper is available to download freely from the AQA website. To navigate around the website, you want QUALIFICATIONS, GCSE, MATHS, MATHEMATICS, MATHEMATICS A (LINEAR), 3301 MATERIALS. Question 1 a). = = 1.25 b) i) Type directly into your calculator ii) Each of the inputs were to 2 dp so 2 dp for our answer would be appropriate Question 2 a) expand 8x 28 = x subtract 3x from both sides 5x 28 = 12 add 28 to both sides 5x = 40 divide both sides by 5 x = 8 b) add 9 to both sides = 11 multiply both sides by 7 y = 77 (we could have multiplied both sides by 7 first to get y 63 = 14, then add 63 to both sides, y = 77) Page 1

2 Question 3 My diagram is NOT to scale First put your compass point on P and use the compass to find two points on the Road that are both the same distance away from P (equidistant from P and marked with arcs). Keeping your compass at the same setting. Now from each of these points (where they meet the road) draw two more arcs, one above the line and one below the line. (P should be the point where the two arcs above the road meet). Join P up to where the two arcs meet below the road. b) the shortest real distance is found by measuring with a ruler the distance from P to where it meets the road. This is 5.8cm which is equivalent to 5.8 x 0.5 = 2.9 km. Question 4 x x 3 + 2x Comment (2 x 3) = = 33 too small (2 x 4) = = 72 too big (2 x 3.5) = too small (2 x 3.8) = too big (2 x 3.7) = too small We can see that the answer lies between 3.7 and 3.8. The answer is 3.7 (as this is away from 60 whereas 3.8 is away from 60) Page 2

3 Question 5 The price we are given is after the reduction has been applied. If we call our original price p then p x 0.85 = (we multiply by 0.85 because 100% - 15% = 85% = = 0.85) divide both sides by 0.85 p = = 14,720 Question 6 a) draw a line of best fit that goes through as many of the points as possible. Then draw a line up from 2007 until it meets the line of best fit. Then take this line across until it meets the y axis. The estimate of the RPI is 202 b) The RPI in 2005 is 192 (from the table). to find the percentage increase, first find the change, divide by the old and multiply by 100 ( ) 100 x 100 = 92% c) RPI in 2000 is 170 (from the table) RPI in 2006 is 197 (from the table) to find the percentage increase, first find the change, divide by the old and multiply by 100 ( ) 170 x 100 = x 100 = 15.9% Page 3

4 Question 7 a) (x + 2)(x 4) = x 2 + 2x 4x 8 = x 2-2x 8 b) 2(m + 5) 3(2m 1) = 2m m + 3 = -4m + 13 Question 8 a) the triangle has a right angle so immediately we should think of either Pythagoras or basic trigonometry. By Pythagoras Theorem a 2 + b 2 = c 2 (where c is the longest side) x = 32 2 x = 1024 subtract 441 from both sides x 2 = 583 square root both sides x = 24.1 cm b) label all the sides from the point of view of the angle. The longest side is the hypotenuse (18cm). y is the adjacent. So we have adj (A) and hyp (H). From SOHCAHTOA we can see that we need cos. cos 35⁰ = multiply both sides by x cos 35⁰ = y y = 14.7cm Question 9 The area of the lawn, where r = 2.2m A = π x = m 2 Bob needs 75g for every square metre of lawn. As he has square metres he will need 75 x = g = 1.14 kg As he needs 1.14kg (which is more than 1kg) he will not have enough lawn feed Page 4

5 Question 10 a) for two lines to be parallel they must have the same gradient. When a line is written in the form y = mx + c, m is the gradient and c is the y intercept. The gradient of line A and line D is 3 so these two lines are parallel to each other. b) when written in the form y = mx + c, c represents the y intercept (where the line crosses the y axis). Line B and line C have the same c of 2 and so both cross the y axis at 2 c) for two lines to be perpendicular their gradients must be the negative reciprocal of each other (or when you multiply the gradients together you get -1) line B has a gradient of -3 and line C has a gradient of ⅓ (-3 x ⅓ = -1) line B and line C are perpendicular to each other Question 11 Density is mass divided by volume (if you didn t know this you could work it out from the units given kg/m 3 ) 1 tonne = 1000kg Volume of cube is = density = = kg/m 3 Question 12 a) The dashed line is x =2 and the part that is shaded is greater than 2. The region is x 2 (this is a strict inequality ( we know this because the line is dashed. If it had been a solid line then our region would be defined by x 2) b) x + y = 5 y = x + 1 R y = Page 5

6 Question 13 First ball Second ball 0.7 green 0.7 green 0.3 yellow 0.3 yellow 0.7 green 0.3 yellow b) the probability that the first ball is green and the second ball is yellow is 07 x 0.3 = 0.21 the probability that the first ball is yellow and the second ball is green is 0.3 x 0.7 = 0.21 so the probability that both balls are different colours will be = 0.42 Question 14 a) Simply put straight into your calculator: (2 x 1.67 x ) + (2.66 x ) = x = 2.99 x (I have given the answer to the same degree of accuracy as the data we were given) b) the above is the mass of 1 molecule of water. To find out how many molecules there are in 1 gram we divide 1 by the mass of 1 molecule. 1 (2.994 x ) = 3.34 x Question 15 As triangle ABC is isosceles the line BM will bisect the angle ABC so angle ABM will equal angle CBM angle BAM = angle BCM (as given in question that triangle is isosceles) We have now defined two angles as being the same, so the third angle must also be the same. All we need to do now is prove that one side is the same. We know that AB = BC as triangle ABC is isosceles We now have ASA so can uniquely define the triangles and they must be congruent Page 6

7 Question 16 a) look for something that goes into 8a 2 and into goes into both so put 2 on the outside of brackets 2(4a 2 25) This will factorise further. A hint in the question was that it said factorise fully. 4a 2 25 is the difference of two squares so can be expressed as (2a 5)(2a + 5) So 8a 2 50 = 2(2a 5)(2a + 5) b) for this type of question you usually need to factorise the top and the bottom and they are bound to have one factor in common which you can cancel working on the numerator: to factorise 12x 2 36x + 15 we can first see that 3 goes into all terms so we can first take that out 3(4x 2 12x + 5) Then work on factorising 4x 2 12x + 5 As we have a number in front of the x 2, it makes the factorisation a bit harder. We must first multiply 4 and 5 together to get 20 then we need to find two numbers that multiply together to give 20 and add together to give -12. This would be -2 and -10. Rewrite 4x 2 12x + 5 as 4x 2 2x 10x + 5 (this is just splitting -12x into the two components -2x and - 10x) Now factorise in pairs 2x(2x 1) 5(2x 1) We should have the same in both brackets (which we do) As (2x 1) goes into both terms we can factorise that out to get (2x 1)(2x 5) Remember we had already factorised the 3 out so the numerator should be 3(2x 1)(2x 5) Working on the denominator: First we can factorise 3 out to give 3(4x 2 1) 4x 2 1 is the difference of two squares so can be expressed as (2x 1)(2x + 1) We have 3(2x 1)(2x + 1) Putting the numerator and denominator back together 3(2x 1)(2x 5) = 3(2x 1)(2x + 1) 3(2x 1)(2x 5) 3(2x 1)(2x + 1) We can see that the 3s will cancel out and so will (2x 1) so we are left with Page 7

8 Question 17 a) Height, h (cm) Number of students (n) Midpoint (m) n x m 140 h h h h h h Total Mean = = cm b) It helps to add two extra rows; class width and frequency density. Frequency density = frequency class width. Here, frequency is the number of adults. Height, h (cm) 140 h h h h h 200 Class width Number of note 2 28 note adults Density Frequency 1.25 note (from the histogram) 5.6 (from the histogram) 1.2 note note 5 note 1: = 1.25 note 2: 3.4 x 5 = 17 note 3: 5.6 x 5 = 28 note 4: = 1.2 note 5: = 0.8 check that the total number of adults makes 100 ( = 100 ) Page 8

9 Question 18 volume of sphere (is given in formulae sheet) = πr3, where r is 3 volume of sphere = π x 33 = 36π = cm 3 volume of cone (is given in formulae sheet) = πr2 h, where r is 6 and h is 8 volume of cone = π x 62 x 8 = 96π = cm 3 I have kept the volumes with the symbol π, as when we find the fraction this will cancel out. If I had used the rounded figures of and then we wouldn t get the simple fraction, but rather just a decimal figure of (but with some rounding error). Fraction taken up is volume of sphere volume of cone Fraction = 36π 96π = Question 19 a)we have to solve this by substitution. If we rearrange 2x + y = 3 to make y the subject 2x + y = 3 subtract 2x from both sides y = 3 2x now substitute this value for y into x 2 + y 2 = 5 x 2 + (3 2x) 2 = 5 x 2 + (3 2x) (3 2x) = 5 x 2 + (3 2x) (3 2x) = 5 expand the brackets x x 2 6x 6x = 5 group terms 5x 2 12x + 9 = 5 subtract 5 from both sides 5x 2 12x + 4 = 0 b) first try to factorise: this is a harder quadratic as there is a number in front of the x 2 so we must first multiply the 5 by the 4 to get 20. Then we need to find two numbers that multiply to give +20 and combine to give -12. These two numbers are -2 and -10. Rewrite the equation splitting the -12x term into -2x and -10x 5x 2 2x 10x + 4 = 0 factorise in pairs x(5x 2) 2(5x 2) = 0 we should have the same factor for both pairs (which we do and it is (5x 2)) Page 9

10 as (5x 2) goes into both terms we can factorise again (5x 2)(x 2) = 0 If you multiply two things together to get 0 then one of them must equal 0 So 5x 2 = 0 or x -2 = 0 5x 2 = 0 add 2 to both sides 5x = 2 divide both sides by 5 x = x 2 = 0 add 2 to both sides x = 0 We have x = 2 or x = = 0.4 Alternatively, we could have used the quadratic formula here (given in formulae sheet) x5x So x = 2 or 0.4 where a = 5, b = -12 and c = Page 10

11 Question 20 In order to be able to work out the area of triangle PQR it would be helpful to have the angle PQR (we could then use the formula ½absinC). In order to work out the angle PQR it would be helpful to have the length of PR (we could then use the cosine rule rearranged). In order to work out the length of PR we need to apply Pythagoras theorem to triangle PSR. First start with triangle PSR. This is right angled so we can apply Pythagoras theorem. PR 2 = PS 2 + SR 2 P 5 PR 2 = = = 169 square root both sides PR = 13 S 12 R Now consider triangle PQR. We have all the sides so can work out any of the angles. The cosine rule (from formulae sheet) is Q 8 14 a 2 = b 2 + c 2 2bccosA add 2bccosA to both sides 2bccosA = b 2 + c 2 a 2 divide both sides by 2bc cosa = b 2 + c 2 a 2 2bc 13 R take the inverse cos (cos -1 ) of both sides A = cos -1 (b 2 + c 2 a 2 ) 2bc Where A is the angle we want (angle PQR), so a (the opposite side) must be 13, b and c can be 8 and 14 (but it is not important which way around, I have taken b = 8 and c = 14) angle PQR = cos -1 ( ) = cos -1 ( ) = cos -1 ( ) = ⁰ (we haven t finished yet 2x8x 14 I have kept a few decimal places) P Now we have two sides and the angle between we can use the formula for the area of a triangle (from formulae sheet) ½absinC where C is ⁰, a and b are 8 and 14 (doesn t matter which way around) Area = ½ x 8 x 14 x sin ⁰ = 51.2 cm Page 11

Question 21 Each of the four pieces of luggage is accurate to the nearest 50g so for example 2kg 400g could really be anywhere between 2kg 375g and 2kg 425g We have to assume the worst case scenario

12 Question 21 Each of the four pieces of luggage is accurate to the nearest 50g so for example 2kg 400g could really be anywhere between 2kg 375g and 2kg 425g We have to assume the worst case scenario (to be sure that Marvin is within his allowance) The worst case scenario is that the four lots weigh 2kg 425g, 2kg 825g, 2kg 775g and 1kg 875 g This total weight is 9kg 900 (9.9kg) At the airport the scales are accurate to the nearest 100g so the allowance of 10kg could be as low as 9kg 950g and as high as 10kg 50g. Again we have to assume the worst case scenario to be sure and take the Airport s reading as 9kg 950g (9.950kg) 9kg 900g (9.9) is less than 9kg 950g (9.950) so Marvin is ok. Question 22 Notice that the probabilities change for the second ball because the question said NOT replaced First ball Second ball 4 7 red red blue blue red blue To have at least one of them red we could have (red, red), (red, blue) or (blue, red) ( x ) + ( x ) + ( x ) = + + = = = Alternatively the only option we don t want is (blue, blue) with probability x = = So probability of at least one red is 1 - = = Page 12

13 Question 23 We can say y x 3 or y = kx 3 also that y or y = (note we need a different constant to k so used m) y = kx 3 and y = putting in the values we have been given we can work out what the constants k and m must be 2 = k x 2 3 and 2 = 2 = 8k divide both sides by 8 k = ¼ 2 = multiply both sides by 16 m = 32 We have y = ¼x 3 and y = so when x = 4, y = ¼ x 4 3 =16 and when y = 16, 16 = multiply both sides by z 16z = 32 divide both sides by 16 z = 2 It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by AQA. In addition these solutions may not necessarily constitute the only possible solutions. If you found this helpful then take a look at where you will find plenty more Page 13

Mathematics Higher Tier, June /2H (Paper 2, calculator)

Mathematics Higher Tier, June /2H (Paper 2, calculator) Link to past paper on AQA website: www.aqa.org.uk The associated question paper is available to download freely from the AQA website. To navigate around the website, choose QUALIFICATIONS, GCSE, MATHS,