Mathematics Higher Tier, November /1H (Paper 1, non calculator)

Transcription

1 Link to past paper on AQA website: The question paper associated with these worked answers is available to download for free from the AQA website. You can navigate around the website in a few ways but one way is to go to QUALIFICATIONS, GCSE, MATHS, MATHEMATICS, MATHEMATICS A (LINEAR), KEY MATERIALS. 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 Each number needs to be rounded to 1 significant figure 52.3 = 50 (1 significant figure) 97.8 = 100 (1 significant figure) 19.4 = 20 (1 significant figure) We have = = 50 x 5 = 250 Question 2 a) angle c is vertically opposite angle a b) angle d is the alternate angle to angle f c) angle g is the corresponding angle to angle c Page 1

2 Question 3 a) and c) b) the points show a strong negative correlation (which means that as the birth rate increases the life expectancy decreases) the graph has a negative gradient d) See the red dashed lines on the graph. Draw a vertical line up from 16 on the x axis until it meets the line of best fit. Then draw a horizontal line from the line of best fit to meet the y axis. 76 years e) all the data that we have is in the range from 8 to 39. We do not know that the relationship we have noted would be the same for higher or smaller birth rates. We would have to extrapolate the data which may not be reliable Page 2

3 Question 4 In order to compare the two shampoos we must either have the same volume for each or the same price I am going to find the equivalent price for both shampoos for 50 ml 1 st shampoo = ml cost 7.50, 50 ml will cost 7.50/5 = nd shampoo = ml cost 12.80, 50 ml will cost 12.80/8 = st shampoo is cheaper so is the better value Question 5 Jenny earns 8 x 4.50 = 36 each weekend She saves one-third of this, so she saves 12 each week We need to know how many 12 there are in x 12 = 108 (8 x 12 = 96) Jenny will have enough money to buy her ipod after 9 weeks Page 3

4 Question 6 a) Marcus has stopped at the petrol station when the graph is a horizontal. He has gone 14 km. b) Marcus is at the petrol station from 0950 to 1002, a total of 12 minutes. c) i) ii) He has travelled 50 km and takes 40 minutes to get home. Halving each of these figures means that he would travel 25 km in 20 minutes. Now multiplying each figure by 3, this also means 75 km in 60 minutes Speed = 75 km/h Page 4

5 Question 7 a) i) front elevation (the view from the front) ii) side elevation (view from the side) b) the total surface area will be the total number of faces that are can be seen from all the angles. Each face is 1 cm 2. From the top (plan view) we can see 5 From the front we can see 5 From the side we can see 3 From the other side we can see 3 From the back we can see 5 From the bottom we can see 5 In total the surface area is 26 cm Page 5

6 Question 8 a) expand the brackets 10w 10 = 15 add 10 to both sides 10w = 25 divide both sides by 10 w = 2.5 b) expand the brackets 5t + 12 = 3t + 15 subtract 3t from both sides 2t + 12 = 15 subtract 12 from both sides 2t = 3 divide both sides by 2 t = 1.5 Question 9 The circumference of a circle is 2πr or πd (where r is the radius and d is the diameter) Here we have a semi-circle so we want half of the circumference of the whole circle The length of the wire is πd/2 = 3.14 x 40/2 = 3.14 x 20 = 62.8 cm There are four sides to a square Each side will be = 15.7 cm x = 15.7 cm Page 6

7 Question 10 Fuel costs Harry will drive 30,000 x 3 = 90,000 kilometres in total 90, = x 5 = 4500 litres of fuel Cost = 4500 x 1.20 = 5400 Road Tax and Insurance 450 x 3 = 1350 Total Servicing costs 500 (no adjustment needed) Total cost = = 7250 Question 11 a) relative frequency is number of blue discs divided by number of trials for example for 10 trials: 5/10 = 0.5 we have 0.4 = number of blue discs/50 multiply both sides by x 0.4 = number of blue discs 20 = total number of blue discs after 50 trials b) the relative frequency with the largest number of trials is the most reliable. This starts to give an idea of the probability of getting a blue disc. 0.3 x 40 = 12 There are about 12 blue discs in the bag Page 7

8 Question 12 a) add t to both sides y + t = swap sides = y + t multiply both sides by w x = w(y + t) b) label both equations eqn 1 and eqn 2 we can solve simultaneous equations by elimination or by substitution elimination method multiply eqn 1 by 2 so that we have 2x in both equations (and relabel as eqn 3) 4y = 2x + 12 eqn 3 y = 2x 3 eqn 2 now that we have 2x in both equations we can eliminate the x terms by subtracting eqn 2 from eqn 3 4y y = (2x + 12) (2x 3) 3y = = = 15 divide both sides by 3 y = 5 put back into eqn 1 to get x (2 x 5) = x = x + 6 subtract 6 from both sides x = 4 x = 4, y = 5 to check, put both back into eqn 2: 5 = (2 x 4) 3 = 8 3 = Page 8

9 substitution method from eqn 2 we have y = 2x 3 substitute this value for y into eqn 1 2 x (2x 3) = x + 6 expand the brackets 4x 6 = x + 6 subtract x from both sides 3x - 6 = 6 add 6 to both sides 3x = 12 divide both sides by 3 x = 4 put back into eqn 2 to get y y = (2 x 4) 3 = 8 3 = 5 x = 4, y = 5 to check, put both back into eqn 1: (2 x 5) = = Page 9

10 Question 13 a) median is between 32 and 33 minutes (say 33) draw a horizontal line across from 300 until it meets the graph then drop a vertical line down to meet the x axis. b) the upper quartile is 39 draw a horizontal line across from 450 (3/4 of 600) until it meets the graph then drop a vertical line down to meet the x axis the lower quartile is 24 draw a horizontal line across from 150 (1/4 of 600) until it meets the graph then drop a vertical line down to meet the x axis interquartile range = = 15 minutes c) draw a vertical line up from 55 to meet the graph then go horizontally across to meet the y axis at 570 this means that 570 students took less than 55 minutes and therefore 30 students took more than 55 minutes to calculate the percentage, turn into a fraction and multiply by 100 x = = 5% Page 10

11 Question 14 a) 5 x 3 = 15 x 4 x x 3 = x 4+3 = x 7 y 2 x y 7 = y 2+7 = y 9 15x 7 y 9 b) in order to add fractions we need a common denominator 6 is a multiple of 2 and 3 so use 6 as the denominator = = We have + = = multiply both sides by 6 5x = = divide both sides by 5 x = = Page 11

12 Question 15 If we first look at triangle PRS we have that tan x = 0.8 P opp S x 5 cm adj R tan x = 0.8 = = multiply both sides by 5 we have PR = 0.8 x 5 = 4 Now looking at triangle PQR we have cos y = 0.9 and now know that PR = 4 cm 4 cm P hyp R y adj Q cos y = 0.9 = = multiply both sides by 4 QR = 0.9 x 4 = 3.6 cm Page 12

13 Question 16 a)i) perimeter of A is = 10 cm perimeter of B is = 20 cm the scale factor of the perimeters will be 2 (20 10) or more simply as the scale factor for the lengths is 2 the scale factor for the perimeter will also be 2 ii) area scale factor will be 2 2 = 4 area of A is 2 x 3 = 6 cm 2, area of B is 4 x 6 = 24 cm = 4 b) we are going from the larger shape to the smaller so the scale factor for the lengths will be.. = the scale factor for the area will be ( )2 = area of P = 54 x = 6 cm2 Question 17 a) divide 3 by b) = = = one tenth of = Using a common denominator of = = = + = Page 13

14 Question 18 A negative power turns the fraction upside down ( )-4 = ( )4 = 2 4 = 16 Question 19 To prove that two triangles are congruent then we must prove that they are the same triangle (that is that they have the same length sides and the same angles). One way of doing this is by showing two sides are the same length and the one angle between them is equal (SAS) (as this uniquely defines a triangle) Length AC = length AC (as it is the same line) Length CR = length CP (as they are two sides of the same square) Angle ACR = angle ACP = 135⁰ (AC is the diagonal of a square so angle ACD = angle ACB = 45⁰, angle DCR = angle BCP = 90⁰ as it is the corner of a square, = 135⁰) We have SAS so the two triangles ACR and ACP must be congruent Question 20 a) the quadratic formula is: 4 2 -b = +5, so b = -5 2a = 6, so a = 3 4ac = 48 and we know a = 3 so 12c = 48, and c = 4 a = 3 b = -5 c = 4 b) i) the discriminant (b 2 4ac) is negative. We cannot square root a negative number so Joanne will not be able to find any solutions. ii) as there are no solutions this means that the graph does not meet the line y = 0. This is graph R Page 14

15 Question 21 a) ( ) 2 = ( ) ( ) = = = x 5 = 12 + (2 x 4 x 5) = 12 + (2 x 2 x 5 ) = b) if this triangle is right angled then Pythagorus s theorem will work Where a = 2 and b = and c is the longest side ( ) a 2 = 4 b 2 = (2 + 5) 2 = (2 + 5) (2 + 5) = = a 2 + b 2 = = we already know from part a) that c 2 = so a 2 + b 2 c 2 Pythagorus s Theorem does not work here so the triangle is not right angled Page 15

16 Question 22 a)(x 3) 2 = (x 3) (x 3) = x x 3x = x 2-6x + 9 b)i) this is a translation of 3 units in the positive x direction ii) we already know from part a) that y = x 2-6x + 9 can be written as y = (x 3) 2 so we want to know how we get from y = x 2 to y = (x 3) 2 we can see from the graph in part i) that this is a translation of 3 units in the positive x direction Page 16

17 Question 23 The only way that this is going to happen is if John takes a Green counter from Bag A, puts this in Bag B and then pulls out a Red counter from Bag B. The probability of John taking a Green counter from Bag A is Let the probability of John taking a Red counter from Bag B be p Then x p = multiply both sides by 6 p = = this means that at the time that John takes a counter from Bag B there were 4 Red and 3 Green. One of these Green counters had come from Bag A so there must have been 2 Green counters in Bag B at the start. Answer is 2 Green counters If you found these solutions helpful and would like to see some more then visit our website 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 Page 17