Mathematics IGCSE Higher Tier, June /3H (Paper 3H)

Transcription

1 Link to examining board: The question paper associated with these solutions is available to download for free from the Edexcel website. The navigation around the website sometimes changes. However one possible route is to follow the above link, then SUBJECTS Mathematics, QUALIFICATIONS (from the LH Panel), under INTERNATIONAL GCSE FROM 2003 choose MATHEMATICS. Otherwise you can order the paper from the Edexcel Publications by phoning them on +44 (0) 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 Apple Fool For people For 1 person note 1 For 15 people note 2 Cooking apples 900 g 150 g 2250 g 750 g Sugar 100 g 1.7 g 250 g Double cream 300 ml 50 ml 750 ml For 5 people note 3 Note 1: we have the ingredients for people so divide each of these figures by Note 2: we now have the ingredients for 1 person so multiply each of these figures by 15 Note 3: from the ingredients for 1 person, multiply by 5 a) for 15 people we need: 2250 g of cooking apples 250 g of sugar 750 ml of double cream b) for 5 people we need: 750 g of cooking apples Question 2 a) i) for this we need to ignore the line PR altogether x = 2⁰ ii) alternate angles are equal b)i) here we need to ignore the line PQ altogether y = 71⁰ ii) corresponding angles are equal Page 1

2 Question 3 a) the three numbers have a median of 4 and as there are only three numbers then the middle number must equal 4 if we put a + 2, b + 2 and c + 2 in order of size, they will have the same order as a, b and c when in size order the median will be the middle number. This is 2 more than the previous median. Median = = b) the range of the numbers will be the largest take away the smallest this will be the same as the original range (7) as we have simply added 2 to the highest and to the lowest. The range is still 7. Question 4 a) 5(n + ) = 5n + 30 b) y c) 4(x 2) = 3 4x 8 = 3 add 8 to both sides 4x = 11 divide both sides by 4 x = 2.75 Question 5 a) we want to find of = x = x = x = b) if are women then this means that the total number must be a multiple of 12 (otherwise would give us a part of a person!) if wear glasses then this means that the total number must be a multiple of 8 (otherwise would give us a part of a person!) so our total number must be a multiple of both 12 and 8. We want the least common multiple. Multiple of 12 are 12, 24, 3, Multiples of 8 are 8, 1, 24, 32 The least common multiple is Page 2

3 Question a) the modal class is the class that has the highest frequency. This is 400 V 500. b) Volume of water (V m 3 ) Frequency Mid point Midpoint x frequency 0 V V V V V V Total Mean = = 395 c) Volume of water (V m 3 ) Cumulative Frequency 0 V V V V V V To get the cumulative frequency you just keep adding the next category to the previous Eg cum freq for 2 nd category is =, cum freq for 3 rd category is + = 12 The final cumulative frequency should be the same as the total frequency it is d) e) There are 80 families so the median will be the 40 th family. Draw a line across from 40 til it meets the curve then drop down to meet the x axis. (shown as a red dashed line). The median is 425 m Page 3

4 Question 7 We have a right angled triangle so we can use SOHCAHTOA Label all the sides from the point of view of the angle hyp.8 cm opp 41⁰ x cm adj We have hyp (H) and we are trying to find Adj (A). We don t have Opp (O) and don t want it. SOHCAHTOA this leaves CAH (cosine) cos 41⁰ = =. multiply both sides by.8.8cos 41⁰ = x x =.8 cos 41⁰ = 5.13 cm (3 significant figures) Question 8 It is really important that you read this question carefully. The $178 represents her income after tax has been deducted. Therefore $178 is equivalent to 7%. We want to work out what income is equivalent to 100%. If we divide $178 by 7 then we will know what 1% is worth. Then we scale this back up by multiplying by 100 to get what 100% is worth x 100 = $ Page 4

5 Question 9 a) this is a reflection in the mirror line y = -x b) c) this is a reflection in the mirror line y = Page 5

6 Question 10 a) -4 x 3 (as there is a full circle above the -4 we use, and as there is an open circle about the 3 we use ) b) i) subtract 9 from both sides 2x -8 divide both sides by 2 x -4 ii) integer means whole number n could be -3, -2, or -1 Question 11 a) area of a circle is (given in formulae sheet) πr 2 if diameter is 1 then radius will be half of this. Radius (r) = 8 cm area = π x 8 2 = 4π = 201 cm 2 (3 significant figures) b) substituting the values for h, x and y we have V = x π x 1.4 x ((3x (.52 )) + ( 3 x 8 2 ) ) x π x 1.4 x ( ) x π x 1.4 x = = 5050 cm3 (3 significant figures) Page

7 Question 12 a) x y note 1 13 note 2 2 note 3-9 note 4-14 note Note 1: (-2) 3 (12 x -2) + 2 = = = 18 Note 2: (-1) 3 (12 x -1) + 2 = = = 13 Note 3: (0) 3 (12 x 0) + 2 = = 2 Note 4: (1) 3 (12 x 1) + 2 = = -9 Note 5: (2) 3 (12 x 2) + 2 = = -14 b) c) i) = 3x2 12x ii) to find the gradient at a particular point, we first differentiate to get, then substitute our particular value for x into the equation for gradient = (3 x 5 2 ) (12 x 5) = 75 0 = 15 Question 13 angle PQS = angle PRS = 3⁰ because angles in the same segment are equal angle PQR = 90⁰ because angles in semicircle make 90⁰ (we know PQR is a semicircle since PR goes through the centre C it is a diameter of the circle) angle RQS = angle PQR angle PQS = 90 3 = 54⁰ Page 7

8 Question 14 a) i) AB is 20 cm, correct to 1 significant figures. It could be as low as 15 or as high as 25 (and still get rounded to 20) lower bound is 15 cm ii) BC is 8.3 cm, correct to 2 significant figures. It could be as low as 8.25 or as high as 8.35 (and still get rounded to 8.3) lower bound is 8.25 cm b) area of a triangle is ½ x base x height for this to be as low as possible we would want the base to be as small as possible and the height to be as small as possible. lower bound for area = ½ x 15 x 8.25 = cm 2 c) tan x⁰ = for this to be as small as possible we need the top (opp) to be as small as possible but the bottom (adj) to be as big as possible the smallest the opp could be is 8.25, the largest the adj could be is 25 lower bound for tan x⁰ = = 0.33 Question 15 a) E 1/r 2 can be written as E = k/r 2 where k is a constant to be determined substituting in the values we have for E and r 4 = k = k 2500 multiply both sides by = k k = 10,000 E = 10,000/r 2 b) substitute r = 20 E = 10, = 10, = 25 c) substitute E = = 10,000/r 2 multiply both sides by r x r 2 = 10,000 divide both sides by 100 r 2 = 10, =.25 square root both sides r =.25 = Page 8

9 Question 1 (3-5)(3-5) = = 14-5 Question 17 The two prisms are mathematically similar which means that all the dimensions are in the same proportion as each other. If we double one length, then all the other lengths are also doubled. We are trying to find the surface area of the larger prism so we want a scale factor with the larger number on top. We have the two lengths in common so our linear scale factor is = we are trying to find an area though which is two dimensional so our area scale factor is the square of the linear scale factor area scale factor = total surface area of Q = 544 x = 1224 cm2 Question 18 factorise the top and the bottom x 2 + x = x(x + ) x 2 3 = (x + )(x ) difference of the square so we have = Page 9

10 Question cent 10 cent 20 cent cent 10 cent 20 cent 5 cent 10 cent 20 cent 5 cent 10 cent 3 20 cent a) probability of two 20 cent coins is x = x = b) of the nine outcomes on the right the 2 nd, 3 rd and th outcome all satisfy this. 5 cent then 10 cent, or 5 cent then 20 cent, or 10 cent then 20 cent ( x ) + ( x ) + ( x ) = + + = Page 10

11 Question 20 Start by looking at triangle ABT (where T is the top of the flagpole). From this we can work out the length of TB (height of the flagpole). T opp A 13⁰ 40 Adj B tan 13⁰ = = multiply both sides by 40 40tan 13⁰ = TB TB = (keep plenty of decimal places as we are still working with this number) now we can look at triangle CBT, and use the fact that TB (height of flagpole) is T opp C 19⁰ Adj B tan 19⁰ = =. multiply both sides by BC BC x tan 19⁰ = divide both sides by tan 19⁰ BC = tan 19⁰ = Page 11

12 now we can go to triangle ABC and use Pythagoras Theorem to work out the length of AC C B 40 A AC 2 = = = square root both sides AC = = 48.2 m (3 significant figures) Page 12

13 Question 21 For both equations we have y as the subject so if they both equal y then they will equal each other 2x 2 = 3x + 14 rearrange to form a quadratic in x and with the quadratic equalling 0 subtract 3x from both sides 2x 2 3x = 14 subtract 14 from both sides 2x 2 3x 14 = 0 Try to factorise first multiply 2 by -14 to get -28, we must find two numbers that multiply to give -28 and combine to give -3 these two numbers are +4 and -7 rewrite the quadratic splitting the middle term into +4x and 7x 2x 2 + 4x 7x 14 = 0 factorise in pairs 2x(x + 2) 7(x + 2) = 0 we should have the same thing in both brackets which we do (x + 2) now factorise again with (x + 2) on the outside (x + 2)(2x 7) = 0 If two things multiply together to give 0 then one or the other of them must be 0 if x + 2 = 0 then x = -2 if 2x 7 = 0 then 2x = 7 and x = 3.5 now put these x values back into either of the two original equations to get y when x = -2, y = 2 x (-2) 2 = 2 x 4 = 8 when x = 3.5, y = 2 x = 24.5 To check, see if you get the same answer for y if put into the second equation Y = (3 x -2) + 14 = = 8, y = (3 x 3.5) + 14 = = 24.5 x = -2, y = 8 or x = 3.5, y = 24.5 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