Mathematics IGCSE Higher Tier, June 2008 ... - Chatterton Tuition

Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Link to examining board: http://www.edexcel.com 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)1623 467467 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 when typing in to your calculator use brackets (3.6 x 4.8) ÷ (5.6 – 3.2) = 7.2 or calculate the top and bottom separately as 17.28 ÷ 2.4 = 7.2

Question 2 there are only red, black and white discs in the bag so the probabilities of each of these will add to 1 we know the probability of red is 0.6 so the probability of black or white is 0.4 1 – 0.6 = 0.4 we know the probability of black and white are the same (as there are the same number of discs) probability of black = ½ x 0.4 = 0.2

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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 3 a) and b)

a) shape Q is an enlargement of shape P by scale factor 2 (all sides are doubled) and the centre of enlargement is (1, 3) (where the red tramlines meet) b) R is a reflection of P in the mirror line y = x (shown by green dashed line) everywhere on this line the x co-ordinate equals the y co-ordinate so y = x

Question 4 Copper 3 210 g note 1

Tin 1

Bronze (total) 4 280 g

Note 1: whatever we did to Bronze to get from 4 to 280 g we do to Copper 280 ÷ 4 = 70 3 x 70 = 210 g


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 5 a) i) means intersection. We want all the members that are in set A that are also in set B. That is the members that appear in both sets. 1, 9, 17 ii) means union. We want all the members that appear in set A or in set B. 1, 5, 9, 13, 17, 25, 33 b) there aren’t any members that are in set A and also in set C. The sets do not overlap at all. means empty

Question 6 label the sides from the point of view of the angle

hyp

3 cm opp

x⁰ 8 cm

adj

we have opp and adj so need to use tan (TOA)

tan x = =

take the inverse tan (tan-1) of both sides

x = tan-1 () = 20.6⁰ (1 decimal place)


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 7 the formula for the circumference of a circle is given in formulae sheet C = 2πr or πd we have diameter of 7.8 cm so I will use C = πd C = π x 7.8 = 24.504 = 24.5 cm (3 significant figures)

Question 8 a) we are told that the number of sticks n in the pth term is 2p + 1 so n = 2p + 1 b) 2p + 1 = n subtract 1 from both sides 2p = n – 1 divide both sides by 2 p=

Question 9 a) expand the brackets 7x – 7 = 5 – 2x add 2x to both sides 9x – 7 = 5 add 7 to both sides 9x = 12 divide both sides by 9

x= b) i) subtract 5 from both sides 4x 16 divide both sides by 4 x4 ii) from part i) we know n 4, also n has to be a positive integer (so more than 0 and a whole number) so n could be 1, 2, 3 or 4 www.chattertontuition.co.uk 0775 950 1629

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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 10 a) $28,250 represents 100% we want to know what percentage $29,832 represents 29,832 – 28,250 = $1582 expressed as a fraction of the original salary we have

we can work out the percentage by multiplying this by 100

x 100 = 5.6%

b) if original salary is 100% and we increase by 5.2% then final salary will be 105.2% which can be written as 1.052 Pedro’s original salary was S then S x 1.052 = 28,141 divide both sides by 1.052 S = $26,750


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 11 a) mode means the “most often”, the modal class is the class that has the highest frequency modal class is 60 p 70 b) to calculate the mean we must first add two more columns to the table. We need mid- point and then mid-point x frequency. We find the total of the mid-point x frequency table and then divide by the total frequency. Pulse rate (p Frequency Mid-point (for pulse Mid-point x frequency beats/min) rate) 50 p 60 7 55 385 60 p 70 21 65 1365 70 p 80 15 75 1125 80 p 90 14 85 1190 90 p 100 3 95 285 Total 60 4350 mean = 4350 ÷ 60 = 72.5 beats/min

c) draw a horizontal line across from a cumulative frequency of 30 until it meets the curve. Then drop this down to meet the pulse rate axis (shown as red dashed line) median pulse rate is 124.5 beats/min d) draw a vertical line up from a pulse rate of 131 until it meets the curve. Then draw this across to meet the cumulative frequency axis at 48. We want the number with a pulse rate more than 131 so we need to subtract 48 from 60 60 – 48 = 12 www.chattertontuition.co.uk 0775 950 1629

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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 12 area of a circle is πr2 (from formulae sheet) If the radius of the small circle is r then the radius of the larger circle will be 3r we have πr2 = 4 we want π x (3r)2 = π x 9r2 = 9 x πr2 = 9 x 4 = 36 cm2

Question 13 the exterior angle of any regular polygon is 360⁰ divided by the number of sides the interior angle will then be 180⁰ minus the exterior angle interior angle of a regular hexagon is 180 – (360 ÷ 6) = 180 – 60 = 120⁰ the interior angle of a square is 90⁰ we know that angles on a point add up to 360⁰ interior angle of the unknown polygon is 360 – (120 + 90) = 360 – 210 = 150⁰ exterior angle of the unknown polygon is 180 – 150 = 30⁰ 360 ÷ 12 = 30⁰ so the unknown polygon has 12 sides 12 sides

Question 14 a) 5 goes into both terms 5(2y – 3) b) 3, p and q go into both terms 3pq(3p + 4q) c) i) we need to find two numbers that multiply to make -16 and add to make 6 the two numbers are +8 and -2 (x + 8)(x – 2) ii) (x + 8)(x – 2) = 0 if two things multiply to make 0 then one or the other of them equals 0 if x + 8 = 0 then x = -8 if x – 2 = 0 then x = 2 x = -8 or 2 www.chattertontuition.co.uk 0775 950 1629

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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 15 a) i) 57.5 kg (anything more would be rounded up to 58 or more kg) ii) 56.5 kg (anything less would be rounded to 56 or less kg) b) Alice’s weight is anywhere from 61.5 kg to 62.5 kg if we want the largest possible difference between the two weights then we want the biggest weight for Alice and the smallest weight for Mia 62.5 – 56.5 = 6 kg

Question 16

a) the probability that he takes one even card is (as there are 5 even cards out of a possible 9 cards) the probability that the second card he takes is also even will be the same (as he has replaced the first card) the probability that he picks two even numbers will be x = b) to get a sum of 43 he could have 20 and 23 21 and 22 22 and 21 23 and 20

Each of these four possibilities has a probability of x =

So the overall probability will be 4 x =


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 17 a) d √! this means d = k x √! where k is a constant to be found we know that h = 225, when d = 54, substituting these values in we get: 54 = k x √225 = k x 15 divide both sides by 15 k = 3.6 d = 3.6 √! b) when h = 64 m d = 3.6 x √64 = 3.6 x 8 = 28.8 km c) when d = 61.2 km 61.2 = 3.6 x √! divide both sides by 3.6 √! = 17 square both sides h = 289 m

Question 18 We have a triangle with angles but no right angles so cannot use basic trigonometry so this question will be about the sine rule or the cosine rule. Sine rule (given in formulae sheet) We are wanting a length so use the version of the sine rule with the sides on top $%& '

(

*

= $%& ) = $%& +

$%& ⁰

-.

= $%& -⁰

multiply both sides by sin 35⁰ a=

-. / $%& ⁰ $%& -⁰

= 4.3395 = 4.34 cm (3 significant figures)


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 19 we don’t want square roots on the bottom so multiply the top and bottom by √8 √

x

√ √ √ = = √

=

√ / / /√ =

= 3√2

Question 20 a) i) and ii) angles at the centre are twice the angle at the circumference 118 ÷ 2 = 59⁰ b) angles on a straight line add up to 180⁰ angle ACB = 180 – (x + 36) = 180 – x – 36 = 144 – x angle at the centre is twice the angle at the circumference x = 2 x (144 – x) x = 288 – 2x add 2x to both sides 3x = 288 divide both sides by 3 x = 96⁰


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 21 a) draw on the graph (at the point (3, 6.5)) a tangent (straight line that just touches the curve) Then find two co-ordinates on that line to work out the gradient

gradient =

122343*3 1 5 * 41637 122343*3 1 8 * 41637

two possible co-ordinates are (4,11) and (2,2) gradient = = = 4.5 b) to solve f(x) = 0 we need to see where the curve y = f(x) meets the line y =0 this happens when x = -1.7 c) draw another line on the graph at the points where x = -1 and x = 1. Join up these two points to make the new line. i)the y intercept of this line is where the line meets the y axis. This is when y =12 so c=12.

ii) the line meets the curve again when x = 4, so this is our third solution.


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) Question 22 If we split the kite down the middle we have two equal triangles

2 cm

2 cm

45⁰

45⁰

y

y

10 cm

10 cm x

x

We can work out the area of each triangle by using ½ab sinC from formulae sheet where C is the angle between the two sides a and b (here shown as angle y) first we need to calculate angle y we can use the sine rule to calculate angle x and then from this work out y $%& 8

=

$%&

multiply both sides by 2 sin x =

$%&

= 0.1414…

take the inverse sin (sin-1) of both sides x = sin-1 (0.1414…) = 8.1301… angles in a triangle add up to 180⁰ y = 180 – (45 + 8.1301) = 126.8699 area of triangle = ½ x 2 x 10 x sin 126.8699 = 8 cm2 area of kite = 2 x 8 = 16 cm2


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Mathematics IGCSE Higher Tier, June 2008 4400/3H (Paper 3H) If you liked this paper and want to see some more then click on this link: http://www.chattertontuition.co.uk/maths-revision-papers

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