Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H)

Mathematics IGCSE Higher Tier, May 2009 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 a) Mathematics Biology 5 2 80 32 note 1

Total 7

Note 1: whatever we did to 5 to get to 80, we do to 2 80 ÷ 5 = 16 2 x 16 = 32 32 candidates for IGCSE biology b) Foundation 1 20 note 2

Higher 3

Total 4 80

Note 2: whatever we did to 4 to get to 80, we do to 1 80 ÷ 4 = 20 1 x 20 = 20 20 candidates for Foundation

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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 2 speed = distance ÷ time speed = 40, time = 13.25 (note that 15 minutes is 0.25 not 0.15) 40 = distance ÷ 13.25 multiply both sides by 13.25 40 x 13.25 = distance distance = 530 km

Question 3

We look at each point from (0,0). The point (4, 4) was 4 along and 4 up (from (0,0)) so it will now be 10 along and 10 up. The point (6,4) was 6 along and 4 up so will now be 15 along and 10 up. The point (6,8) was 6 along and 8 up so will now be 15 along and 20 up. The dashed tram lines help us to be sure that we have enlarged the triangle correctly.


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H)

Question 4 If we work out 0.35 of 10 we get 0.35 x 10 = 3.5. As this is not a whole number we know that this probability must be wrong because we can’t have 3.5 red beads (or 3.5 beads of any colour).

Question 5 a) p(p + 7) b) add 5x to both sides 4 = 5x + 2 rewrite 5x + 2 = 4 subtract 2 from both sides 5x = 2 divide both sides by 5 x= c) t3+6 = t9 d)3(4y + 5) – 5(2y + 3) 12y + 15 – 10y – 15 group terms 2y

Question 6

a) x 100 = 35% b) $204 represents 30% 204 ÷ 30 = $6.80 represents 1% 6.80 x 100 = $680 represents 100% Weekly pay is $680


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 7 label all the sides (opp, adj, hyp) from the point of view of the angle

hyp 3.6 cm 7.9 cm

opp

x⁰ adj

From SOHCAHTOA, we don’t have adj so we are left with SOHCAHTOA . sin x⁰ = = = 0.455696…

.

take the inverse sin (sin-1) of both sides x = sin-1(0.455696…) = 27.1⁰


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 8 A = 1, 3, 9, 27 B = 1, 3, 9 C = 2, 4, 6, 8 a) A B is everything that is in A together with everything that is in B

1, 3, 9, 27 b)A C is everything that is in both of A and B there is nothing that is in both of A and B so this set is empty the statement is true A C = c) B is completely contained within A. B A. There is no overlap with C and A or with C and B

A B

C

Question 9 Because AB is a tangent to the circle it will meet the radius (OA) at a right angle. We have a right angled triangle so can apply Pythagorus’ Theorem to calculate OB. a2 + b2 = c2 where c is the longest side 4.72 + 5.92 = OB2 OB2 = 56.9 square root both sides OB = 7.543 BC = OB – OC BC = 7.453 – 4.7 = 2.84 cm (3 significant figures)


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 10 a) Distance (d km)

Frequency

Midpoint

0 d 20 20 d 40 40 d 60 60 d 80 80 d 100 total

8 24 5 2 1 40

10 30 50 70 90

Midpoint x frequency 80 720 250 140 90 1280

Estimate of mean = 1280 ÷ 40 = 32 km

b) draw a vertical line up from 25 on the x axis to meet the curve then go across to meet the y axis. This meets at 12.5 so could take 12 or 13 (see red dashed line) 13 people walked less than 25 km c) draw horizontal lines across from the y axis at 30 and at 10. Where these meet the curve drop down to meet the x axis (see red dashed lines) the upper quartile is 38, the lower quartile is 22 interquartile range = 38 – 22 = 16 km


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 11 a)We can solve these by elimination or by substitution By elimination We want the same number of x or the same number of y Multiply the top equation by 5 and the bottom equation by 2 then we will have 10x in both 10x – 15y = 45 10x + 8y = 22 we need to subtract the first equation from the second equation to eliminate the x terms 8y - - 15y = 22 – 45 8y + 15y = -23 23y = -23 divide both sides by 23 y = -1 now put this back into the first equation to get x 2x – (3 x -1) = 9 2x - -3 = 9 2x + 3 = 9 subtract 3 from both sides 2x = 6 divide both sides by 2 x=3 We have x = 3 and y = -1 To check: put back into the second equation (5 x 3) + (4 x -1) = 15 + -4 = 15 – 4 = 11 By substitution rearrange the first equation to make y the subject add 3y to both sides 2x = 3y + 9 rewrite 3y + 9 = 2x subtract 9 from both sides 3y = 2x – 9 divide both sides by 3 y=

substitute this value for y into the second equation 5x + (4 x ) = 11 multiply by 3 15x + 4(2x – 9) = 33 expand 15x + 8x – 36 = 33 group terms www.chattertontuition.co.uk 0775 950 1629

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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) 23x – 36 = 33 add 36 to both sides 23x = 69 divide both sides by 23 x=3 Now substitute this value for x back into y = y=

"

= = =

-1

We have x = 3 and y = -1 To check: put back into the second equation (5 x 3) + (4 x -1) = 15 + -4 = 15 – 4 = 11 b) this is what we have just done (3, -1)

Question 12 a) 150 million = 150,000,000 = 1.5 x 108 b) 108 milliion ÷ 150 million = 108 ÷ 150 = 0.72 = 7.2 x 10-1 astronomical units


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 13 a) 70 = 2(4L + (4 x 2) + 2L) 70 = 2(6L + 8) divide both sides by 2 35 = 6L + 8 rewrite 6L + 8 = 35 subtract 8 from both sides 6L = 27 divide both sides by 6 L = 4.5 b) A = 2LW + 2HW + 2HL subtract 2WH from both sides A – 2HW = 2LW + 2HL rewrite 2LW + 2HL = A – 2HW factorise 2L(W + H) = A – 2HW divide both sides by 2 and by (W + H) L=

#$% %&$"

Question 14 a) i) angle AED = 47⁰ ii) this uses the fact that DE and AT are parallel to each other. The line AE goes through both of the parallel lines and “alternate angles are equal” b) ABCD forms a cyclic quadrilateral so angle BCD will be 180 – 56 = 124⁰ opposite angles in a cyclic quadrilateral add to give 180⁰ c)i) angle ADB will be 47⁰ ii) this is because of the “alternate segment theorem”


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 15 a) use the f(x) graph to see where this meets the line x = 3, draw this across to meet the x axis at 12 f(3) = 12 b) the graphs f(x) and g(x) meet in 3 places so there are three solutions x = 0.2, 3.6 and 6.1 c) g(1) = 5 f(5) = 0 so f(g(1)) = 0 d) draw by eye a tangent at the point (1, 16). On this tangent find two nice points ((1.5, 20) and (0, 8.5) in my case) gradient = difference in y coordinates divided by difference in x coordinates '. ((. gradient = (. = (. = 7.7


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 16 a) total surface area will be curved surface area plus area of base curved surface area given in formulae sheet is πrl where r = 4, l = 9 π x 4 x 9 = 36π = 113.097… area of base this is area of a circle given in formulae sheet is πr2 where r = 4 π x 14 = 16π = 50.265… total surface area = 113.097 + 50.265 = 163.362 = 163 cm2 (3 significant figures) b) Mathematically similar means that the cones are exactly the same shape and in the same proportions just that one might be bigger than the other. )

The scale factor for the lengths is =

The scale factor for the volume will be x x =

'


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 17 Second counter

First counter

3 4 1 5

1 4

1

3 5 1 5

3

3 5

1 4 1 4

1 4

5 1

2 4

5 1

3 4

3

3

i) x ) = = ( ii) we will get a sum of 6 if we have (1, 5), (3, 3) or (5, 1) ( (

( ( ( ( ' ) x ) + x )) + x )) = + + = = ( = )


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 18 We need to factorise the top and the bottom and hope that something will cancel To factorise the top we need to first multiply 5 by -3 to get -15. We need to find two numbers that multiply to get -15 but add to get +14. These two numbers are +15 and -1. Rewrite the expression splitting the 14x into +15x and -1x 5x2 + 15x – 1x – 3 factorise in pairs 5x(x + 3) – 1(x + 3) we should have the same thing in both brackets, we do factorise again (x + 3)(5x – 1) To factorise the bottom, we need to first factorise out the 2 2(25x2 – 1) then recognise that this is the difference of the squares 2(5x – 1)(5x + 1) putting the fraction back together again in factorised form & " (" (" &("

we can see that (5x – 1) is on the top and the bottom so can cancel each other out we are left with & " &("


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 19 We want the arc length AB plus the chord length AB Arc length AB the circumference of a full circle is 2πr we need to scale this down as we don’t have a full circle ' ' 2πr x = 2 x π x 6 x = 8.168… = 8.168 (keeping several decimal places as not finished yet) Chord length AB we can use cosine rule (given in formulae sheet) a2 = b2 + c2 – 2bccosA where A is the angle opposite the side a let a = AB, b = c = 6, A = 78⁰ AB2 = 62 + 62 – (2 x 6 x 6 x cos 78⁰) AB2 = 36 + 36 – 14.9696… = 57.030… square root both sides AB = 7.5518… = 7.552 (keeping several decimal places as not finished yet) Perimeter of shaded segment APB = 8.168 + 7.552 = 15.720 = 15.7 cm (3 significant figures)

Question 20 230 is only correct to 2 significant figures so could be as low as 225 (and would still get rounded to 230 to 2 sf) If the length of the side of the square is x then x2 = 225 square root both sides x = √225 = 15 lower bound for perimeter will be 4 x 15 = 60 cm


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Mathematics IGCSE Higher Tier, May 2009 4400/3H (Paper 3H) Question 21 We can use the cosine rule (given in formulae sheet) a2 = b2 + c2 – 2bccosA where A is the angle opposite the side a let a = x + 4, b = x + 6, c = x, A = 60⁰ (x + 4)2 = (x + 6)2 + x2 – (2 x (x + 6) x x x cos 60⁰) (x + 4)(x + 4) = (x + 6) (x + 6) + x2 – (2x (x + 6) x ½) (x + 4)(x + 4) = (x + 6) (x + 6) + x2 – (2x (x + 6) x ½)

x2 + 4x + 4x + 16 = x2 + 6x + 6x + 36 + x2 – (x(x + 6)) group terms and continue to expand x2 + 8x + 16 = 2x2 + 12x + 36– x2 – 6x x2 + 8x + 16 = x2 + 6x + 36 subtract x2 from both sides 8x + 16 = 6x + 36 subtract 6x from both sides 2x + 16 = 36 subtract 16 from both sides 2x = 20 divide both sides by 2 x = 10 cm

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