Solved 0580 - m16 - QP - 22..''..typed [PDF]

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Cambridge



IGCSE



Cambridge International Examinations Cambridge International General Certificate of Secondary Education



CANDIDATE NAME CENTRE NUMBER



CANDIDATE NUMBER



* 7 5 8 5 9 5 5 1 0 7 *



0580/22



MATHEMATICS



February/March 2016



Paper 2 (Extended)



1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials:



Electronic calculator Tracing paper (optional)



Geometrical instruments



READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. If working is needed for any question it must be shown below that question. Electronic calculators should be used. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For π, use either your calculator value or 3.142. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 70.



The syllabus is approved for use in England, Wales and Northern Ireland as a Cambridge International Level 1/Level 2 Certificate.



This document consists of 12 printed pages. DC (LEG/FD) 112489/3 © UCLES 2016



CAMBRIDGE International Examinations



[Turn over



2 1



Solve (x – 7)(x + 4) = 0. Solution:



x2 + 4x - 7x - 28



x( x + 4 ) - 7( x + 4 ) (x+4)(x-7) either x + 4 = 0 or x - 7 = 0 x = 4 , -7



 2



4



-7



x= ................................. or x= .................................[1]



Factorise 2x – 4xy. Solution: 2x(1 - 2y)



2x(1 - 2y)



................................................... [2] 3



C



NOT TO SCALE



3.5 m



B



0.9 m



A



Calculate angle BAC. Solution:



By trignometrical ratios ; cos = adjacent / hypotenuse cos BAC = 0.9 / 3.5 = 0.2571 BAC = cos-1 0.2571 BAC = 75.10 0



75.1 Angle BAC= .................................................. [2]



4



Solve the inequality. Solution:



6n + 3 > 8n



=> 3 > 8n - 6n



=>



3 > 2n



=>



n 0.444 = 4 / 9



4/9







................................................... [2]



7



22.3 cm



NOT TO SCALE



25° 27.6 cm



Calculate the area of this triangle. Solution:



Area of Triangle



=



½ × AC × AB × sin A



= ½ × 22.3 × 27.6 × sin25 2 = 130.1 cm



130.1



 © UCLES 2016



0580/22/F/M/16



............................................cm2 [2] [Turn over



4 8



3 -2 m. Find the inverse of the matrix c -8 7 Solution:



1 / |det|



d a c b



( ) (



=> 1 / (3 × 7) - (-2 × -8)



=> 1 / 21 - 16 2 3



7 8



)



(



=>



7 8



2 3



1/5



)



(



7 8



2 3



)



f



7/5



8/5



9



7 Without using your calculator, work out 1 12 . + 13 20







You must show all your working and give your answer as a mixed number in its simplest form.



2/5



3/5



p



[2]



Solution: 19 / 12



+



13 / 20



95 / 60



+



39 / 60



95 + 39 / 60



134 / 60 67 / 30



67 / 30







................................................... [3]



10 The scale on a map is 1 : 20 000. The area of a lake on the map is 1.6 square centimetres.



Calculate the actual area of the lake. Give your answer in square metres. Solution: 1 cm2 = 40000 m2



1.6 cm2 = 1.6 × 40000 = 64000 m2



64000



..............................................m2 [3]



© UCLES 2016



0580/22/F/M/16



5 11



A



O



NOT TO SCALE



38° 25 cm B







The diagram shows a sector of a circle, centre O, radius 25 cm. The sector angle is 38°.







Calculate the length of the arc AB. Give your answer correct to 4 significant figures. Solution: Lenght of the arc AB = (x0 / 360) × 2 p r = (380 / 360) × 2 × p × 25 AB = 16.58 cm







16.58 AB= ............................................ cm [3]



12 A metal pole is 500 cm long, correct to the nearest centimetre. The pole is cut into rods each of length 5.8 cm, correct to the nearest millimetre.



Calculate the largest number of rods that the pole can be cut into. Solution: Upper bound of pole = 500 + 0.5 = 500.5 cm Lower bound of rod = 5.8 - 0.05 = 5.75 cm



Hence , larger number of rods = 500.5 / 5.75 cm = 87.04 cm



© UCLES 2016



0580/22/F/M/16



87.04 cm ................................................... [3]



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6 13 (a) Write 2016 as the product of prime factors. Solution:



2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 = 25 × 32 × 7



2 × 3 ×7 ................................................... [3] 5







2



(b) Write 2016 in standard form. 3



2.016 × 10 ................................................... [1]



14 Simplify.



(a) x 3 y 4 # x 5 y 3 Solution: 4 3 x3 x5 × y y



3+5



x



x8 y7 ................................................... [2]



4+3



y



8



x y7







3 (b) ^3p 2 m 5h



Solution:



= 33 p2×3 m5×3 = 27 p6 m15



6



© UCLES 2016



15



27 p m ................................................... [2]







0580/22/F/M/16



7 15 p° 2.8 cm



3.6 cm



NOT TO SCALE



5.3 cm



Find the value of p. Solution:



c2 = a2 + b2 – 2ab cos C,



cos p =( 2.82 + 3.62 - 5.32 ) / { 2(2.8)(3.6) } = - 7.29 / 20.16 = - 0.3616 0 p = cos-10.3616



p0 = 111.20 0



111.2 p= .................................................. [4]



 16 Raj measures the height, hcm, of 70 plants. The table shows the information. Height (hcm) Frequency



10  h 20



20  h 40



40  h 50



50  h 60



60  h 90



7



15



27



13



8



Calculate an estimate of the mean height of the plants. Solution:



mean = Sfx / Sf = {( 7×15 )+( 15×30 )+( 27×45 )+( 13×55 )+( 8+75 )} / 70 = ( 105 + 450 + 1215 + 715 + 600 ) / 70 = 3085 / 70 mean = 44.1 cm



44.1 ............................................................ cm [4]



 © UCLES 2016



0580/22/F/M/16



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8 17 Solve the equation 3x 2 - 11x + 4 = 0 . Show all your working and give your answers correct to 2 decimal places. Solution:



x = [ -b (under square root b2 - 4ac ) / 2a = [ -(-11) under square root (-11) - 4× 3 × 4 / 2(3) = 11 (under square root 121 - 48) / 6 = 11 (under square root 73) / 6 = 11 (+/-8.544004) / 6 2



either x = (11+8.544004)/6 x = 3.26 or 0.41



or



x=(11-8.544004)/6



3.26







© UCLES 2016



0.41



x=............................ or x= ............................[4]



0580/22/F/M/16



9 18 (a) x° NOT TO SCALE 47°







Find the value of x. Solution:







0



47 x= .................................................. [1]



X = 470 ( alternate angles )



(b) 85° 115°



NOT TO SCALE y°



97°



Find the value of y. Solution:



3600 = 850 + 1150 + 970 + x0 (sum of quadrilateral) x0 = 630 0 0 0 \ y = 180 - 63 (supplementary angles) 0 = 117







0



117 y= .................................................. [2]



(c) 58° NOT TO SCALE



z° O







The diagram shows a circle, centre O.







Find the value of z. Solution:







© UCLES 2016



x0 = 580 × 2 ( angle at center = 2 × angle at circumference) = 1160 z0 = 3600 - 1160 = 2440



0580/22/F/M/16



244 z= .................................................. [2] 0



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10 19



y 4



3



2



1



–3



–2



–1



0



1



2



3



4



x



–1



Find the four inequalities that define the region that is not shaded. y and =) –2 ................................................... y (> and =) 1/2x + 1 ................................................... y (< and =) -x + 3 ................................................... [5]



© UCLES 2016



0580/22/F/M/16



11 20 The nth term of a sequence is an 2 + bn .



(a) Write down an expression, in terms of aand b, for the 3rd term. a(3)2 + 3b 9a + 3b 9a + 3b ................................................... [1]







(b) The 3rd term of this sequence is 21 and the 6th term is 96. Find the value of aand the value of b. You must show all your working. Solution: (6)2a + 6b = 96 ----equation i 36a + 6b = 96 6a + b = 16 b = 16 - 6a



and



9a + 3b = 21----- equation ii



put b = 16 - 6a in equation ii 9a + 3(16 - 6a) = 21 9a + 48 - 18a =21 9a = 27 a=3 And b = 16 - 6a = 16 - 18 b = -2







hence a = 3 b = -2



3 a= ..................................................







-2 b= .................................................. [4]



Question 21 is printed on the next page.



© UCLES 2016



0580/22/F/M/16



[Turn over



12 21 Dan either walks or cycles to school. The probability that he cycles to school is 13 .



(a) Write down the probability that Dan walks to school. 2/3 ................................................... [1]







(b) When Dan cycles to school the probability that he is late is 18 .











When Dan walks to school the probability that he is late is 38 .











Complete the tree diagram. 1 8 1 3



Cycles



7/8 ..........



3 8 2/3 ..........



5/8



(c) Calculate the probability that











Not late Late



Walks



..........



Late



Not late



[2]



(i) Dan cycles to school and is late, Solution: 1/3 × 1/8 = 1/24



1/24 ................................................... [2]











(ii) Dan is not late. Solution:



1/3 × 7/8 or 2/3 × 5/8 (1/3 × 7/8) + (2/3 × 5/8) 7/24 + 10/24 = 17/24



17/24 ................................................... [3]



Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.



© UCLES 2016



0580/22/F/M/16