(A/s) Mathematics Oct/Nov 2012 paper 4 variant 2
Mark scheme found on page 7 / 8 pages total, pdf
s engine = KE gain + PE gain + WD against resistance Work done is 1000 000 J (1000 kJ) A1ft 3 ft incorrect answer in (i) Special Ruling applying to part (i) for candidates who omit the weight component in applying Newton’s second law. (Max 3 out of 5) (i) v bottom = 30000/(1250 × 4 + 1
(A/s) Mathematics Oct/Nov 2009 paper 4 variant 2
Mark scheme found on page 6 / 6 pages total, pdf
s 1 = 0.6 × 50 2 – 0.004 × 50 3 (= 1000) B1 = + + 5. 27 50 1000 2 2 t s M1 For using ‘average speed = total distance /total time’ [1000 + 30t 2 = 27.5(50 + t 2 )] M1 For substituting s
(A/s) Mathematics May/June 2011 paper 1 variant 1
Mark scheme found on page 5 / 6 pages total, pdf
Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 11 © University of Cambridge International Examinations 2011 5 (i) θ θ θ 2 2 2 sin 1 sin sin 2 − = 1 0 1 sin sin 2 2 4 = − + θ θ AG (ii) 0 ) 1 )(sin 1 sin 2 ( 2 2 =
(A/s) Mathematics Oct/Nov 2019 paper 3 variant 1
Mark scheme found on page 8 / 17 pages total, pdf
9709/31 Cambridge International A Level – Mark Scheme PUBLISHED Question 4(i) Answer State October/November 2019 Marks Guidance B1 OE ( −10 = k × 1× 1000 ) dN = ke −0.02t N and show k = – 0.01 dt 1 4(ii) Separate variables correctly and integrate at least one sideB1 Obtain term ln NB1 OE Obtain term 0.
(A/s) Mathematics Oct/Nov 2008 paper 4
Mark scheme found on page 6 / 7 pages total, pdf
s second law to A (3 terms): (can be scored in (ii) by applying Newton’s second law to B instead) 0.5g – T = 0.5x2.5 A1ft Tension is 3.75N A1 [3] (ii) T – mg = 2.5m or 0.5g – mg = 0.5x2.5 + 2.5m B1ft ft from T – 0.5g
(A/s) Mathematics Oct/Nov 2004 paper 4
Mark scheme found on page 7 / 9 pages total, pdf
s second law; equation must contain F (or P / v ) and ma terms Equation F – 1130 + 1250 g sin3 o = 1250 x 0.2 contains not more than one error A1 Equation is correct A1 22000 = 725.8 v M1 For using P = Fv Speed is 30.3 ms -1 A1 5 4 (i) Gain
(A/s) Mathematics Oct/Nov 2019 paper 1 variant 3
Question paper found on page 7 / 20 pages total, pdf
s−1 . Find the rate of increase of the volume at this instant. [4] .....................................................................................................................
(A/s) Mathematics May/June 2016 paper 4 variant 1
Mark scheme found on page 5 / 7 pages total, pdf
s second law with 3 terms to the car a = 132/1000 = 0.132 ms – 2 A1 [3] 4 P cos θ = 48 cos α – 14 sin α and/or P sin θ = 50 – 48 sin α –14 cos α M1 For resolving forces horizontally and/or vertically P cos θ = 48(24/25) – 14(
(A/s) Mathematics Oct/Nov 2009 paper 7 variant 1
Mark scheme found on page 6 / 6 pages total, pdf
Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9709 71 © UCLES 2009 7 (i) 3 C ~ N(990, 5.2 2 × 3) (= N(990, 81.12)) 2 W ~ N(1000, 7.1 2 × 2 (= N(1000, 100.82)) 3 C
(A/s) Mathematics May/June 2017 paper 6 variant 1
Question paper found on page 9 / 12 pages total, pdf
9 (b) The lengths of metal rods have a normal distribution with mean 16 cm and standard deviation 0.2 cm. Rods which are shorter than 15.75 cm or longer than 16.25 cm are not usable. Find the expected number of usable rods in a batch of 1000 rods. [4] ...............................
(A/s) Mathematics May/June 2017 paper 6 variant 1
Mark scheme found on page 10 / 12 pages total, pdf
9709/61 Question 6(a)(i) May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Answer Marks B1 rounding to ±0.674 or 0.675 z = 0.674 0.674 = Guidance M1 standardising, no cc, no sq rt, no sq, σ may still be present on RHS 6.8 − µ 0.25 µ M1 subst and
(A/s) Mathematics Oct/Nov 2004 paper 3
Mark scheme found on page 10 / 10 pages total, pdf
Page 5 Mark Scheme Syllabus Paper A AND AS LEVEL – NOVEMBER 2004 9709 3 © University of Cambridge International Examinations 2005 10 (i) State or imply t h t V d d d d 1000 = B1 State or imply h m t h h k t V − = − = 03 . 0 30 d d or d d B1 Show that k
(A/s) Mathematics May/June 2019 paper 6 variant 1
Question paper found on page 6 / 16 pages total, pdf
6 4 The Mathematics and English A-level marks of 1400 pupils all taking the same examinations are shown in the cumulative frequency graphs below. Both examinations are marked out of 100. 1500 1400 1300 1200 English 1100 Mathematics 1000 900 800 700 600 500 400 300 200 100 0 0 10 20
(A/s) Mathematics Oct/Nov 2003 paper 6
Question paper found on page 2 / 4 pages total, pdf
2 1 A computer can generate random numbers which are either 0 or 2. On a particular occasion, it generates a set of numbers which consists of 23 zeros and 17 twos. Find the mean and variance of this set of 40 numbers. [4] 2 The floor areas, x m 2 , of 20 factories are as follows. 150 350 450 578 595 64
(A/s) Mathematics May/June 2020 paper 4 variant 2
Question paper found on page 9 / 12 pages total, pdf
s−1 . The car travels up this section of the road at constant speed with the engine working at 60 kW. Find this constant speed. [5] ...........................................................................................
(A/s) Mathematics May/June 2020 paper 4 variant 2
Mark scheme found on page 10 / 11 pages total, pdf
9709/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED Question 5(a)(i) Answer May/June 2020 Marks B1 DF = 750 Power = their (750) × 32 = 24kW B1 FT 2 5(a)(ii) 16000 = DF × 32 DF = 500M1 500 − 750 = 1250 × aM1 a =
(A/s) Mathematics May/June 2013 paper 4 variant 3
Mark scheme found on page 4 / 7 pages total, pdf
s Speed is 16.9 m s -1 A1 [4] Alternative Scheme 1 WD against friction = 0.02 W cos α × 50 B1 PE loss = W × 50 sin α B1 M1 For using Gain in KE = Loss in PE – WD against friction Speed is 16.9 m s -1 A1 [4] 2 (i) M1 PE loss = B’s loss
(A/s) Mathematics Oct/Nov 2019 paper 3 variant 1
Question paper found on page 6 / 20 pages total, pdf
6 4 The number of insects in a population t weeks after the start of observations is denoted by N . The population is decreasing at a rate proportional to N e−0.02t . The variables N and t are treated as dN continuous, and it is given that when t = 0, N = 1000 and = −10. dt (i) Show that N and t satisfy the
(A/s) Mathematics Oct/Nov 2016 paper 6 variant 1
Mark scheme found on page 4 / 6 pages total, pdf
Page 4 1 2 Mark Scheme Cambridge International AS/A Level – October/November 2016 M1 ±0.674 seen M1 Standardising no cc, no sq, no sq rt k = 24.7 A1 diff 0 1 2 prob 6/36 10/36 8/36 3 4 5 6/36 4/36 2/36 Expectation = (0+10+16+18+16+10)/36
(A/s) Mathematics Oct/Nov 2022 paper 3 variant 3
Mark scheme found on page 14 / 15 pages total, pdf
Cambridge International AS & A Level – Mark Scheme PUBLISHED 9709/33 Question 10(b) Answer Marks Separate variables and integrate one side Guidance M1 Obtain terms −100ln (30 − 0.01V ) and t, or equivalent A1 FT FT their a and b. + A1 FT Evaluate a constant, or use t = 0, V = 0 as limits, in a solution containing
(A/s) Mathematics Feb/March 2019 paper 1 variant 2
Mark scheme found on page 9 / 15 pages total, pdf
9709/12 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question 5(ii) Answer Marks March 2019 Guidance M1 Correct method for scalar product 0 8 u = 2 v = −1 u.v = ‒2 ‒ 42 6 −7 |u| × |v| =22 + 62
(A/s) Mathematics May/June 2014 paper 4 variant 1
Mark scheme found on page 5 / 6 pages total, pdf
s second law for both particles and eliminating T, or using (M + m)a = (M – m)g Acceleration is 5 ms – 2 A1 M1 For using s = 0 + 2 1 at 2 Distance is 0.9 m A1 [4] (ii) 2 1 0.6 × V = 0.9 → V = 3 B1 ft distance in (i) M1 For using 0
(A/s) Mathematics Feb/March 2016 paper 1 variant 2
Mark scheme found on page 5 / 6 pages total, pdf
Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9709 12 © Cambridge International Examinations 2016 6 (i) 2 22 ππ =+ Arrh 2 2 1000 1000 π π =→= rhh r Sub for h into A → 2 2000 2 π =+ Ar r AG B1 M1 A1 [3] (ii) 2 d2
(A/s) Mathematics Oct/Nov 2022 paper 3 variant 3
Question paper found on page 14 / 20 pages total, pdf
14 10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time t minutes after filling begins the volume of water in the pool is V litres. The pool has a small leak and loses water at a rate of 0.01V litres per minute. The differential equation satisfied by V and t
(A/s) Mathematics May/June 2019 paper 4 variant 2
Question paper found on page 10 / 16 pages total, pdf
s−1 , where v > 2, the resistance to motion of the car is Av + B N, where A and B are constants. The car can travel along a horizontal road at a steady speed of 18 m s−1 when its engine is working at 36 kW. The car can travel up a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.0
(A/s) Mathematics May/June 2007 paper 4
Mark scheme found on page 5 / 6 pages total, pdf
s Q = 1.5t 2 – 0.1t 3 (+ C) s Q (10) = 50 (or s Q (5) = 25) Greatest velocity is 10 ms -1 M1 A1 M1 A1ft M1 A1 6 For using ∫ = dt v s Q Q For using limits 0 to 10 or equivalent (or 0 to 5 if the candidate states or implies
(A/s) Mathematics May/June 2019 paper 4 variant 2
Mark scheme found on page 11 / 13 pages total, pdf
1000 g × 1/20]M1 Use DF = resistance + weight component (case 2) 12A + B = 1250 oeA1 Correct equation, unsimplified DM1 Solve two simultaneous equations in A and B only for A or B Dependent on both previous M1’s A = 125, B = –250 A1 Both correct 7 © UCLES 2019 Page 11 of 13
(A/s) Mathematics May/June 2011 paper 1 variant 1
Question paper found on page 3 / 4 pages total, pdf
S O 2 q r In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = 2 θ radians. Arc PST has centre O and radius r , and the line ASB is a tangent to the arc PST at S . (i) Find the total area of the shaded regions in terms of r and θ . [4] (ii) In the case where θ =
(A/s) Mathematics Oct/Nov 2013 paper 4 variant 1
Mark scheme found on page 6 / 6 pages total, pdf
t ) = –0.2t 2 + 50 A1 At A, –0.2t 2 + 50 = 0 t = √250 B1 M1 For integrating v(t) and using s(10) = 100 s(t) = –t 3 /15 + 50t – 1000/3 A1 M1 For finding s(√250) Distance OA is 194 m A1 7
(A/s) Mathematics Oct/Nov 2017 paper 1 variant 3
Mark scheme found on page 4 / 15 pages total, pdf
9709/13 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question 1 Answer Marks ½ n − 24 + ( n − 1 ) 6 ~ 3000 October/November 2017 Guidance M1 Use correct formula with RHS ≈ 3000 (e.g. 3010). Note: ~ denotes any inequality or equality ( 3 ) ( n 2 − 5 n −
(A/s) Mathematics Oct/Nov 2022 paper 4 variant 3
Mark scheme found on page 9 / 12 pages total, pdf
s = 12t − 8t dt s = 4t − 4t B1 Solves v X = vY . 10 3 *M1 Uses s = v dt . s X = 6t 2 + 12t dt s X = 2t 3 + 6t 2 2 3 3 10 10 sX = 2 + 6 3 3
(A/s) Mathematics Oct/Nov 2019 paper 1 variant 3
Mark scheme found on page 7 / 15 pages total, pdf
S 3 V= 1 1 30 dV 3 2 SOI when S = 700 → = = ×S × 7 7 7 7 14 dS 2 S2 DM1 dS 1 dS dV dV = × OE used with dt = 2 and their 14 d t dS dt 30 A1 OE 30 or 4.29 7
(A/s) Mathematics Oct/Nov 2008
Examiner report found on page 10 / 17 pages total, pdf
s second law to A and to B . This leads to a pair of simultaneous equations in three unknowns, a , T and m . Most such candidates were able to find the route to a which is independent of T and m and, after finding the value of a , were able to find T and m . Answers : (i)(a) 2.5 ms − 2 , (b) 3
(A/s) Mathematics Feb/March 2018 paper 6 variant 2
Mark scheme found on page 6 / 12 pages total, pdf
s cales starting at (0 0,0) and axes labe elled cf and time in i mins, all l points correct; (a allow straight line es or curves) 600 400 n in median attem mpt on increasing CF C graph M1 450 seen (indepen ndent); 200 0 0 5 10 15 TIME, IN M MINUTES t 0 3 4 5 6 8 10 0
(A/s) Mathematics May/June 2012 paper 1 variant 1
Mark scheme found on page 4 / 7 pages total, pdf
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 11 © University of Cambridge International Examinations 2012 1 tan 2 x = 2 2 x = 63.4 or 243.4 x = 31.7 or 121.7 (allow 122) M1 A1 A1A1 [4] 1 solution sufficient For 2 nd A1 allow
(A/s) Mathematics Oct/Nov 2016 paper 6 variant 1
Question paper found on page 2 / 4 pages total, pdf
2 1 The random variable X is such that X ∼ N 20, 49 . Given that P X > k = 0.25, find the value of k . [3] 2 Two fair six-sided dice with faces numbered 1, 2, 3, 4, 5, 6 are thrown and the two scores are noted. The difference between the two scores is defined as follows. • If the scores
(A/s) Mathematics May/June 2013 paper 7 variant 3
Question paper found on page 3 / 4 pages total, pdf
3 6 Calls arrive at a helpdesk randomly and at a constant average rate of 1.4 calls per hour. Calculate the probability that there will be (i) more than 3 calls in 2 1 2 hours, [3] (ii) fewer than 1000 calls in four weeks (672 hours). [4] 7 In the past the weekly profit at a store had mean $34 600 and
(A/s) Mathematics May/June 2016 paper 4 variant 2
Mark scheme found on page 6 / 8 pages total, pdf
s second law to the car (3 terms) a = –0.5 ms – 2 or d = 0.5 ms – 2 A1 3 (ii) DF = 1100 g sin 8 + 1550 [= 3081] M1 For stating the equilibrium of the three forces 80000 = 3081 v M1 For using P = Fv with F involving a weight and a
(A/s) Mathematics May/June 2002
Examiner report found on page 33 / 43 pages total, pdf
s must be clearly done with an inequality statement and/or a diagram with both values clearly shown. The conclusion must then be drawn with no contradictions. Answers : (i) 14.2; (ii) 50.3; (iii) Reject exam board’s claim. Question 5 This was a poorly attempted question. Many candidates indicated that they knew what was meant by a type I and type II error but were
(A/s) Mathematics Oct/Nov 2002 paper 4
Question paper found on page 2 / 4 pages total, pdf
s 1 .[3] 2 A man runs in a straight line. He passes through a fixed point A with constant velocity 7 m s 1 at time t 0. At time t s his velocity is v ms 1 . The diagram shows the graph of v against t for the period 0 t 40. (i) Show that the man runs more than 154 m in the first 24
(A/s) Mathematics Oct/Nov 2004 paper 3
Question paper found on page 4 / 4 pages total, pdf
s − 1 . At the same time, water begins to flow out at a rate proportional to √ h ,where h m is the depth of the water at time t s. When h = 1, d h d t = 0.02. (i) Show that h satisfies the differential equation d h d t = 0.01 ( 3 − √ h ) . [ 3 ] It is
(A/s) Mathematics Oct/Nov 2016 paper 6 variant 2
Mark scheme found on page 5 / 6 pages total, pdf
s) (5×40+10×30+17.5×18+32.5×34+62.5×8)/130 = 2420/130 = 18.6 thousand (iii) median group = 8 – 12 thousand LQ group = 3 – 7 thousand [5] Labels fd and capacity (thousands) Correct horizontal scale required. Vertical scale linear from 0 Σ fx/130 where x is mid point attempt
(A/s) Mathematics Oct/Nov 2013 paper 4 variant 1
Mark scheme found on page 4 / 6 pages total, pdf
s 2 nd law, F = μR and R = mg Decelerations of P and Q are 2 ms –2 and 2.5 ms –2 . A1 2 (ii) M1 For using s = ut + ½ at 2 and s P = s Q + 5 8t – t 2 = 3t – 1.25t 2 +5 A1 t = √120 – 10 (=0.
(A/s) Mathematics Oct/Nov 2013 paper 1 variant 1
Mark scheme found on page 6 / 6 pages total, pdf
S formulae Substitute or divide Ignore any other solns for r and a 10 (i) () [] [] 2 2 3 3 d d 2 − × − = x x y At 24 d d , 2 1 − = = x y x − − = − 2 1 24 8 x y 20 24 + − = x y (ii)
(A/s) Mathematics Oct/Nov 2009 paper 4 variant 1
Question paper found on page 2 / 4 pages total, pdf
s kinetic energy as it moves from A to B . [4] 2 A B C D E 6 m s –1 0.65 m A smooth narrow tube AE has two straight parts, AB and DE , and a curved part BCD . The part AB is vertical with A above B , and DE is horizontal. C is the lowest point of the tube and is 0.65 m below
(A/s) Mathematics Oct/Nov 2013 paper 1 variant 1
Question paper found on page 4 / 4 pages total, pdf
clearance have unwit tingly been included, the publisher will be pleased to make amends at the earliest possible opportuni ty. University of Cambridge International Examinations is par t of the Cambridge Assessment Group. Cambridge Assessment i s the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is i tself a department of the University of Cambridge. © UCLES 2013 9709/11/O/N/13
(A/s) Mathematics May/June 2010 paper 6 variant 2
Question paper found on page 2 / 4 pages total, pdf
2 1 The times in minutes for seven students to become proficient a t a new computer game were measured. The results are shown below. 15 10 48 10 19 14 16 (i) Find the mean and standard deviation of these times. [2] (ii) State which of the mean, median or mode you consider would be m ost appropriate to use as a measure of central tendency
(A/s) Mathematics Oct/Nov 2010 paper 7 variant 2
Mark scheme found on page 4 / 6 pages total, pdf
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9709 72 © UCLES 2010 1 0.605 ± z × 1000 ) 605 . 0 1 ( 605 . 0 − × M1 z = 1.645 seen [0.580, 0.630] B1 A1 [3] Allow [0.58, 0.63
(A/s) Mathematics Oct/Nov 2010 paper 7 variant 1
Mark scheme found on page 4 / 6 pages total, pdf
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9709 71 © UCLES 2010 1 0.605 ± z × 1000 ) 605 . 0 1 ( 605 . 0 − × M1 z = 1.645 seen [0.580, 0.630] B1 A1 [3] Allow [0.58, 0.63
(A/s) Mathematics May/June 2011 paper 6 variant 1
Question paper found on page 3 / 4 pages total, pdf
3 6 There are 5000 schools in a certain country. The cumulative f requency table shows the number of pupils in a school and the corresponding number of schools. Number of pupils in a school ≤ 100 ≤ 150 ≤ 200 ≤ 250 ≤ 350 ≤ 450 ≤ 600 Cumulative frequency 200 800 1600 2100 4100 4700 5
(A/s) Mathematics Oct/Nov 2011 paper 4 variant 1
Mark scheme found on page 5 / 5 pages total, pdf
s = 2/5 × 2 5 16 . 0 t – 0.016 3 t /3 M1 For using s = ∫ vdt A1 Distance is 1070 m A1 3 (iv) 0 ) 016 . 0 192 . 0 ( 3 1 2 5 = − t t M1 For attempting to solve s( t ) = 0 Value of t is 144 A1
(A/s) Mathematics May/June 2012 paper 6 variant 1
Mark scheme found on page 5 / 6 pages total, pdf
s.f. (Accept 0.0878 to 0.0889) 6 (i) σ μ − = − 6 253 . 1 0.648 = σ μ − 12 μ = 9.9 σ = 3.15 or 3.16 (ii) need P(z < –1 or z > 1) = 1 – Ф 1) + Ф (–1) = 2 – 2 × 0.
(A/s) Mathematics May/June 2016 paper 4 variant 1
Question paper found on page 2 / 4 pages total, pdf
s. The lift then travels for 6 s at constant speed and finally slows down, with a constant decele ration, stopping in a further 4 s. (i) Sketch a velocity-time graph for the motion. [3] (ii) Find the total distance travelled by the lift. [2] 2 A box of mass 25 kg is pulled, at a constant speed, a distance of 36 m up a
(A/s) Mathematics Feb/March 2016 paper 1 variant 2
Question paper found on page 2 / 8 pages total, pdf
2 1 (i) Find the coefficients of x 4 and x 5 in the expansion of 1 − 2 x 5 . [2] (ii) It is given that, when 1 + px 1 − 2 x 5 is expanded, there is no term in x 5 . Find the value of the constant p . [2] 2 A curve for which d y d x = 3
(A/s) Mathematics Oct/Nov 2009 paper 4 variant 1
Mark scheme found on page 4 / 6 pages total, pdf
. . . . . (Third alternative) M1 For force diagram showing P, Q and –R, and values of any 2 angles at ‘O’ shown. Angle between P and Q is 120 o and between P and –R is 140 o A1 [Q/sin140 o = 12/sin120 o ] M1 For using Lami’s theorem Q = 8.91 A1 4
(A/s) Mathematics Oct/Nov 2010 paper 7 variant 2
Question paper found on page 2 / 4 pages total, pdf
2 1 In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a 90% confidence interval for the proportion of adults in the whole populatio n who use email. [3] 2 People arrive randomly and independently at a supermarket c heckout at an average rate of 2 people every 3 minutes. (i) Find the probability that exactly 4 people arrive in a 5
(A/s) Mathematics Oct/Nov 2010 paper 7 variant 1
Question paper found on page 2 / 4 pages total, pdf
2 1 In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a 90% confidence interval for the proportion of adults in the whole populatio n who use email. [3] 2 People arrive randomly and independently at a supermarket c heckout at an average rate of 2 people every 3 minutes. (i) Find the probability that exactly 4 people arrive in a 5
(A/s) Mathematics May/June 2013 paper 4 variant 2
Mark scheme found on page 6 / 7 pages total, pdf
s(20) s(20) = ½(8 × 8) + ½(8 + 2) × 12 (= 92) A1 M1 For using the gradient property to find acceleration in 3 rd phase a = (6.5 – 2)/6 (= 0.75) A1 [s(t) = 92 + 2(t – 20) + 0.375(t – 20) 2 M1 Displacement
(A/s) Mathematics May/June 2015 paper 7 variant 2
Mark scheme found on page 5 / 6 pages total, pdf
Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9709 72 © Cambridge International Examinations 2015 5 (i) 14800/50 or 296 2 504390000 '296' 4950 − (= 187.755) = 188 (3 sf) B1 M1 A1 3 Oe (ii) 2
(A/s) Mathematics Oct/Nov 2012
Examiner report found on page 36 / 59 pages total, pdf
s second law and P = Fv are necessary, and many were successful in obtaining the values of v at the bottom and the top of the hill. Unfortunately a few candidates did not use Newton’s second law and used the given values of the accelerations as the values of the velocities. A very considerable proportion of candida tes, who used Newton’s second law and P = Fv , failed to
(A/s) Mathematics May/June 2006
Examiner report found on page 11 / 19 pages total, pdf
s second law. The equation thus derived contains four terms, one of which requires as a preliminary the calculation of the acceleration of the block. Not surprisingly the equati on rarely contained all four terms, the component of weight and the mass-acceleration term being the most frequent absentees. Another error using this method reflects a confusion fo r some candidates between the angle of inclination of the hill and the angle α
(A/s) Mathematics Oct/Nov 2012 paper 4 variant 2
Question paper found on page 3 / 4 pages total, pdf
s engine is constant and equal to 30 000 W. The car’s acceleration is 4 ms − 2 at the bottom of the hill and is 0.2 ms − 2 at the top. The resistance to the car’s motion is 1000 N. Find (i) the car’s gain in kinetic energy, [5] (ii) the work done by the car’s engine. [3] 7
(A/s) Mathematics Oct/Nov 2003 paper 4
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2003 paper 2
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2003 paper 1
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics May/June 2006 paper 3
Question paper found on page 2 / 4 pages total, pdf
2 1 Given that x = 4 ( 3 − y ) ,express y in terms of x .[3] 2 Solve the inequality 2 x > | x − 1 | .[4] 3 The parametric equations of a curve are x = 2 θ + sin 2 θ , y = 1 − cos 2 θ . Show that d y d x = tan θ .[5] 4(i) Express
(A/s) Mathematics Oct/Nov 2003 paper 6
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2003 paper 3
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2003
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2008 paper 4
Question paper found on page 3 / 4 pages total, pdf
s. The journey has three stages. In the first stage the train starts from rest at A and accelerates uniformly until its speed is V ms − 1 .Inthe second stage the train travels at constant speed V ms − 1 for 600 s. During the third stage of the journey the train decelerates uniformly, coming to rest at B . (i) Sketch the velocity-time graph for the train’s
(A/s) Mathematics Oct/Nov 2003 paper 7
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics Oct/Nov 2003 paper 5
Mark scheme found on page 28 / 33 pages total, pdf
Page 1 Mark Scheme Syllabus Paper AICE AND A AND AS LEVEL – NOVEMBER 2003 9709/0390 6 © University of Cambridge Local Examinations Syndicate 2003 1 x 0 2 freq 23 17 OR P(0) = 23/40, P(2) = 17/40 Mean = 34/40 = 0.850 Variance = 2 ) 85 . 0 ( 40 / )
(A/s) Mathematics May/June 2004
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics May/June 2004 paper 5
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics May/June 2004 paper 3
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics Oct/Nov 2004 paper 5
Question paper found on page 2 / 4 pages total, pdf
s − 1 and 40 m s − 1 respectively. (ii) Find the time taken for the car to travel from A to B .[4] 4 A particle is projected from a point O on horizontal ground with speed 50 m s − 1 at an angle θ to the horizontal. Given that the speed of the particle when it is at its highest point is 40 m s −
(A/s) Mathematics May/June 2012 paper 6 variant 1
Question paper found on page 3 / 4 pages total, pdf
s tanding next to his or her partner? [3] (ii) How many different arrangements are there if the 7 friends al l stand together and the 7 partners all stand together? [2] (b) A group of 9 people consists of 2 boys, 3 girls and 4 adults. In h ow many ways can a team of 4 be chosen if (i) both boys are in the team, [
(A/s) Mathematics May/June 2004 paper 6
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics May/June 2004 paper 7
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics Oct/Nov 2013 paper 6 variant 1
Mark scheme found on page 4 / 6 pages total, pdf
s dollars B1 B1 B1 B1 4 Linear scale or 5 values shown and labels or in heading, need thousands of dollars, Correct median Correct quartiles Correct end points of whiskers not through box (ii) 1.5 × 170 = 255 Expensive houses above 690 + 170 × 1.5 = 945 i.e. 957 and 986 thousands of dollars M1 A1 2
(A/s) Mathematics May/June 2004 paper 2
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics May/June 2004 paper 1
Mark scheme found on page 21 / 34 pages total, pdf
s ( t ) (iii) ( t 2 – 36)(1 – 0.25 t 2 ) = 0 M1 For attempting to solve s = 0 (depends on both method marks in (i)) or ∫ t vdt 0 =36 (but not –36) for t 2 by factors or formula method Roots of quadratic are 4, 36 A1 t = 2, 6 A1 ft 3
(A/s) Mathematics Oct/Nov 2013 paper 7 variant 2
Question paper found on page 2 / 4 pages total, pdf
2 1 Each computer made in a factory contains 1000 components. On average, 1 in 30 000 of these components is defective. Use a suitable approximate distri bution to find the probability that a randomly chosen computer contains at least 1 faulty compone nt. [4] 2 Heights of a certain species of animal are known to be normall y distributed with standard deviation 0.17 m. A conservationist
(A/s) Mathematics Oct/Nov 2013 paper 7 variant 1
Question paper found on page 2 / 4 pages total, pdf
2 1 Each computer made in a factory contains 1000 components. On average, 1 in 30 000 of these components is defective. Use a suitable approximate distri bution to find the probability that a randomly chosen computer contains at least 1 faulty compone nt. [4] 2 Heights of a certain species of animal are known to be normall y distributed with standard deviation 0.17 m. A conservationist
(A/s) Mathematics May/June 2016 paper 7 variant 3
Mark scheme found on page 4 / 6 pages total, pdf
Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 73 © Cambridge International Examinations 2016 Qu Answer Marks Notes 1 192.4 ± z 43.6 150 z = 2.326 to 2.329 191 to 194 (3 sf) M1 B1 A1 [3] Allow 43.6 150 Allow one side for M1
(A/s) Mathematics Oct/Nov 2013
Examiner report found on page 32 / 61 pages total, pdf
s = ut + ½ gt 2 with s the distance that A falls from the time the string breaks until A reaches the floor, and u = –1.6. Answer: (i) 4.2 N (ii) 0.6 s Question 7 (i) This part was reasonably well attempted by t hose who recognised the need to use calculus. (ii) Very few candidates recognised the need to use v(1
(A/s) Mathematics Oct/Nov 2004 paper 4
Question paper found on page 2 / 4 pages total, pdf
s − 2 .[5] 4 A lorry of mass 16 000 kg climbs from the bottom to the top of a straight hill of length 1000 m at a constant speed of 10 m s − 1 . The top of the hill is 20 m above the level of the bottom of the hill. The driving force of the lorry is constant and equal to 5000 N
(A/s) Mathematics May/June 2011
Examiner report found on page 49 / 63 pages total, pdf
Cambridge International Advanced Subsidiary Level and Advanced Level 9709 Mathematics June 2011 Principal Examiner Report for Teachers © 2011 Question 4 The candidates who made a serious attempt at th is question recognised the need for the use of combinations in parts (i) and (iii). (i) The three possible team options were often ov erlooked and some candidates gave 1 option of 3 combinations either correctly multiplied
(A/s) Mathematics May/June 2014 paper 7 variant 3
Question paper found on page 3 / 4 pages total, pdf
s that the digit 5 is appearing less often than it should. In order to test this claim the manufacturer u ses the machine to generate 25 digits and finds that exactly 1 of these digits is a 5. (i) Carry out a test of Max’s claim at the 2.5% significance level. [5] (ii) Max carried out a similar hypothesis test by generating 1000 digits between
(A/s) Mathematics May/June 2013 paper 1 variant 2
Mark scheme found on page 5 / 6 pages total, pdf
S π π + = → r r S π π 500 2 2 + = (ii) r d d S 2 500 4 r r π π − = = 0 when r ³ = 125 → r = 5 → S = 150 π (iii) 2 2 d d r S 3 1000 4 r π π + = This is positive → Minimum M1
(A/s) Mathematics May/June 2013 paper 4 variant 2
Question paper found on page 3 / 4 pages total, pdf
s speed increases to 8ms − 1 with constant acceleration. During the next 12 seconds P ’s speed decreases to 2ms − 1 with constant deceleration. P then moves with constant acceleration for 6 seconds, reaching A with speed 6.5 m s − 1 . (i) Sketch the velocity-time graph for P ’s motion. [2] The displacement of P from O , at time t seconds after