(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 1
Question paper found on page 17 / 24 pages total, pdf
17 (a) Show how the letter “E” can be represented by the eight 6-bit registers (four registers have been done for you). f e d c b a 0 0 0 0 0 0 1 2 3 1 2 3 4 5 6 7 8 0 0 0 0 0 0 4 0 0 0 0 0 0 5 6 7 0 0 0 0 0 0 8 [4]
(A/s) Computer Science Oct/Nov 2015 paper 3 variant 3
Question paper found on page 10 / 12 pages total, pdf
10 (b) The truth table for a logic circuit with four inputs is given below: INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1
(A/s) Computer Science Oct/Nov 2015 paper 3 variant 1
Question paper found on page 10 / 12 pages total, pdf
10 (b) The truth table for a logic circuit with four inputs is given below: INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1
(A/s) Computer Science Oct/Nov 2018 paper 3 variant 3
Question paper found on page 9 / 16 pages total, pdf
9 (b) A logic circuit with four inputs produces the following truth table. INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0
(A/s) Computer Science May/June 2018 paper 3 variant 1
Question paper found on page 9 / 16 pages total, pdf
9 (b) A logic circuit with four inputs produces the following truth table. INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 1 0
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 2
Question paper found on page 17 / 24 pages total, pdf
17 (a) Show how the letter “E” can be represented by the eight 6-bit registers (four registers have been done for you). f e d c b a 0 0 0 0 0 0 1 2 3 1 2 3 4 5 6 7 8 0 0 0 0 0 0 4 0 0 0 0 0 0 5 6 7 0 0 0 0 0 0 8 [4]
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 2
Mark scheme found on page 9 / 12 pages total, pdf
Page 9 Mark Scheme Cambridge IGCSE – October/November 2014 (d) A B Syllabus 0420 C D 1 1500 2 18 3 60 4 5 Paper 12 45 40 6 1800 1 mark profit 1 mark 1 mark [3] 12 (a) 1 mark for each of four rows shown in bold below; there are two possible ways of doing this – one set
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 1
Mark scheme found on page 9 / 12 pages total, pdf
Page 9 Mark Scheme Cambridge IGCSE – October/November 2014 (d) A B Syllabus 0420 C D 1 1500 2 18 3 60 4 5 Paper 11 45 40 6 1800 1 mark profit 1 mark 1 mark [3] 12 (a) 1 mark for each of four rows shown in bold below; there are two possible ways of doing this – one set
(A/s) Mathematics - Further May/June 2016 paper 1 variant 2
Mark scheme found on page 8 / 10 pages total, pdf
Page 8 Mark Scheme Cambridge International A Level – May/June 2016 Qu 10 Syllabus 9231 Solution 1 1 0 − 2 0 0 P = 0 1 1 , D = 0 − 1 0 or equivalent (in correct order) 0 0 1 0 0 1 1 − 1 1
(A/s) Mathematics - Further May/June 2016 paper 1 variant 1
Mark scheme found on page 8 / 10 pages total, pdf
Page 8 Mark Scheme Cambridge International A Level – May/June 2016 Qu 10 Syllabus 9231 Solution 1 1 0 − 2 0 0 P = 0 1 1 , D = 0 − 1 0 or equivalent (in correct order) 0 0 1 0 0 1 1 − 1 1
(A/s) Computer Science May/June 2017 paper 1 variant 1
Question paper found on page 11 / 16 pages total, pdf
11 (b) (i) A second data block is received as shown in the following table. There are errors in this data block. Identify and then circle two bits in the table which must be changed to remove the errors. 7 1 0 0 1 1 0 0 6 0 0 0 1 1 0 0 Bit position 5 4 3 2 0 0 1 1 1 0 0 0 1 1
(A/s) Computer Science May/June 2017 paper 1 variant 1
Question paper found on page 10 / 16 pages total, pdf
10 5 A computer receives data from a remote data logger. Each data block is a group of 8 bytes. A block is made up of seven data bytes and a parity byte. Each data byte has a parity bit using odd parity. The parity byte also uses odd parity. The following table shows a data block before transmission. Bit position 0 is the parity bit. 7 1 0 1 1
(IGCSE) Physics May/June 2018 paper 2 variant 1
Question paper found on page 12 / 16 pages total, pdf
12 32 The diagram shows two voltmeters P and Q connected to a potential divider. X V voltmeter P V voltmeter Q The sliding connection at point X is moved towards the top of the diagram. What happens to the reading on P and to the reading on Q? reading on P reading on Q A decreases decreases B decreases increases C increases decreases D increases increases 33 The diagram represents a
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 2
Question paper found on page 16 / 24 pages total, pdf
16 12 An advertising sign uses large LED characters controlled by a microprocessor. Each letter is formed from a grid made up of eight rectangles numbered 1 to 8: 1 2 3 4 5 6 7 8 For example, the letter “Z” is formed as follows: 1 2 4 5 7 8 Each rectangle has six LEDs that can light up; these LEDs are labelled “a” to “f”:
(A/s) Computer Science May/June 2015 paper 3 variant 3
Question paper found on page 14 / 16 pages total, pdf
14 Each greenhouse has eight sensors (numbered 1–8). • The byte at address 150 is used to store eight 1-bit flags. • A flag is set to indicate whether its associated sensor reading is waiting to be processed. • More than one sensor reading may be waiting to be processed at any particular moment. • Data received from the sensors is stored in a block of eight consecutive bytes
(A/s) Computer Science Oct/Nov 2015 paper 3 variant 2
Question paper found on page 10 / 12 pages total, pdf
10 (b) The truth table for a logic circuit with four inputs is given below: INPUT (i) OUTPUT P Q R S Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 1
(A/s) Computer Science Oct/Nov 2018 paper 3 variant 2
Question paper found on page 7 / 16 pages total, pdf
7 (c) The truth table for a logic circuit with four inputs is shown. INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1
(A/s) Computer Science Oct/Nov 2018 paper 3 variant 1
Question paper found on page 9 / 16 pages total, pdf
9 (b) A logic circuit with four inputs produces the following truth table. INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0
(A/s) Computer Science May/June 2018 paper 3 variant 3
Question paper found on page 9 / 16 pages total, pdf
9 (b) A logic circuit with four inputs produces the following truth table. INPUT (i) OUTPUT A B C D X 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 1 0
(IGCSE) Computer Studies Oct/Nov 2013 paper 1 variant 3
Mark scheme found on page 9 / 13 pages total, pdf
Page 9 Mark Scheme IGCSE – October/November 2013 Syllabus 0420 Paper 13 (b) A B C X 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 } } } } 1 mark 1 mark 1 mark 1 mark [4] © Cambridge International Examinations 2013
(A/s) Computer Science Oct/Nov 2017 paper 3 variant 3
Mark scheme found on page 6 / 8 pages total, pdf
9608/33 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question Answer 5(a)(ii) A B C X 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 5(b)(i) 1 3 R Q Q Initially 1 0 0 1 R changed to 1 1 1
(A/s) Computer Science For examination from 2015 paper 1
Specimen question paper found on page 5 / 16 pages total, pdf
5 2 When data is transmitted, it may become corrupted. (a) Explain how a parity check can be used to detect a possible error in a transmitted byte. [3] (b) Describe how parity can be used to identify and correct the single error in this transmitted data block. 0 1 1 0 1 1 0 1 1 0 0 1 0 1 1 1 0 1 0 1 0 1
(IGCSE) Co-ordinated Sciences Oct/Nov 2012 paper 5 variant 2
Ir found on page 6 / 8 pages total, pdf
40 0 10 20 30 60 180 170 1 20 10 0 40 30 180 170 1 60 15 0 0 14 0 14 15 0 20 10 0 20 10 0 40 30 60 180 170 1 40 30 60 0654/52/CI/O/N/12 180 170 1 0 14
(A/s) Mathematics - Further May/June 2020 paper 2 variant 2
Mark scheme found on page 13 / 13 pages total, pdf
9231/22 Cambridge International AS & A Level – Mark Scheme PUBLISHED Question 8(c) Answer Eigenvalues of A are 3, 6 and −2 . λ = 3:i j k −4 1 0 1 1 = 0 0 0 3 −1 0 0 λ = 6:i j k −1
(A/s) Mathematics - Further Oct/Nov 2016 paper 1 variant 1
Mark scheme found on page 5 / 15 pages total, pdf
Page 5 3 Mark Scheme Cambridge International A Level – October/November 2016 Syllabus 9231 Paper 11 1 1 0 − 1 0 0 , P = 0 1 1 D = 0 1 0 soi 0 0 1 0 0 2 B1B1 P − 1 AP = D ⇒
(A/s) Mathematics Feb/March 2021 paper 5 variant 2
Mark scheme found on page 15 / 16 pages total, pdf
9709/52 Question 7(b)(i) March 2021 Cambridge International AS & A Level – Mark Scheme PUBLISHED Answer Marks Guidance Method 1 [1 – P(0,1,2)] = 1 – (10C0 0∙30 0∙710 + 10C1 0∙31 0∙79 + 10C2 0∙32 0∙78)M1 = 1 – (0∙028248 + 0∙
(A/s) Computer Science May/June 2017 paper 1 variant 2
Mark scheme found on page 6 / 7 pages total, pdf
9608/12 Question 5(a)(iv) Answer 1 Op code 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 1 1 1 1 14 5E 2 14 5E 1 1 LDR #77 2 LDR #77 © UCLES 2017 3 Operand Both correct op codes Operand
(IGCSE) Computer Science Oct/Nov 2016 paper 1 variant 2
Question paper found on page 4 / 12 pages total, pdf
4 4 Nine bytes of data are transmitted from one computer to another. Even parity is used. An additional parity byte is also sent. The ten bytes arrive at the destination computer as follows: parity bit bit 2 bit 3 bit 4 bit 5 bit 6 bit 7 bit 8 byte 1 1 1 1 0 1 1 1 0 byte 2 0 0 0 0 0 1 0 1 byte 3 0
(A/s) Computer Science May/June 2017 paper 1 variant 3
Question paper found on page 11 / 16 pages total, pdf
11 (b) (i) A second data block is received as shown in the following table. There are errors in this data block. Identify and then circle two bits in the table which must be changed to remove the errors. 7 1 0 0 1 1 0 0 6 0 0 0 1 1 0 0 Bit position 5 4 3 2 0 0 1 1 1 0 0 0 1 1
(IGCSE) Computer Science May/June 2015 paper 1 variant 2
Question paper found on page 6 / 16 pages total, pdf
6 5 Parity checks are often used to check for errors that may occur during data transmission. (a) A system uses even parity. Tick (✓) to show whether the following three bytes have been transmitted correctly or incorrectly. Received byte Byte transmitted correctly Byte transmitted incorrectly 1 1 0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 1 0 0 1 [3]
(A/s) Computer Science May/June 2018 paper 1 variant 2
Question paper found on page 6 / 16 pages total, pdf
6 3 Parity bits can be used to verify data. (a) The following binary number is transmitted using odd parity. Add the missing parity bit. Parity bit 0 1 0 0 0 0 0 [1] (b) In the following data transmitted, the first column contains the parity bits, and the last row contains the parity byte. A device transmits the data using even parity. Circle the error in the
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 3
Question paper found on page 19 / 24 pages total, pdf
19 (b) In order to prevent errors, an 8-bit register is used. The 8th bit will contain: • 0 – if the first 7 bits add up to an even number • 1 – if the first 7 bits add up to an odd number Complete the 8th bit for each register. The first register has been completed for you. 1 2 3 4 5 6 7 8 Reg
(IGCSE) Co-ordinated Sciences Oct/Nov 2012 paper 5 variant 3
Ir found on page 6 / 8 pages total, pdf
40 0 10 20 30 60 180 170 1 20 10 0 40 30 180 170 1 60 15 0 0 14 0 14 15 0 20 10 0 20 10 0 40 30 60 180 170 1 40 30 60 0654/53/CI/O/N/12 180 170 1 0 14
(IGCSE) Physics Oct/Nov 2005 paper 5
Question paper found on page 3 / 8 pages total, pdf
(f) A student carried out this experiment using a 360 ° protractor, as shown in Fig. 1.2. Explain how the student could use this protractor to measure the angle between the metre rule and the bench. You may draw a diagram if you wish. Fig. 1.2 (g) The range of angles measured in this experiment may be quite small. Using the same apparatus, with the masses
(A/s) Computer Science Oct/Nov 2015 paper 3 variant 2
Mark scheme found on page 5 / 6 pages total, pdf
Page 5 5 Mark Scheme Cambridge International A Level – October/November 2015 Syllabus 9608 Paper 32 (a) (i) Z = P. Q . R + P . Q .R + P.Q.R [1] [1] [1] (ii) PQ 00 01 11 10 0 0 0 0 1 1 0 0 1 1 R [1] (iii) 1 mark each loop PQ
(A/s) Computer Science May/June 2017 paper 1 variant 3
Question paper found on page 10 / 16 pages total, pdf
10 5 A computer receives data from a remote data logger. Each data block is a group of 8 bytes. A block is made up of seven data bytes and a parity byte. Each data byte has a parity bit using odd parity. The parity byte also uses odd parity. The following table shows a data block before transmission. Bit position 0 is the parity bit. 7 1 0 1 1
(A/s) Computer Science May/June 2017 paper 1 variant 3
Mark scheme found on page 5 / 8 pages total, pdf
9608/13 May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question Answer Marks 4(a)(i) 500 1 4(a)(ii) 496 1 4(a)(iii) 502 1 4(a)(iv) 86 1 4(b) 3 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 0
(A/s) Computer Science May/June 2017 paper 1 variant 1
Mark scheme found on page 5 / 8 pages total, pdf
9608/11 May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question Answer Marks 4(a)(i) 500 1 4(a)(ii) 496 1 4(a)(iii) 502 1 4(a)(iv) 86 1 4(b) 3 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 0
(A/s) Mathematics - Further May/June 2021 paper 2 variant 2
Mark scheme found on page 10 / 13 pages total, pdf
9231/22Cambridge International AS & A Level – Mark Scheme PUBLISHED QuestionAnswer 6(a) Marks Eigenvalues of A are 5, −6 and 1. λ = 5: M1 Uses vector product (or equations) to find corresponding Eigenvectors. A1 A1 A1 for each correct Eigenvector. A1 − 154 3 i j k 2 22 λ = −6 :
(A/s) Mathematics - Further May/June 2021 paper 2 variant 1
Mark scheme found on page 10 / 13 pages total, pdf
9231/21Cambridge International AS & A Level – Mark Scheme PUBLISHED QuestionAnswer 6(a) Marks Eigenvalues of A are 5, −6 and 1. λ = 5: M1 Uses vector product (or equations) to find corresponding Eigenvectors. A1 A1 A1 for each correct Eigenvector. A1 − 154 3 i j k 2 22 λ = −6 :
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 3
Mark scheme found on page 11 / 13 pages total, pdf
Page 11 Mark Scheme Cambridge IGCSE – October/November 2014 14 (a) Row number: Syllabus 0420 1 2 3 4 5 6 7 Reg 1: 0 1 1 0 0 0 0 Reg 2: 1 0 0 1 0 0 1 Reg 3: 1 0 0 1 0 1 0 Reg 4: 1 0 0 1 1 0 0 Reg 5: 0 1 1 0 0 0
(A/s) Mathematics - Further May/June 2020 paper 2 variant 1
Mark scheme found on page 13 / 13 pages total, pdf
9231/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED Question 8(c) Answer Eigenvalues of A are 3, 6 and −2 . λ = 3:i j k −4 1 0 1 1 = 0 0 0 3 −1 0 0 λ = 6:i j k −1
(IGCSE) Computer Science May/June 2015 paper 1 variant 2
Question paper found on page 9 / 16 pages total, pdf
9 (b) Complete the truth table for the safety system. Workspace G C L 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 X [4] (c) Complete the truth table for the XOR gate: A B 0 0 0 1 1 0 1 1 C [1] © UCLES 2015 0478/12/M
(A/s) Computer Science May/June 2017 paper 3 variant 1
Mark scheme found on page 4 / 7 pages total, pdf
9608/31 Question 3(a) May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Answer Marks X = A.(B + (B . C)) B.C B + B.C A. 3(b) 3 1 1 1 A B C Working Space X 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0
(IGCSE) Computer Studies May/June 2011 paper 1 variant 1
Question paper found on page 16 / 24 pages total, pdf
16 10 (a) Two logic gates are the AND gate and the OR gate. Complete the truth tables for these two gates: OR gate AND gate A B 0 For Examiner's Use X A B 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1 X [2] (b) Complete the truth table for the following logic circuit: A AND OR B X AND C
(IGCSE) Computer Studies Oct/Nov 2012 paper 1 variant 3
Mark scheme found on page 11 / 12 pages total, pdf
Page 11 Mark Scheme IGCSE – October/November 2012 Syllabus 0420 Paper 13 15 (a) A B C X 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1 1 mark 1 mark 1 mark 1 mark [4] (b) 1 mark for gate name + 1
(A/s) Mathematics - Further May/June 2019 paper 1 variant 1
Mark scheme found on page 14 / 20 pages total, pdf
9231/11 Cambridge International A Level – Mark Scheme PUBLISHED Question 9(i) Answer Marks A 2e = A ( Ae ) = λ Ae = λ 2e May/June 2019 Guidance M1 A1 AG 2 9(ii) Eigenvalues of A are n, 2 n and 3n . B1 M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. i j k −
(A/s) Computer Science May/June 2015 paper 1 variant 3
Question paper found on page 14 / 16 pages total, pdf
14 7 The table shows assembly language instructions for a processor which has one general purpose register, the Accumulator (ACC). Instruction Explanation Op code Operand LDD <address> Direct addressing. Load contents of given address to ACC STO <address> Store the contents of ACC at the given address LDI <address> Indirect addressing. The address to be used is at the given address. Load the contents of this second address
(A/s) Biology May/June 2019 paper 2 variant 3
Question paper found on page 6 / 16 pages total, pdf
6 (d) Fig. 2.2 shows the mean transpiration rate of a xerophyte between 08:00 and 19:00. 5.0 4.5 4.0 3.5 3.0 mean transpiration rate / arbitrary units 2.5 2.0 1.5 1.0 0.5 0 0 0 0 0 0 0 0 0 0 0 0 0 :0 :0 :0 :0 :0 :0 :0 :0 :0
(A/s) Computer Science Oct/Nov 2016 paper 3 variant 2
Question paper found on page 10 / 12 pages total, pdf
10 5 (a) (i) A half adder is a logic circuit with the following truth table. Input Output X Y A B 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 The following logic circuit is constructed. P X Q HALF ADDER Y B A J X A HALF ADDER Y B R K Complete the following truth table for this logic circuit. Input
(A/s) Computer Science Oct/Nov 2017 paper 3 variant 2
Question paper found on page 14 / 16 pages total, pdf
14 (d) There are eight temperature sensors numbered 1 to 8. Readings from these sensors are stored in four 16-bit memory locations. The memory locations have addresses from 4000 to 4003. Each memory location stores two sensor readings as two unsigned binary integers. Sensor 1 reading is stored in bits 8 to 15 of address 4000; Sensor 2 reading is stored in bits 0 to
(A/s) Mathematics Oct/Nov 2020 paper 5 variant 1
Mark scheme found on page 8 / 16 pages total, pdf
9709/51 Cambridge International A Level – Mark Scheme PUBLISHED Question 2(a) Answer Marks October/November 2020 Guidance M1 Equation of form 0·6 × a + 0·4 × b = 0·58; a = 0·3, 0·7, b = x, (1 – x) 0·6 × 0·7 + 0·4(1 – x) = 0·58 ≡ 0·42 + 0
(A/s) Mathematics - Further Oct/Nov 2012 paper 1 variant 2
Mark scheme found on page 10 / 12 pages total, pdf
Page 10 Qu No 10 Mark Scheme GCE A LEVEL – October/November 2012 Commentary States eigenvalues. Finds eigenvectors. States P and D. Finds inverse of P. Finds A n . Syllabus 9231 Solution Marks Part Total Mark B1 1 Eigenvalues are 1, 2, 3. i j k 28 1 λ = 1 e 1 = 0 4 −
(A/s) Mathematics - Further Oct/Nov 2012 paper 1 variant 1
Mark scheme found on page 10 / 12 pages total, pdf
Page 10 Qu No 10 Mark Scheme GCE A LEVEL – October/November 2012 Commentary States eigenvalues. Finds eigenvectors. States P and D. Finds inverse of P. Finds A n . Syllabus 9231 Solution Marks Part Total Mark B1 1 Eigenvalues are 1, 2, 3. i j k 28 1 λ = 1 e 1 = 0 4 −
(IGCSE) Computer Studies May/June 2014 paper 1 variant 2
Mark scheme found on page 12 / 14 pages total, pdf
Page 12 Mark Scheme IGCSE – May/June 2014 Syllabus 0420 Paper 12 (b) one mark – letter “Y” or 25 th letter One mark – the binary number 0 0 0 0 1 1 0 0 1 0 0 0 has been shifted (to the left) 3 places – so the binary number becomes 0 0 0 0 0 0 0 1 1 0 0 1 –
(IGCSE) Computer Studies Oct/Nov 2014 paper 1 variant 1
Question paper found on page 16 / 24 pages total, pdf
16 12 An advertising sign uses large LED characters controlled by a microprocessor. Each letter is formed from a grid made up of eight rectangles numbered 1 to 8: 1 2 3 4 5 6 7 8 For example, the letter “Z” is formed as follows: 1 2 4 5 7 8 Each rectangle has six LEDs that can light up; these LEDs are labelled “a” to “f”:
(A/s) Computer Science Oct/Nov 2020 paper 3 variant 2
Question paper found on page 2 / 16 pages total, pdf
2 1 In a particular computer system, real numbers are stored using floating-point representation, with: • • • 12 bits for the mantissa 4 bits for the exponent two’s complement form for both mantissa and exponent. (a) Calculate the denary value for the following floating-point number. Show your working. Mantissa 0 1 0 1 0 0 0 Exponent 0 0 0 0 0 0 1 1 0 Working
(IGCSE) Physics May/June 2018 paper 2 variant 2
Question paper found on page 11 / 16 pages total, pdf
11 33 The circuit shown contains two gates. P R Q Which truth table describes the operation of the circuit? A B C D P Q R P Q R P Q R P Q R 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 0 1 0 1 1
(IGCSE) Computer Science For examination from 2020 paper 1
Specimen question paper found on page 6 / 14 pages total, pdf
6 4 A digital alarm clock is controlled by a microprocessor. It uses the 24-hour clock system (i.e. 6 pm is 18:00). Each digit in a typical display is represented by a 4-digit binary code. For example: 0 0 0 0 1st digit (0) 1 0 0 0 2nd digit (8) 0 0 1 1 3rd digit (3) 0 1 0 1
(IGCSE) Physics Oct/Nov 2005 paper 6
Question paper found on page 3 / 12 pages total, pdf
(c) A student carries out this experiment using the 360 ° protractor shown in Fig. 1.2. Fig. 1.2 Explain how the student could use this protractor to measure the angle between the metre rule and the bench. You may draw a diagram if you wish. ...............................................
(A/s) Mathematics - Further Oct/Nov 2018 paper 1 variant 2
Mark scheme found on page 9 / 20 pages total, pdf
9231/12 Cambridge International A Level – Mark Scheme PUBLISHED Question Answer 5(i) 3 2 0 1 3 6 5 − 1 3 → 0 9 8 − 2 5 0 − 3 − 2 0 − 1 0 3 0 → 0 0 2 0 1 −
(IGCSE) Computer Science May/June 2018 paper 1 variant 2
Mark scheme found on page 5 / 10 pages total, pdf
0478/12 Cambridge IGCSE – Mark Scheme PUBLISHED Question 3(a) 3(b) Answer 3 Hours 0 0 0 0 0 0 1 0 Minutes 0 0 0 1 1 1 1 1 Seconds 0 0 1 1 1 0 1 0 1 mark for each correct section: 3 5 2 Hours 6 5 Minutes Question 5 Seconds Answer Marks 1 mark for each correct section: 1 1 0 1 mark
(A/s) Mathematics - Further Oct/Nov 2016 paper 1 variant 2
Mark scheme found on page 5 / 15 pages total, pdf
Page 5 3 Mark Scheme Cambridge International A Level – October/November 2016 Syllabus 9231 Paper 12 1 1 0 − 1 0 0 , P = 0 1 1 D = 0 1 0 soi 0 0 1 0 0 2 B1B1 P − 1 AP = D ⇒
(A/s) Computer Science May/June 2017 paper 3 variant 3
Mark scheme found on page 4 / 7 pages total, pdf
9608/33 Question 3(a) May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Answer Marks X = A.(B + (B . C)) B.C B + B.C A. 3(b) 3 1 1 1 A B C Working Space X 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0
(A/s) Mathematics - Further May/June 2011 paper 1 variant 3
Mark scheme found on page 13 / 15 pages total, pdf
Page 13 Qu No Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2011 Commentary Solution EITHER (Alternative) (i) Uses Ae = λe (3 times) Forms 3 linear equations (3 times) Syllabus 9231 a b d e g h c 0 0 f 1 = −
(IGCSE) Physics May/June 2018 paper 2 variant 3
Question paper found on page 13 / 16 pages total, pdf
13 33 The diagram shows a logic circuit with inputs X and Y. X Y Q The output is Q. Which truth table is correct? A B X Y Q X Y Q 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 C D X Y Q X Y Q 0 0 1 0 0 0 0 1
(IGCSE) Computer Science Oct/Nov 2016 paper 1 variant 3
Question paper found on page 8 / 16 pages total, pdf
8 (d) When eight bytes of data have been collected, they are transmitted to a computer 100 km away. Parity checks are carried out to identify if the data has been transmitted correctly. The system uses even parity and column 1 is the parity bit. The eight bytes of data are sent together with a ninth parity byte: parity bit column 2 column 3 column 4 column 5 column 6 column
(IGCSE) Computer Science For examination from 2020 paper 1
Specimen mark scheme found on page 3 / 8 pages total, pdf
3 3 (a) 1 mark for each logic gate correctly connected A AND NOT Y OR T AND S NOT [5] (b) A T S Y 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 } ] ] ] ] 1 mark 1 mark 1 mark 1 mark [4
(IGCSE) Computer Science For examination from 2016 paper 1
Specimen mark scheme found on page 3 / 8 pages total, pdf
3 3 (a) 1 mark for each logic gate correctly connected A AND NOT Y OR T AND S NOT [5] (b) A T S Y 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 } ] ] ] ] 1 mark 1 mark 1 mark 1 mark [4
(IGCSE) Computer Studies Oct/Nov 2003 paper 1
Question paper found on page 10 / 16 pages total, pdf
10 12 Two 7 segment displays are used on a car dashboard to give information to the driver. Each segment is numbered as shown. 1 6 2 7 5 3 4 (1) (2) For example, the information 1P shown above is represented by: (1) 7 6 5 4 3 2 1 0 0 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 and
(IGCSE) Computer Science For examination from 2020 paper 1
Specimen question paper found on page 5 / 14 pages total, pdf
5 (b) Complete the truth table for this alarm system. A T S 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Y [4] © UCLES 2017 0478/01/SP/20 [Turn over
(IGCSE) Computer Science For examination from 2015 paper 1
Specimen question paper found on page 5 / 14 pages total, pdf
5 (b) Complete the truth table for this alarm system. A T S 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Y [4] © UCLES 2012 0478/01/SP/15 [Turn over
(A/s) Mathematics - Further May/June 2022 paper 2 variant 1
Mark scheme found on page 12 / 13 pages total, pdf
9231/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED Question 8(a) Answer Marks 3 Guidance M1 Sets determinant equal to zero. 3 a 0 5 1 0 0 10a 6 0 1 May/June 2022 2 a 53 A1 2 8(b) Eigenvalues of A are 3, 1 and 2. 3:i j
(IGCSE) Computer Studies Oct/Nov 2013 paper 1 variant 2
Mark scheme found on page 10 / 16 pages total, pdf
Page 10 Mark Scheme IGCSE – October/November 2013 Syllabus 0420 Paper 12 (10) (a) (i) A B X 0 0 0 1 1 1 1 0 1 1 } } 1 0 1 mark 1 mark [2] (ii) NAND gate (if truth table above is incorrect, allow follow through in part (ii)) [1] (b) A B C X 0 0
(A/s) Computer Science May/June 2018 paper 3 variant 3
Mark scheme found on page 6 / 8 pages total, pdf
9608/33 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question Answer May/June 2018 Marks 4(a)(i) 2 marks all products correct, 1 mark 2 or 3 products correct X = A.B.C + A.B.C + A.B.C + A.B.C 2 4(a)(ii) 1 mark for all correct bits 1 AB 0 1 C 4(a)(iii) 00
(A/s) Computer Science Oct/Nov 2018 paper 3 variant 1
Mark scheme found on page 6 / 8 pages total, pdf
9608/31 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question 4(b)(i) October/November 2018 Answer Marks 1 mark per bullet point max 2: • • 4 Correct column headings and row headings – values only Correct column headings and row headings – order 1 mark for 2 correct rows or columns, 2 marks for 4 correct rows or columns (based on headings) AB CD 4
(A/s) Computer Science May/June 2021 paper 3 variant 1
Question paper found on page 2 / 16 pages total, pdf
2 1 In a particular computer system, two real numbers, A and B, are stored using floating-point representation with: • • • 12 bits for the mantissa 4 bits for the exponent two’s complement form for both mantissa and exponent. Number A Mantissa 1 1 0 0 0 Number B 0 0 0 0 0 0 Mantissa 0 (a) (i) 0 Exponent 1 1 1 0 0 0
(A/s) Computer Science May/June 2017 paper 3 variant 2
Mark scheme found on page 4 / 7 pages total, pdf
9608/32 Question 3(a) May/June 2017 Cambridge International AS/A Level – Mark Scheme PUBLISHED Answer Marks 4 S = ( P + ( Q + R ) ) . R P ( Q + R ) ( P +( Q + R ) ) . R (must be outside final brackets) 1 1 1 1 Or P ( Q + R ) P +( Q +
(IGCSE) Physics May/June 2016 paper 4 variant 1
Question paper found on page 19 / 20 pages total, pdf
Suggest a modification to the circuit in Fig. 10.2 to produce the output Z in the truth table below. It may help you to compare this truth table with the truth table in (b) . (c) output Z input D input B input A 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1
(A/s) Mathematics - Further Oct/Nov 2021 paper 2 variant 2
Mark scheme found on page 11 / 14 pages total, pdf
9231/22 Cambridge International AS & A Level – Mark Scheme PUBLISHED Question 6(b) Answer Marks October/November 2021 Guidance M1 Applies A = PDP −1. 4 0 0 A = P 0 5 0 P −1 0 0 6 1 6 6 4 −12 −16 4 30 36 1
(IGCSE) Arabic - Foreign Language May/June 2014 paper 2 variant 3
Question paper found on page 10 / 20 pages total, pdf
10 Section 2 Exercise 1 Questions 17–26 . 3$&8 (! 1 + 5') E ! . C '8) C $) -3 3 - + + & 0 . $ ! ' + E) = $ / *F , ( #0 : - *0 ) . , + + & 3 ) & + 1 41 : 3! * 0 + ! 3! I H
(IGCSE) Computer Science Feb/March 2017 paper 1 variant 2
Mark scheme found on page 9 / 9 pages total, pdf
0478 / 12 Cambridge IGCSE – Mark Scheme PUBLISHED Question 14(b) March 2017 Answer Marks 4 marks for 8 correct bits 3 marks for 6 correct bits 2 marks for 4 correct bits 1 mark for 2 correct bits © UCLES 2017 4 A B C X 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1
(A/s) Mathematics - Further For examination from 2017 paper 1
Specimen mark scheme found on page 7 / 16 pages total, pdf
9231/01 [Further Maths 1] AS & A Level Mathematics – Mark Scheme SPECIMEN Question 6 Answer Partial Marks Marks 6 1 1 : 10 8 10 0 ~ 0 oe 7 5 7 6 1 2 M1A1 0 0
(A/s) Computer Science May/June 2018 paper 1 variant 2
Mark scheme found on page 5 / 9 pages total, pdf
9608/12 Cambridge International AS/A Level – Mark Scheme PUBLISHED Question 3(a) May/June 2018 Answer Marks 1 mark for correct parity bit 1 Parity bit 0 3(b) 0 1 0 0 0 0 1 mark for the correct bit circled. 1 Parity bit Parity byte 3(c) 0 Data 1 0 1 0 1 1 1 1 0 1 1 0 0 1 1 0 1
(A/s) Computer Science Oct/Nov 2017 paper 3 variant 3
Question paper found on page 9 / 16 pages total, pdf
9 5 (a) (i) Complete the truth table for this 2-input NAND gate: A 0 0 1 1 A X B B 0 1 0 1 X [1] (ii) Complete the truth table for this 3-input NAND gate: A 0 0 0 0 1 1 1 1 A X B C B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0
(IGCSE) Computer Science May/June 2016 paper 1 variant 3
Mark scheme found on page 5 / 10 pages total, pdf
Page 5 5 Mark Scheme Cambridge IGCSE – May/June 2016 Syllabus 0478 Paper 13 (a) 1 mark for each correct gate, with correct source of input(s) [6] (b) D S T Working Space 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 4
(IGCSE) Computer Studies Oct/Nov 2012 paper 1 variant 2
Mark scheme found on page 8 / 12 pages total, pdf
Page 8 Mark Scheme IGCSE – October/November 2012 Syllabus 0420 Paper 12 11 (a) P NOT 1 mark AND OR T X 1 mark 1 mark NOT 1 mark AND W 1 mark NOT Note: accept answers using MIL symbols e.g. AND [5] (b) P T W X 0 0 0 1 0 0 1 0 0 1 0 1 1 mark 1 mark 0