Data representation

1(a) Working: one mark for calculation of the mantissa and one mark for 3 calculation or use of the exponent Exponent: one from: = 0.11101 × 23 // 0.11101 × 211 // 0.11101 × 103 // 0.11101 × 1011 = 1.00011 × 23 // 1.00011 × 211 // 1.00011 × 103 // 1.00011 × 1011 = appropriate shifting of binary point for +7.25 Mantissa: one from: = 111.01 (conversion to binary +7.25 – 10 bits) = 0111010000 (mantissa 10 bits for +7.25 = 1000101111(one’s complement mantissa for –7.25) = 1000110000 (two’s complement mantissa for –7.25) Correct Answer (Max 1) Mantissa Exponent 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1(b) One mark for working out the exponent 3 One mark for working out the mantissa One mark for the correct answer Example answers • =1.011000111 × 27 (exponent is 7) • =10110001.11 // –128 + 32 + 16 + 1 + 0.5 + 0.25 // convert to positive 01001110.01 (and add a minus sign to the answer) • –78.25 1(c) One mark for working 3 One mark for correct mantissa One mark for correct exponent Example answers Number of places added to exponent for normalisation −6 for number to retain its value // mantissa moved 6 places left Mantissa 0 1 1 1 0 0 0 0 0 0 Exponent 1 0 0 0 0 1 1(d)(i) One mark for each correct marking point (Max 3) 3 • Requires 11 bits / more than 10 bits to store (accurately) / reference to maximum (positive) number that can be stored = 511 • Denary 513 in binary is 1000000001 // Normalised: 0.1000000001 • Results in overflow © UCLES 2021 Page 3 of 10 1(d)(ii) One mark for each correct marking point (Max 2) 2 • The number of bits for the mantissa must be increased • 11/12 bits mantissa and 5/4 bits exponent Question Answer Marks

1(a) Working: one mark for calculation of the mantissa and one mark for 3 calculation or use of the exponent Exponent: one from: = 0.11101 × 23 // 0.11101 × 211 // 0.11101 × 103 // 0.11101 × 1011 = 1.00011 × 23 // 1.00011 × 211 // 1.00011 × 103 // 1.00011 × 1011 = appropriate shifting of binary point for +7.25 Mantissa: one from: = 111.01 (conversion to binary +7.25 – 10 bits) = 0111010000 (mantissa 10 bits for +7.25 = 1000101111(one’s complement mantissa for –7.25) = 1000110000 (two’s complement mantissa for –7.25) Correct Answer (Max 1) Mantissa Exponent 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1(b) One mark for working out the exponent 3 One mark for working out the mantissa One mark for the correct answer Example answers • =1.011000111 × 27 (exponent is 7) • =10110001.11 // –128 + 32 + 16 + 1 + 0.5 + 0.25 // convert to positive 01001110.01 (and add a minus sign to the answer) • –78.25 1(c) One mark for working 3 One mark for correct mantissa One mark for correct exponent Example answers Number of places added to exponent for normalisation −6 for number to retain its value // mantissa moved 6 places left Mantissa 0 1 1 1 0 0 0 0 0 0 Exponent 1 0 0 0 0 1 1(d)(i) One mark for each correct marking point (Max 3) 3 • Requires 11 bits / more than 10 bits to store (accurately) / reference to maximum (positive) number that can be stored = 511 • Denary 513 in binary is 1000000001 // Normalised: 0.1000000001 • Results in overflow © UCLES 2021 Page 3 of 10 1(d)(ii) One mark for each correct marking point (Max 2) 2 • The number of bits for the mantissa must be increased • 11/12 bits mantissa and 5/4 bits exponent Question Answer Marks

1(a) Working: one mark for calculation of the mantissa and one mark for 3 calculation or use of the exponent Exponent: one from: = 0.11101 × 23 // 0.11101 × 211 // 0.11101 × 103 // 0.11101 × 1011 = 1.00011 × 23 // 1.00011 × 211 // 1.00011 × 103 // 1.00011 × 1011 = appropriate shifting of binary point for +7.25 Mantissa: one from: = 111.01 (conversion to binary +7.25 – 10 bits) = 0111010000 (mantissa 10 bits for +7.25 = 1000101111(one’s complement mantissa for –7.25) = 1000110000 (two’s complement mantissa for –7.25) Correct Answer (Max 1) Mantissa Exponent 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1(b) One mark for working out the exponent 3 One mark for working out the mantissa One mark for the correct answer Example answers • =1.011000111 × 27 (exponent is 7) • =10110001.11 // –128 + 32 + 16 + 1 + 0.5 + 0.25 // convert to positive 01001110.01 (and add a minus sign to the answer) • –78.25 1(c) One mark for working 3 One mark for correct mantissa One mark for correct exponent Example answers Number of places added to exponent for normalisation −6 for number to retain its value // mantissa moved 6 places left Mantissa 0 1 1 1 0 0 0 0 0 0 Exponent 1 0 0 0 0 1 1(d)(i) One mark for each correct marking point (Max 3) 3 • Requires 11 bits / more than 10 bits to store (accurately) / reference to maximum (positive) number that can be stored = 511 • Denary 513 in binary is 1000000001 // Normalised: 0.1000000001 • Results in overflow © UCLES 2021 Page 3 of 10 1(d)(ii) One mark for each correct marking point (Max 2) 2 • The number of bits for the mantissa must be increased • 11/12 bits mantissa and 5/4 bits exponent Question Answer Marks

1(a)(i) One mark for each correct marking point (Max 2) 2 • 010111000110 (correct mantissa) • 0111 (correct exponent) 1(a)(ii) One mark for each correct consequence 2 One mark for each correct justification Consequence • The precision/accuracy of the number would be reduced Justification • … because the least significant bits of the original number have been truncated/lost // the original number had 13 bits / 14 bits with sign but the mantissa can only store 12 bits 1(b) One mark for each correct marking point (Max 3) 3 • To store the maximum range of numbers in the minimum number of bytes / bits • Normalisation minimises the number of leading zeros/ones represented • Maximising the number of significant bits // maximising the (potential) precision / accuracy of the number for the given number of bits • … enables very large / small numbers to be stored with accuracy. • Avoids the possibility of many numbers having multiple representations. © UCLES 2021 Page 3 of 10

1(a)(i) One mark for each correct marking point (Max 2) 2 • 010111000110 (correct mantissa) • 0111 (correct exponent) 1(a)(ii) One mark for each correct consequence 2 One mark for each correct justification Consequence • The precision/accuracy of the number would be reduced Justification • … because the least significant bits of the original number have been truncated/lost // the original number had 13 bits / 14 bits with sign but the mantissa can only store 12 bits 1(b) One mark for each correct marking point (Max 3) 3 • To store the maximum range of numbers in the minimum number of bytes / bits • Normalisation minimises the number of leading zeros/ones represented • Maximising the number of significant bits // maximising the (potential) precision / accuracy of the number for the given number of bits • … enables very large / small numbers to be stored with accuracy. • Avoids the possibility of many numbers having multiple representations. © UCLES 2021 Page 3 of 10

1(a) Two marks for working 3 One mark for correct answer Working: Conversion to binary + 202 = 11001010 // repeated division by 2 // 128 + 64 + 8 + 2 Appropriate shifting of binary point for + 202 = 0.1100101  28 // exponent = 8 Answer: = 01100101 00001000 (stored as mantissa and exponent) 1(b) Two marks for working 3 One mark for correct answer Working: • Appropriate method of conversion e.g. = 10011010 (one’s complement of 8-bit mantissa) = 10011011 (two’s complement of 8-bit mantissa) –256 + 32 + 16 + 4 +2 • Realisation that the exponent doesn’t change // value of exponent = 8 // appropriate shifting of binary point Answer: = 10011011 00001000 (stored as mantissa and exponent) 1(c)(i) The mantissa does not begin with 01/10 (as its most significant bits) 1 // the mantissa begins with 00 // first two digits are the same. 1(c)(ii) One mark for each point: 2 • Correct mantissa • Correct exponent Mantissa Exponent 0 1 1 1 1 0 0 0 0 0 0 1 0 1 1 0 © UCLES 2022 Page 4 of 15

1(a) Mantissa Exponent 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1(b) Two marks for working 3 • correct calculation of exponent seen • correct application of exponent to mantissa seen One mark for correct answer Working: = 1.0110010011  29 //exponent = 9 = 1011001001.1 (moving bp 9 places to right) // evaluate two’s complement For example: –512 + 128 + 64 + 8 + 1 + 0.5 Answer: –310.5 // –3101/

1(a) Two marks for working 3 One mark for correct answer Working: Conversion to binary + 202 = 11001010 // repeated division by 2 // 128 + 64 + 8 + 2 Appropriate shifting of binary point for + 202 = 0.1100101  28 // exponent = 8 Answer: = 01100101 00001000 (stored as mantissa and exponent) 1(b) Two marks for working 3 One mark for correct answer Working: • Appropriate method of conversion e.g. = 10011010 (one’s complement of 8-bit mantissa) = 10011011 (two’s complement of 8-bit mantissa) –256 + 32 + 16 + 4 +2 • Realisation that the exponent doesn’t change // value of exponent = 8 // appropriate shifting of binary point Answer: = 10011011 00001000 (stored as mantissa and exponent) 1(c)(i) The mantissa does not begin with 01/10 (as its most significant bits) 1 // the mantissa begins with 00 // first two digits are the same. 1(c)(ii) One mark for each point: 2 • Correct mantissa • Correct exponent Mantissa Exponent 0 1 1 1 1 0 0 0 0 0 0 1 0 1 1 0 © UCLES 2022 Page 4 of 15

1(a) One mark per mark point (Max 4) 4  conversion of 113.75 to binary seen 1110001.11  exponent for normalisation 7 converted to binary 111 // evidence of binary point moved 7 places // evidence of finding exponent = 7  system 1 answer  system 2 answer showing correct version from system 1 System 1 Mantissa Exponent 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 System 2 Mantissa Exponent 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 1(b) One mark per mark point (Max 2) 2  the mantissa in system 2 does not have enough bits to store the whole binary number // 10 bits required and only 8 bits available  so precision is lost / the number is truncated Question Answer Marks

1(a) One mark per mark point 2  correct mantissa  correct exponent with associated working Answer Mantissa Exponent 0 1 0 1 0 1 0 1 1 1 0 0 0 1 1 0 Working exponent = 6 (movement of 6 bicimal places seen to find what exponent should be) calculation of denary 6 to binary (000)110 1(b) One mark per mark point (Max 3) 3 MP1 the mantissa of the number would need to be 0.101011111001 / 13 bits / digits MP2 … it can only store 10 bits / digits MP3 The 3 least significant digits would be truncated MP4 …causing a loss of precision Question Answer Marks

1(a) One mark per mark point (Max 4) 4  conversion of 113.75 to binary seen 1110001.11  exponent for normalisation 7 converted to binary 111 // evidence of binary point moved 7 places // evidence of finding exponent = 7  system 1 answer  system 2 answer showing correct version from system 1 System 1 Mantissa Exponent 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 System 2 Mantissa Exponent 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 1(b) One mark per mark point (Max 2) 2  the mantissa in system 2 does not have enough bits to store the whole binary number // 10 bits required and only 8 bits available  so precision is lost / the number is truncated Question Answer Marks

1(a) One mark for working (Max 1) 3 • conversion of 65.25 to binary seen e.g. 1000001.01 = 65.25 // 64 + 1 + 0.25 / ¼ One mark per mark point (Max 2) • correct mantissa • correct exponent Mantissa Exponent 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 1 1(b) One mark per mark point (Max 2) 2 MP1 the decimal fraction 0.20 cannot be represented exactly (the closest is 0.25 / 0.1875) MP2 therefore, there will be a loss of precision due to a rounding error/truncation Question Answer Marks

1(a) One mark per mark point (Max 1) 3 • conversion of −96.75 to binary e.g., positive 96.75, flip the bits + 1 to give 10011111.01 // –128 + 16 + 8 + 4 + 2 + 1 + 0.25 / ¼ seen One mark per mark point (Max 2) • correct mantissa • correct exponent Mantissa Exponent 1 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1(b) One mark per mark point (Max 3) 3 MP1 Real numbers (can) have a fractional part (such as 1/3 and ½) / (such as 0.4 and 0.25) MP2 The fixed length of the storage means that you can’t store very large / very small numbers MP3 Binary numbers represent numbers based on powers of 2, with limited fractional representations such as 1/2, 1/4, 1/8, 1/16, etc. MP4 It isn’t possible to store all fractions with the level of precision provided by this system MP5 …the fractional part of the number is as close as possible within these constraints. Question Answer Marks

1(a) One mark for working (Max 1) 3 • conversion of 65.25 to binary seen e.g. 1000001.01 = 65.25 // 64 + 1 + 0.25 / ¼ One mark per mark point (Max 2) • correct mantissa • correct exponent Mantissa Exponent 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 1 1(b) One mark per mark point (Max 2) 2 MP1 the decimal fraction 0.20 cannot be represented exactly (the closest is 0.25 / 0.1875) MP2 therefore, there will be a loss of precision due to a rounding error/truncation Question Answer Marks

1(a) One mark per mark point (Max 3) 3 MP1 conversion of exponent 001001 to 9 MP2 application of exponent to mantissa to go from 0.100111100 to 100111100 // 256 + 32 + 16 + 8 + 4 seen // 64/128 + 8/128 + 4/128 + 2/128 + 1/128 = 79/128 // 1/2 + 1/16 + 1/32 + 1/64 + 1/128 = 79/128 MP3 correct answer = 316 1(b) One mark per mark point (Max 3) 3 MP1 number converted to binary 10011001.01 // number converted to positive 102.75, reversed bits and 1 added. (0)1100110.11  10011001.00  10011001.01 // -128 + 16 + 8 + 1 + 0.25 = –102.75 MP2 exponent = 7 // Moving binary point the correct number of places MP3 correct answer Mantissa Exponent 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 1 © Cambridge University Press & Assessment 2024 Page 4 of 14

1(a) One mark per mark point (Max 2) 2 MP1 When the number of bits in the mantissa is raised, the precision / accuracy of the number represented increases // when the number of bits in the mantissa is lowered, the precision / accuracy of the number represented reduces. MP2 When the number of bits in the exponent is reduced, the range of numbers that can be represented is reduced // when the number of bits in the exponent is increased, the range of possible numbers that can be represented increases. MP3 When the range increases the accuracy decreases // When the range decreases the accuracy increases. 1(b) One mark per mark point (Max 3) 3  number converted to binary e.g. 54.8125 = 00110110.1101 // Fractions method 1/2 + 1/4 + 1/16 + 1/32 + 1/128 + 1/256 + 1/1024 = 877/1024 // 32 + 16 + 4 + 2 + 0.5 + 0.25 + 0.0625 / (1/2 + 1/4 + 1/16)  exponent = 6 // Moving binary point the correct number of places  correct answer Mantissa Exponent 0 1 1 0 1 1 0 1 1 0 1 0 0 1 1 0 Question Answer Marks

1(a) One mark per mark point (Max 3) 3 MP1 conversion of exponent 001001 to 9 MP2 application of exponent to mantissa to go from 0.100111100 to 100111100 // 256 + 32 + 16 + 8 + 4 seen // 64/128 + 8/128 + 4/128 + 2/128 + 1/128 = 79/128 // 1/2 + 1/16 + 1/32 + 1/64 + 1/128 = 79/128 MP3 correct answer = 316 1(b) One mark per mark point (Max 3) 3 MP1 number converted to binary 10011001.01 // number converted to positive 102.75, reversed bits and 1 added. (0)1100110.11  10011001.00  10011001.01 // -128 + 16 + 8 + 1 + 0.25 = –102.75 MP2 exponent = 7 // Moving binary point the correct number of places MP3 correct answer Mantissa Exponent 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 1 © Cambridge University Press & Assessment 2024 Page 4 of 14

1(a) One mark per mark point (Max 1) 3 • correct answer • statement regarding number losing precision/rounding error One mark per mark point for working (Max 2) • number converted to binary 201.125 = 11001001.001 // 128 + 64 + 8 + 1 + 0.125 / 1/ seen 8 • use of the exponent e.g. moving the binary point 8 places /  28. Mantissa Exponent 0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 0 1(b) One mark per mark point (Max 2) 3 • application of exponent to go from 1.010110011 to 101011.0011 // x 25 // movement of binary point 5 places seen • –32 + 8 + 2 + 1 + .125 + .0625 // –32 + 8 + 2 + 1 + 1/ + 1/ seen 8 16 // –1 + ¼ + 1/ + 1/ + 1/ + 1/ // –1 + 179/ // –333/ 16 32 256 512 512 512 One mark for correct answer (Max 1) • –20.8125 // –2013/ 16 © Cambridge University Press & Assessment 2024 Page 4 of 17

4(a) One mark for working 2 • application of exponent to mantissa to go from 0.10001110111 to 01000111.0111 // moving the binary point 7 places // multiplying by 27/128 in the fractions method // 64 + 4 + 2 +1 + .25+ .125 + .0625 seen One mark for correct answer • 71.4375 // 717/ 16 4(b) One mark per mark point (Max 2) 4 • correct mantissa – exact answer only • correct exponent – exact answer only Mantissa Exponent 1 0 0 1 1 1 0 1 1 0 1 0 0 1 1 0 One mark per mark point for working (Max 2) • number converted to binary e.g., positive binary version of 49.1875 = 0110001.0011 // two’s complement version bits flipped and 1 added = 1001110.1101 // –64 + 8 + 4 + 2 + .5 + .25 + .0625 // –64 + 14.8125 • use of the exponent e.g. moving the binary point 6 places /  26. © Cambridge University Press & Assessment 2024 Page 6 of 15

1(a) One mark per mark point (Max 1) 3 • correct answer • statement regarding number losing precision/rounding error One mark per mark point for working (Max 2) • number converted to binary 201.125 = 11001001.001 // 128 + 64 + 8 + 1 + 0.125 / 1/ seen 8 • use of the exponent e.g. moving the binary point 8 places /  28. Mantissa Exponent 0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 0 1(b) One mark per mark point (Max 2) 3 • application of exponent to go from 1.010110011 to 101011.0011 // x 25 // movement of binary point 5 places seen • –32 + 8 + 2 + 1 + .125 + .0625 // –32 + 8 + 2 + 1 + 1/ + 1/ seen 8 16 // –1 + ¼ + 1/ + 1/ + 1/ + 1/ // –1 + 179/ // –333/ 16 32 256 512 512 512 One mark for correct answer (Max 1) • –20.8125 // –2013/ 16 © Cambridge University Press & Assessment 2024 Page 4 of 17

2(a) One mark per mark point 2 MP1 Correct mantissa MP2 Correct exponent Mantissa Exponent 0 1 1 0 1 0 1 1 1 0 1 1 1 0 1 0 2(b) One mark per mark point for working (Max 2) 4 • number converted to binary e.g., positive binary version of 25.3125 = (0)11001.0101 • two’s complement version bits flipped and 1 added = 100110.1011 • -32 + 4 + 2 + 0.5 + 0.125 + 0.0625 // -32 + 4 + 2 + 1/2 + 1/8 + 1/16 • movement of binary point seen (5 places) One mark per mark point • correct mantissa • correct exponent Mantissa Exponent 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 1 © Cambridge University Press & Assessment 2025 Page 7 of 15

2(a) Two marks for working 3 • number converted to binary (0)1111100.0111 • use of exponent = 7 // Moving binary point the correct number (7) of places One mark for correct answer Mantissa Exponent 0 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 © Cambridge University Press & Assessment 2025 Page 4 of 12 2(b) Two marks for working 3 • correct use of exponent seen • correct conversion method from binary to denary One mark for correct answer Working: 1010001.01011 // moving bp 6 places to right Evaluation of two’s complement −64 +16 + 1 + 0.25 + 0.0625 + 0.03125 // −64 + 16 + 1 + 1/4 + 1/16 + 1/32 // Converting two’s complement back and evaluating positive binary number 32 + 8 + 4 + 2 + 0.5 + 0.125 + 0.03125 // 32 + 8 + 4 + 2 + 1/2 + 1/8 + 1/32 Fractions methods - award both working marks for either (– 2048 + 512 + 32 + 8 + 2 + 1) / 2048 x 26 = -1493 / 32 OR – 1 + 1/4 + 1/64 + 1/ 256 + 1/1024 + 1/2048 x 26 = -1493 / 32 Answer: −46.65625 // −4621/ 32 Question Answer Marks

2(a) One mark for two’s complement version and one mark for denary version 2 Mantissa Exponent 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 127  2120 // 0.9921875 x 2127 // 127/128 x 2127 2(b) One mark per mark point for working (Max 2) 4 • number converted to binary e.g., positive binary version of 3.59375 = (0)11.10011 • negative two’s complement version - bits flipped and 1 added = 100.01101 • -4 + 1/4 + 1/8 + 1/32 // -4 + 025 +0.125 + 0.03125 // -(64 + 32 + 16 + 2 + 1)/32 • -22 + 2-2 + 2-3 + 2-5 • 22 (- 20 + 2-4 + 2-5 + 2-7) One mark per mark point • correct mantissa • correct exponent, with working seen. Mantissa Exponent 1 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 Question Answer Marks

2(a) One mark per mark point (Max 2) 2 MP1 Correct mantissa MP2 Correct exponent Mantissa Exponent 0 1 1 1 0 1 0 1 1 0 1 0 1 0 0 1 © Cambridge University Press & Assessment 2025 Page 6 of 15 2(b) One mark per mark point (Max 4) 4 MP1 correct method to find the binary number MP2 additional working towards binary number MP3 correct use of exponent MP4 correct answer in the space provided Two from: e.g. 76.1875 = 64+8+4+0.125+0.0625 (0)1001100.0011 –76.1875 = –128+32+16+2+1+0.5+0.25+0.0625 10110011.11 01 One mark movement of binary point by 7 places // 1.011001111 01 x 27 One mark Mantissa Exponent 1 0 1 1 0 0 1 1 1 1 0 1 0 1 1 1 Question Answer Marks Guidance

2(a) Two marks for working 3 • correct calculation/application of exponent seen • correct method to find the final answer Working: Exponent = 8 + 2 + 1 = 11 // =0.111100101 x 211 // =11110010100.0 (moving bp 11 places to right) Method to find the answer // 1024 + 512 + 256 + 128 + 16 + 4 One mark for correct answer Denary value: 1940 Example of solution using fractions: 2(b) One mark per mark point (Max 3) 3 MP1 correct method to find the binary number MP2 correct use of exponent MP3 correct answer in the space provided Working: 26.6875 converted to binary (0)11010.1011 // 16+8+2+0.5+0.125+0.0625 movement of binary point by 5 places Mantissa Exponent 0 1 1 0 1 0 1 0 1 1 0 0 0 1 0 1 © Cambridge University Press & Assessment 2025 Page 7 of 17

2(a) One mark per mark point (Max 3) 3 MP1 the precision of the number stored in the mantissa will be reduced MP2 the number of bits available for the exponent will increase to 6 MP3 … this will increase the range of numbers that can be stored. 2(b) One mark per mark point (Max 3) 3 MP1 a process involving a calculation / the multiplication of two large numbers could take place MP2 the result might be outside of the range of values possible to store in the given system MP3 … leading to the most significant bits of the mantissa/exponent being lost (which is an overflow). Question Answer Marks Guidance