GCSE Computer Science: Data Representation

Data representation is the part of the course where marks are lost to arithmetic rather than to understanding, so every question here is worked and every wrong option is a number a real slip produces.

The byte 10110101 comes up twice, once as a conversion to denary and once as a conversion to hexadecimal, and the two answers check each other: 128 plus 32 plus 16 plus 4 plus 1 is 181, and B5 is eleven sixteens plus five, which is also 181. Among the wrong answers is 173, which is what you get by reading the sixteens column as eights — the single most common place-value error on this topic, and one an automated checker made on this very question while the bank was being verified.

Seventy-six goes the other way, into eight bits, with the three wrong options being the numbers you actually land on by taking the wrong powers of two. A left shift turns 00001101 into 00011010 and thirteen into twenty-six, with the answer required to say both. And an eight-bit register holding 11111111 has one added to it, where the point is not that something goes wrong but what goes wrong: the result needs nine bits, the ninth cannot be kept, and the overflow is flagged.

Two questions are the file size calculations that carry the marks. An image twenty pixels by thirty at three bits a pixel is six hundred pixels, eighteen hundred bits, two hundred and twenty-five bytes — and 1800 is offered as an answer for anyone who stops one step early. A recording of four seconds sampled eight thousand times a second at eight bits a sample is thirty-two kilobytes, with the version that forgets the duration and the version that forgets to divide by eight both available.

The last two are about what is really being stored. A character set is approached through the codes for capital A and lower-case a, where the answer has to be that there is no letter in memory at all, only a number, and the table saying which number means which character is agreed in advance so that two machines can read each other's files. And compression is set as a choice between a spreadsheet of accounts and a photograph, with the second half of the answer doing the work: lossy compression permanently discards detail, and uncompressing does not bring it back.

Every conversion and calculation has been checked by hand. Nothing is reproduced from any exam board specification, past paper or mark scheme.

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  • Convert between binary, denary and hexadecimal without a place-value slip
  • Say what hexadecimal is for, and who it is for
  • Perform a binary shift and state its effect on the value
  • Explain what overflow is in terms of the width of a register
  • Say what a character set is and what is actually stored when text is saved
  • Calculate an image file size from dimensions and colour depth
  • Calculate a sound file size from duration, sample rate and bit depth
  • Choose between lossy and lossless compression and say what lossy removes

Ten worked conversions and calculations: 10110101 into denary and into hex, 76 into eight bits, why a programmer writes B5 rather than 10110101, 00001101 shifted one place left, one added to a full eight-bit register, the codes 65 and 97 for the same letter in two cases, a twenty by thirty image at three bits a pixel, four seconds sampled eight thousand times at eight bits, and a spreadsheet and a photograph to be made smaller. Twelve flashcards carry the vocabulary — bit, byte, binary, denary, hexadecimal, most significant bit, binary shift, overflow, character set, pixel, colour depth, sample rate.

Sample question

A student is asked to convert the binary number 10110101 into denary. What is the correct denary value?

See the answer

181

To convert 10110101, add the values of the positions where the bit is 1: 128 + 32 + 16 + 4 + 1 = 181. A common error is miscalculating the sum by missing the bit in the 16s place or misaligning the powers of two.

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