Edexcel GCSE Computer Science (1CP2) Exam Questions

Award one mark for:

• 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Award one mark for any of the following up to a maximum of two marks:

• A set/collection of characters (1)
• (Characters such as) letters, digits and symbols/punctuation (1)
• Each character is represented by a unique code/binary value (1)
• A standard set of characters so that text can be exchanged between computer systems (1)

Allow 'a list of characters that a computer can represent' for the first mark.

Do not accept a program/software that converts text into binary.

Do not accept 'a set of binary codes' on its own.

Award up to two marks for a linked description such as:

• 6 bits give only 64 different codes (1) which is fewer than the number of letters, digits and symbols that must be represented (1)
• 7 bits give 128 different codes (1) which is enough to represent every letter, digit and symbol that is needed (1)
• Adding a bit doubles the number of different codes from 64 to 128 (1) so that every character needed can have its own code (1)
• There are more than 64 characters that must be represented (letters, digits and symbols) (1) so 6 bits cannot give each one a different code (1)

A statement of the number of codes without a development can earn one mark only; the development must be linked to the statement for the second mark.

Do not accept responses that only say computers use binary (1s and 0s).

Do not accept that 7 bits are needed because 7 is a bigger number than 6, with no reference to the number of codes or characters.

Do not accept responses about 8 bits, extended ASCII or Unicode.

Award up to two marks for a linked explanation such as:

• The computer looks up the character in the ASCII character set/table (1) to find the code that represents it (1)
• Each character has its own unique ASCII code (1) so the letter A is converted to and stored as a 7-bit binary pattern (1)
• ASCII uses 7 bits for each character (1) so A is stored as its ASCII code 1000001 (binary) / 65 (denary) (1)

The second mark requires a linked expansion, so a point on its own (e.g. 'A has the code 65') can only earn one mark.

For both marks, the expansion must follow/associate with the statement.

Accept an 8-bit pattern with a leading 0 (e.g. 01000001) as the binary code for A.

Do not accept 'the character is stored as text' without saying it is changed into a code/binary.

Do not accept a binary value on its own (e.g. 1000001) without saying that it is the ASCII code/that the character is converted to it.