A car has many embedded systems - Edexcel - GCSE Computer Science - Question 2 - 2018 - Paper 1
Question 2
A car has many embedded systems.
(a) An embedded system processes binary numbers.
(i) The speed limit for some roads is 60 miles per hour.
Convert the denary numbe... show full transcript
Worked Solution & Example Answer:A car has many embedded systems - Edexcel - GCSE Computer Science - Question 2 - 2018 - Paper 1
Step 1
Convert the denary number 60 to 8-bit binary.
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Answer
The denary number 60 can be converted to binary by dividing the number by 2 and recording the remainders:
60 ÷ 2 = 30 → remainder 0
30 ÷ 2 = 15 → remainder 0
15 ÷ 2 = 7 → remainder 1
7 ÷ 2 = 3 → remainder 1
3 ÷ 2 = 1 → remainder 1
1 ÷ 2 = 0 → remainder 1
Reading the remainders from bottom to top gives us 111100. However, we need an 8-bit representation, so we add leading zeros:
Answer: 00111100
Step 2
Convert the 8-bit binary number 0010 0011 to denary.
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Answer
To convert the binary number 0010 0011 to denary, we calculate the value based on the position of each bit:
0 × 2^7 = 0
0 × 2^6 = 0
1 × 2^5 = 32
0 × 2^4 = 0
0 × 2^3 = 0
1 × 2^2 = 4
1 × 2^1 = 2
1 × 2^0 = 1
Adding these values together gives us:
total = 32 + 4 + 2 + 1 = 35.
Answer: 35
Step 3
Complete the table to show the result of 0010 0011 + 0100 1010.
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Answer
To add the two binary numbers:
0010 0011
+ 0100 1010
--------------
0110 1101
The result of adding 0010 0011 and 0100 1010 is:
Answer: 0110 1101
Step 4
Identify one reason why encryption is used.
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Answer
One reason why encryption is used is:
To keep data secure.
Step 5
Complete the table using a Caesar Cipher algorithm.
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Answer
Using a Caesar Cipher algorithm with a shift of +4:
Plain Text
Shift
Cipher Text
brakes
+4
fveliw
clutch
+4
fylxq
Both entries represent the characters shifted by 4 positions in the alphabet.
Step 6
Explain one special condition a Caesar Cipher algorithm must handle.
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Answer
One special condition a Caesar Cipher algorithm must handle is:
It must wrap around the alphabet. For example, if the shift extends past 'z', it should loop back to the beginning of the alphabet (so 'z' shifted by 1 becomes 'a').
