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_ _ a _ _ _ _ _ ` _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ : Question 1 [June 1993]
(a) In ASCII C=code, the letters of the alphabet are stored using consecutive codes. If B is represented by 1000010, what is the representation of G?
(b) Give the binary value of 2310 + 11012.
Suggested Response
(a) B is 1000010
G is 5 letters to the right from B
The binary value of 5 is 101
Therefore, G is:
10000102+101210001112
(b) The binary value of 2310 is 101112
Therefore,
1111101112+110121001002
Question 2 [June 1994]
(a) In a certain character coding system, each character occupies 7 bits and the letters of the alphabet are assigned consecutive codes. If M is represented by 1010001, what is the representation of H?
(b) What is the value of X in the following equation?
5210 X2 = 111112
Suggested Response
(a) M is 1010001
H is 5 letters to the left of M
The binary value of 5 is 101
Therefore, H is:
102210100012_101210011002
(b) 52 10 X2 = 111112
X2 = 5210 111112
5210 is 1101002
21210020221 1101002_111112101012
The value of X is 101012
Question 3 [June 1995]
(a) Convert the decimal number 55 to its binary equivalent.
(b) Convert the binary number 1011010 to its decimal equivalent.
(c) Add the binary numbers 11010 and 10011.
(d) The ASCII code for the letter H has a decimal equivalent of 72. What is the ASCII code for the letter P?
Suggested Response
(a)
255Remainder2271213126123021101
5510 = 1101112
(b)
10110106432168421
64 + 16 + 8 + 2 = 90
Therefore, 10110102 = 9010
1110102+1001121011012 (c)
(d) H is 72
H is 8 letters to the right of H
72 + 8 = 80
280Remainder24002200210025022121001
8010 = 10100002
Therefore, ASCII code for P is 1010000
Question 4 [June 1996]
(a) Convert the decimal number 5710 to a binary number.
(b) Convert the binary number 110111012 to a decimal number.
(c) Explain the term parity, as it relates to the coding of characters.
(d) Add the binary numbers 1101012 and 1011102.
Suggested Response
(a)
257Remainder2281214027023121101
5710 = 1110012
(b)
110111011286432168421
128 + 64 + 16 + 8 + 4 + 1 = 221
Therefore, 110111012 = 22110
Parity is as an error checking system using a dummy value within the coding of characters. It is used to detect whether the transmission or coding of data is complete or accurate.
1111101012+101110211000112
Question 5 [June 1997]
(a) Convert the decimal number 83 to its binary equivalent.
(b) Convert the binary number 11001101 to its decimal equivalent.
(c) Add the two binary numbers 1101 and 100101.
(d) Explain the difference between discrete and continuous data.
Suggested Response
(a)
283Remainder24112201210025022121001
8310 = 10100112
(b)
110011011286432168421
128 + 64 + 8 + 4 + 1 = 205
Therefore, 110011012 = 20510
(c)
11111012+10010121100102
(d) Discrete data is based on counts in whole numbers and the values cannot be meaningfully subdivided. For example, you cannot have 2.5 people.
Continuous data is data that can be measured and broken down into smaller parts and still have meaning. For example, measurement is continuous data as it can be expressed as 22mm, 25.2mm, 27.5mm, etc and you actually get more information from what is being measured.
Question 6 [June 1998]
(a) Convert 8710 to its binary equivalent.
(b) Add the two binary numbers 1101012 and 111012, giving your answer in binary form.
(c) Give the BCD representation of 5910.
(d) Give the BCD representation of -5910 .
Suggested Response
(a)
287Remainder24312211210125022121001
8710 = 10101112
11111101012+11101210100102(b)
(c) 5 = 0101 in BCD
9 = 1001 in BCD
Therefore, 5910 = 0101 1001 in BCD
(d) (-) = 1011 in BCD
5910 = 0101 1001 in BCD
Therefore, -5910 = 1011 0101 1001 in BCD
Question 7 [June 1999]
(a) Convert the decimal number 9010 to a binary number.
(b) Convert the binary number 110112 to a decimal number.
(c) Add the two binary numbers 11012 and 101012, giving your answer as a binary number.
(d) Determine the BCD representation of 8710.
Suggested Response
(a)
290Remainder24502221211025122121001
9010 = 10110102
(b)
11011168421
16 + 8 + 2 + 1 = 27
Therefore, 110112 = 2710
(c)
11111012+1010121000102
(d) 8 = 1000 in BCD
7 = 0111 in BCD
Therefore, 8710 = 1000 0111 in BCD
Question 8 [June 2000]
(a) Convert 9910 to its binary equivalent.
(b) Add the two binary numbers 1101012 and 110112, giving your answer in decimal form.
(c) Explain how integer can be represented using BCD.
(d) Give the BCD representation of -6510.
Suggested Response
(a)
299Remainder24912241212026023021101
9910 = 11000112
(b)
111111101012+11011210100002
10100006432168421
64 + 16 = 80
Therefore, 10100002 = 8010
(c) The 4 digit binary representation for each digit is used. A positive sign is indicated by the pattern 1010 and a negative sign by 1011.
(d) (-) = 1011 in BCD
6 = 0110 in BCD
5 = 0101 in BCD
Therefore, -6510 = 1011 0110 0101 in BCD
Question 9 [June 2001]
(a) Convert the binary number 01010101 to a decimal number.
(b) Convert the decimal number 7 (seven) to a binary number.
Suggested Response
(a)
010101011286432168421
64 + 16 + 4 + 1 = 85
Therefore, 010101012 = 8510
(b)
27Remainder23121101
710 = 1112
Question 10 [June 2002 Specimen]
(a) Explain the difference between odd parity and even parity. (1 mark)
(b) The ASCII representation for the letter D is 1000100
(i) What is the ASCII representation for the letter C? (1 mark)
(ii) What is the ASCII representation for the letter G? (1 mark)
(c) Give the decimal equivalent for the binary number 01001001. (1 mark)
(d) Give the binary equivalent for the decimal number 25. (1 mark)
(e) Add the two binary numbers 00110011 and 00100110 giving the answer as a binary number. (1 mark)
Suggested Response
(a) Under odd parity, the number of ones in a byte is odd. An extra (parity) bit is added and set to 1 if the number of ones isnt odd or it is set to 0 if it is already odd. Upon retrieval of the byte after data transmission, if the number of ones including the parity bit isnt odd, then an error must have occurred.
Under even parity, the number of ones in a byte is even. An extra (parity) bit is added and set to 1 if the number of ones isnt even or it is set to 0 if it is already even. Upon retrieval of the byte after data transmission, if the number of ones including the parity bit isnt even, then an error must have occurred.
(b) (i) C is 1 letter to the left of D
Therefore, C is:
1022110001002_1210000112
(ii) G is 3 letters to the right of D
The binary value of 3 is 11
10001002+11210001112 Therefore, G is:
(c)
010010011286432168421
64 + 8 + 1 = 73
Therefore, 010010012 = 7310
(d)
225Remainder212126023021101
2510 = 110012
(e)
111001100112+001001102010110012
Question 11 [June 2002]
Integers are good for representing discrete data. Real numbers and floating point values are good for representing continuous data. Explain these statements.
Suggested Response
Integers are good for representing discrete data because discrete data is based on counts in which only a finite number of values is possible as the values cannot be subdivided meaningfully. Subdividing a whole number will create a fractional or decimal component. Integers are whole numbers that cannot be written with a fractional or decimal component.
On the other hand, real numbers and floating point values are good for representing continuous data because continuous data is data that can be measured and broken down into smaller parts and still have meaning. Real numbers and floating point values are used to represent fractional and decimal components of numbers which provides additional meaning.
Question 12 [June 2002]
(a) Using an 8-bit representation, give the
i) binary representation for the decimal number 10
ii) ones complement representation for the decimal number NEGATIVE 10
twos complement representation for the decimal number NEGATIVE 10
sign-and-magnitude representation for the decimal number NEGATIVE 10
(b) Using even parity with an 8-bit data byte, where the leftmost bit is to be used as the parity bit, which of the following are valid bit patterns?
00000000
00000011
00000010
10000001
Suggested Response
(a) i)
210Remainder25022121001
1010 = 000010102 (Note: Four 0 bits are added to the leftmost end to increase number of bits to 8)
ii) The 8-bit representation of 10 is 000010102
-10 in ones complement will therefore be:
11110101 (Note: Bits are flipped to create the ones complement value)
iii) The ones complement of -10 is 11110101
Therefore, the twos complement representation is:
111110101+111110110
iv) 10001010 (Note: This simply requires changing the leftmost bit in
00001010 to 1, which is the sign bit representing negative)
(b) The valid bit patterns using even parity are:
i, ii and iv
Question 13 [June 2003]
The eight-bit representation of the decimal number 25 is 00011001.
(a) What is the twos complement representation of NEGATIVE 25? (1 mark)
(b) What are the sign and magnitude representations of NEGATIVE 25? (1 mark)
(c) What is the decimal equivalent of 00110111? (1 mark)
(d) The ASCII representation of H is 01001000. What is the ASCII representation of D? (1 mark)
Suggested Response
(a) The ones complement of 00011001 is 11100110
Therefore, the twos complement representation is:
11100110+111100111
(b) 10011001
(c)
001101111286432168421
32 + 16 + 4 + 2 + 1 = 55
(d) H is 01001000
D is 4 letters to the left of H
The binary value of 4 is 100
02010010002_1002010001002Therefore, D is:
Question 14 [June 2004]
(a) State the binary and BCD representation for 187. (2 marks)
(b) Using eight-bit, twos complement binary numbers, compute 11 7. (2 marks)
Suggested Response
(a)
2187Remainder293124612230211125122121001
18710 = 101110112
1 = 0001 in BCD
8 = 1000 in BCD
7 = 0111 in BCD
Therefore, 18710 = 0001 1000 0111 in BCD
(b) 1110 = 000010112 (8-bit binary)
710 = 000001112 (8-bit binary)
11111000 (flip bits to give -7 in ones complement)
+ 1
11111001 (-7 in twos complement)
00001011 (1110)
+ 11111001 (-710)
100000100 (410)
Ans. = 00000100 (i.e. drop left-most bit as answer should be 8-bits)
Note: Brackets contain explanations.
Question 15 [June 2005]
(a) Convert the decimal number 25 to binary, using an eight-bit representation. (1 mark)
(b) The binary number 1100101101 uses 10 bits to store sign and magnitude. What is the decimal number equivalent? (2 marks)
(c) State the Binary Coded Decimal (BCD) representation for the decimal number 358. (2 marks)
(d) State the eight-bit twos complement representation of the decimal number 63. (2 marks)
Suggested Response
(a)
225Remainder212126023021001
25 = 10001 (5 bits)
Place zeros (three in all) to the left of the number to make it up to 8 bits.
Therefore, 25 = 000100012
(b)
1100101101_2561286432168421
Note: The leftmost bit is a sign-bit (1 represents the negative sign) and should not be included in the addition of the weights.
Therefore, 256 + 32 + 8 + 4 + 1 = -301
(c) 3 = 0011
5 = 0101
8 = 1000
Therefore, 358 = 0011 0101 1000
(d)
263Remainder2311215127123121101
63 = 111111
The 8-bits twos complement representation would be:
00111111
Question 16 [June 2006]
Perform calculations to determine which of the following examples represents binary coded decimal, 8-bit sign and magnitude or even parity. Make sure to state your answer in each example.
(a) 1011000 1
(b) -19 = 1111 0001 1001
(c) -4 = 10000100 (6 marks)
Suggested Response
(a) The number of 1-bit is even, therefore it can only be concluded that this example represents even parity. No calculation can be performed.
(b) Based on the organization of the bits in groups of four in the binary component, the example represents binary coded decimal and is proven as follows:
84210001
= 1
84211001
8 + 1 = 9
The negative sign (-) in BCD is represented by 1111
Therefore, -19 = 1111 0001 1001
(c) The binary equivalent of 4 is 100 while a 1 bit is used to represent a negative, thus making this example an 8-bit sign and magnitude number. It can be proven as follows:
-643216842110000100
= -4
Therefore, -4 = 10000100
DATA REPRESENTATION AND MANIPULATION: Questions and Answers
PAGE
Questions are the copyright material of the Caribbean Examinations Council (CXC).
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