Number Systems and Data Representation in Computers

The correct answer is D, E. Hexadecimal has base 16, and because 2 raised to the power 4 equals 16, exactly four binary bits make one hexadecimal digit. The given number 1110 is already a group of four bits, so it converts in one step. Its place values are 8, 4, 2 and 1, and the bits are 1, 1, 1 and 0, so the decimal value is 8 + 4 + 2 + 0, that is 14. In hexadecimal the decimal values 10 to 15 are written as the letters A to F, so 10 is A, 11 is B, 12 is C, 13 is D, 14 is E and 15 is F. Fourteen is therefore E. Option A is wrong because F stands for 15, whose binary form is 1111. Option B is wrong because B stands for 11, that is 1011. Option C is wrong because A stands for 10, that is 1010.

The correct answer is C, 110010111. Octal has base 8, and because 2 raised to the power 3 equals 8, every octal digit expands into exactly three binary bits. Take the digits of 627 one at a time. The digit 6 becomes 110, the digit 2 becomes 010 and the digit 7 becomes 111. Writing them in the same order gives 110 010 111, that is 110010111. A quick check works the other way: grouping 110010111 in threes from the right gives 110, 010 and 111, which read back as 6, 2 and 7. Option A is wrong because 111010110 reads as 7, 2 and 6, which is octal 726. Option B is wrong because 110010110 ends in 110 and so reads as 626. Option D is wrong because 011010111 reads as 3, 2 and 7, that is octal 327.

The correct answer is A, (321)8. To turn a binary number into octal, group the bits in threes starting from the rightmost bit, adding leading zeros to complete the last group, because three bits cover the eight octal digits. The number 11010001 splits into 11, 010 and 001, and the first group is written as 011. The group 011 has the value 3, the group 010 has the value 2 and the group 001 has the value 1, so the octal number is 321. Checking in decimal confirms it: 11010001 is 128 + 64 + 16 + 1, which equals 209, and octal 321 is 3 times 64 plus 2 times 8 plus 1, again 209. Option B is wrong because 123 is the same digits reversed. Option C is wrong because 641 would need the bits 110100001. Option D is wrong because 146 in octal equals 102 in decimal.

The correct answer is C, 10111. The safest way to handle a binary subtraction in an examination is to convert both numbers to decimal, subtract, and convert the answer back. The first number 10010000 has ones in the places worth 128 and 16, so it equals 144. The second number 1111001 has ones in the places worth 64, 32, 16, 8 and 1, so it equals 121. The difference is 144 minus 121, which is 23. Now convert 23 to binary by repeated division by 2, which gives the remainders 1, 1, 1, 0 and 1 read upwards, that is 10111. Option A is wrong because 11101 equals 29. Option B is wrong because 11011 equals 27. Option D is wrong because 10011 equals 19. Only 10111 equals 23, so C is the answer.

The correct answer is B, 4. A bit, short for binary digit, is a single 0 or 1 and is the smallest unit of data a computer handles. Four bits together are called a nibble, which is exactly half of a byte, and the playful name comes from a nibble being a small bite. The nibble matters in practice because one hexadecimal digit is stored in exactly four bits, which is why hexadecimal is such a convenient shorthand for binary. Option A is wrong because two bits form no named unit and can represent only four combinations. Option C is wrong because eight bits make one byte, the unit in which a single character is normally stored. Option D is wrong because sixteen bits make two bytes, often called a word on older machines. Remember the ladder: bit, nibble of 4 bits, byte of 8 bits.

The correct answer is B, 1024. Storage units in computing grow in powers of two rather than powers of ten, because addresses are binary. One kilobyte is 2 raised to the power 10 bytes, which equals 1024 bytes. The same step of 1024 continues upward: 1024 kilobytes make one megabyte, 1024 megabytes make one gigabyte and 1024 gigabytes make one terabyte. Option A is wrong because 1000 bytes is the decimal kilobyte used by disc manufacturers in advertising, and it is exactly the trap this question sets. Option C is wrong because 2048 bytes is two kilobytes. Option D is wrong because 512 bytes is half a kilobyte, a figure students recall from the classic disc sector size. In competitive examinations always take one kilobyte as 1024 bytes unless the question itself says otherwise.

The correct answer is B, 128. The octal system has base 8, so it uses only the eight digits from 0 to 7. The general rule is that the largest digit of any number system is one less than its base, which means the digits 8 and 9 can never appear in an octal number. The number 128 contains the digit 8, so it cannot be an octal number, although it is a perfectly valid decimal or hexadecimal number. Option A is wrong as a choice because 377 uses only 3 and 7, both of which are allowed. Option C is wrong because 745 uses 7, 4 and 5, all below 8. Option D is wrong because 66 uses only the digit 6. Apply the same test to binary, where only 0 and 1 are allowed, and to hexadecimal, where 0 to 9 and A to F are allowed.

The correct answer is C, 25. To convert a binary number to decimal, write the place values from the right as 1, 2, 4, 8, 16 and so on, multiply each bit by its place value and add the results. The number 11001 has five bits, so the place values from left to right are 16, 8, 4, 2 and 1, and the bits are 1, 1, 0, 0 and 1. The sum is 16 + 8 + 0 + 0 + 1, which equals 25. Option A is wrong because 21 in binary is 10101. Option B is wrong because 23 in binary is 10111. Option D is wrong because 27 in binary is 11011. A useful check is that a binary number ending in 1 is always odd and one ending in 0 is always even, which immediately rules out any even option here.

The correct answer is B, 7. ASCII stands for American Standard Code for Information Interchange. In its standard form it is a seven-bit code, so it can represent 2 raised to the power 7, that is 128 different characters: the English capital and small letters, the ten digit symbols, punctuation marks and a set of control codes. In this code the capital letter A is 65, the small letter a is 97 and the digit character 0 is 48. Option A is wrong because six bits would allow only 64 characters, too few for both cases of the alphabet. Option C is wrong because eight bits describe Extended ASCII and EBCDIC, which reach 256 characters. Option D is wrong because sixteen bits describe the older fixed-width form of Unicode, the standard created so that Devanagari and every other script could be represented.

The correct answer is B, 0110. The 2's complement is found in two steps. First take the 1's complement by changing every 0 to 1 and every 1 to 0; for 1010 this gives 0101. Then add 1 to that result: 0101 plus 1 equals 0110. So the 2's complement of 1010 is 0110. Computers store negative numbers in 2's complement form because the same adder circuit can then perform both addition and subtraction, which saves hardware. Option A is wrong because 0101 is only the 1's complement, the first step, and forgetting to add one is the commonest mistake in this question. Option C is wrong because 1011 is simply 1010 with the last bit changed. Option D is wrong because 1001 is neither complement of the given number. Remember: 2's complement equals 1's complement plus one.

The correct answer is C, 13. Hexadecimal has base 16 and therefore needs sixteen different symbols. The digits 0 to 9 serve for the first ten values, and because no single digit exists for the values 10 to 15, the letters A to F are borrowed. The mapping runs A for 10, B for 11, C for 12, D for 13, E for 14 and F for 15. The letter D is the fourth of these letters, so it stands for 13, whose binary form is 1101. Option A is wrong because 11 is B. Option B is wrong because 12 is C. Option D is wrong because 14 is E. A simple memory aid is that the letters continue the count from ten, so the position of the letter in the alphabet after A, plus ten, gives its value.

The correct answer is C, 255. One byte holds eight bits, and each bit can be either 0 or 1, so the number of different patterns is 2 raised to the power 8, that is 256. Those 256 patterns represent the numbers from 0 to 255 when the byte is unsigned, because counting starts at zero. The largest pattern, 11111111, adds up to 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1, which equals 255. Option A is wrong because 127 is the largest positive value of a signed byte, where the leftmost bit is kept as the sign bit and only seven bits remain for the value. Option B is wrong because 128 is the place value of the leftmost bit alone. Option D is wrong because 256 is the count of patterns, not the largest value, a distinction examiners test often.