### Previous IES Questions on Number Systems, Binary Codes and 1's and 2's Complements

1. Find the value of ‘r’, if √(41)r = (7)10. (1988)

2. Find the number of 1’s in an unsigned binary representation of given number. 163*9 + 162*7 + 16*5 + 3. (1991)

3. Find the number of 1’s in an unsigned binary representation of given number. 3 * 4096 + 15 * 256 + 5 * 16 + 3. (1995)

4. Find the value of ‘r’, if √(224)r = (13)r. (1997)
a. 10
b. 8
c. 5
d. 6

5. Convert the following: (1993)
a. 110.011(2) = X(10)
b. 1118(10) = Y(H)
Answer: X = 6.375 & Y = 45E

6. Consider a 3 bit error detection and 1 bit error correction hamming code for 4 bit data. The extra parity bits required would be ________________, and to detect 3 bit error, the minimum distance of the code should be ____________. (1992)
If two codewords are Hamming distance d apart, it will take d one-bit errors to convert one into the other.
To detect (but not correct) up to d errors per length n, you need a coding scheme where codewords are at least (d+1) apart in Hamming distance. Then d errors can't change into another legal code, so we know there's been an error.
To correct d errors, need codewords (2d+1) apart. Then even with d errors, bitstring will be d away from original and (d+1) away from nearest legal code. Still closest to original. Original can be reconstructed.

7. In 2’s compliment arithmetic, explain the condition for overflow. (1992)
The rules for detecting overflow in a two's complement sum are simple:
If the sum of two positive numbers yields a negative result, the sum has overflowed.
If the sum of two negative numbers yields a positive result, the sum has overflowed.
Otherwise, the sum has not overflowed.
It is important to note the overflow and carry out can each occur without the other. In unsigned numbers, carry out is equivalent to overflow. In two's complement, carry out tells you nothing about overflow.
The reason for the rules is that overflow in two's complement occurs, not when a bit is carried out out of the left column, but when one is carried into it. That is, when there is a carry into the sign. The rules detect this error by examining the sign of the result. A negative and positive added together cannot overflow, because the sum is between the addends. Since both of the addends fit within the allowable range of numbers, and their sum is between them, it must fit as well.

8. Following 7 bit single error correcting Hamming coded message is received as 1000110. Determine if the message is correct, by assuming that at most 1-bit error could be corrected. If the message contains an error, find the bit which is erroneous and give the correct message. (1994)
Answer: incorrect message. message is corrupted at 6th position, to get the correct word reverse the bit in 6th position. so the correct word is 1100110.

9. What is the distance of the following code? If the code contains 000000, 010101, 000111, 011001 and 111111. (1995)
a. 2
b. 3
c. 4
d. 1

10. The octal representation of an integer is 342(8). If this were to be treated as an 8 bit integer in an 8085 based computer, its decimal representation is ________. (1998)
a. 226
b. – 98
c. 76
d. – 30

11. Zero has two representations in___________ (1999)
a. Sign magnitude
b. 1’s complement
c. 2’s complement
d. Both a and b

12. The number 43 in 2’s complement representation is (2000)
a. 01010101
b. 11010101
c. 00101011
d. 10101011

13. 2’s complement representation of (– 539)10 in hexadecimal is (2001)
a. ABE
b. DBC
c. DE5
d. 9E7

14. What is equivalent value for the decimal value 0.25 (2002)
a. 0.1(2)
b. 0.01(2)
c. 0.00111…(2)
d. Can’t be represented precisely

15. 2’s complement representation of decimal value (– 15) is (2002)
a. 1111
b. 11111
c. 111111
d. 10001

16. Assume all numbers are in 2’s complement form, which of the following numbers is divisible by 11111011? (2003)
a. 11100111
b. 11100100
c. 11010111
d. 11011011

17. If 73(x) = 54(y) then the possible values of x and y are (2004 & 2011)
a. 8, 16
b. 10,12
c. 9,13
d. 8,11

18. Let A = 11111010 and B = 00001010 be two 8-bit 2’s complement numbers. Their product in 2’complement is (2004)
a. 11000100
b. 10011100
c. 10100101
d. 11010101

19. An 8085 microprocessor executes the following instructions: Two numbers are represented in signed 2’s complement form as P = 11101101 and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2’s complement form is (2012)
a. 100000111
b. 00000111
c. 11111001
d. 011111001

20. If (11X1Y)8 = (12C9)16 then the values of X and Y are (2012)
a. 3, 1
b. 5, 7
c. 7, 5
d. 1, 5

21. Binary data is being represented in size of byte and in 2’s complement form. The number of 0’s present in representation of (– 127)10 is (2012)
a. 8
b. 7
c. 6
d. 5

22. The decimal equivalent of binary number 10110.11 is (2013)
a. 16.75
b. 20.75
c. 16.50
d. 22.75

23. Hexadecimal conversion of decimal number 227 will be (2013)
a. A3
b. E3
c. CC
d. C3

24. A seven bit Hamming code is received as 1111101. What is the correct code? (2013)
a. 1101111
b. 1011111
c. 1111111
d. 1111011

25. The number of 1’s present in the binary representation of 15*256 + 5*16 + 3 are (2014)
a. 8
b. 9
c. 10
d. 11

26. Given (135)X + (144)X = (323)X. what is the value of x? (2005 & 2014)
a. 5
b. 3
c. 12
d. 6

27. Which one of the following is correct answer when 11011(2) is subtracted from 11101(2) by using 1’s complement method? (2015)
a. 01001
b. 10001
c. 00011
d. 00010

28. An Excess-3 code arithmetic operation is used to perform (2015)
b. Binary subtraction
d. BCD subtraction

29. Convert the decimal 41.6875 into octal (2015)
a. 51.54
b. 51.13
c. 54.13
d. 51.51

30. The decimal equivalent of binary 110.001 is (2015)
a. 6.25
b. 6.125
c. 62.5
d. 0.612

31. Given (125)r = (203)5. The value of r will be (2015)
a. 16
b. 10
c. 8
d. 6

32. The 9’s complement of (25.639)10 is (2015)
a. 74.360
b. 0.6732
c. 6.732
d. 7.436

33. If (2.3)4 + (1.2)4 = (Y)4, then what is the value of Y? (2005)
a. 10.1
b. 10.01
c. 10.2
d. 1.02

34. The number of 1’s in 8-bit representation of – 127 in 2’s complement form is ‘m’ and that in 1’s complement form is ‘n’. What is the value of m:n? (2005)
a. 2:1
b. 1:2
c. 3:1
d. 1:3

35. Assertion (A): There is no overflow after an addition, if one number is positive and the other is negative.
Reason (R): Adding a positive number to a negative number always produces a result, which is smaller than the larger of the two. (2005)
a. Both A and R are individually true and R is the correct explanation of A
b. Both A and R are individually true and R is not the correct explanation of A
c. A is true but R is false
d. A is false but R is true

36. A Gray code is a/an___________ (2005)
a. Binary weight code
b. Arithmetic code
c. Code which exhibits a single bit change between two consecutive codes
d. Alphanumeric code

37. Which of the following subtraction operations do not result in F16?
1. (BA)16 – (AB)16
2. (BC)16 – (CB)16
3. (CB)16 – (BC)16
Select the correct answer using the codes given below: (2006)
a. Only 1 and 2
b. Only 1 and 3
c. Only 2 and 3
d. All, 1 , 2 and 3

38. What is the Gray code word for the binary number 101011? (2006)
a. 101011
b. 110101
c. 011111
d. 111110

39. What is the addition of (– 64)10 and (80)16? (2007)
a. (– 16)10
b. (16)16
c. (1100000)2
d. (01000000)2

40. (24)8 is expressed in Gray code. Which one of the following is correct? (2008)
a. 11000
b. 10100
c. 11110
d. 11111

41. Two 2’s complement numbers having sign bits ‘x’ and ‘y’ are added and the sign bit of the result is ‘z’. which Boolean function indicates overflow condition? (2008)
a. xyz
b. x’y’z’
c. x’y’z + xyz’
d. xy + yz + zx

42. Consider the following statements: (2009)
1. Taking 2’s complement is equivalent to sign change
2. In the 2’s complement representation, the MSB is zero for a positive number
3. In a 4 bit binary representation of a binary number A, A + 1’s complement of A = 24.
Which of the above statements are correct?
a. 1 and 2 only
b. 1 and 3 only
c. 2 and 3 only
d. 1, 2 and 3

43. The hexadecimal representation of (657)8 is (2010)
a. 1AFH
b. D78H
c. D71H
d. 32FH

44. Find radix ‘r’, if √(41)r = 5(10).

45. Find radix, if 302/20 = 12.1

46. In a positional weight system, x and y are two successive digits and xy = 25(10) and yx = 31(10). Determine radix, x and y.

1. 2. 3. Oh, I didn't know about that!.. Yesterday I've read something similar at https://qanda.typicalstudent.org/

4. in question no 16 answer will be 'A'

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