1. What is the binary equivalent of (-3/8)?
a) (10011)2
b) (0011)2
c) (1100)2
d) (1101)2
Explanation: The number (-3/8) is stored in the computer as the 2’s complement of (3/8)
We know that the binary equivalent of (3/8)=0011
Thus the twos complement of 0011=1101.
2. Which of the following is the correct representation of a floating point number X?
a) 2E
b) M.2E(1/2<M<1)
c) 2M.2E(1/2<M<1)
d) None of the mentioned
Explanation: The binary floating point representation commonly used in practice, consists of a mantissa M, which is the fractional part of the number and falls in the range 1/2<M<1, multiplied by the exponential factor 2E, where the exponent E is either a negative or positive integer. Hence a number X is represented as X= M.2E(1/2<M<1).
3. What is the mantissa and exponent respectively obtained when we add 5 and 3/8 in binary float point representation?
a) 0.101010,011
b) 0.101000,011
c) 0.101011,011
d) 0.101011,101
Explanation: We can represent the numbers in binary float point as
5=0.101000(2011)
3/8=0.110000(2101)=0.000011(2011)
=>5+3/8=(0.101000+0.000011)(2011)=(0.101011)(2011)
Therefore mantissa=0.101011 and exponent=011
4. What is the largest floating point number that can be represented using a 32-bit word?
a) 3*1038
b) 1.7*1038
c) 0.2*1038
d) 0.3*1038
Explanation: Let the mantissa be represented by 23 bits plus a sign bit and let the exponent be represented by 7 bits plus a sign bit
5. If E=0 and M=0, then which of the following statement is true about X?
a) Not a number
b) Infinity
c) Defined
d) Zero
Explanation: According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)s.2E-127(M).
From the above equation we can interpret that,
If E=0 and M=0, then the value of X is 0.
6. The truncation error for the sign magnitude representation is symmetric about zero.
a) True
b) False
Explanation: The truncation error for the sign magnitude representation is symmetric about zero and falls in the range
-(2-b-2-bm) ≤ Et ≤ (2-b-2-bm)
7. . What is the range of round-off error for a foxed point representation?
a) [-0.5(2-b+2-bm), 0.5(2-b+2-bm)]
b) [0, (2-b+2-bm)]
c) [0, (2-b-2-bm)]
d) [-0.5(2-b-2-bm), 0.5(2-b-2-bm-bm)]
Explanation: The round-off error is independent of the type of fixed point representation. The maximum error that can be introduced through rounding is 0.5(2-b+2-bm) and this can be either positive or negative, depending on the value of x. Therefore, the round-off error is symmetric about zero and falls in the range
[-0.5(2-b-2-bm), 0.5(2-b-2-bm-bm)]
8. What is the 2’s complement of (1100)2?
a) (0100)2
b) (0011)2
c) (0111)2
d) (1100)2
Explanation: The ones complement of (1100)2 is (0011)2. Thus the two complement of this number is obtained as (0011)2+(0001)2=(0100)2.
9. The binary digit b-A is called as ______________
a) LSB
b) Total value
c) MSB
d) None of the mentioned
Explanation: Since the binary digit b-A is the first bit in the representation of the real number, it is called as the most significant bit(MSB) of the number
10. If E=255 and M≠0, then which of the following statement is true about X?
a) Not a number
b) Infinity
c) Defined
d) Zero
Explanation: According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)s.2E-127(M).
From the above equation we can interpret that,
If E=255 and M≠0, then X is not a number.