1.Perform binary addition of 1101 + 0010 is ________
a) 1110
b) 1111
c) 0111
d) 1,1101
Answer: b
Explanation: The addition is performed as :
1101
+ 0010
_______
1111
Therefore, the result is 1111.
2.The addition 1+1 gives 0 as a result.
a) True
b) False
Answer: a
Explanation: The two result obtained is 0 with a carry of 1. This carry is transferred to the next higher column.
3. The result of 0*1 in binary is ____________
a) 0
b) 1
c) invalid
d) 10
Answer: a
Explanation: The binary multiplication of any number with 0 will give the result 0 itself. Any binary number when multiplied by 0 gives 0 only. e.g. 1101 * 0000 = 0000.
4.The multiplication of 110 * 111 is performed. What is a general term used for 111?
a) Dividend
b) Quotient
c) Multiplicand
d) Multiplier
Answer: d
Explanation: 111 is called the multiplier.
Whenever a multiplication is performed the second term is called the multiplier whereas the first term is called the multiplicand.
5.The result obtained on binary multiplication of 1010 * 1100 is _____________
a) 0001111
b) 0011111
c) 1111100
d) 1111000
Answer: d
Explanation: The solution is as follows :
1010
* 1100
________
0000
0000
1010
1010
_______
1111000
6. Which of the following is often called the double precision format?
a) 64-bit
b) 8-bit
c) 32-bit
d) 128-bit
Answer: a
Explanation: The 64-bit format is referred to as the double precision format. It has 1 sign bit, 8 exponent bits and 23 bits for the mantissa.
7.What do you call the intermediate terms in binary multiplication?
a) Multipliers
b) Mid terms
c) Partial Products
d) Multiplicands
Answer: c
Explanation: The intermediate terms are called partial terms. The mid terms obtained in the binary multiplications are the partial ones whereas the answer obtained is called the final product.
8.The result that is smaller than the smallest number obtained is referred to as ___________
a) NaN
b) Underflow
c) Smallest
d) Mantissa
Answer: b
Explanation: It is referred to as underflow. Nan stands for not a number. Mantissa is the part after the decimal.
9.The number of sign bits in a 32-bit IEEE format is _______
a) 1
b) 11
c) 9
d) 23
Answer: a
Explanation: There is only 1 sign bit in all the standards. In a 32-bit format, there is 1 sign bit, 8 bits for the exponent and 23 bits for the mantissa.
10.Express the decimal format of the signed binary number (101010)2 .
a) 10
b) 12
c) -12
d) -10
Answer: d
Explanation: The first bit is the sign bit whereas the rest of the bits are magnitude bits. So the number is: 01010 =23 * 1 + 21 * 1 =8+2 =10.
But , the sign bit is 1,
Therefore the answer is : (-2)10.
11.__________________ is a straightforward method of representing positive and negative numbers.
a) Radix
b) Complement
c) Sign Magnitude
d) Encode
Answer: c
Explanation: Sign Magnitude is used for the representation of positive and negative numbers. If the leftmost digit is 0, the number is positive. If the leftmost digit is 1, the number is negative.
12.The additive inverse of a number is the number which when added to the original number gives 1 as a result.
a) True
b) False
Answer: b
Explanation: Additive Inverse of a number is the number which gives 0 and not 1 when added to the original number. e.g. number=45, additive inverse =-45, after addition they give 0.
13.The 1’s complement of 1 in 4 bits is __________
a) 0001
b) 0
c) 1001
d) 1110
Answer: d
Explanation: 1’s complement is obtained by reversing the bits from 0 to 1 and vice-versa. Binary of 1 is : 0001 and 1’s complement is : 1110.
14.The binary number 111 in its 2’s complement form is ____________
a) 010
b) 001
c) 000
d) 111
Answer: b
Explanation: 2’s complement is obtained by adding 1 to the 1’s complement. 1’s complement of 111: 000 and 2’s complement:001.
15. The sign magnitude representation of -9 is ___________
a) 00001001
b) 11111001
c) 10001001
d) 11001
Answer: c
Explanation: In case of a negative number, the leftmost digit is 1 if the number is negative. Therefore, +9=00001001 and -9=10001001. Similarly for all other negative numbers.
16.If you are given a word of size n bits, the range of 2’s complement of binary numbers is ________
a) -2n+1 to +2n+1
b) -2n-1 to +2n-1
c) -2n-1 to +2n+1
d) -2n-1 to +2n-1-1
Answer: d
Explanation: 2’s complement is obtained by adding 1 to the 1’s complement. For e.g. 5 :0101 and 1’s complement=1010 and 2’s complement=1011.
17.In both signed magnitude and 2’s complement , positive and negative numbers are separated using ______________
a) LSB
b) MSB
c) 0
d) 1
Answer: b
Explanation: The positive and negative numbers are separated using the MSB. MSB is the Most Significant Bit. MSB is the leftmost bit. e.g. If 1000 is the number then 1 is the most significant bit.
18.Single Precision format comprises of _________ bits.
a) 4
b) 8
c) 16
d) 32
Answer: d
Explanation: The single precision format comprises of 32-bits. It has 1 sign bit, 8 bits for exponent and 23 for the mantissa.
19.If m is the 2’s complement and n is the binary number, then ______________
a) m=n’
b) m=n’+1
c) m=n’-1
d) m=n
Answer: b
Explanation: 2’s complement is simply obtained by addition of 1. So if n is the number and we take the 2’s complement, add 1 to it, we get the 2’s complement. Therefore, m=n’+1.
20.The possible number of bit patterns with 8 bits ________________
a) 128
b) 8
c) 24
d) 256
Answer: d
Explanation: The total number of patterns that can be formed using n-bits are 2n. Here, possible patterns are: 28=256.
21.Which of the following is used for binary multiplication?
a) Restoring Multiplication
b) Booth’s Algorithm
c) Pascal’s Rule
d) Digit-by-digit multiplication
Answer: b
Explanation: The Booth’s Algorithm is used for the multiplication of binary numbers. It is basically used for the multiplication of 2 signed numbers. This is a very important algorithm in binary arithmetic.
22.One extra bit is added on the left of a binary number, in case of Binary Multiplication using Booth’s Algorithm.
a) True
b) False
Answer: a
Explanation: The statement is true as an extra bit is added when we multiply 2 binary numbers by using Booth’s.
Let us take an example if we multiply 2 * – 3.
The first step is to obtain the binary equivalents.
Hence, 2=10 and -3=01.
Now after adding the extra bit 2=010 and -3=101.
We add 0 in case of positive numbers whereas 1 in negative numbers.
23.The binary number obtained after applying RSC on 11010 will be ___________
a) 11101
b) 10100
c) 01101
d) 01000
Answer: c
Explanation: RSC stands for Right-Shift Circulant. So, whenever the numbers are shifted to the right an extra 0 bit is added to the left. Here, after the right shift of 11010, the number obtained will be 01101.
24.The result of >> of 11001 by 3-bits will be ______________
a) 01000
b) 01111
c) 00011
d) 11111
Answer: a
Explanation: >> is the bitwise left shift operator in binary arithmetic.
Applying >>(left-shift) by 3-bits on the number 11001 will result in 3 zeroes on the right, i.e., 01000.
25.Booth’s Algorithm is applied on _____________
a) decimal numbers
b) binary numbers
c) hexadecimal numbers
d) octal Numbers
Answer: b
Explanation: Booth’s Algorithm is applied only on signed and unsigned binary numbers.
Although, the values of other number systems can be converted to binary, and then the multiplication could be performed.
26.If Booth’s Multiplication is performed on the numbers 22*3, then what is 3 referred to as __________
a) accumulator
b) multiplicand
c) quotient
d) multiplier
Answer: d
Explanation: It is referred to as the multiplier. Multiplier is denoted by Q in booth’s algorithm. 22 is called the multiplicand. These numbers are first converted to their binary equivalents and further the multiplication is performed.
27.What is the default value of accumulator in booth’s multiplication of two 4-bit binary numbers?
a) 0
b) 1
c) 0000
d) 00000
Answer: d
Explanation: The correct answer is d because in case of Booth’s algorithm an extra bit is always added to the binary numbers. The 4-bit binary numbers become 5-bit numbers after adding the extra bit. Accumulator is always assigned 0 bits of the order of the binary numbers whose multiplication is to be performed.
28.What is the value of n in multiplication of 110* 1000?
a) 2
b) 3
c) 4
d) 0
Answer: c
Explanation: In Booth’s, n denotes the number of bits that the higher binary number has when multiplication is performed. Here, since there are 4 bits in 1000, the answer is n=4.
29.What will be the value obtained after multiplication of (-2) * (-3) using Booth’s Algorithm?
a) 6
b) -6
c) -2
d) -3
Answer: a
Explanation: After applying the procedure of Booth’s Algorithm, the value obtained will be 6. Even though the result is obtained in its 2’s complement for but then it is reconverted to its normalized form. Also, the value obtained after decimal multiplication is the same as the value obtained after binary multiplication.
30.What does the data transfer instruction STA stand for?
a) Store Accumulator
b) Send Accumulator
c) Send Action
d) Store Action
Answer: a
Explanation: STA is a data transfer instruction given whenever a value is to be copied on the accumulator. It is used for the fetch operation.
31.What is the 1’s complement of 11010?
a) 11010
b) 11011
c) 00110
d) 00101
Answer: d
Explanation: The 1’s complement of a number is obtained by converting all the 0 bits to 1 and all 1’s to 0’s. Here, 11010 gets converted to 00101.
32.2’s complement is obtained by adding 1 to 1’s complement of a number.
a) True
b) False
Answer: a
Explanation: The statement is true. Let us take an example: Consider the number 110000.
1’s complement of the number is 001111. 2’s complement=001111+1=010000.
33.The 10’s complement of 562 is __________
a) 4
b) 3
c) 7
d) 8
Answer: b
Explanation: 10’s complement is obtained by adding 1 to the 9’s complement of a number.
Here, 9’s complement=999-562=436.
Therefore, 10’s complement=436+1=437.
34.The 9’s complement of 6578 is ___________
a) 1234
b) 3421
c) 3124
d) 3420
Answer: b
Explanation: 9’s complement is obtained by subtracting 9 from each bit of the number.
Here, 9999-6578=3421.
Therefore, the 9’s complement is 3421.
35. >> operator is used to denote _________
a) left shift
b) right shift
c) greater than
d) less than
Answer: b
Explanation: >> operator denotes the right shift in binary arithmetic.
E.g. If we say, >>110 by 2-bits, the value obtained will be 001.
Similarly, << denotes the right shift operator.
36.The subtraction using 1’s complement of 110 – 100 will give the result ___________
a) -011
b) 011
c) 010
d) -010
Answer: c
Explanation: To subtract using 1’s complement: Take the 1’s complement of the 2nd number (here, 011) and add it to the first number. (110+011=1,001).
If the carry is generated, it is added to the result (001+1=010).
If a carry is not generated then it is converted to its 1’s complement form.
37.The result obtained on subtraction using 2’s complement of 1111-0010 will be _________
a) 11101
b) 11011
c) 1101
d) 1011
Answer: c
Explanation: Firstly, the 2’s complement of the 2nd number is obtained.
Then, if carry is generated, it is discarded and the number is positive.
Otherwise, 2’s complement of the result is obtained and the answer is negative.
Here, 1111+1110=1, 1101. Since carry is generated, it is discarded and the answer is positive.
38.The 1’s complement of 1111111110.101 is _______________
a) 1.010
b) 0000000000.010
c) 1111111110.101
d) 1.101
Answer: a
Explanation: The 1’s complement of a number is obtained by reversing the bits with value 1 to 0 and the bits with value 0 to 1. Here, 11111111110.101 gets converted to 000000001.010 in its 1’s complement format.
39.7’s complement of 432 is _________________
a) 432
b) 543
c) 345
d) 777
Answer: c
Explanation: To find the 7’s complement of any number, we simply subtract 7 from each of the bits.
Here, 777-432=345. So, the 7’s complement is 345.
40.The 16’s complement of 74E will be __________
a) 8B2
b) F8B2
c) 2B8
d) 8C2
Answer: b
Explanation: The 16’s complement is obtained by subtracting 16 from each of the bits of that number.
Here, 16’s complement of 074E will be F8B2.
41. Binary addition of 1 + 1 gives the result _____________
a) 0
b) 1
c) 2
d) 10
Answer: a
Explanation: The result obtained is 0 with a carry of 1. This carry obtained is added to the next higher column. Binary system can have only two numbers 0 and 1.
42.4-bit is a valid IEEE- format.
a) True
b) False
Answer: b
Explanation: The two main IEEE formats are 16-bit and 32-bit. No format of 4-bit exists. Further, 8-bit format also exists.
In IEEE format, there are certain bits allotted to sign, exponent and mantissa.
43.The result of 0 – 1 in binary is ______________
a) 0
b) 1
c) 11
d) 10
Answer: b
Explanation: The binary subtraction 0 – 1 gives the result 1. A borrow of 1 is although generated and is removed from the next higher column.
44.The complement of any number can be given by a general term _______________
a) Bn – 1 – N
b) Bn -1 + N
c) Bn +1 – N
d) Bn +1 + N
Answer: a
Explanation: Complement of a number can be given by the term :
C= Bn – 1 – N.
Here,
B= base
n= number of digits
N= the number.
45.The IEEE standard followed by almost all the computers for floating point arithmetic _____
a) IEEE 260
b) IEEE 488
c) IEEE 754
d) IEEE 610
Answer: c
Explanation: Most computers follow the IEEE 754 bit format for floating point arithmetic. IEEE stands for electrical and electronic engineers. It has all the floating point arithmetic specifications.
46.Which of the following is often called the double precision format?
a) 64-bit
b) 8-bit
c) 32-bit
d) 128-bit
Answer: a
Explanation: The 64-bit format is referred to as the double precision format. It has 1 sign bit, 8 exponent bits and 23 bits for the mantissa.
47.What is NaN in IEEE standards?
a) Not arithmetic
b) Not a negation
c) Not a number
d) Not a Node
Answer: c
Explanation: It stands for not a number in IEEE standards. A NaN is obtained whenever a result id indeterminate like whenever anything is divided by 0.
48.The result that is smaller than the smallest number obtained is referred to as ___________
a) NaN
b) Underflow
c) Smallest
d) Mantissa
Answer: b
Explanation: It is referred to as underflow. Nan stands for not a number. The mantissa is the part after the decimal.
49.The number of sign bits in a 32-bit IEEE format _________
a) 1
b) 11
c) 9
d) 23
Answer: a
Explanation: There is only 1 sign bit in all the standards. In a 32-bit format, there is 1 sign bit, 8 bits for the exponent and 23 bits for the mantissa.
50.Express the decimal format of the signed binary number (101010)2 .
a) 10
b) 12
c) -12
d) -10
Answer: d
Explanation: The first bit is the sign bit whereas the rest of the bits are magnitude bits. So the number is: 01010 =23 * 1 + 21 * 1 =8+2 =10.
But, the sign bit is 1 ,
Therefore the answer is : (-2)10.
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