McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 1. |
Any signed negative binary number is recognised by its |
| A. | msb |
| B. | lsb |
| C. | byte |
| D. | nibble |
| Answer» A. msb | |
| Explanation: any negative number is recognized by its msb (most significant bit). | |
| 2. |
The parameter through which 16 distinct values can be represented is known as |
| A. | bit |
| B. | byte |
| C. | word |
| D. | nibble |
| Answer» C. word | |
| Explanation: it can be represented up to 16 different values with the help of a word. nibble is a combination of four bits and byte is a combination of 8 bits. it is “word” which is said to be a collection of 16-bits on most of the systems. | |
| 3. |
If the decimal number is a fraction then its binary equivalent is obtained by the number continuously by 2. |
| A. | dividing |
| B. | multiplying |
| C. | adding |
| D. | subtracting |
| Answer» B. multiplying | |
| Explanation: on multiplying the decimal number continuously by 2, the binary equivalent is obtained by the collection of the integer part. however, if it’s an integer, then it’s binary equivalent is determined by dividing the number by 2 and collecting the remainders. | |
| 4. |
The representation of octal number (532.2)8 in decimal is |
| A. | (346.25)10 |
| B. | (532.864)10 |
| C. | (340.67)10 |
| D. | (531.668)10 |
| Answer» A. (346.25)10 | |
| Explanation: octal to decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index position. | |
| 5. |
The decimal equivalent of the binary number (1011.011)2 is |
| A. | (11.375)10 |
| B. | (10.123)10 |
| C. | (11.175)10 |
| D. | (9.23)10 |
| Answer» A. (11.375)10 | |
| Explanation: binary to decimal conversion is obtained by multiplying 2 to the power of base index along with the value at that index position. | |
| 6. |
An important drawback of binary system is |
| A. | it requires very large string of 1’s and 0’s to represent a decimal number |
| B. | it requires sparingly small string of 1’s and 0’s to represent a decimal number |
| C. | it requires large string of 1’s and small string of 0’s to represent a decimal number |
| D. | it requires small string of 1’s and large string of 0’s to represent a decimal number |
| Answer» A. it requires very large string of 1’s and 0’s to represent a decimal number | |
| Explanation: the most vital drawback of binary system is that it requires very large string of 1’s and 0’s | |
| 7. |
The decimal equivalent of the octal number (645)8 is |
| A. | (450)10 |
| B. | (451)10 |
| C. | (421)10 |
| D. | (501)10 |
| Answer» C. (421)10 | |
| Explanation: octal to decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index position. | |
| 8. |
The largest two digit hexadecimal number is |
| A. | (fe)16 |
| B. | (fd)16 |
| C. | (ff)16 |
| D. | (ef)16 |
| Answer» C. (ff)16 | |
| Explanation: (fe)16 is 254 in decimal system, while (fd)16 is 253. (ef)16 is 239 in decimal system. and, | |
| 9. |
What is the addition of the binary numbers 11011011010 and 010100101? |
| A. | 0111001000 |
| B. | 1100110110 |
| C. | 11101111111 |
| D. | 10011010011 |
| Answer» C. 11101111111 | |
| Explanation: the rules for binary addition are : 0 + 0 = 0 | |
| 10. |
Representation of hexadecimal number (6DE)H in decimal: |
| A. | 6 * 162 + 13 * 161 + 14 * 160 |
| B. | 6 * 162 + 12 * 161 + 13 * 160 |
| C. | 6 * 162 + 11 * 161 + 14 * 160 |
| D. | 6 * 162 + 14 * 161 + 15 * 160 |
| Answer» A. 6 * 162 + 13 * 161 + 14 * 160 | |
| Explanation: hexadecimal to decimal conversion is obtained by multiplying 16 to the power of base index along with the value at that index position. | |
| 11. |
100101 × 0110 = ? |
| A. | 1011001111 |
| B. | 0100110011 |
| C. | 101111110 |
| D. | 0110100101 |
| Answer» C. 101111110 | |
| Explanation: the rules for binary multiplication are: 0 * 0 = 0 | |
| 12. |
Perform multiplication of the binary numbers: 01001 × 01011 = ? |
| A. | 001100011 |
| B. | 110011100 |
| C. | 010100110 |
| D. | 101010111 |
| Answer» A. 001100011 | |
| Explanation: the rules for binary multiplication are: 0 * 0 = 0 | |
| 13. |
Divide the binary numbers: 111101 ÷ 1001 and find the remainder |
| A. | 0010 |
| B. | 1010 |
| C. | 1100 |
| D. | 0011 |
| Answer» D. 0011 | |
| Explanation: binary division is accomplished using long division method. | |
| 14. |
Divide the binary number (011010000) by (0101) and find the quotient |
| A. | 100011 |
| B. | 101001 |
| C. | 110010 |
| D. | 010001 |
| Answer» B. 101001 | |
| 15. |
Binary coded decimal is a combination of |
| A. | two binary digits |
| B. | three binary digits |
| C. | four binary digits |
| D. | five binary digits |
| Answer» C. four binary digits | |
| Explanation: binary coded decimal is a combination of 4 binary digits. for example-8421. | |
| 16. |
The decimal number 10 is represented in its BCD form as |
| A. | 10100000 |
| B. | 01010111 |
| C. | 00010000 |
| D. | 00101011 |
| Answer» C. 00010000 | |
| Explanation: the decimal number 10 is represented in its bcd form as 0001 0000, in accordance to 8421 for each of the two digits. | |
| 17. |
Carry out BCD subtraction for (68) – (61) using 10’s complement method. |
| A. | 00000111 |
| B. | 01110000 |
| C. | 100000111 |
| D. | 011111000 |
| Answer» A. 00000111 | |
| Explanation: first the two numbers are converted into their respective bcd form using 8421 sequence. then binary subtraction is carried out. | |
| 18. |
When numbers, letters or words are represented by a special group of symbols, this process is called |
| A. | decoding |
| B. | encoding |
| C. | digitizing |
| D. | inverting |
| Answer» B. encoding | |
| Explanation: when numbers, letters or words are represented by a special group of symbols, this process is called encoding. encoding in the sense of fetching the codes or words in a computer. it is done to secure the transmission of information. | |
| 19. |
A three digit decimal number requires for representation in the conventional BCD format. |
| A. | 3 bits |
| B. | 6 bits |
| C. | 12 bits |
| D. | 24 bits |
| Answer» C. 12 bits | |
| Explanation: the number of bits needed to represent a given decimal number is always greater than the number of bits required for a straight binary encoding of the same. hence, a three digit decimal number requires 12 bits for representation in bcd format. | |
| 20. |
How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes? |
| A. | 12 |
| B. | 14 |
| C. | 16 |
| D. | 18 |
| Answer» B. 14 | |
| Explanation: total number of decimals to be represented = 10000 = 104 = 2n (where n is the number of bits required) = 213.29. therefore, the number of bits required for straight binary encoding = 14. | |
| 21. |
The excess-3 code for 597 is given by |
| A. | 100011001010 |
| B. | 100010100111 |
| C. | 010110010111 |
| D. | 010110101101 |
| Answer» A. 100011001010 | |
| Explanation: the addition of ‘3’ to each digit yields the three new digits ‘8’, ’12’ and ’10’. hence, the corresponding four-bit binary equivalents are 100011001010, in accordance to 8421 format. | |
| 22. |
The decimal equivalent of the excess-3 number 110010100011.01110101 is |
| A. | 970.42 |
| B. | 1253.75 |
| C. | 861.75 |
| D. | 1132.87 |
| Answer» A. 970.42 | |
| Explanation: the conversion of binary numbers into digits ‘1100’, ‘1010’, ‘0011’, ‘0111’ and ‘0101’ gives | |
| 23. |
In boolean algebra, the OR operation is performed by which properties? |
| A. | associative properties |
| B. | commutative properties |
| C. | distributive properties |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the expression for associative property is given by a+(b+c) = (a+b)+c & a*(b*c) = (a*b)*c. | |
| 24. |
The expression for Absorption law is given by |
| A. | a + ab = a |
| B. | a + ab = b |
| C. | ab + aa’ = a |
| D. | a + b = b + a |
| Answer» A. a + ab = a | |
| Explanation: the expression for absorption law is given by: a+ab = a. | |
| 25. |
DeMorgan’s theorem states that |
| A. | (ab)’ = a’ + b’ |
| B. | (a + b)’ = a’ * b |
| C. | a’ + b’ = a’b’ |
| D. | (ab)’ = a’ + b |
| Answer» A. (ab)’ = a’ + b’ | |
| Explanation: the demorgan’s law states that (ab)’ = a’ + b’ & (a + b)’ = a’ * b’, as per the dual property. | |
| 26. |
(A + B)(A’ * B’) = ? |
| A. | 1 |
| B. | 0 |
| C. | ab |
| D. | ab’ |
| Answer» B. 0 | |
| Explanation: the demorgan’s law states that (ab)’ = a’ + b’ & (a + b)’ = a’ * b’, as per the dual property. | |
| 27. |
According to boolean law: A + 1 = ? |
| A. | 1 |
| B. | a |
| C. | 0 |
| D. | a’ |
| Answer» A. 1 | |
| Explanation: a + 1 = 1, as per 1’s property. | |
| 28. |
The involution of A is equal to |
| A. | a |
| B. | a’ |
| C. | 1 |
| D. | 0 |
| Answer» A. a | |
| Explanation: the involution of a means double inversion of a (i.e. a”) and is equal to a. | |
| 29. |
Simplify Y = AB’ + (A’ + B)C. |
| A. | ab’ + c |
| B. | ab + ac |
| C. | a’b + ac’ |
| D. | ab + a |
| Answer» A. ab’ + c | |
| Explanation: y = ab’ + (a’ + b)c = ab’ + (ab’)’c | |
| 30. |
The boolean function A + BC is a reduced form of |
| A. | ab + bc |
| B. | (a + b)(a + c) |
| C. | a’b + ab’c |
| D. | (a + c)b |
| Answer» B. (a + b)(a + c) | |
| Explanation: (a + b)(a + c) = aa + ac + ab + bc | |
| 31. |
In which of the following gates the output is 1 if and only if at least one input is 1? |
| A. | and |
| B. | nor |
| C. | nand |
| D. | or |
| Answer» D. or | |
| Explanation: in or gate we need at least one bit to be equal to 1 to generate the output as 1 because or means any of the condition out of two is equal to 1 which means if at least one input is 1 then it shows output as 1. | |
| 32. |
The time required for a gate or inverter to change its state is called |
| A. | rise time |
| B. | decay time |
| C. | propagation time |
| D. | charging time |
| Answer» C. propagation time | |
| Explanation: the time required for a gate or inverter to change its state is called propagation time. | |
| 33. |
What is the minimum number of two input NAND gates used to perform the function of two input OR gates? |
| A. | one |
| B. | two |
| C. | three |
| D. | four |
| Answer» C. three | |
| Explanation: y = a + b. this is the equation of or gate. we require 3 nand gates to create or gate. we can also write, | |
| 34. |
Odd parity of word can be conveniently tested by |
| A. | or gate |
| B. | and gate |
| C. | nand gate |
| D. | xor gate |
| Answer» D. xor gate | |
| Explanation: odd parity of word can be conveniently tested by xor gate, since, xor outputs 1 only when the input has odd number of 1’s. | |
| 35. |
The number of full and half adders are required to add 16-bit number is |
| A. | 8 half adders, 8 full adders |
| B. | 1 half adders, 15 full adders |
| C. | 16 half adders, 0 full adders |
| D. | 4 half adders, 12 full adders |
| Answer» B. 1 half adders, 15 full adders | |
| Explanation: half adder has two inputs and two outputs whereas full adder has 3 inputs and 2 outputs. one half adder can add the least significant bit of the two numbers whereas full adders are required to add the remaining 15 bits as they all involve adding carries. | |
| 36. |
Which of the following will give the sum of full adders as output? |
| A. | three point major circuit |
| B. | three bit parity checker |
| C. | three bit comparator |
| D. | three bit counter |
| Answer» D. three bit counter | |
| Explanation: counters are used for counting purposes in ascending or descending order. three bit counter will give the sum of full adders as output. | |
| 37. |
An OR gate can be imagined as |
| A. | switches connected in series |
| B. | switches connected in parallel |
| C. | mos transistor connected in series |
| D. | bjt transistor connected in series |
| Answer» B. switches connected in parallel | |
| Explanation: or gate means addition of two inputs, which outputs when any of the input is high. due to this reason, it is imagined as switches connected in parallel. | |
| 38. |
How many full adders are required to construct an m-bit parallel adder? |
| A. | m/2 |
| B. | m |
| C. | m-1 |
| D. | m+1 |
| Answer» C. m-1 | |
| Explanation: we need adder for every bit. so we should need m bit adders. a full adder adds a carry bit to two inputs and produces an output and a carry. but the most significant bits can use a half adder which differs from the full adder as in that it has no carry input, so we need m-1 full adders and 1 half adder in m bit parallel adder. | |
| 39. |
Boolean Function is of the form of |
| A. | truth values |
| B. | k=f(x,y,x) |
| C. | algebraic expression |
| D. | truth table |
| Answer» A. truth values | |
| Explanation: the boolean function is of the form of algebraic expressions or truth table. a boolean function is of the form as that of option 2. the result obtained from a boolean function can be a truth value or a fallacy. | |
| 40. |
The result of X+X.Y is X. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: this is a boolean expression. it can be evaluated as : | |
| 41. |
In the boolean function w=f(X,Y,Z), what is the RHS referred to as |
| A. | right hand side |
| B. | expression |
| C. | literals |
| D. | boolean |
| Answer» B. expression | |
| Explanation: the rhs is commonly referred to as the expression. the symbols x, y and z are commonly referred to as the literals of the function. | |
| 42. |
The general form for calculating the number of rows in a truth table is |
| A. | 2n |
| B. | 2n+1 |
| C. | 2n |
| D. | 2n+1 |
| Answer» C. 2n | |
| Explanation: the general form for the calculation of rows in a truth table is 2n . here, n= number of literals in the function. | |
| 43. |
The number of literals in the expression F=X.Y’ + Z are |
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
| Answer» B. 3 | |
| Explanation: literals the generally the number of variables used in any boolean expression. here, since there are 3 variables x, y and z, therefore the answer is 3. | |
| 44. |
The complement term for X’.Y’.Z + X.Y will be |
| A. | xyz’+x’y’ |
| B. | (x+y+z’)(x’+y’) |
| C. | (x+y+z’)(x’+y) |
| D. | (x+y+z’)(x’+y) |
| Answer» B. (x+y+z’)(x’+y’) | |
| Explanation: the or and and operators are interchanged. the complement terms are reduced to normal terms and the result is obtained. | |
| 45. |
What is the complement of X’Y’Z? |
| A. | x+yz |
| B. | x’+y+’z’ |
| C. | x+y+z’ |
| D. | xyz’ |
| Answer» C. x+y+z’ | |
| Explanation: the complement is obtained by converting the complement terms to normal terms and vice versa. also, the or and and operators are | |
| 46. |
The minterm of any expression is denoted by |
| A. | mt |
| B. | m |
| C. | m |
| D. | min |
| Answer» B. m | |
| Explanation: the variables forming an and term is generally the minterm. it is denoted by m. (lower case letter). | |
| 47. |
The min term when X=Y=Z=0 is |
| A. | x’+y’+z’ |
| B. | xyz |
| C. | x’y’z’ |
| D. | x+y+z |
| Answer» C. x’y’z’ | |
| Explanation: there minterm expression is obtained using the and term. here, when x=y=z=0, the expression obtained is x’y’z’. | |
| 48. |
The max term when X=Y=Z=1 is |
| A. | x’+y’+z’ |
| B. | xyz |
| C. | x’y’z’ |
| D. | x+y+z |
| Answer» A. x’+y’+z’ | |
| Explanation: the max term consists of variables forming an or term. here, when x=y=z=1, the expression is x’+y’+z’. | |
| 49. |
A product term containing all K variables of the function in either complemented or uncomplemented form is called a |
| A. | minterm |
| B. | maxterm |
| C. | midterm |
| D. | ∑ term |
| Answer» A. minterm | |
| Explanation: a product term containing all k variables of the function in either complemented or uncomplemented form is called a minterm. a sum term containing all k variables of the function in either complemented or uncomplemented form is called a maxterm. | |
| 50. |
The canonical sum of product form of the function y(A,B) = A + B is |
| A. | ab + bb + a’a |
| B. | ab + ab’ + a’b |
| C. | ba + ba’ + a’b’ |
| D. | ab’ + a’b + a’b’ |
| Answer» B. ab + ab’ + a’b | |
| Explanation: a + b = a.1 + b.1 = a(b + b’) + b(a + a’) = ab + ab’ + ba +ba’ = ab + ab’ + a’b = ab | |
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