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Home » Technical » Electronics » Aptitude Test - 3
1.Convert (214)8 into decimal?
A. (140)10
B. (141)10
C. (142)10
D. (140)10
2.Convert (0.345)10 into an octal number?
A. (0.16050)8
B. (0.26050)8
C. (0.19450)8
D. (0.16050)8
3.Convert the binary number (01011.1011)2 into decimal?
A. (11.6875)10
B. (11.5874)10
C. (10.9876)10
D. (11.6875)10
4.Octal to binary conversion: (24)8 =?
A. (111101)2
B. (010100)2
C. (111100)2
D. (111101)2
5.Convert binary to octal: (110110001010)2 =?
A. (5512)8
B. (6612)8
C. (4532)8
D. (5512)8
6.What is the addition of the binary numbers 11011011010 and 010100101?
A. 0111001000
B. 1100110110
C. 11101111111
D. 0111001000
7.Perform binary addition: 101101 + 011011 = ?
A. 011010
B. 1010100
C. 101110
D. 011010
8.Perform binary subtraction: 101111 – 010101 = ?
A. 100100
B. 010101
C. 011010
D. 100100
9.Binary subtraction of 100101 – 011110 is?
A. 000111
B. 111000
C. 010101
D. 000111
10.Perform multiplication of the binary numbers: 01001 × 01011 = ?
A. 001100011
B. 110011100
C. 010100110
D. 001100011
11.100101 × 0110 = ?
A. 1011001111
B. 0100110011
C. 101111110
D. 1011001111
12.On multiplication of (10.10) and (01.01), we get?
A. 101.0010
B. 0010.101
C. 011.0010
D. 101.0010
13.Divide the binary numbers: 111101 ÷ 1001 and find the remainder?
A. 0010
B. 1010
C. 1100
D. 0010
14.Divide the binary number (011010000) by (0101) and find the quotient?
A. 100011
B. 101001
C. 110010
D. 100011
15.Binary subtraction of 101101 – 001011 = ?
A. 100010
B. 010110
C. 110101
D. 100010
16.1’s complement of 1011101 is?
A. 0101110
B. 1001101
C. 0100010
D. 0101110
17.2’s complement of 11001011 is?
A. 01010111
B. 11010100
C. 00110101
D. 01010111
18.On subtracting (01010)2 from (11110)2 using 1’s complement, we get?
A. 01001
B. 11010
C. 10101
D. 01001
19.On subtracting (010110)2 from (1011001)2 using 2’s complement, we get?
A. 0111001
B. 1100101
C. 0110110
D. 0111001
20.On subtracting (001100)2 from (101001)2 using 2’s complement, we get?
A. 1101100
B. 011101
C. 11010101
D. 1101100
21.On addition of 28 and 18 using 2’s complement, we get ?
A. 00101110
B. 0101110
C. 00101111
D. 00101110
22.On addition of +38 and -20 using 2’s complement, we get?
A. 11110001
B. 100001110
C. 010010
D. 11110001
23.On addition of -46 and +28 using 2’s complement, we get?
A. -10010
B. -00101
C. 01011
D. -10010
24.On addition of -33 and -40 using 2’s complement, we get?
A. 1001110
B. -110101
C. 0110001
D. 1001110
25.On subtracting +28 from +29 using 2’s complement, we get?
A. 11111010
B. 111111001
C. 100001
D. 11111010
26.If the number of bits in the sum exceeds the number of bits in each added numbers, it results in?
A. Successor
B. Overflow
C. Underflow
D. Successor
27.An overflow is a?
A. Hardware problem
B. Software problem
C. User input problem
D. Hardware problem
28.An overflow occurs in?
A. MSD position
B. LSD position
C. Middle position
D. MSD position
29.Logic circuitry is used to detect?
A. Underflow
B. MSD
C. Overflow
D. Underflow
30.1’s complement can be easily obtained by using?
A. Comparator
B. Inverter
C. Adder
D. Comparator
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