IMPORTANT COMPUTER APTITUDE PRACTICE QUESTIONS FOR SBI PO & CLERK MAINS

Hope all are ready for SBI Clerk & Po exams.Here is some of the important computer aptitude questions, you can practice very well for your mains examination and can score more marks in the upcoming exams.

1) Convert the given Decimal number into Binary form -173

a) 001011112

b) 001010012

c) 101011112

d) 101010012

e) 101011012

2) Convert the given Decimal into octal form -130

a) 3018

b) 3038

c) 2028

d) 1048

e) 2038

3) Convert the given base 10 into base 16 – 310

a) D216

b) C216

c) C316

d) A316

e) D316

4) Convert the given Decimal number into Binary number – 195

a) 001111102

b) 111110002

c) 110001112

d) 001110002

e) 001010002

5) Convert the binary number into octal number – 100100110

a) 5448

b) 4468

c) 5648

d) 4548

e) 5548

6) Convert the given binary number into octal number – 111100110

a) 6568

b) 7868

c) 5568

d) 7468

e) 7568

7) Convert the given base 10 to base 16 – 200

a) B816

b) D216

c) A316

d) C616

e) C816

8) Convert the given Binary number into Decimal number – 1000112

a) 30

b) 31

c) 32

d) 34

e) 35

9) Convert the given binary number into Decimal number – 0011102

a) 17

b) 12

c) 14

10) Convert the given Decimal into octal – 135

a) 2168

b) 2098

c) 2058

d) 2078

e) 2068

11) Convert the given Binary number into octal number 111001001

a) 7228

b) 7118

c) 6558

d) 6118

e) 7668

12) What will be the 1’s complement of decimal number 44

a) 111000

b) 100110

c) 010011

d) 110010

e) 001100

13) What will be the 1’s complement of decimal number 22 ?

a) 01000

b) 01101

c) 01010

d) 01011

e) 01001

14) What will be the 2’s complement of 1001002

a) 11001

b) 01100

c) 10101

d) 11100

e) 11111

15) What will be the 2’s complement of 1110112 ?

a) 000101

b) 110001

c) 110000

d) 100011

e) 110011

16) What will be the 2’s complement of 1000112 ?

a) 111101

b) 010101

c) 011100

d) 111101

e) 011101

17) What will be the 1’s complement of decimal number 45 ?

a) 111000

b) 010010

c) 011110

d) 111000

e) 011101

18) Convert the given hexadecimal to decimal number (base 10) – BC2

a) 304010

b) 301010

c) 201810

d) 201010

e) 401010

19) Convert the given hexadecimal to decimal number (base 10) – AC2

a) 375410

b) 275310

c) 265410

d) 274410

e) 275410

20) Convert the given hexadecimal to decimal number (base 10) – CC2

a) 324110

b) 331110

c) 326610

d) 326110

e) 321110

1) e

173/2 – Remainder 1

86/2 – Remainder 0

43/2 – Remainder 1

21/2 – Remainder 1

10/2 – Remainder 0

5/2 – Remainder 1

2/2 – Remainder 0

1

173=> 101011012

2) c

130/8 – Remainder 2

16/8 – Remainder 0

2

130 => 2028

3) a

310 / 16 -remainder 2

13 – D

(A = 10 , B = 11 , C = 12 , D = 13)

4) c

110001112

195 /2 – Remainder 1

97/2 – Remainder 1

48/2 – Remainder 1

24/2 – Remainder 0

12/2 – Remainder 0

6/2 – Remainder 0

3/2 – Remainder 1

1

195 -> 110001112

5) b

4468

100100110 = > 100/100/110

4 / 4 / 6

100100110 => 4468

6) d

7468

111100110 => 111/100/110

7 / 4 / 6

111100110 => 7468

7) e

C816

200 / 16 – Remainder 8

12 – C => C816

8) d

34

1000112 = 25 + 0 + 0 + 0 + 1 + 1

32 + 0 + 0 + 0 + 2 = > 34

9) c

0011102

0 + 0 + 23 + 22 + 21 + 0

0 + 0 + 8 + 4 + 2 + 0 => 14

10) d

135/8 – Remainder 7

16/8 – Remainder 0

2

135 => 2078

11) b

1110010012

111/001/001

7/ 1/ 1

1110010012 => 711

12) c

44/2 – Remainder – 0

22/2 – Remainder 0

11/2 – Remainder 0

5/2 – Remainder 1

2/2 – Remainder 0

1

44 => 010011

13) e

22/2 – Remainder 0

11/2 – Remainder 1

5/2 – Remainder 1

2/2 – Remainder 0

1

22 => 01001

14) d

1001002 = 011011

(+1)

= 011100

15) a

1110112 = 000100

(+1)

= 000101

16) c

1000112 = 011100

(+1)

= 011101

17) b

45/2 – Remainder 1

22/2 – Remainder 0

11/2 – Remainder 1

5/2 – Remainder 1

2/2 – Remainder 0

1

45 => 101101

=> 010010

18) b

BC2 => 11 12 2

11 x 162 + 12 x 161 + 2 x 160

2816 + 192 + 2

301010

19) e

AC2 => 10 12 2

10 x 162 + 12 x 161 + 2 x 160

2560 + 192 + 2

275410

20) c

CC2 = 12 12 2

12x 162 + 12 x 161 + 2 x 160

3072 + 192 + 2

326610